On (special) relativity: what’s relative?

This is my third and final post about special relativity. In the previous posts, I introduced the general idea and the Lorentz transformations. I present these Lorentz transformations once again below, next to their Galilean counterparts. [Note that I continue to assume, for simplicity, that the two reference frames move with respect to each other along the x- axis only, so the y- and z-component of u is zero. It is not all that difficult to generalize to three dimensions (especially not when using vectors) but it makes an intuitive understanding of what’s relativity all about more difficult.]

CaptureAs you can see, under a Lorentz transformation, the new ‘primed’ space and time coordinates are a mixture of the ‘unprimed’ ones. Indeed, the new x’ is a mixture of x and t, and the new t’ is a mixture as well. You don’t have that under a Galilean transformation: in the Newtonian world, space and time are neatly separated, and time is absolute, i.e. it is the same regardless of the reference frame. In Einstein’s world – our world – that’s not the case: time is relative, or local as Hendrik Lorentz termed it, and so it’s space-time – i.e. ‘some kind of union of space and time’ as Minkowski termed it  that transforms. In practice, physicists will use so-called four-vectors, i.e. vectors with four coordinates, to keep track of things. These four-vectors incorporate both the three-dimensional space vector as well as the time dimension. However, we won’t go into the mathematical details of that here.

What else is relative? Everything, except the speed of light. Of course, velocity is relative, just like in the Newtonian world, but the equation to go from a velocity as measured in one reference frame to a velocity as measured in the other, is different: it’s not a matter of just adding or subtracting speeds. In addition, besides time, mass becomes a relative concept as well in Einstein’s world, and that was definitely not the case in the Newtonian world.

What about energy? Well… We mentioned that velocities are relative in the Newtonian world as well, so momentum and kinetic energy were relative in that world as well: what you would measure for those two quantities would depend on your reference frame as well. However, here also, we get a different formula now. In addition, we have this weird equivalence between mass and energy in Einstein’s world, about which I should also say something more.

But let’s tackle these topics one by one. We’ll start with velocities.

Relativistic velocity

In the Newtonian world, it was easy. From the Galilean transformation equations above, it’s easy to see that

v’ = dx’/dt’ = d(x – ut)/dt = dx/dt – d(ut)/dt = v – u

So, in the Newtonian world, it’s just a matter of adding/subtracting speeds indeed: if my car goes 100 km/h (v), and yours goes 120 km/h, then you will see my car falling behind at a speed of (minus) 20 km/h. That’s it. In Einstein’s world, it is not so simply. Let’s take the spaceship example once again. So we have a man on the ground (the inertial or ‘unprimed’ reference frame) and a man in the spaceship (the primed reference frame), which is moving away from us with velocity u.

Now, suppose an object is moving inside the spaceship (along the x-axis as well) with a (uniform) velocity vx’, as measured from the point of view of the man inside the spaceship. Then the displacement x’ will be equal to x’vx’ t’. To know how that looks from the man on the ground, we just need to use the opposite Lorentz transformations: just replace u by –u everywhere (to the man in the spaceship, it’s like the man on the ground moves away with velocity –u), and note that the Lorentz factor does not change because we’re squaring and (–u)2 u2. So we get:

Capture

Hence, x’ = vx’ t’ can be written as x = γ(vx’ t’ + ut’). Now we should also substitute t’, because we want to measure everything from the point of view of the man on the ground. Now, t = γ(t’ + uvx’ t’/c2). Because we’re talking uniform velocities, v(i.e. the velocity of the object as measured by the man on the ground) will be equal to x divided by t (so we don’t need to take the time derivative of x), and then, after some simplifying and re-arranging (note, for instance, how the t’ factor miraculously disappears), we get:

Capture

What does this rather complicated formula say? Just put in some numbers:

  • Suppose the object is moving at half the speed of light, so 0.5c, and that the spaceship is moving itself also at 0.5c, then we get the rather remarkable result that, from the point of view of the observer on the ground, that object is not going as fast as light, but only at vx = (0.5c + 0.5c)/(1 + 0.5·0.5) = 0.8c.
  • Or suppose we’re looking at a light beam inside the spaceship, so something that’s traveling at speed c itself in the spaceship. How does that look to the man on the ground? Just put in the numbers: vx = (0.5c + c)/(1 + 0.5·1) = ! So the speed of light is not dependent on the reference frame: it looks the same – both to the man in the ship as well as to the man on the ground. As Feynman puts it: “This is good, for it is, in fact, what the Einstein theory of relativity was designed to do in the first place–so it had better work!”

It’s interesting to note that, even if u has no y– or z-component, velocity in the direction will be affected too. Indeed, if an object is moving upward in the spaceship, then the distance of travel of that object to the man on the ground will appear to be larger. See the triangle below: if that object travels a distance Δs’ = Δy’ = Δy = v’Δt’ with respect to the man in the spaceship, then it will have traveled a distance Δs = vΔt to the man on the ground, and that distance is longer.

CaptureI won’t go through the process of substituting and combining the Lorentz equations (you can do that yourself) but the grand result is the following:

vy = (1/γ)vy’ 

1/γ is the reciprocal of the Lorentz factor, and I’ll leave it to you to work out a few numeric examples. When you do that, you’ll find the rather remarkable result that vy is actually less than vy’. For example, for u = 0.6c, 1/γ will be equal to 0.8, so vy will be 20% less than vy’. How is that possible? The vertical distance is what it is (Δy’ = Δy), and that distance is not affected by the ‘length contraction’ effect (y’ = y). So how can the vertical velocity be smaller?  The answer is easy to state, but not so easy to understand: it’s the time dilation effect: time in the spaceship goes slower. Hence, the object will cover the same vertical distance indeed – for both observers – but, from the point of view of the observer on the ground, the object will apparently need more time to cover that distance than the time measured by the man in the spaceship: Δt > Δt’. Hence, the logical conclusion is that the vertical velocity of that object will appear to be less to the observer on the ground.

How much less? The time dilation factor is the Lorentz factor. Hence, Δt = γΔt’. Now, if u = 0.6c, then γ will be equal to 1.25 and Δt = 1.25Δt’. Hence, if that object would need, say, one second to cover that vertical distance, then, from the point of view of the observer on the ground, it would need 1.25 seconds to cover the same distance. Hence, its speed as observed from the ground is indeed only 1/(5/4) = 4/5 = 0.8 of its speed as observed by the man in the spaceship.

Is that hard to understand? Maybe. You have to think through it. One common mistake is that people think that length contraction and/or time dilation are, somehow, related to the fact that we are looking at things from a distance and that light needs time to reach us. Indeed, on the Web, you can find complicated calculations using the angle of view and/or the line of sight (and tons of trigonometric formulas) as, for example, shown in the drawing below. These have nothing to do with relativity theory and you’ll never get the Lorentz transformation out of them. They are plain nonsense: they are rooted in an inability of these youthful authors to go beyond Galilean relativity. Length contraction and/or time dilation are not some kind of visual trick or illusion. If you want to see how one can derive the Lorentz factor geometrically, you should look for a good description of the Michelson-Morley experiment in a good physics handbook such as, yes :-), Feynman’s Lectures.

visual effect 2

So, I repeat: illustrations that try to explain length contraction and time dilation in terms of line of sight and/or angle of view are useless and will not help you to understand relativity. On the contrary, they will only confuse you. I will let you think through this and move on to the next topic.

Relativistic mass and relativistic momentum

Einstein actually stated two principles in his (special) relativity theory:

  1. The first is the Principle of Relativity itself, which is basically just the same as Newton’s principle of relativity. So that was nothing new actually: “If a system of coordinates K is chosen such that, in relation to it, physical laws hold good in their simplest form, then the same laws must hold good in relation to any other system of coordinates K’ moving in uniform translation relatively to K.” Hence, Einstein did not change the principle of relativity – quite on the contrary: he re-confirmed it – but he did change Newton’s Laws, as well as the Galilean transformation equations that came with them. He also introduced a new ‘law’, which is stated in the second ‘principle’, and that the more revolutionary one really:
  2. The Principle of Invariant Light Speed: “Light is always propagated in empty space with a definite velocity [speed] c which is independent of the state of motion of the emitting body.”

As mentioned above, the most notable change in Newton’s Laws – the only change, in fact – is Einstein’s relativistic formula for mass:

mv = γm0

This formula implies that the inertia of an object, i.e. its mass, also depends on the reference frame of the observer. If the object moves (but velocity is relative as we know: an object will not be moving if we move with it), then its mass increases. This affects its momentum. As you may or may not remember, the momentum of an object is the product of its mass and its velocity. It’s a vector quantity and, hence, momentum has not only a magnitude but also a direction:

 pv = mvv = γm0v 

As evidenced from the formula above, the momentum formula is a relativistic formula as well, as it’s dependent on the Lorentz factor too. So where do I want to go from here? Well… In this section (relativistic mass and momentum), I just want to show that Einstein’s mass formula is not some separate law or postulate: it just comes with the Lorentz transformation equations (and the above-mentioned consequences in terms of measuring horizontal and vertical velocities).

Indeed, Einstein’s relativistic mass formula can be derived from the momentum conservation principle, which is one of the ‘physical laws’ that Einstein refers to. Look at the elastic collision between two billiard balls below. These balls are equal – same mass and same speed from the point of view of an inertial observer – but not identical: one is red and one is blue. The two diagrams show the collision from two different points of view: left, we have the inertial reference frame, and, right, we have a reference frame that is moving with a velocity equal to the horizontal component of the velocity of the blue ball.

Relcollision

The points to note are the following:

  1. The total momentum of such elastic collision before and after the collision must be the same.
  2. Because the two balls have equal mass (in the inertial reference frame at least), the collision will be perfectly symmetrical. Indeed, we may just turn the diagram ‘upside down’ and change the colors of the balls, as we do below, and the values w, u and v (as well as the angle α) are the same.

Elastic collision

As mentioned above, the velocity of the blue and red ball and, hence, their momentum, will depend on the frame of reference. In the diagram on the left, we’re moving with a velocity equal to the horizontal component of the velocity of the blue ball and, therefore, in this particular frame of reference, the velocity (and the momentum) of the blue ball consists of a vertical component only, which we refer to as w.

From this point of view (i.e. the reference frame moving with, the velocity (and, hence, the momentum) of the red ball will have both a horizontal as well as a vertical component. If we denote the horizontal component by u, then it’s easy to show that the vertical velocity of the red ball must be equal to sin(α)v. Now, because u = cos(α)v, this vertical component will be equal to tan(α)u. But so what is tan(α)u? Now, you’ll say, that is quite evident: tan(α)u must be equal to w, right?

No. That’s Newtonian physics. The red ball is moving horizontally with speed u with respect to the blue ball and, hence, its vertical velocity will not be quite equal to w. Its vertical velocity will be given by the formula which we derived above: vy = (1/γ)vy’, so it will be a little bit slower than the w we see in the diagram on the right which is, of course, the same w as in the diagram on the left. [If you look carefully at my drawing above, then you’ll notice that the w vector is a bit longer indeed.]

Huh? Yes. Just think about it: tan(α)= (1/γ)w. But then… How can momentum be conserved if these speeds are not the same? Isn’t the momentum conservation principle supposed to conserve both horizontal as well as vertical momentum? It is, and momentum is being conserved. Why? Because of the relativistic mass factor.

Indeed, the change in vertical momentum (Δp) of the blue ball in the diagram on the left or – which amounts to the same – the red ball in the diagram on the right (i.e. the vertically moving ball) is equal to Δpblue = 2mww. [The factor 2 is there because the ball goes down and then up (or vice versa) and, hence, the total change in momentum must be twice the mwamount.] Now, that amount must be equal to Δpred, which is equal to Δpblue = 2mv(1/γ)w. Equating both yields the following grand result:

mv/m= γ ⇔ mv = γmw

What does this mean? It means that mass of the red ball in the diagram on the left is larger than the mass of the blue ball. So here we have actually derived Einstein’s relativistic mass formula from the momentum conservation principle !

Of course you’ll say: not quite. This formula is not the mu = γmformula that we’re used to ! Indeed, it’s not. The blue ball has some velocity w itself, and so the formula links two velocities v and w. However, we can derive  mv = γmformula as a limit of mv = γmw for w going to zer0. How can w become infinitesimally small? If the angle α becomes infinitesimally small. It’s obvious, then, that v and u will be practically equal. In fact, if w goes to zero, then mw will be equal to m0 in the limiting case, and mv will be equal to mu. So, then, indeed, we get the familiar formula as a limiting case:

mu = γm

Hmm… You’ll probably find all of this quite fishy. I’d suggest you just think about it. What I presented above, is actually Feynman’s presentation of the subject, but with a bit more verbosity. Let’s move on to the final.

Relativistic energy

From what I wrote above (and from what I wrote in my two previous posts on this topic), it should be obvious, by now, that energy also depends on the reference frame. Indeed, mass and velocity depend on the reference frame (moving or not), and both appear in the formula for kinetic energy which, as you’ll remember, is

K.E. = mc– m0c= (m – m0)c= γm0c– m0c= m0c2(γ – 1).

Now, if you go back to the post where I presented that formula, you’ll see that we’re actually talking the change in kinetic energy here: if the mass is at rest, it’s kinetic energy is zero (because m = m0), and it’s only when the mass is moving, that we can observe the increase in mass. [If you wonder how, think about the example of the fast-moving electrons in an electron beam: we see it as an increase in the inertia: applying the same force does no longer yield the same acceleration.]

Now, in that same post, I also noted that Einstein added an equivalent rest mass energy (E= m0c2) to the kinetic energy above, to arrive at the total energy of an object:

E = E+ K.E. = mc

Now, what does this equivalence actually mean? Is mass energy? Can we equate them really? The short answer to that is: yes.

Indeed, in one of my older posts (Loose Ends), I explained that protons and neutrons are made of quarks and, hence, that quarks are the actual matter particles, not protons and neutrons. However, the mass of a proton – which consists of two up quarks and one down quark – is 938 MeV/c(don’t worry about the units I am using here: because protons are so tiny, we don’t measure their mass in grams), but the mass figure you get when you add the rest mass of two u‘s and one d, is 9.6 MeV/conly: about one percent of 938 ! So where’s the difference?

The difference is the equivalent mass (or inertia) of the binding energy between the quarks. Indeed, the so-called ‘mass’ that gets converted into energy when a nuclear bomb explodes is not the mass of quarks. Quarks survive: nuclear power is binding energy between quarks that gets converted into heat and radiation and kinetic energy and whatever else a nuclear explosion unleashes.

In short, 99% of the ‘mass’ of a proton or an electron is due to the strong force. So that’s ‘potential’ energy that gets unleashed in a nuclear chain reaction. In other words, the rest mass of the proton is actually the inertia of the system of moving quarks and gluons that make up the particle. In such atomic system, even the energy of massless particles (e.g. the virtual photons that are being exchanged between the nucleus and its electron shells) is measured as part of the rest mass of the system. So, yes, mass is energy. As Feynman put it, long before the quark model was confirmed and generally accepted:

“We do not have to know what things are made of inside; we cannot and need not justify, inside a particle, which of the energy is rest energy of the parts into which it is going to disintegrate. It is not convenient and often not possible to separate the total mc2 energy of an object into (1) rest energy of the inside pieces, (2) kinetic energy of the pieces, and (3) potential energy of the pieces; instead we simply speak of the total energy of the particle. We ‘shift the origin’ of energy by adding a constant m0c2 to everything, and say that the total energy of a particle is the mass in motion times c2, and when the object is standing still, the energy is the mass at rest times c2.” (Richard Feynman’s Lectures on Physics, Vol. I, p. 16-9)

 So that says it all, I guess, and, hence, that concludes my little ‘series’ on (special) relativity. I hope you enjoyed it.

Post scriptum:

Feynman describes the concept of space-time with a nice analogy: “When we move to a new position, our brain immediately recalculates the true width and depth of an object from the ‘apparent’ width and depth. But our brain does not immediately recalculate coordinates and time when we move at high speed, because we have had no effective experience of going nearly as fast as light to appreciate the fact that time and space are also of the same nature. It is as though we were always stuck in the position of having to look at just the width of something, not being able to move our heads appreciably one way or the other; if we could, we understand now, we would see some of the other man’s time—we would see “behind”, so to speak, a little bit. Thus, we shall try to think of objects in a new kind of world, of space and time mixed together, in the same sense that the objects in our ordinary space-world are real, and can be looked at from different directions. We shall then consider that objects occupying space and lasting for a certain length of time occupy a kind of a “blob” in a new kind of world, and that when we look at this “blob” from different points of view when we are moving at different velocities. This new world, this geometrical entity in which the “blobs” exist by occupying position and taking up a certain amount of time, is called space-time.”

If none of what I wrote could convey the general idea, then I hope the above quote will. 🙂 Apart from that, I should also note that physicists will prefer to re-write the Lorentz transformation equations by measuring time and distance in so-called equivalent units: velocities will be expressed not in km/h but as a ratio of c and, hence, = 1 (a pure number) and so u will also be a pure number between 0 and 1. That can be done by expressing distance in light-seconds ( a light-second is the distance traveled by light in one second or, alternatively, by expressing time in ‘meter’. Both are equivalent but, in most textbooks, it will be time that will be measured in the ‘new’ units. So how do we express time in meter?

It’s quite simple: we multiply the old seconds with c and then we get: timeexpressed in meters = timeexpressed in seconds multiplied by 3×10meters per second. Hence, as the ‘second’ the first factor and the ‘per second’ in the second factor cancel out, the dimension of the new time unit will effectively be the meter. Now, if both time and distance are expressed in meter, then velocity becomes a pure number without any dimension, because we are dividing distance expressed in meter by time expressed in meter, and it should be noted that it will be a pure number between 0 and 1 (0 ≤ u ≤ 1), because 1 ‘time second’ = 1/(3×108) ‘time meters’. Also, c itself becomes the pure number 1. The Lorentz transformation equations then become:

Capture

They are easy to remember in this form (cf. the symmetry between x ut and t  ux) and, if needed, we can always convert back to the old units to recover the original formulas.

I personally think there is no better way to illustrate how space and time are ‘mere shadows’ of the same thing indeed: if we express both time and space in the same dimension (meter), we can see how, as result of that, velocity becomes a dimensionless number between zero and one and, more importantly, how the equations for x’ and t’ then mirror each other nicely. I am not sure what ‘kind of union’ between space and time Minkowski had in mind, but this must come pretty close, no?

Final note: I noted the equivalence of mass and energy above. In fact, mass and energy can also be expressed in the same units, and we actually do that above already. If we say that an electron has a rest mass of 0.511 MeV/c(a bit less than a quarter of the mass of the u quark), then we express the mass in terms of energy. Indeed, the eV is an energy unit and so we’re actually using the m = E/c2 formula when we express mass in such units. Expressing mass and energy in equivalent units allows us to derive similar ‘Lorentz transformation equations’ for the energy and the momentum of an object as measured under an inertial versus a moving reference frame. Hence, energy and momentum also transform like our space-time four-vectors and – likewise – the energy and the momentum itself, i.e. the components of the (four-)vector, are less ‘real’ than the vector itself. However, I think this post has become way too long and, hence, I’ll just jot these four equations down – please note, once again, the nice symmetry between (1) and (2) – but then leave it at that and finish this post. 🙂

Capture

On (special) relativity: the Lorentz transformations

I just skyped to my kids (unfortunately, we’re separated by circumstances) and they did not quite get the two previous posts (on energy and (special) relativity). The main obstacle is that they don’t know much – nothing at all actually – about integrals. So I should avoid integrals. That’s hard but I’ll try to do so in this post, in which I want to introduce special relativity as it’s usually done, and so that’s not by talking about Einstein’s mass-energy equivalence relation first.

Galilean/Newtonian relativity

A lot of people think they understand relativity theory but they often confuse it with Galilean (aka Newtonian) relativity and, hence, they actually do not understand it at all. Indeed, Galilean or Newtonian relativity is as old as Galileo and Newton (so that’s like 400 years old), who stated the principle of relativity as a corollary to the laws of motion: “The motions of bodies included in a given space are the same amongst themselves, whether that space is at rest or moves uniformly forward in a straight line.”

The Galilean or Newtonian principle of relativity is about adding and subtracting speeds: if I am driving at 120 km/h on some highway, but you overtake me at 140 km/h, then I will see you go past me at the rather modest speed of 20 km/h. That’s all what there is to it.

Now, that’s not what Einstein‘s relativity theory is about. Indeed, the relationship between your and my reference frame (yours is moving with respect to mine, and mine is moving with respect to yours but with opposite velocity) is very simple in this example. It involves a so-called Galilean transformation only: if my coordinate system is (x, y, z, t), and yours is (x‘, y, z‘, t‘), then we can write:

(1) x’ = x – ut (or x = x’ + ut),  (2) y’ = y, (3) z’ = z and (4) t’ = t

To continue the example above: if we start counting at t = t’ = 0 when you are overtaking me, and if we both consider ourselves to be at the center of our reference frame (i.e. x = 0 where I am and x’ = 0 where you are), then you will be at x = 10 km after 30 minutes from my point of view, and I will be at x’ = –10 km (so that’s 10 km behind) from your point of view. So x’ = x – ut indeed, with u = 20 km/h.

Again, that’s not what Einstein’s principle of relativity is about. They knew that very well in the 17th century already. In fact, they actually knew that much earlier but Descartes formalized his Cartesian coordinate system only in the first half of the 17th century and, hence, it’s only from that time onwards that scientists such as Newton and Huygens started using it to transform the laws of physics from one frame of reference to another. What they found is that those laws remained invariant.

For example, the conservation law for momentum remains valid even if, as illustrated below, an inertial observer will see an elastic collision, such as the one illustrated, differently than a observer who’s moving along: for the observer who’s moving along, the (horizontal) speed of the blue ball will be zero, and the (horizontal) speed of the red ball will be twice the speed as observed by the inertial observer. That being said, both observers will find that momentum (i.e. the product of mass and velocity: p = mv) is being conserved in such collisions.

RelcollisionBut, again, that’s Galilean relativity only: the laws of Newton are of the same form in a moving system as in a stationary system and, therefore, it is impossible to tell, by making experiments, whether our system is moving or not. In other words: there is no such thing as ‘absolute speed’. But, so – let me repeat it again – that is not what Einstein’s relativity theory is about.

Let me give a more interesting example of Galilean relativity, and then we can see what’s wrong with it. The speed of a sound wave is not dependent on the motion of the source: the sound of a siren of an ambulance or a noisy car engine will always travel at a speed of 343 meter per second, regardless of the motion of the ambulance. So, while we’ll experience a so-called Doppler effect when the ambulance is moving – i.e. a higher pitch when it’s approaching than when it’s receding – this Doppler effect does not have any impact on the speed of the sound wave. It only affects the frequency as we hear it. The speed of the wave depends on the medium only, i.e. air in this case.

Indeed, the speed of sound will be different in another gas, or in a fluid, or in a solid, and there’s a surprisingly simple function for that – the so-called Newton-Laplace equation: vsound = (k/ρ)2. In this equation, k is a coefficient of ‘stiffness’ of the medium (even if ‘stiffness’ sounds somewhat strange as a concept to apply to gases), and ρ is the density of the medium (so lower or higher air density will increase/decrease the speed of sound).

Dopplerfrequenz

This has nothing to do with speed being absolute. No. The Galilean relativity principle does come into play, as one would expect: it is actually possible to catch up with a sound wave (or with any wave traveling through some medium). In fact, that’s what supersonic planes do: they catch up with their own sound waves. However, in essence, planes are not any different from cars in terms of their relationship with the sound that they produce. It’s just that they are faster: the sound wave they produce also travels at a speed of 1,235 km/h, and so cars can’t match that, but supersonic planes can!

[As for the shock wave that is being produced as these planes accelerate and actually ‘break’ the ‘sound barrier’, that has to do with the pressure waves the plane creates in front of itself (just like a traveling compresses the air in front of it). These pressure waves also travel at the speed of sound. Now, as the speed of the object increases, the waves are forced together, or compressed, because they cannot get out of the way of each other. Eventually they merge into one single shock wave, and so that’s what happens and creates the ‘sonic boom’, which also travels at the speed of sound. However, that should not concern us here. For more information on this, I’d refer to Wikipedia, as I got these illustrations from that source, and I quite like the way they present the topic.] 

Dopplereffectsourcemovingrightatmach1.4

The Doppler effect looks somewhat different (it’s illustrated above) but so, once again, this phenomenon has nothing to do with Einstein’s relativity theory. Why not? Because we are still talking Galilean relativity here. Indeed, let’s suppose our plane travels at twice the speed of sound (i.e. Mach 2 or almost 2,500 km/h). For us, as inertial observers, the speed of the sound wave originating at point 0 in the illustration above (i.e. the reference frame of the inertial observer) will be equal to dx/dt = 1235 km/h. However, for the pilot, the speed of that wave will be equal to

dx’/dt = d(x – ut)/dt = dx/dt – d(ut)/dt = dx/dt – d(ut)/dt = 1235 km/h – u

= 1235 km/h – u = 1235 km/h – 2470 km/h = – 1235 km/h

In short, from the point of view of the pilot, he sees the wave front of the wave created at point 0 traveling away from him (cf. the negative value) at 1235 km/h, i.e. the speed of sound. That makes sense obviously, because he travels twice as fast. However – I cannot repeat it enough – this phenomenon has nothing to do with Einstein’s theory of relativity: if they could have imagined supersonic travel, Galileo, Newton and Huygens would have predicted that too.

So what’s Einstein’s theory of (special) relativity about?

Einstein’s principle of relativity

In 1865, the Scottish mathematical physicist James Clerk Maxwell –  I guess it’s important to note he’s Scottish with that referendum coming 🙂 – finally discovered that light was nothing but electromagnetic radiation – so radio waves, (visible) light, X-rays, gamma rays,… It’s all the same: electromagnetic radiation, also known as light tout court.

Now, the equations that describe how electromagnetic radiation (i.e. light) travels through space are beautiful but involve operators which you may not recognize and, hence, I will not write them down. The point to note is that Maxwell’s equations were very elegant but… There were two major difficulties with them:

  1. They did not respect Galilean relativity: if we transform them using the above-mentioned Galilean transformation (x’ = x – uty’ = y, z’ = z and t’ = t) then we do not get some relative speed of light. On the contrary, according to Maxwell’s equations, from whatever reference frame you look at light, it should always travel at the same (absolute) speed of light c = 299,792 km/h. So c is a constant, and the same constant, ALWAYS.
  2. Scientists did not have any clue about the medium in which light was supposed to travel. The second half of the 19th century saw lots of experiments trying to discover evidence of a hypothetical ‘luminiferous aether’ in which light was supposed to travel, and which should also have some ‘stiffness’ and ‘density’, but so they could not find any trace of it. No one ever did, and so now we’ve finally accepted that light can actually travel in a vacuum, i.e. in plain nothing.

So what? Well… Let’s first look at the first point. Just like a sound wave, the motion of the source does not have any impact on the speed of light: it goes out in all directions at the same speed c, whether it is emitted from a fast-moving car or from some beacon near the sea. However, unlike sound waves, Maxwell’s equations imply that we cannot catch up with them. That’s troublesome, very troublesome, because, according to the above-mentioned Galilean transformation rules,

i.e. v’ = dx’/dt = dx/dt – u = v – u,   

some light beam that is traveling at speed vc past a spaceship that itself is traveling at speed u – let’s say u = 0.2c for example – should have a speed of c‘ = c – 0.2c = 0.8c = = 239,834 km/h only with respect to the spaceship. However, that’s not what Maxwell’s equations say when you substitute x, y, z and t for x‘, y‘, z‘ and t‘ using those four simple equations x’ = x – uty’ = yz’ = z and t’ = t. After you do the substitution, the transformed Maxwell equations will once again yield that c’ = c = 299,792 km/h, and not c’ = 0.8×299,792 km/h = 239,834 km/h.

That’s weird ! Why? Well… If you don’t think that this is weird, then you’re actually not thinking at all ! Just compare it with the example of our sound wave. There is just no logic to it !

The discovery startled all scientists because there could only be possible solutions to the paradox:

  1. Either Maxwell’s equations were wrong (because they did not observe the principle of (Galilean relativity) or, else,
  2. Newton’s equations (and the Galilean transformation rules – i.e. the Galilean relativity principle) are wrong.

Obviously, scientists and experimenters first tried to prove that Maxwell had it all wrong – if only because no experiment had ever shown Newton’s Laws to be wrong, and so it was probably hard – if not impossible – to try to come up with one that would ! So, instead, experimenters invented all kinds of wonderful apparatuses trying to show that the speed of the light was actually not absolute.

Basically, these experiments assumed that the speed of the Earth, as it rotates around the Sun at a speed of 108,000 km per hour, would result in measurable differences of c that would depend on the direction of the apparatus. More specifically, the speed of the light beam, as measured, would be different if the light beam would be traveling parallel to the motion of the Earth, as opposed to the light beam traveling at right angle to the motion of the Earth. Why? Well… It’s the same idea as the car chasing its own light beams, but I’ll refer to you to other descriptions of the experiment, because explaining these set-ups would take too much time and space. 🙂 I’ll just say that, because 108,000 km/h (on average) is only about 30 km per second (i.e. 0.0001 times c), these experiments relied on (expected) interference effects. The technical aspect of these experiments is really quite interesting. However, as mentioned above, I’ll refer you to Wikipedia or other sources if you’d want more detail.

Just note the most famous of those experiments: the 1887 Michelson-Morley experiment, also known as ‘the most famous failed experiment in history’ because, indeed, it failed to find any interference effects: the speed of light always was the speed of light, regardless of the direction of the beam with respect to the direction of motion of the Earth.

The Lorentz transformations

Once the scientists had recovered from this startling news (Michelson himself suffered from a nervous breakdown for a while, because he really wanted to find that interference effect in order to disprove Maxwell’s Laws), they suggested solutions.

The math was solved first. Indeed, just before the turn of the century, the Dutch physicist Hendrik Antoon Lorentz suggested that, if material bodies would contract in the direction of their motion with a factor (1 – u2/c2)1/2 and, in addition, if time would also be dilated with a factor (1 – u2/c2)–1/2, then the Michelson-Morley results could be explained. Of course, scientists objected to this ‘explanation’ as being very much ‘ad hoc’.

So then came Einstein. He just took the math for granted, so Einstein basically accepted the so-called Lorentz transformations that resulted from it, and corrected Newton’s Law in order to set physics right again.

And so that was it. As it turned out, all that was needed in fact, was to do away with the assumption that the inertia (or mass) of an object is a constant and, hence, that it does not vary with its velocity. For us, today, it seems obvious: mass also varies, and the factor involved is the very same Lorentz factor that we mentioned above: γ = (1 – u2/c2)–1/2. Hence, the m in Newton’s Second Law (F = d(mv)/dt) is not a constant but equal to m = γm0. For all speeds that we, human beings, can imagine (including the astronomical speed of the Earth in orbit around the Sun), the ‘correction’ is too small to be noticeable, or negligible, but so it’s there, as evidenced by the Michelson-Morley experiment, and, some hundred years later, we can actually verify it in particle accelerators.

As said, for us, today, it’s obvious (in my previous post, I mention a few examples: I explain how the mass of electrons in an electron beam is impacted by their speed, and how the lifetime of muon increases because of their speed) but one hundred years ago, it was not. Not at all – and so that’s why Einstein was a genius: he dared to explore and accept the non-obvious.

Now, what then are the correct transformations from one reference frame to another? They are referred to as the Lorentz transformations, and they can be written down (in a simplified form, assuming relative motion in the x direction only) as follows:

Capture

Now, I could point out many interesting implications, or come up with examples, but I will resist the temptation. I will only note two things about them:

1. These Lorentz transformations actually re-establish the principle of relativity: the Laws of Nature – including the Laws of Newton as corrected by Einstein’s relativistic mass formula – are of the same form in a moving system as in a stationary system, and therefore it is impossible to tell, by making experiments, whether the system is moving or not.

2. The second thing I should note is that the equations above imply that the idea of absolute time is no longer valid: there is no such thing as ‘absolute’ or ‘universal’ time. Indeed, Lorentz’ concept of ‘local time’ is a most profound departure from Newtonian mechanics that is implicit in these equations.

Indeed, space and time are entangled in these equations as you can see from the –ut and –ux/c2 terms in the equation for x’ and t’ respectively and, hence, the idea of simultaneity has to be abandoned: what happens simultaneously in two separated places according to one observer, does not happen at the same time as viewed by an observer moving with respect to the first. Let me quickly show how.

Suppose that in my world I see two events happening at the same time t0 but so they happen at two different places x1 and x2. Now, if you are movingaway from me at a (uniform) speed u, then equation (4) tells us that you will see these two events happen at two different times t1‘ and t2‘, with the time difference t1‘ – t2‘ equal to t1‘ – t2‘ = γ[u(x1 – x2)/c2], with γ the above-mentioned Lorentz factor. [Just do the calculation for yourself using equation 4.]

Of course, the effect is negligible for most speeds that we, as human beings, can imagine, but it’s there. So we do not have three separate space coordinates and one time coordinates, but four space-time coordinates that transform together, fully entangled, when applying those four equations above. 

That observation led the German mathematician Hermann Minkowski, who helped Einstein to develop his theory of four-dimensional space-time, to famously state that “Space of itself, and time of itself, will sink into mere shadows, and only a kind of union between them shall survive.”

Post scriptum: I did not elaborate on the second difficulty when I mentioned Maxwell’s equations: the lack of a need for a medium for light to travel through. I will let that rest for the moment (or, else, you can just Google some stuff on it). Just note that (1) it is kinda convenient that electromagnetic radiation does not need any medium (I can’t see how one would incorporate that in relativity theory) and (2) that light does seem to slow down in a medium. However, the explanation for that (i.e. for light to have an apparently lower speed in a medium) is to be found in quantum mechanics and so we won’t touch upon that complex matter here (for now that is). The point to note is that this slowing down is caused by light interacting with the matter it encounters as it travels through the medium. It does not actually go slower. However, I need to stop here as this is, yet again, a post which has become way too long. On the other hand, I am hopeful my kids will actually understand this one, because it does not involve integrals. 🙂

Another post for my kids: introducing (special) relativity

In my previous post, I talked about energy, and I tried to keep it simple – but also accurate. However, to be completely accurate, one must, of course, introduce relativity at some point. So how does that work? What’s ‘relativistic’ energy? Well… Let me try to convey a few ideas here.

The first thing to note is that the energy conservation law still holds: special theory or not, the sum of the kinetic and potential energies in a (closed) system is always equal to some constant C. What constant? That doesn’t matter: Nature does not care about our zero point and, hence, we can add or subtract any (other) constant to the equation K.E. + P.E. = T + U = C.

That being said, in my previous post, I pointed out that the constant depends on the reference point for the potential energy term U: we will usually take infinity as the reference point (for a force that attracts) and associate it with zero potential (U = 0). We then get a function U(x) like the one below: for gravitational energy we have U(x) = –GMm/x, and for electrical charges, we have U(x) = q1q2/4πε0x. The mathematical shape is exactly the same but, in the case of the electromagnetic forces, you have to remember that likes repel, and opposites attract, so we don’t need the minus sign: the sign of the charges takes care of it.

Capture

Minus sign? In case you wonder why we need that minus sign for the potential energy function, well… I explained that in my previous post and so I’ll be brief on that here: potential energy is measured by doing work against the force. That’s why. So we have an infinite sum (i.e. an integral) over some trajectory or path looking like this: U = – ∫F·ds.

For kinetic energy, we don’t need any minus sign: as an object picks up speed, it’s the force itself that is doing the work as its potential energy is converted into kinetic energy, so the change in kinetic energy will equal the change in potential energy, but with opposite sign: as the object loses potential energy, it gains kinetic energy. Hence, we write ΔT = –ΔU = ∫F·ds..

That’s all kids stuff obviously. Let’s go beyond this and ask some questions. First, why can we add or subtract any constant to the potential energy but not to the kinetic energy? The answer is… Well… We actually can add or subtract a ‘constant’ to the kinetic energy as well. Now you will shake your head: Huh? Didn’t we have that T = mv2/2 formula for kinetic energy? So how and why could one add or subtract some number to that?

Well… That’s where relativity comes into play. The velocity v depends on your reference frame. If another observer would move with and/or alongside the object, at the same speed, that observer would observe a velocity equal to zero and, hence, its kinetic energy – as that observer would measure it – would also be zero. You will object to that, saying that a change of reference frame does not change the force, and you’re right: the force will cause the object to accelerate or decelerate indeed, and if the observer is not subject to the same force, then he’ll see the object accelerate or decelerate indeed, regardless of his reference frame is a moving or inertial frame. Hence, both the inertial as well as the moving observer will see an increase (or decrease) in its kinetic energy and, therefore, both will conclude that its potential energy decreases (or increases) accordingly. In short, it’s the change in energy that matters, both for the potential as well as for the kinetic energy. The reference point itself, i.e. the point from where we start counting so to say, does not: that’s relative. [This also shows in the derivation for kinetic energy which I’ll do below.]

That brings us to the second question. We all learned in high school that mass and energy are related through Einstein’s mass-energy relation, E = mc2, which establishes an equivalence between the two: the mass of an object that’s picking up speed increases, and so we need to look at both speed and mass as a function of time. Indeed, remember Newton’s Law: force is the time rate of change of momentum: F = d(mv)/dt. When the speed is low (i.e. non-relativistic), then we can just treat m as a constant and write that  F = mdv/dt = ma (the mass times the acceleration). Treating m as a constant also allows us to derive the classical (Newtonian) formula for kinetic energy:

Capture

So if we assume that the velocity of the object at point O is equal to zero (so vo = 0), then ΔT will be equal to T and we get what we were looking for: the kinetic energy at point P will be equal to T = mv2/2.

Now, you may wonder why we can’t do that same derivation for a non-constant mass? The answer to that question is simple: taking the m factor out of the integral can only be done if we assume it is a constant. If not, then we should leave it inside. It’s similar to taking a derivative. If m would not be constant, then we would have to apply the product rule to calculate d(mv)/dt, so we’d write d(mv)/dt = (dm/dt)v + m(dv/dt). So we have two terms here and it’s only when m is constant that we can reduce it to d(mv)/dt = m(dv/dt).

So we have our classical kinetic energy function. However, when the velocity gets really high – i.e. if it’s like the same order of magnitude as the velocity of light – then we cannot assume that mass is constant. Indeed, the same high-school course in physics that taught you that E = mc2 equation will probably also have taught you that an object can never go faster than light, regardless of the reference frame. Hence, as the object goes faster and faster, it will pick up more momentum, but its rate of acceleration should (and will) go down in such way that the object can never actually reach the speed of light. Indeed, if Newton’s Law is to remain valid, we need to correct it such a way that m is no longer constant: m itself will increase as a function of its velocity and, hence, as a function of time. You’ll remember the formula for that:

Capture

This is often written as m = γm0, with m0 denoting the mass of the object at rest (in your reference frame that is) and γ = (1 – v2/c2)–1/2 the so-called Lorentz factor. The Lorentz factor is named after a Dutch physicist who introduced it near the end of the 19th century in order to explain why the speed of light is always c, regardless of the frame of reference (moving or not), or – in other words – why the speed of light is not relative. Indeed, while you’ll remember that there is no such thing as an absolute velocity according to the (special) theory of relativity, the velocity of light actually is absolute ! That means you will always see light traveling at speed c regardless of your reference frame. To put it simply, you can never catch up with light and, if you would be traveling away from some star in a spaceship with a velocity of 200,000 km per second, and a light beam from that star would pass you, you’d measure the speed of that light beam to be equal to 300,000 km/s, not 100,000 km/s. So is an absolute speed that acts as an absolute speed limit regardless of your reference frame. [Note that we’re talking only about reference frames moving at a uniform speed: when acceleration comes into play, then we need to refer to the general theory of relativity and that’s a somewhat different ball game.]

The graph below shows how γ varies as a function of v. As you can see, the mass increase only becomes significant at speeds of like 100,000 km per second indeed. Indeed, for v = 0.3c, the Lorentz factor is 1.048, so the increase is about 5% only. For v = 0.5c, it’s still limited to an increase of some 15%. But then it goes up rapidly: for v = 0.9c, the mass is more than twice the rest mass: m ≈ 2.3m0; for v = 0.99c, the mass increase is 600%: m ≈ 7m0; and so on. For v = 0.999c – so when the speed of the object differs from c only by 1 part in 1,000 – the mass of the object will be more than twenty-two times the rest mass (m ≈ 22.4m0).

Lorentz_factor

You probably know that we can actually reach such speeds and, hence, verify Einstein’s correction of Newton’s Law in particle accelerators: the electrons in an electron beam in a particle accelerator get usually pretty close to c and have a mass that’s like 2000 times their rest mass. How do we know that? Because the magnetic field needed to deflect them is like 2000 times as great as their (theoretical) rest mass. So how fast do they go? For their mass to be 2000 times m0, 1 – v2/c2 must be equal to 1/4,000,000. Hence, their velocity v differs from c only by one part in 8,000,000. You’ll have to admit that’s very close.

Other effects of relativistic speeds

So we mentioned the thing that’s best known about Einstein’s (special) theory of relativity: the mass of an object, as measured by the inertial observer, increases with its speed. Now, you may or may not be familiar with two other things that come out of relativity theory as well:

  1. The first is length contraction: objects are measured to be shortened in the direction of motion with respect to the (inertial) observer. The formula to be used incorporates the reciprocal of the Lorentz factor: L = (1/γ)L0. For example, a meter stick in a space ship moving at a velocity v = 0.6c will appear to be only 80 cm to the external/inertial observer seeing it whizz past… That is if he can see anything at all of course: he’d have to take like a photo-finish picture as it zooms past ! 🙂
  2. The second is time dilation, which is also rather well known – just like the mass increase effect – because of the so-called twin paradox: time will appear to be slower in that space ship and, hence, if you send one of two twins away on a space journey, traveling at such relativistic speed, he will come back younger than his brother. The formula here is a bit more complicated, but that’s only because we’re used to measure time in seconds. If we would take a more natural unit, i.e. the time it takes light to travel a distance of 1 m, then the formula will look the same as our mass formula: t = γt0 and, hence, one ‘second’ in the space ship will be measured as 1.25 ‘seconds’ by the external observer. Hence, the moving clock will appear to run slower – to the external (inertial) observer that is.

Again, the reality of this can be demonstrated. You’ll remember that we introduced the muon in previous posts: muons resemble electrons in the sense that they have the same charge, but their mass is more than 200 times the mass of an electron. As compared to other unstable particles, their average lifetime is quite long: 2.2 microseconds. Still, that would not be enough to travel more than 600 meters or so – even at the speed of light (2.2 μs × 300,000 km/s = 660 m). But so we do detect muons in detectors down here that come all the way down from the stratosphere, where they are created when cosmic rays hit the Earth’s atmosphere some 10 kilometers up. So how do they get here if they decay so fast? Well, those that actually end up in those detectors, do indeed travel very close to the speed of light and, hence, while from their own point of view they live only like two millionths of a second, they live considerably longer from our point of view.

Relativistic energy: E = mc2

Let’s go back to our main story line: relativistic energy. We wrote above that it’s the change of energy that matters really. So let’s look at that.

You may or may not remember that the concept of work in physics is closely related to the concept of power. In fact, you may actually remember that power, in physics at least, is defined as the work done per second. Indeed, we defined work as the (dot) product of the force and the distance. Now, when we’re talking a differential distance only (i.e. an infinitesimally small change only), then we can write dT = F·ds, but when we’re talking something larger, then we have to do that integral: ΔT = ∫F·ds. However, we’re interested in the time rate of change of T here, and so that’s the time derivative dT/dt which, as you easily verify, will be equal to dT/dt = (F·ds)/dt = F·(ds/dt) = F·and so we can use that differential formula and we don’t need the integral. Now, that (dot) product of the force and the velocity vectors is what’s referred to as the power. [Note that only the component of the force in the direction of motion contributes to the work done and, hence, to the power.]

OK. What am I getting at? Well… I just want to show an interesting derivation: if we assume, with Einstein, that mass and energy are equivalent and, hence, that the total energy of a body always equals E = mc2, then we can actually derive Einstein’s mass formula from that. How? Well… If the time rate of change of the energy of an object is equal to the power expended by the forces acting on it, then we can write:

dE/dt = d(mc2)/dt = F·v

Now, we cannot take the mass out of those brackets after the differential operator (d) because the mass is not a constant in this case (relativistic speeds) and, hence, dm/dt ≠ 0. However, we can take out c2 (that’s an absolute constant, remember?) and we can also substitute F using Newton’s Law (F = d(mv)/dt), again taking care to leave m between the brackets, not outside. So then we get:

d(mc2)/dt = c2dm/dt = [d(mv)/dt]·v = d(mv)/dt

In case you wonder why we can replace the vectors (bold face) v and d(mv) by their magnitudes (or lengths) v and d(mv): v and mv have the same direction and, hence, the angle θ between them is zero, and so v·v =v││v│cosθ =v2. Likewise, d(mv) and v also have the same direction and so we can just replace the dot product by the product of the magnitudes of those two vectors.

Now, let’s not forget the objective: we need to solve this equation for m and, hopefully, we’ll find Einstein’s mass formula, which we need to correct Newton’s Law. How do we do that? We’ll first multiply both sides by 2m. Why? Because we can then apply another mathematical trick, as shown below:

c2(2m)·dm/dt = 2md(mv)/dt ⇔ d(m2c2)/dt = d(m2v2)/dt

However, if the derivatives of two quantities are equal, then the quantities themselves can only differ by a constant, say C. So we integrate both sides and get:

m2c= m2v+ C

Be patient: we’re almost there. The above equation must be true for all velocities v and, hence, we can choose the special case where v = 0 and call this mass m0, and then substitute, so we get m0c= m00+ C = C. Now we put this particular value for C back in the more general equation above and we get:

mc= mv+ m0c⇔ m = mv2/c2 +m⇔ m(1 – v2/c2) = m⇔ m = m0/(1 – v2/c2)–1/2

So there we are: we have just shown that we get the relativistic mass formula (it’s on the right-hand side above) if we assume that Einstein’s mass-energy equivalence relation holds.

Now, you may wonder why that’s significant. Well… If you’re disappointed, then, at the very least, you’ll have to admit that it’s nice to show how everything is related to everything in this theory: from E = mc2, we get m0/(1 – v2/c2)–1/2. I think that’s kinda neat!

In addition, let us analyze that mass-energy relation in another way. It actually allows us to re-define kinetic energy as the excess of a particle over its rest mass energy, or – it’s the same expression really – or the difference between its total energy and its rest energy.

How does that work? Well… When we’re looking at high-speed or high-energy particles, we will write the kinetic energy as:

K.E. = mc– m0c= (m – m0)c= γm0c– m0c= m0c2(γ – 1). 

Now, we can expand that Lorentz factor γ = (1 – v2/c2)–1/2 into a binomial series (the binomial series is an infinite Taylor series, so it’s not to be confused with the (finite) binomial expansion: just check it online if you’re in doubt). If we do that, we we can write γ as an infinite sum of the following terms:

γ = 1 + (1/2)v2/c+ (3/8)v4/c+ (5/16)v6/c+ …

Now, when we plug this back into our (relativistic) kinetic energy equation, we can scrap a few things (just do it) to get where I wanted to get:

K.E. = (1/2)m0v+ (3/8)m0v4/c+ (5/16)m0v6/c+ …

Again, you’ll wonder: so what? Well… See how the non-relativistic formula for kinetic energy (K.E. = m0v2/2) appears here as the first term of this series and, hence, how the formula above shows that our ‘Newtonian’ formula is just an approximation. Of course, at low speeds, the second, third etcetera terms represent close to nothing and, hence, then our Newtonian ‘approximation is obviously pretty good of course !

OK… But… Now you’ll say: that’s fine, but how did Einstein get inspired to write E = mc2 in the first place? Well, truth be told, the relativistic mass formula was derived first (i.e. before Einstein wrote his E = mc2 equation), out of a derivation involving the momentum conservation law and the formulas we must use to convert the space-time coordinates from one reference frame to another when looking at phenomena (i.e. the so-called Lorentz transformations). And it was only afterwards that Einstein noted that, when expanding the relativistic mass formula, that the increase in mass of a body appeared to be equal to the increase in kinetic energy divided by c2 (Δm = Δ(K.E.)/c2). Now, that, in turn, inspired him to also assign an equivalent energy to the rest mass of that body: E0 = m0c2. […] At least that’s how Feynman tells the story in his 1965 Lectures… But so we’ve actually been doing it the other way around here!

Hmm… You will probably find all of this rather strange, and you may also wonder what happened to our potential energy. Indeed, that concept sort of ‘disappeared’ in this story: from the story above, it’s clear that kinetic energy has an equivalent mass, but what about potential energy?

That’s a very interesting question but, unfortunately, I can only give a rather rudimentary answer to that. Let’s suppose that we have two masses M and m. According to the potential energy formula above, the potential energy U between these two masses will then be equal to U = –GMm/r. Now, that energy is not interpreted as energy of either M or m, but as energy that is part of the (M, m) system, which includes the system’s gravitational field. So that energy is considered to be stored in that gravitational field. If the two masses would sit right on top of each other, then there would be no potential energy in the (M, m) system and, hence, the system as a whole would have less energy. In contrast, when we separate them further apart, then we increase the energy of the system as a whole, and so the system’s gravitational field then increases. So, yes, the potential energy does impact the (equivalent) mass of the system, but not the individual masses M and m. Does that make sense?

For me , it does, but I guess you’re a bit tired by now and, hence, I think I should wrap up here. In my next (and probably last) post on relativity, I’ll present those Lorentz transformations that allow us to ‘translate’ the space and time coordinates from one reference frame to another, and in that post I’ll also present the other derivation of Einstein’s relativistic mass formula, which is actually based on those transformations. In fact, I realize I should have probably started with that (as mentioned above, that’s how Feynman does it in his Lectures) but, then, for some reason, I find the presentation above more interesting, and so that’s why I am telling the story starting from another angle. I hope you don’t mind. In any case, it should be the same, because everything is related to everything in physics – just like in math. That’s why it’s important to have a good teacher. 🙂

A post for my kids: on energy and potential

We’ve been juggling with a lot of advanced concepts in the previous post. Perhaps it’s time I write something that my kids can understand too. One of the things I struggled with when re-learning elementary physics is the concept of energy. What is energy really? I always felt my high school teachers did a poor job in trying to explain it. So let me try to do a better job here.

A high-school level course usually introduces the topic using the gravitational force, i.e. Newton’s Third Law: F = GmM/r2. This law states that the force of attraction is proportional to the product of the masses m and M, and inversely proportional to the square of the distance r between those two masses. The factor of proportionality is equal to G, i.e. the so-called universal gravitational constant, aka the ‘big G’ (G ≈ 6.674×10-11 N(m/kg)2), as opposed to the ‘little g’, which is the gravity of Earth (g ≈ 9.80665 m/s2). As far as I am concerned, it is at this point where my high-school teacher failed.

Indeed, he would just go on and simplify Newton’s Third Law by writing F = mg, noting that g = GM/rand that, for all practical purposes, this g factor is constant, because we are talking small distances as compared to the radius of the Earth. Hence, we should just remember that the gravitational force is proportional to the mass only, and that one kilogram amounts to a weight of about 10 newton (9.80665 kg·m/s2 (N) to be precise). That simplification would then be followed by another simplification: if we are lifting an object with mass m, we are doing work against the gravitational force. How much work? Well, he’d say, work is – quite simply – the force times the distance in physics, and the work done against the force is the potential energy (usually denoted by U) of that object. So he would write U = Fh = mgh, with h the height of the object (as measured from the surface of the Earth), and he would draw a nice linear graph like the one below (I set m to 10 kg here, and h ranges from 0 to 100 m).

Potential energy uniform gravitation field

Note that the slope of this line is slightly less than 45 degrees (and also note, of course, that it’s only approximately 45 degrees because of our choice of scale: dU/dh is equal to 98.0665, so if the x and y axes would have the same scale, we’d have a line that’s almost vertical).

So what’s wrong with this graph? Nothing. It’s just that this graph sort of got stuck in my head, and it complicated a more accurate understanding of energy. Indeed, with examples like the one above, one tends to forget that:

  1. Such linear graphs are an approximation only. In reality, the gravitational field, and force fields in general, are not uniform and, hence, g is not a constant: the graph below shows how g varies with the height (but the height is expressed in kilometer this time, not in meter).
  2. Not only is potential energy usually not a linear function but – equally important – it is usually not a positive real number either. In fact, in physics, U will usually take on a negative value. Why? Because we’re indeed measuring and defining it by the work done against the force.

Erdgvarp

So what’s the more accurate view of things? Well… Let’s start by noting that potential energy is defined in relation to some reference point and, taking a more universal point of view, that reference point will usually be infinity when discussing the gravitational (or electromagnetic) force of attraction. Now, the potential energy of the point(s) at infinity – i.e. the reference point – will, usually, be equated with zero. Hence, the potential energy curve will then take the shape of the graph below (y = –1/x), so U will vary from zero (0) to minus infinity (–∞) , as we bring the two masses closer together. You can readily see that the graph below makes sense: its slope is positive and, hence, as such it does capture the same idea as that linear mgh graph above: moving a mass from point 1 to point 2 requires work and, hence, the potential energy at point 2 is higher than at point 1, even if both values U(2) and U(1) are negative numbers, unlike the values of that linear mgh curve.

Capture

How do you get a curve like that? Well… I should first note another convention which is essential for making the sign come out alright: if the force is gravity, then we should write F = –GmMr/r3. So we have a minus sign here. And please do note the boldface type: F and r are vectors, and vectors have both a direction and magnitude – and so that’s why they are denoted by a bold letter (r), as opposed to the scalar quantities G, m, M or r).

Back to the minus sign. Why do we have that here? Well… It has to do with the direction of the force, which, in case of attraction, will be opposite to the so-called radius vector r. Just look at the illustration below, which shows, first, the direction of the force between two opposite electric charges (top) and then (bottom), the force between two masses, let’s say the Earth and the Moon.

Force and radius vector

So it’s a matter of convention really.

Now, when we’re talking the electromagnetic force, you know that likes repel and opposites attract, so two charges with the same sign will repel each other, and two charges with opposite sign will attract each other. So F12, i.e. the force on q2 because of the presence of q1, will be equal to F12 = q1q2r/r3. Therefore, no minus sign is needed here because qand q2 are opposite and, hence, the sign of this product will be negative. Therefore, we know that the direction of F comes out alright: it’s opposite to the direction of the radius vector r. So the force on a charge q2 which is placed in an electric field produced by a charge q1 is equal to F12 = q1q2r/r3. In short, no minus sign needed here because we already have one. Of course, the original charge q1 will be subject to the very same force and so we should write F21 = –q1q2r/r3. So we’ve got that minus sign again now. In general, however, we’ll write Fij = qiqjr/r3 when dealing with the electromagnetic force, so that’s without a minus sign, because the convention is to draw the radius vector from charge i to charge j and, hence, the radius vector r in the formula F21 would point in the other direction and, hence, the minus sign is not needed.

In short, because of the way that the electromagnetic force works, the sign always come out right: there is no need for a minus sign in front. However, for gravity, there are no opposite charges: masses are always alike, and so likes actually attract when we’re talking gravity, and so that’s why we need the minus sign when dealing with the gravitational force: the force between a mass i and another mass j will always be written as Fij = –mimjr/r3, so here we do have to put the minus sign, because the direction of the force needs to be opposite to the direction of the radius vector and so the sign of the ‘charges’ (i.e. the masses in this case), in the case of gravity, does not take care of that.

One last remark here may be useful: always watch out to not double-count forces when considering a system with many charges or many masses: both charges (or masses) feel the same force, but with opposite direction. OK. Let’s move on. If you are confused, don’t worry. Just remember that (1) it’s very important to be consistent when drawing that radius vector (it goes from the charge (or mass) causing the force field to the other charge (or mass) that is being brought in), and (2) that the gravitational and electromagnetic forces have a lot in common in terms of ‘geometry’ – notably that inverse proportionality relation with the square of the distance between the two charges or masses – but that we need to put a minus sign when we’re dealing with the gravitational force because, with gravitation, likes do not repel but attract each other, as opposed to electric charges.

Now, let’s move on indeed and get back to our discussion of potential energy. Let me copy that potential energy curve again and let’s assume we’re talking electromagnetics here, and that we’re have two opposite charges, so the force is one of attraction.

Capture

Hence, if we move one charge away from the other, we are doing work against the force. Conversely, if we bring them closer to each other, we’re working with the force and, hence, its potential energy will go down – from zero (i.e. the reference point) to… Well… Some negative value. How much work is being done? Well… The force changes all the time, so it’s not constant and so we cannot just calculate the force times the distance (Fs). We need to do one of those infinite sums, i.e. an integral, and so, for point 1 in the graph above, we can write:

Capture

Why the minus sign? Well… As said, we’re not increasing potential energy: we’re decreasing it, from zero to some negative value. If we’d move the charge from point 1 to the reference point (infinity), then we’d be doing work against the force and we’d be increasing potential energy. So then we’d have a positive value. If this is difficult, just think it through for a while and you’ll get there.

Now, this integral is somewhat special because F and s are vectors, and the F·ds product above is a so-called dot product between two vectors. The integral itself is a so-called path integral and so you may not have learned how to solve this one. But let me explain the dot product at least: the dot product of two vectors is the product of the magnitudes of those two vectors (i.e. their length) times the cosine of the angle between the two vectors:

F·d=│F││ds│cosθ

Why that cosine? Well… To go from one point to another (from point 0 to point 1, for example), we can take any path really. [In fact, it is actually not so obvious that all paths will yield the same value for the potential energy: it is the case for so-called conservative forces only. But so gravity and the electromagnetic force are conservative forces and so, yes, we can take any path and we will find the same value.] Now, if the direction of the force and the direction of the displacement are the same, then that angle θ will be equal to zero and, hence, the dot product is just the product of the magnitudes (cos(0) = 1). However, if the direction of the force and the direction of the displacement are not the same, then it’s only the component of the force in the direction of the displacement that’s doing work, and the magnitude of that component is Fcosθ. So there you are: that explains why we need that cosine function.

Now, solving that ‘special’ integral is not so easy because the distance between the two charges at point 0 is zero and, hence, when we try to solve the integral by putting in the formula for F and finding the primitive and all that, you’ll find there’s a division by zero involved. Of course, there’s a way to solve the integral, but I won’t do it here. Just accept the general result here for U(r):

U(r) = q1q2/4πε0r

You can immediately see that, because we’re dealing with opposite charges, U(r) will always be negative, while the limit of this function for r going to infinity is equal to zero indeed. Conversely, its limit equals –∞ for r going to zero. As for the 4πεfactor in this formula, that factor plays the same role as the G-factor for gravity. Indeed, εis an ubiquitous electric constant: ε≈ 8.854×10-12 F/m, but it can be included in the value of the charges by choosing another unit and, hence, it’s often omitted – and that’s what I’ll also do here. Now, the same formula obviously applies to point 2 in the graph as well, and so now we can calculate the difference in potential energy between point 1 and point 2:

Capture

Does that make sense? Yes. We’re, once again, doing work against the force when moving the charge from point  1 to point 2. So that’s why we have a minus sign in front. As for the signs of qand q2, remember these are opposite. As for the value of the (r2 – r1) factor, that’s obviously positive because  r2 > r1. Hence, ΔU = U(1) – U(2) is negative. How do we interpret that? U(2) and U(1) are negative values, the difference between those two values, i.e. U(1) – U(2), is negative as well? Well… Just remember that ΔU is minus the work done to move the charge from point 1 to point 2. Hence, the change in potential energy (ΔU) is some negative value because the amount of work that needs to be done to move the charge from point 1 to point 2 is decidedly positive. Hence, yes, the charge has a higher energy level (albeit negative – but that’s just because of our convention which equates potential energy at infinity with zero) at point 2 as compared to point 1.

What about gravity? Well… That linear graph above is an approximation, we said, and it also takes r = h = 0 as the reference point but it assigns a value of zero for the potential energy there (as opposed to the –∞ value for the electromagnetic force above). So that graph is actually an linearization of a graph resembling the one below: we only start counting when we are on the Earth’s surface, so to say.

Capture

However, in a more advanced physics course, you will probably see the following potential energy function for gravity: U(r) = –GMm/r, and the graph of this function looks exactly the same as that graph we found for the potential energy between two opposite charges: the curve starts at point (0, –∞) and ends at point (∞, 0).

OK. Time to move on to another illustration or application: the covalent bond between two hydrogen atoms.

Application: the covalent bond between two hydrogen atoms

The graph below shows the potential energy as a function of the distance between two hydrogen atoms. Don’t worry about its exact mathematical shape: just try to understand it.

potential-energy-curve-H2-moleculecovalent_bond_hydrogen_3

 

 

Natural hydrogen comes in Hmolecules, so there is a bond between two hydrogen atoms as a result of mutual attraction. The force involved is a chemical bond: the two hydrogen atoms share their so-called valence electron, thereby forming a so-called covalent bond (which is a form of chemical bond indeed, as you should remember from your high-school courses). However, one cannot push two hydrogen atoms too close, because then the positively charged nuclei will start repelling each other, and so that’s what is depicted above: the potential energy goes up very rapidly because the two atoms will repel each other very strongly.

The right half of the graph shows how the force of attraction vanishes as the two atoms are separated. After a while, the potential energy does not increase any more and so then the two atoms are free.

Again, the reference point does not matter very much: in the graph above, the potential energy is assumed to be zero at infinity (i.e. the ‘free’ state) but we could have chosen another reference point: it would only shift the graph up or down. 

This brings us to another point: the law of energy conservation. For that, we need to introduce the concept of kinetic energy once again.

The formula for kinetic energy

In one of my previous posts, I defined the kinetic energy of an object as the excess energy over its rest energy:

K.E. = T = mc– m0c= γm0c– m0c= (γ–1)m0c2

γ is the Lorentz factor in this formula (γ = (1–v2/c2)-1/2), and I derived the T = mv2/2 formula for the kinetic energy from a Taylor expansion of the formula above, noting that K.E. = mv2/2 is actually an approximation for non-relativistic speeds only, i.e. speeds that are much less than c and, hence, have no impact on the mass of the object: so, non-relativistic means that, for all practical purposes, m = m0. Now, if m = m0, then mc– m0c2 is equal to zero ! So how do we derive the kinetic energy formula for non-relativistic speeds then? Well… We must apply another method, using Newton’s Law: the force equals the time rate of change of the momentum of an object. The momentum of an object is denoted by p (it’s a vector quantity) and is the product of its mass and its velocity (p = mv), so we can write

F = d(mv)/dt (again, all bold letters denote vectors).

When the speed is low (i.e. non-relativistic), then we can just treat m as a constant and so we can write F = mdv/dt = ma (the mass times the acceleration). If m would not be constant, then we would have to apply the product rule: d(mv) = (dm/dt)v + m(dv/dt), and so then we would have two terms instead of one. Treating m as a constant also allows us to derive the classical (Newtonian) formula for kinetic energy:

Capture

So if we assume that the velocity of the object at point O is equal to zero (so vo = 0), then ΔT will be equal to T and we get what we were looking for: the kinetic energy at point P will be equal to T = mv2/2.

Energy conservation

Now, the total energy – potential and kinetic – of an object (or a system) has to remain constant, so we have E = T + U = constant. As a consequence, the time derivative of the total energy must equal zero. So we have:

E = T + U = constant, and dE/dt = 0

Can we prove that with the formulas T = mv2/2 and U = q1q2/4πε0r? Yes, but the proof is a bit lengthy and so I won’t prove it here. [We need to take the derivatives ∂T/∂t and ∂U/∂t and show that these derivatives are equal except for the sign, which is opposite, and so the sum of those two derivatives equals zero. Note that ∂T/∂t = (dT/dv)(dv/dt) and that ∂U/∂t = (dU/dr)(dr/dt), so you have to use the chain rule for derivatives here.] So just take a mental note of that and accept the result:

(1) mv2/2 + q1q2/4πε0= constant when the electromagnetic force is involved (no minus sign, because the sign of the charges makes things come out alright), and
(2) mv2/2 – GMm/= constant when the gravitational force is involved (note the minus sign, for the reason mentioned above: when the gravitational force is involved, we need to reverse the sign).

We can also take another example: an oscillating spring. When you try to compress a (linear) spring, the spring will push back with a force equal to F = kx. Hence, the energy needed to compress a (linear) spring a distance x from its equilibrium position can be calculated from the same integral/infinite sum formula: you will get U = kx2/2 as a result. Indeed, this is an easy integral (not a path integral), and so let me quickly solve it:

Capture

While that U = kx2/2 formula looks similar to the kinetic energy formula, you should note that it’s a function of the position, not of velocity, and that the formula does not involve the mass of the object we’re attaching to the string. So it’s a different animal altogether. However, because of the energy conservation law, the graph of both the potential and kinetic energy will obviously reflect each other, just like the energy graphs of a swinging pendulum, as shown below. We have:

T + U = mv2/2 + kx2/2 = C

energy - pendulum - 2 energy - pendulum

Note: The graph above mentions an ‘ideal’ pendulum because, in reality, there will be an energy loss due to friction and, hence, the pendulum will slowly stop, as shown below. Hence, in reality, energy is conserved, but it leaks out of the system we are observing here: it gets lost as heat, which is another form of kinetic energy actually.

energy - pendulum - 3

Another application: estimating the radius of an atom

very nice application of the energy concepts introduced above is the so-called Bohr model of a hydrogen atom. Feynman introduces that model as an estimate of the size (or radius) of an atom (see Feynman’s Lectures, Vol. III, p. 2-6). The argument is the following.

The radius of an atom is more or less the spread (usually denoted by Δ or σ) in the position of the electron, so we can write that Δx = a. In words, the uncertainty about the position is the radius a. Now, we know that the uncertainty about the position (x) also determines the uncertainty about the momentum (p = mv) of the electron because of the Uncertainty Principle ΔxΔp ≥ ħ/2 (ħ ≈ 6.6×10-16 eV·s). The principle is illustrated below, and in a previous posts I proved the relationship. [Note that k in the left graph actually represents the wave number of the de Broglie wave, but wave number and momentum are related through the de Broglie relation p = ħk.]

example of wave packet

Hence, the order of magnitude of the momentum of the electron will – very roughly – be p ≈ ħ/a. [Note that Feynman doesn’t care about factors 2 or π or even 2π (h = 2πħ): the idea is just to get the order of magnitude (Feynman calls it a ‘dimensional analysis’), and that he actually equates p with p = h/a, so he doesn’t use the reduced Planck constant (ħ).]

Now, the electron’s potential energy will be given by that U(r) = q1q2/4πε0formula above, with q1= e (the charge of the proton) and q2= –e (i.e. the charge of the electron), so we can simplify this to –e2/a. 

The kinetic energy of the electron is given by the usual formula: T = mv2/2. This can be written as T = mv2/2 = m2v2/2m = p2/2m =  h2/2ma2. Hence, the total energy of the electron is given by

E = T + U = h2/2ma– e2/a

What does this say? It says that the potential energy becomes smaller as a gets smaller (that’s because of the minus sign: when we say ‘smaller’, we actually mean a larger negative value). However, as it gets closer to the nucleus, it kinetic energy increases. In fact, the shape of this function is similar to that graph depicting the potential energy of a covalent bond as a function of the distance, but you should note that the blue graph below is the total energy (so it’s not only potential energy but kinetic energy as well).

CaptureI guess you can now anticipate the rest of the story. The electron will be there where its total energy is minimized. Why? Well… We could call it the minimum energy principle, but that’s usually used in another context (thermodynamics). Let me just quote Feynman here, because I don’t have a better explanation: “We do not know what a is, but we know that the atom is going to arrange itself to make some kind of compromise so that the energy is as little as possible.”

He then calculates, as expected, the derivative dE/da, which equals dE/da = –h2/ma3 e2/a2. Setting dE/da equal to zero, we get the ‘optimal’ value for a: 

ah2/me=0.528×10-10 m = 0.528 Å (angstrom)

Note that this calculation depends on the value one uses for e: to be correct, we need to put the 4πε0 factor back in. You also need to ensure you use proper and compatible units for all factors. Just try a couple of times and you should find that 0.528 value.

Of course, the question is whether or not this back-of-the-envelope calculation resembles anything real? It does: this number is very close to the so-called Bohr radius, which is the most probable distance between the proton and and the electron in a hydrogen atom (in its ground state) indeed. The Bohr radius is an actual physical constant and has been measured to be about 0.529 angstrom. Hence, for all practical purposes, the above calculation corresponds with reality. [Of course, while Feynman started with writing that we shouldn’t trust our answer within factors like 2, π, etcetera, he concludes his calculation by noting that he used all constants in such a way that it happens to come out the right number. :-)]

The corresponding energy for this value for a can be found by putting the value aback into the total energy equation, and then we find:

E= –me4/2h= –13.6 eV

Again, this corresponds to reality, because this is the energy that is needed to kick an electron out of its orbit or, to use proper language, this is the energy that is needed to ionize a hydrogen atom (it’s referred to as a Rydberg of energy). By way of conclusion, let me quote Feynman on what this negative energy actually means: “[Negative energy] means that the electron has less energy when it is in the atom than when it is free. It means it is bound. It means it takes energy to kick the electron out.”

That being said, as we pointed out above, it is all a matter of choosing our reference point: we can add or subtract any constant C to the energy equation: E + C = T + U + C will still be constant and, hence, respect the energy conservation law. But so I’ll conclude here and – of course – check if my kids understand any of this.

And what about potential?

Oh – yes. I forgot. The title of this post suggests that I would also write something on what is referred to as ‘potential’, and it’s not the same as potential energy. So let me quickly do that.

By now, you are surely familiar with the idea of a force field. If we put a charge or a mass somewhere, then it will create a condition such that another charge or mass will feel a force. That ‘condition’ is referred to as the field, and one represents a field by field vectors. For a gravitational field, we can write:

F = mC

C is the field vector, and F is the force on the mass that we would ‘supply’ to the field for it to act on. Now, we can obviously re-write that integral for the potential energy as

U = –∫F·ds = –m∫C·ds = mΨ with Ψ (read: psi) = ∫C·d= the potential

So we can say that the potential Ψ is the potential energy of a unit charge or a unit mass that would be placed in the field. Both C (a vector) as well Ψ (a scalar quantity, i.e. a real number) obviously vary in space and in time and, hence, are a function of the space coordinates x, y and z as well as the time coordinate t. However, let’s leave time out for the moment, in order to not make things too complex. [And, of course, I should not say that this psi has nothing to do with the probability wave function we introduced in previous posts. Nothing at all. It just happens to be the same symbol.]

Now, U is an integral, and so it can be shown that, if we know the potential energy, we also know the force. Indeed, the x-, y and z-component of the force is equal to:

F= – ∂U/∂x, F= – ∂U/∂y, F= – ∂U/∂z or, using the grad (gradient) operator: F = –∇U  

Likewise, we can recover the field vectors C from the potential function Ψ:

C= – ∂Ψ/∂x, C= – ∂Ψ/∂y, C= – ∂Ψ/∂z, or C = –∇Ψ

That grad operator is nice: it makes a vector function out of a scalar function.

In the ‘electrical case’, we will write:

F = qE

 And, likewise,

U = –∫F·ds = –q∫E·ds = qΦ with Φ (read: phi) = ∫E·d= the electrical potential.

Unlike the ‘psi’ potential, the ‘phi’ potential is well known to us, if only because it’s expressed in volts. In fact, when we say that a battery or a capacitor is charged to a certain voltage, we actually mean the voltage difference between the parallel plates of which the capacitor or battery consists, so we are actually talking the difference in electrical potential ΔΦ = Φ– Φ2., which we also express in volts, just like the electrical potential itself.

Post scriptum:

The model of the atom that is implied in the above derivation is referred to as the so-called Bohr model. It is a rather primitive model (Wikipedia calls it a ‘first-order approximation’) but, despite its limitations, it’s a proper quantum-mechanical view of the hydrogen atom and, hence, Wikipedia notes that “it is still commonly taught to introduce students to quantum mechanics.” Indeed, that’s Feynman also uses it in one of his first Lectures on Quantum Mechanics (Vol. III, Chapter 2), before he moves on to more complex things.

Time reversal and CPT symmetry (III)

Although I concluded my previous post by saying that I would not write anything more about CPT symmetry, I feel like I have done an injustice to Val Fitch, James Cronin, and all those other researchers who spent many man-years to painstakingly demonstrate how the weak force does not always respect the combined charge-parity (C-P) symmetry. Indeed, I did not want to denigrate their efforts when I noted that:

  1. These decaying kaons (i.e. the particles that are used to demonstrate the CP symmetry-breaking phenomenon) are rather exotic and very short-lived particles; and
  2. Researchers have not been able to find many other traces of non-respect of CP symmetry, except when studying a heavier version of these kaons (the so-called B- and D-mesons) as soon as these could be produced in higher volumes in newer (read: higher-energy) particle colliders (so that’s in the last ten or fifteen years only), but so these B- and D-mesons are even more rare and even less stable.

CP violation is CP violation: it’s plain weird, especially when Fermilab and CERN experiments observed direct CP violation in kaon decay processes. [Remember that the original 1964 Fitch-Cronin experiment could not directly observe CP violation: in their experiment, CP violation in neutral kaon decay processes could only be deduced from other (unexpected) decay processes.]

Why? When one reverses all of the charges and other variables (such as parity which – let me remind you – has to do with ‘left-handedness’ and ‘right-handedness’ of particles), then the process should go in the other direction in an exactly symmetric way. Full stop. If not, there’s some kind of ‘leakage’ so to say, and such ‘leakage’ would be ‘kind-of-OK’ when we’d be talking some kind of chemical or biological process, but it’s obviously not ‘kind-of-OK’ when we’re talking one of the fundamental forces. It’s just not ‘logical’.

Feynman versus ‘t Hooft: pro and contra CP-symmetry breaking

A remark that is much more relevant than the two comments above is that one of the most brilliant physicists of the 20th century, Richard Feynman, seemed to have refused to entertain the idea of CP-symmetry breaking. Indeed, while, in his 1965 Lectures, he devotes quite a bit of attention to Chien-Shiung Wu’s 1956 experiment with decaying cobalt-60 nuclei (i.e. the experiment which first demonstrated parity violation, i.e. the breaking of P-symmetry), he does not mention the 1964 Fitch-Cronin experiment, and all of his writing in these Lectures makes it very clear that he not only strongly believes that the combined CP symmetry holds, but that it’s also the only ‘symmetry’ that matters really, and the only one that Nature truly respects–always.

So Feynman was wrong. Of course, these Lectures were published less than a year after the 1964 Fitch-Cronin experiment and, hence, you might think he would have changed his ideas on the possibility of Nature not respecting CP-symmetry–just like Wolfgang Pauli, who could only accept the reality of Nature not respecting reflection symmetry (P-symmetry) after repeated experiments re-confirmed the results of Wu’s original 1956 experiment.

But – No! – Feynman’s 1985 book on quantum electrodynamics (QED) –so that’s five years after Fitch and Cronin got a Nobel Prize for their discovery– is equally skeptical on this point: he basically states that the weak force is “not well understood” and that he hopes that “a more beautiful and, hence, more accurate understanding” of things will emerge.

OK, you will say, but Feynman passed away shortly after (he died from a rare form of cancer in 1988) and, hence, we should now listen to the current generation of physicists.

You’re obviously right, so let’s look around. Hmm… Gerard ‘t Hooft? Yes ! He is 67 now but – despite his age – it is obvious that he surely qualifies as a ‘next-generation’ physicist. He got his Nobel Prize for “elucidating the quantum structure of electroweak interactions” (read: for clarifying how the weak force actually works) and he is also very enthusiastic about all these Grand Unified Theories (most notably string and superstring theory) and so, yes, he should surely know, shouldn’t he?

I guess so. However, even ‘t Hooft writes that these experiments with these ‘crazy kaons’ – as he calls them – show ‘violation’ indeed, but that it’s marginal: the very same experiments also show near-symmetry. What’s near-symmetry? Well… Just what the term says: the weak force is almost symmetrical. Hence, CP-symmetry is the norm and CP-asymmetry is only a marginal phenomenon. That being said, it’s there and, hence, it should be explained. How?

‘t Hooft himself writes that one could actually try to interpret the results of the experiment by adding some kind of ‘fifth’ force to our world view – a “super-weak force” as he calls it, which would interfere with the weak force only.

To be fair, he immediately adds that introducing such ‘fifth force’ doesn’t really solve the “mystery” of CP asymmetry, because, while we’d restore the principle of CP symmetry for the weak force interactions, we would then have to explain why this ‘super-weak’ force does not respect it. In short, we cannot just reason the problem away. Hence, ‘t Hooft’s conclusion in his 1996 book on The Ultimate Building Blocks of the universe is quite humble: “The deeper cause [of CP asymmetry] is likely to remain a mystery.” (‘t Hooft, 1996, Chapter 7: The crazy kaons)

What about other explanations? For example, you might be tempted to think these two or three exceptions to a thousand cases respecting the general rule must have something to do with quantum-mechanical uncertainty: when everything is said and done, we’re dealing with probabilities in quantum mechanics, aren’t we? Hence, exceptions do occur and are actually expected to occur.

No. Quantum indeterminism is not applicable here. While working with probability amplitudes and probabilities is effectively equivalent to stating some general rules involving some average or mean value and then some standard deviation from that average, we’ve got something else going on here: Fitch and Cronin took a full six months indeed–repeating the experiment over and over and over again–to firmly establish a statistically significant bias away from the theoretical average. Hence, even if the bias is only 0.2% or 0.3%, it is a statistically significant difference between the probability of a process going one way, and the probability of that very same process going the other way.

So what? There are so many non-reversible processes and asymmetries in this world: why don’t we just accept this?Well… I’ll just refer to my previous post on this one: we’re talking a fundamental force here – not some chemical reaction – and, hence, if we reverse all of the relevant charges (including things such as left-handed or right-handed spin), the reaction should go the other way, and with exactly the same probability. If it doesn’t, it’s plain weird. Full stop.

OK. […] But… Perhaps there is some external phenomenon affecting these likelihoods, like these omnipresent solar neutrinos indeed, which I mentioned in a previous post and which are all left-handed. So perhaps we should allow these to enter the equation as well. […] Well… I already said that would make sense–to some extent at least– because there is some flimsy evidence of solar flares affecting radioactive decay rates (solar flares and neutrino outbursts are closely related, so if solar flares impact radioactive decay, we could or should expect them to meddle with any beta decay process really). That being said, it would not make sense from other, more conventional, points of view: we cannot just ‘add’ neutrinos to the equation because then we’d be in trouble with the conservation laws, first and foremost the energy conservation law! So, even if we would be able to work out some kind of theoretical mechanism involving these left-handed solar neutrinos (which are literally all over the place, bombarding us constantly even if they’re very hard to detect), thus explaining the observed P-asymmetry, we would then have to explain why it violates the energy conservation law! Well… Good luck with that, I’d say!

So it is a conundrum really. Let me sum up the above discussion in two bullet points:

  1. While kaons are short-lived particles because of the presence of the second-generation (and, hence, unstable) s-quark, they are real particles (so they are not some resonance or some so-called virtual particle). Hence, studying their behavior in interactions with any force field (and, most notably, their behavior in regard to the weak force) is extremely relevant, and the observed CP asymmetry–no matter how small–is something which should really grab our attention.
  2. The philosophical implications of any form of non-respect of the combined CP symmetry for our common-sense notion of time are truly profound and, therefore, the Fitch-Cronin experiment rightly deserves a lot of accolades.

So let’s analyze these ‘philosophical implications’ (which is just a somewhat ‘charged’ term for the linkage between CP- and time-symmetry which I want to discuss here) somewhat more in detail.

Time reversal and CPT symmetry

In the previous posts, I said it’s probably useful to distinguish (a) time-reversal as a (loosely defined) philosophical concept from (b) the mathematical definition of time-reversal, which is much more precise and unambiguous. It’s the latter which is generally used in physics, and it amounts to putting a minus sign in front of all time variables in any equation describing some situation, process or system in physics. That’s it really. Nothing more.

The point that I wanted to make is that true time reversal – i.e. time-reversal in the ‘philosophical’ or ‘common-sense’ interpretation – also involves a reversal of the forces, and that’s done through reversing all charges causing those forces. I used the example of the movie as a metaphor: most movies, when played backwards, do not make sense, unless we reverse the forces. For example, seeing an object ‘fall back’ to where it was (before it started falling) in a movie playing backwards makes sense only if we would assume that masses repel, instead of attract, each other. Likewise, any static or dynamic electromagnetic phenomena we would see in that backwards playing movie would make sense only if we would assume that the charges of the protons and electrons causing the electromagnetic fields involved would be reversed. How? Well… I don’t know. Just imagine some magic.

In such world view–i.e. a world view which connects the arrow of time with real-life forces that cause our world to change– I also looked at the left- and right-handedness of particles as some kind of ‘charge’, because it co-determines how the weak force plays out. Hence, any phenomenon in the movie having to do with the weak force (such as beta decay) could also be time-reversed by making left-handed particles right-handed, and right-handed particles left-handed. In short, I said that, when it comes to time reversal, only a full CPT-transformation makes sense–from a philosophical point of view that is.

Now, reversing left- and right-handedness amounts to a P-transformation (and don’t interrupt me now by asking why physicists use this rather awkward word ‘parity’ for what’s left- and right-handedness really), just like a C-transformation amounts to reversing electric and ‘color’ charges (‘color’ charges are the charges involved in the strong nuclear force).

Now, if only a full CPT transformation makes sense, then CP-reversal should also mean T-reversal, and vice versa. Feynman’s story about “the guy in the ‘other’ universe” (see my previous post) was quite instructive in that regard, and so let’s look at the finer points of that story once again.

Is ‘another’ world possible at all?

Feynman’s assumption was that we’ve made contact (don’t ask how: somehow) with some other intelligent being living in some ‘other’ world somewhere ‘out there’, and that there are no visual or other common references. That’s all rather vague, you’ll say, but just hang in there and try to see where we’re going with this story. Most notably, the other intelligent being – but let’s call ‘it’ a she instead of ‘a guy’ or ‘a Martian’ – cannot see the universe as we see it: we can’t describe, for instance, the Big and Small Dipper and explain to her what ‘left’ and ‘right’ is referring to such constellations, because she’s sealed off somehow from it (so she lives in a totally different corner of the universe really).

In contrast, we would be able, most probably, to explain and share the concept of ‘upward’ and ‘downwards’ by assuming that she is also attracted by some center of gravity nearby, just like we are attracted downwards by our Earth. Then, after many more hours and days, weeks, months or even years of tedious ‘discussions’, we would probably be able to describe electric currents and explain electromagnetic phenomena, and then, hopefully, she would find out that the laws in her corner of the universe are exactly the same, and so we could thus explain and share the notion of a ‘positive’ and a ‘negative’ charge, and the notion of a magnetic ‘north’ and ‘south’ pole.

However, at this point the story becomes somewhat more complicated, because – as I tried to explain in my previous post – her ‘positive’ electric charge (+) and her magnetic ‘north’ might well be our ‘negative’ electric charge (–) and our magnetic ‘south’. Why? It’s simple: the electromagnetic force does respect charge and also parity symmetry and so there is no way of defining any absolute sense of ‘left’ and ‘right’ or (magnetic) ‘north’ and (magnetic) ‘south’ with reference to the electromagnetic force alone. [If you don’t believe, just look at my previous post and study the examples.]

Talking about the strong force wouldn’t help either, because it also fully respects charge symmetry.

Huh? Yes. Just go through my previous post which – I admit – was probably quite confusing but made the point that a ‘mirror-image’ world would work just as well… except when it comes to the weak force. Indeed, atomic decay processes (beta decay) do distinguish between ‘left-handed’ and ‘right-handed’ particles (as measured by their spin) in an absolute sense that is (see the illustration of decaying muons and their mirror-image in the previous post) and, hence, it’s simple: in order to make sure her ‘left’ and her ‘right’ is the same as ours, we should just ask her to perform those beta decay experiments demonstrating that parity (or P-symmetry) is not being conserved and, then, based on our common definition of what’s ‘up’ and ‘down’ (the commonality of these notions being based on the effects of gravity which, we assume, are the same in both worlds), we could agree that ‘right’ is ‘right’ indeed, and that ‘left’ is ‘left’ indeed.

Now, you will remember there was one ‘catch’ here: if ever we would want to set up an actual meeting with her (just assume that we’ve finally figured out where she is and so we (or she) are on our way to meet each other), we would have to ask her to respect protocol and put out her right hand to greet us, not her left. The reason is the following: while ‘right-handed’ and ‘left-handed’ matter behave differently when it comes to weak force interactions (read: atomic decay processes)–which is how we can distinguish between ‘left’ and ‘right’ in the first place, in some kind of absolute sense that is–the combined CP symmetry implies that right-handed matter and left-handed anti-matter behave just the same–and, of course, the same goes for ‘left-handed’ matter and ‘right-handed’ anti-matter. Hence, after we would have had a painstakingly long exchange on broken P-symmetry to ensure we are talking about the same thing, we would still not know for sure: she might be living in a world of anti-matter indeed, in which case her ‘right’ would actually be ‘left’ for us, and her ‘left’ would be ‘right’.

Hence, if, after all that talk on P-symmetry and doing all those experiments involving P-asymmetry, she actually would put out her left hand when meeting us physically–instead of the agreed-upon right hand… Then… Well… Don’t touch it. 🙂

There is a way out of course. And, who knows, perhaps she was just trying to be humorous and so perhaps she smiled and apologized for the confusion in the meanwhile. But then… […] Hmm… I am not sure if such bad joke would make for a good start of a relationship, even if it would obviously demonstrate superior intelligence. 🙂

Indeed, the Fitch-Cronin experiment brings an additional twist to this potentially romantic story between two intelligent beings from two ‘different’ worlds. In fact, the Fitch-Cronin experiment actually rules out this theoretical possibility of mutual destruction and, therefore, the possibility of two ‘different’ worlds.

The argument goes straight to the heart of our philosophical discussion on time reversal. Indeed, whatever you may or may not have understood from this and my previous posts on CPT symmetry, the key point is that the combined CPT symmetry cannot be violated.

Why? Well… That’s plain logic: the real world does not care about our conventions, so reversing all of our conventions, i.e.

  1. Changing all particles to antiparticles by reversing all charges (C),
  2. Turning all right-handed particles into left-handed particles and vice versa (P), and
  3. Changing the sign of time (T),

describes a world truly going back in time.

Now, ‘her’ world is not going back in time. Why? Well… Because we can actually talk to her, it is obvious that her ‘arrow of time’ points in the same direction as ours, so she is not living in a world that is going back in time. Full stop. Therefore, any experiment involving a combined CP asymmetry (i.e. C-P violation) should yield the same results and, hence, she should find the same bias, i.e. a bias going in the very same direction of the equation, i.e. from left to right, or from right to left – whatever (what we label it, depends on our conventions, which we ‘re-set’ as we talked to her, and, hence, which we share, based on the results of all these beta decay experiments we did to ensure we’re really talking about the ‘same’ direction, and not its opposite).

Is this confusing? It sure is. But let me rephrase the logic. Perhaps it helps.

  1. Combined CPT symmetry implies that if the combined CP-symmetry is broken, then T-symmetry is also broken. Hence, the experimentally established fact of broken CP symmetry (even if it’s only 2 or 3 times per thousand) ensures that the ‘arrow of time’ points in one direction, and in one direction only. To put it simply: we cannot reverse time in a world which does not (fully) respect the principle of CP symmetry.
  2. Now, if you and I can exchange meaningful signals (i.e. communicate), then your and my ‘arrow of time’ obviously point in the same direction. To put it simply, we’re actors in the same movie, and whether or not it is being played backwards doesn’t matter anymore: the point is that the two of us share the same arrow of time. In other words, God did not do any combined CPT-transformation trick on your world as compared to mine, and vice versa.
  3. Hence, ‘your’ world is ‘my’ world and vice versa. So we live in the same world with the very same symmetries and asymmetries.

Now apply this logic to our imaginary new friend (‘she’) and (I hope) you’ll get the point.

To make a long story short, and also to conclude our philosophical digressions here on a pleasant (romantic) note: the fact that we would be able to communicate with her, implies that she’d be living in the same world as ours. We know that now, for sure, because of the broken CP symmetry: indeed, if her ‘time arrow’ points in the same direction, then CP symmetry will be broken in just the very same way in ‘her’ world (i.e. the ‘bias’ will have the same direction, in an absolute sense) as it it is broken in ‘our’ world.

In short, there are only two possible worlds: (1) this world and (2) one and only one ‘other’ world. This ‘other’ world is our world under a full CPT-transformation: the whole movie played backwards in other words, but with all ‘charges’ affecting forces – in whatever form and shape they come (electric charge, color charge, spin, and what have you) reversed or – using that awful mathematical term – ‘negated’.

In case you’d wonder (1): I consider the many-worlds interpretation of quantum mechanics as… Well… Nonsense. CPT symmetry allows for two worlds only. Maximum two. 🙂

In case you’d wonder (2): An oscillating-universe theory, or some kind of cyclic thing (so Big Bangs followed by Big Crunches) are not incompatible with my ‘two-possible-worlds’ view of things. However, this ‘oscillations’ would all take place in the same world really, because the arrow of time isn’t being reversed really, as Big Bangs and Big Crunches do not reverse charges and parities–at least not to my knowledge.

But, of course, who knows?

Postscripts:

1. You may wonder what ‘other’ asymmetries I am hinting at in this post here. It’s quite simple. It’s everything you see around you, including the works of the increasing entropy law. However, if I would have to choose one asymmetry in this world (the real world), as an example of a very striking and/or meaningful asymmetry, it’s the the preponderance of matter over anti-matter, including the preponderance of (left-handed) neutrinos over (right-handed) antineutrinos. Indeed, I can’t shake off that feeling that neutrino physics is going to spring some surprises in the coming decades.

[When you’d google a bit in order to get some more detail on neutrinos (and solar neutrinos in particular, which are the kind of neutrinos that are affecting us right now and right here), you’ll probably get confused by a phenomenon referred to as neutrino oscillation (which refers to a process in which neutrinos change ‘flavor’) but so the basic output of the Sun’s nuclear reactor is neutrinos, not anti-neutrinos. Indeed, the (general) reaction involves two protons combining to form one (heavy) hydrogen atom (i.e. deuterium, which consists of one neutron, one proton and one electron), thereby ejecting one positron (e+) and one (electron) neutrino (ve). In any case, this is not the place to develop the point. I’ll leave that for my next post.]

2. Whether or not you like the story about ‘her’ above, you should have noticed something that we could loosely refer to as ‘degrees of freedom’ is playing some role:

  1. We know that T-symmetry has not been broken: ‘her’ arrow of time points in the same direction.
  2. Therefore, the combined CP-symmetry of ‘her’ world is broken in the same way as in our world.
  3. If the combined CP-symmetry in ‘her’ world is broken in the same way as in ‘our’ world, the individual C and P symmetries have to be broken in the very same way. In other words, it’s the same world indeed. Not some anti-matter world.

As I am neither a physicist nor a mathematician, and not a philosopher either, please do feel free to correct any logical errors you may identify in this piece. Personally, I feel the logic connecting CP violation and individual C- and P-violation needs further ‘flesh on the bones’, but the core argument is pretty solid I think. 🙂

3. What about the increasing entropy law in this story? What happens to it if we reverse time, charge and parity? Well… Nothing. It will remain valid, as always. So that’s why an actual movie being played backwards with charges and parities reversed will still not make any sense to us: things that are broken don’t repair themselves and, hence, at the system level, there’s another type of irreducible ‘arrow of time’ it seems. But you’ll have to admit that the character of that entropy ‘law’ is very different from these ‘fundamental’ force laws. And then just think about it, isn’t it extremely improbable how we human beings have evolved in this universe? And how we are seemingly capable to understand ourselves and this universe? We don’t violate the entropy law obviously (on the contrary: we’re obviously messing up our planet), but I feel we do negate it in a way that escapes the kind of logical thinking that underpins the story I wrote above. But such remarks have nothing to do with math or physics and, hence, I will refrain from them.

4. Finally, for those who’d feel like some kind of ‘feminist’ remark on my use of ‘us’ and ‘her’, I think the use of ‘her’ is explained to underline the idea of ‘other’ and, hence, as a male writer, using ‘her’ to underscore the ‘other’ dimension comes naturally and shouldn’t be criticized. The element which could/should bother a female reader of such ‘through experiments’ is that we seem to assume that the ‘other’ intelligent being is actually somewhat ‘dumber’ than us, because the story above assumes we are actually explaining the experiments of the Wu and Fitch-Cronin team to ‘her’, instead of the other way around. That’s why I inserted the possibility of ‘her’ pulling a practical joke on us by offering us her left hand: if ‘she’ is equally or even more intelligent than us, then she’d surely have figured out that there’s no need to be worried about the ‘other’ being made of anti-matter. 🙂

Time reversal and CPT symmetry (II)

My previous post touched on many topics and, hence, I feel I was not quite able to exhaust the topic of parity violation (let’s just call it mirror asymmetry: that’s more intuitive). Indeed, I was rather casual in stating that:

  1. We have ‘right-handed’ and ‘left-handed’ matter, and they behave differently–at least with respect to the weak force–and, hence, we have some kind of absolute distinction between left and right in the real world.
  2. If ‘right-handed’ matter and ‘left-handed’ matter are not the same, then ‘right-handed’ antimatter and ‘left-handed’ antimatter are not the same either.
  3. CP symmetry connects the two: right-handed matter behaves just like left-handed antimatter, and right-handed antimatter behaves just like left-handed matter.

There are at least two problems with this:

  1. In previous posts, I mentioned the so-called Fitch-Cronin experiment which, back in 1964, provided evidence that ‘Nature’ also violated the combined CP-symmetry. In fact, I should be precise here and say the weak force, instead of ‘Nature’, because all these experiments investigate the behavior of the weak force only. Having said that, it’s true I mentioned this experiment in a very light-hearted manner–too casual really: I just referred to my simple diagrams illustrating what true time reversal entails (a reversal of the forces and, hence, of the charges causing those forces) and that was how I sort of shrugged it all of.
  2. In such simplistic world view, the question is not so much why the weak force violates mirror symmetry, but why gravity, electromagnetism and the strong force actually respect it!

Indeed, you don’t get a Nobel Prize for stating the obvious and, hence, if Val Fitch and James Cronin got one for that CP-violation experiment, C/P or CP violation cannot be trivial matters.

P-symmetry revisited

So let’s have another look at mirror symmetry–also known as reflection symmetry– by following Feynman’s example: let us actually build a ‘left-hand’ clock, and let’s do it meticulously, as Feynman describes it: “Every time there is a screw with a right-hand thread in one, we use a screw with a left-hand thread in the corresponding place of the other; where one is marked ‘IV’ on the face, we mark a ‘VI’ on the face of the other; each coiled spring is twisted one way in one clock and the other way in the mirror-image clock; when we are all finished, we have two clocks, both physical, which bear to each other the relation of an object and its mirror image, although they are both actual, material objects. Now the question is: If the two clocks are started in the same condition, the springs wound to corresponding tightnesses, will the two clocks tick and go around, forever after, as exact mirror images?”

The answer seems to be obvious: of course they will! Indeed, we do observe that P symmetry is being respected, as shown below:

P symmetry

You may wonder why we have to go through the trouble of building another clock. Why can’t we just take one of these transparent ‘mystery clocks’ and just go around it and watch its hand(s) move standing behind it? The answer is simple: that’s not what mirror symmetry is about. As Feynman puts its: a mirror reflection “turns the whole space inside out.” So it’s not like a simple translation or a rotation of space. Indeed, when we would move around the clock to watch it from behind, then all we do is rotating our reference frame (with a rotation angle equal to 180 degrees). That’s all. So we just change the orientation of the clock (and, hence, we watch it from behind indeed), but we are not changing left for right and right for left.

Rotational symmetry is a symmetry as well, and the fact that the laws of Nature are invariant under rotation is actually less obvious than you may think (because you’re used to the idea). However, that’s not the point here: rotational symmetry is something else than reflection (mirror) symmetry. Let me make that clear by showing how the clock might run when it would not respect P-symmetry.

P asymmetry

You’ll say: “That’s nonsense.” If we build that mirror-image clock and also wind it up in the ‘other’ direction (‘other’ as compared to our original clock), then the mirror-image clock can’t run that way. Is that nonsense? Nonsensical is actually the word that Wolfgang Pauli used when he heard about Chien-Shiung Wu’s 1956 experiment (i.e. the first experiment that provided solid evidence for the fact that the weak force – in beta decay for instance – does not respect P-symmetry), but so he had to retract his words when repeated beta decay experiments confirmed Wu’s findings.

Of course, the mirror-image clock above (i.e. the one running clockwise) breaks P-symmetry in a very ‘symmetric’ way. In fact, you’ll agree that the hands of that mirror-image clock might actually turn ‘clockwise’ if its machinery would be completely reversible, so we could wind up its springs in the same way as the original clock. But that’s cheating obviously. However, it’s a relevant point and, hence, to be somewhat more precise I should add that Wu’s experiment (and the other beta decay experiments which followed after hers) actually only found a strong bias in the direction of decay: not all of the beta rays (beta rays consist of electrons really – check the illustration in my previous post for more details) went ‘up’ (or ‘down’ in the mirror-reversed arrangement), but most of them did. 

Wu_experiment

OK. We got that. Now how do we explain it? The key to explaining the phenomenon observed by Wu and her team, is the spin of the cobalt-60 nuclei or, in the muon decay experiment described in my previous post, the spin of the muons. It’s the spin of these particles that makes them ‘left-handed’ or ‘right-handed’ and the decay direction is (mostly) in the direction of the axial vector that’s associated with the spin direction (this axial vector is the thick black arrow in the illustration below).

Axial vector

Hmm… But we’ve got spinning things in (mechanical) clocks as well, don’t we? Yes. We have flywheels and balance wheels and lots of other spinning stuff in a mechanical clock, but these wheels are not equivalent to spinning muons or other elementary particles: the wheels in a clock preserve and transfer angular momentum.

OK… But… […] But isn’t that what we are talking about here? Angular momentum?

No. Electrons spinning around a nucleus have angular momentum as well – referred to as orbital angular momentum – but it’s not the same thing as spin which, somewhat confusingly, is often referred to as intrinsic angular momentum. In short, we could make a detailed analysis of how our clock and its mirror image actually work, and we would find that all of the axial vectors associated with flywheels, balance wheels and springs in a clock would effectively be reversed in the mirror-image clock but, in contrast with the weak decay example, their reversed directions would actually explain why the mirror-image clock is turning counter-clockwise (from our point of view that is), just like the image of the original clock in the mirror does, and, therefore, why a ‘left-handed’ mechanical clock actually respects P-symmetry, instead of breaking it.

Axial and polar vectors in physics

In physics, we encounter such axial vectors everywhere. They show the axis of spin, and their direction is determined by the direction of spin through one of two conventions: the ‘right-hand screw rule’, or the ‘left-hand screw rule’. Physicists have settled on the former, so let’s work with that for the time being.

The other type of vector is a polar vector. That’s an ‘honest’ vector as Feynman calls it–depicting ‘real’ things such as, for example, a step in space, or some force acting in some direction. The figures below (which I took from Feynman’s Lectures) illustrate the idea (and please do note the care with which Feynman reversed the direction of the arrows above the r and ω in the mirror image):

  1. When mirrored, a polar vector “changes its head, just as the whole space turns inside out.”
  2. An axial vector behaves differently when mirrored. It changes too, but in a very different way: it is usually reversed in respect to the geometry of the whole space, as illustrated in the muon decay image above. However, in the illustration below, that is not the case, because the angular velocity ‘vector’ is not reversed when mirrored. So it’s all quite subtle and one has to carefully watch what’s going on really when we do such mirror reflections.

Axial vectors

What’s the third figure about? Well… While it’s not that difficult to visualize all of the axial vectors in a mechanical clock, it’s a different matter when discussing electromagnetic forces, and then to explain why these electromagnetic forces also respect mirror symmetry, just like the mechanical clock. But let’s me try.

When an electric current goes through a solenoid, the solenoid becomes a magnet, especially when wrapped around an iron core. The direction and strength of the magnetic field is given by the magnetic field vector B, and the force on an electrically charged particle moving through such magnetic field will be equal to F = qv×B. That’s a so-called vector cross product and we’ve seen it before: a×b = na││b│sinθ, so we take (1) the magnitudes of a and b, (2) the sinus of the angle between them, and (3) the unit vector (n) perpendicular to (the plane containing) a and b; multiply it all; and there we are: that’s the result. But – Hey! Wait a minute! – there are two unit vectors perpendicular to a and b. So how does that work out?

Well… As you might have guessed, there is another right-hand rule here, as shown below.

2000px-Right_hand_rule_cross_product

Now how does that work out for our magnetic field? If we mirror the set-up and let an electron move through the field? Well… Let’s do the math for an electron moving into this screen, so in the direction that you are watching.

In the first set-up, the B vector points upwards and, hence, the electron will deviate in the direction given by that cross product above: qv×B. In other words, it will move sideways as it moves away from you, into the field. In which direction? Well… Just turn that hand above about 90 degrees and you have the answer: right. Oh… No. It’s left, because q is negative. Right.

In the mirror-image set-up, we have a B’ vector pointing in the opposite direction so… Hey ! Mirror symmetry is not being respected, is it?

Well… No. Remember that we must change everything, including our conventions, so the ‘right-hand rules’ above becomes ‘left-hand rules’, as shown below for example. Surely you’re joking, Mr. Feynman!

P-parity for screw rules

Well… No. F and v are polar vectors and, hence, “their head might change, just as the whole space turns inside out”, but that’s not the case now, because they’re parallel to the mirror. In short, the force F on the electron will still be the same: it will deviate leftwards. I tried to draw that below, but it’s hard to make that red line look like it’s a line going away from you.

Capture

But that can’t be true, you’ll say. The field lines go from north to south, and so we have that B’ vector pointing downwards now.

No, we don’t. Or… Well… Yes. It all depends on our conventions. 🙂  

Feynman’s switch to ‘left-hand rules’ also involves renaming the magnetic poles, so all magnetic north poles are now referred to as ‘south’ poles, and all magnetic south poles are now referred to as ‘north’ poles, and so that’s why he has a B’ vector pointing downwards. Hence, he does not change the convention that magnetic field lines go from north to south, but his ‘north’ pole (in the mirror-image drawing) is actually a ‘south’ pole. Capito? 🙂

[…] OK. Let me try to explain it once again. In reality, it does not matter whether or not a solenoid is wound clockwise or counterclockwise (or, to use the terminology introduced above, whether our solenoid is left-handed or right-handed). The important thing is that the current through the solenoid flows from the top to the bottom. We can only reverse the poles – in reality – if we reverse the electric current, but so we don’t do that in our mirror-image set-up. Therefore, the force F on our charged particle will not change, and B’ is an axial vector alright but this axial vector does not represent the actual magnetic field.

[…] But… If we change these conventions, it should represent the magnetic field, shouldn’t it? And how do we calculate that force then?

OK. If you insist. Here we go:

  1. So we change ‘right’ to ‘left’ and ‘left’ to ‘right’, and our cross-product rule becomes a ‘left-hand’ rule.
  2. But our electrons still go from ‘top’ to ‘bottom’. Hence, the (magnetic) force on a charged particle won’t change.
  3. But if the result has to be the same, then B needs to become –B, or so that’s B’ in our ‘left-handed’ coordinate system.
  4. We can now calculate F using the ‘left-handed’ cross product rule and – because we did not change the convention that field lines go from north to south, we’ll also rename our poles.
  5. Yippee ! All comes out all right: our electron goes left. Sorry. Right. Huh? Yes. Because we’ve agreed to replace ‘left’ by ‘right’, remember? 🙂

[…]

If you didn’t get anything of this, don’t worry. There is actually a much more comprehensible illustration of the mirror symmetry of electromagnetic forces. If we would hang two wires next to each other, as below, and we send a current through them, they will attract if the two currents are in the same direction, and they will repel when the currents are opposite. However, it doesn’t matter if the current goes from left to right or from right to left. As long as the two currents have the same direction (left or right), it’s fine: there will be attraction. That’s all it takes to demonstrate P-symmetry for electromagnetism.

Wires attracting

The Fitch-Cronin experiment

I guess I caused an awful lot of confusion above. Just forget about it all and take one single message home: the electromagnetic force does not care about the axial vector of spinning particles, but the weak force does.

Is that shocking?

No. There are plenty of examples in the real world showing that the direction of ‘spin’ does matter. For instance, to unlock a right-hinged door, you turn the key to the right (i.e. clockwise). The other direction doesn’t work. While I am sure physicists won’t like such simplistic statements, I think that accepting that Nature has similar ‘left-handed’ and ‘right-handed’ mechanisms is not the kind of theoretical disaster that Wolfgang Pauli thought it was. If anything, we just should marvel at the fact that gravity, electromagnetism and the strong force are P- and C-symmetric indeed, and further investigate why the weak force does not have such nice symmetries. Indeed, it respects the combined CPT symmetry, but that amounts to saying that our world sort of makes sense, so that ain’t much.

In short, our understanding of that weak force is probably messy and, as Feynman points out: “At the present level of understanding, you can still see the “seams” in the theories; they have not yet been smoothed out so that the connection becomes more beautiful and, therefore, probably more correct.” (QED, 1985, p. 142). However, let’s stop complaining about our ‘limited understanding’ and so let’s work with what we do understand right now. Hence, let’s have a look at that Fitch-Cronin experiment now and see how ‘weird’ or, on the contrary, how ‘understandable’ it actually is.

To situate the Fitch-Cronin experiment, we first need to say something more about that larger family of mesons, of which the kaons are just one of the branches. In fact, in case you’d not be interested in this story as such, then I’d suggest you just read it as a very short introduction to the Standard Model as such, as it gives a nice short overview of all matter-particles–which is always useful I’d think.

Hadrons, mesons and baryons

You may or may not remember that mesons are unstable particles consisting of one quark and one anti-quark (so mesons consist of two quarks, but one of them should be an anti-quark). As such, mesons are to be distinguished from the ‘other’ group within the larger group of hadrons, i.e. the baryons, which are made of three quarks. [The term ‘hadrons’ itself is nothing but a catch-all for all particles consisting of quarks.]

The most prominent representatives of the baryon family are the (stable) neutron and proton, i.e. the nucleons, which consist of u and d quarks. However, there are unstable baryons as well. These unstable baryons involve the heavier (second-generation) or quarks, or the super-heavy (third-generation) b quark. [As for the top quark (t), that’s so high-energy (and, hence, so short-lived) that baryons made of a t quark (so-called ‘top-baryons’) are not expected to exist but, then, who knows really?]

But kaons are mesons, and so I won’t say anything more about baryons The two illustrations below should be sufficient to situate the discussion.

98E-pic-first-classification-particles

Standard_Model_of_Elementary_Particles

Kaons are just one branch of the meson family. There are, for instance, heavier versions of the kaons, referred to as B- and D-mesons. Let me quickly introduce these:

  1. The ‘B’ in ‘B-meson’ refers to the fact that one of the quarks in a B-meson is a b-quark: a b (bottom) quark is a much heavier (third-generation) version of the (second-generation) s-quark.
  2. As for the ‘D’ in D-meson, I have no idea. D-mesons will always consist of a c-quark or anti-quark, combined with a lighter d, u or s (anti-)quark, but so there’s no obvious relationship between a D-meson and a d-quark. Sorry.
  3. If you look at the quark table above, you’ll wonder whether there are any top-mesons, i.e. mesons consisting of a t quark or anti-quark. The answer to that question seems to be negative: t quarks disintegrate too fast, it is said. [So that resembles the remark on the possiblity of t-baryons.] If you’d google a bit on this, you’ll find that, in essence, we haven’t found any t-mesons as yet but their potential existence should not be excluded.

Anything else? Yes. There’s a lot more around actually. Besides (1) kaons, (2) B-mesons and (3) D-mesons, we also have (4) pions (i.e. a combination of a u and a d, or their anti-matter counterpart), (5) rho-mesons (ρ-mesons can be thought of as excited (higher-energy) pions(6) eta-mesons (η-mesons a rapidly decaying mixture of ud and s quarks or their anti-matter counterparts), as well as a whole bunch of (temporary) particles consisting of a quark and its own anti-matter counterpart, notably the (7) phi (a φ consists of a s and an anti-s), psi (a ψ consists of an c and an anti-c) and upsilon (a φ consists of a b and an anti-b) particles (so all these particles are their own anti-particles).

So it’s quite a zoo indeed, but let’s zoom in on those ‘crazy’ kaons. [‘Crazy kaons’ is the epithet that Gerard ‘t Hooft reserved for them in his In Search of the Ultimate Building Blocks (1996).] What are they really? 

Crazy kaons

Kaons, also know as K-mesons, are, first of all, mesons, i.e. particles made of one quark and one anti-quark (as opposed to baryons, which are made of three quarks, e.g. protons and neutrons). All mesons are unstable: at best, they last a few hundredths of a microsecond, but kaons have much shorter lifetimes than that. Where do we find them? We usually create them in those particle colliders and other sophisticated machinery (the experiment used kaon beams) but we can also find them as a decay product in (secondary) cosmic rays (cosmic rays consist of very high-energy particles and they produce ‘showers’ of secondary particles as they hit our atmosphere).

They come in three varieties: neutral and positively or negatively charged, so we have a K0, a K+, and a K, in principle that is (the story will become more complicated later). What they have in common is that one of the quarks is the rather heavy s-quark (s stands for ‘strange’ but you know what Feynman – and others – think of that name: it’s just a strange name indeed, and so don’t worry too much about it). An s-quark is a so-called second-generation matter-particle and that’s why the kaon is unstable: all second-generation matter-particles are unstable. The second quark is just an ordinary u- or d-quark, i.e. the type of quark you’d find in the (stable) proton or neutron.

But what about the electric charge? Well… I should be complete. The quarks might be anti-quarks as well. That’s nothing to worry about as you’ll remember: anti-matter is just matter but with the charges reversed. So a Kconsists of an s quark and an anti-d quark or –and this is the key to understanding the experiment actually– a K0 can also consist of an anti-s quark and a (normal) d-quark. Note that the s and d quarks have a charge of 1/3 and so the total charge comes out alright. [As for the other kaons, a Kconsists of a u and anti-s quark (the u quark has charge 2/3 and so we have +1 as the total charge), and the K– consists of an anti-u and an s quark (and, hence, we have –1 as the charge), but we actually don’t need them any more for our story.]

So that’s simple enough. Well… No. Unfortunately, the story is, indeed, more complicated than that. The actual kaons in a neutral kaon beam come in two varieties that are a mix of the two above-mentioned neutral K states: a K-long (KL) has a lifetime of about 9×10–11 s, while a K-short (KS) has a lifetime of about 5.2×10–8 s. Hence, at the end of the beam, we’re sure to find Kkaons only.

Huh? mix of two particle states… You’re talking superposition here? Well… Yes. Sort of. In fact, as for what KL and Kactually are, that’s a long and complex story involving what is referred to as a neutral particle oscillation process. In essence, neutral particle oscillation occurs when a (neutral) particle and its antiparticle are different but decay into the same final state. It is then possible for the decay and its time reversed process to contribute to oscillations indeed, that turn the one into the other, and vice versa, so we can write A → Δ → B → Δ → A → etcetera, where A is the particle, B is the antiparticle, and Δ is the common set of particles into which both can decay. So there’s an oscillation phenomenon from one state to the other here, and all the things I noted about interference obviously come into play.

In any case, to make a very long and complicated story short, I’ll summarize it as follows: if CP symmetry holds, then one can show that this oscillation process should result in a very clear-cut situation: a mixed beam of long-lived and short-lived kaons, i.e. a mix of KL and KS. Both decay differently: a K-short particle decays into two pions only, while a K-long particle decays into three pions.

That is illustrated below: at the end of the 17.4 m beam, one should only see three-pion decay events. However, that’s not what Fitch and Cronin measured: the actually saw a one two-pion decay event into every 500 (on average that is)! [I have introduced the pion species in the more general discussion on mesons: you’ll remember they consist of first-generation quarks only, but so don’t worry about it: just note the K-long and K-short particles decay differently. Don’t be confused by the π notation below: it has nothing to do with a circle or so, so 2π just means two pions.]

Kaon beam

That means that the kaon decay processes involved do not observe the assumed CP symmetry and, because it’s the weak force that’s causing those decays, it means that the weak force itself does not respect CP symmetry.

Why is that so?

You may object that these lifetimes are just averages and, hence, perhaps we see these two-pion decays at the end of the beam because some of the K-short particles actually survived much longer !

No. That’s to be ruled out. The short-lived particle cannot be observable more than a few centimeters down the beam line. To show that, one can calculate the time required to drop to 1/500 of the original population of K-short particles. With the stated lifetime (9×10–11 s), the half-life calculation gives a time of 5.5 x 10-10 seconds. At nearly the speed of light, this would give a distance of about 17 centimeters, and so that’s only 1/100 the length of Cronin and Fitch’s beam tube.

But what about the fact that particles live longer when they’re going fast? You are right: the number above ignores relativistic time dilation: the lifetime as seen in the laboratory frame is ‘dilated’ indeed by the relativity factor γ. At 0.98c (i.e. the speed of these kaons, γ =5, and, hence, this “time dilation effect” is very substantial. However, re-calculating the distance gives a revised distance equal to 17γ cm, i.e. 85 cm. Hence, even with kaons speeding at 0.98c, the population would be down by a factor of 500 by the time they got a meter down the beam tube. So for any particle velocity really, all of these K-short particles should have decayed long before they get to the end of the beam line.

Fitch and Cronin did not see that, however: they saw one two-pion decay event for every 500 decay events, so that’s two per thousand (0.2%) and, hence, that is very significant. While the reasoning is complex (these oscillations and the quantum-mechanical calculations involved are not easy to understand), the results clearly shows the kaon decay process does not observe CP symmetry.

OK. So what? How does this violate charge and parity symmetry? Well… That’s a complicated story which involves a deeper understanding of how initial and final states of such processes incorporate CP values, and then showing how these values differ. That’s a story that requires a master’s degree in physics, I must assume, and so I don’t have that. But I can sort of sense the point and I would suggest we just accept it here. [To be precise, the Fitch-Cronin experiment is an indirect ‘proof’ of CP violation only: as mentioned below, only in 1999 would experiments be able to demonstrate direct CP violation.]

OK. So what? Do we see it somewhere else? Well… Fitch and Cronin got a Nobel Prize for this only sixteen years later, i.e. in 1980, and then it took researchers another twenty years to find CP violation in some other process. To be very precise, only in 1999 (i.e. 35 years after the Fitch-Cronin findings), Fermilab and CERN could conclude a series of experiments demonstrating direct CP violation in (neutral) kaon decay processes (as mentioned above, the Fitch-Cronin experiment only shows indirect CP violation), and that then set the stage for a ‘new’ generation of experiments involving B-mesons and D-mesons, i.e. mesons consisting of even heavier quarks (c or b quarks)–so these are things that are even less stable than kaons. So… Well… Perhaps you’re right. There’s not all that many examples really.

Aha ! So what?

Well… Nothing. That’s it. These ‘broken symmetries’ exist, without any doubt, but–you’re right–they are a marginal phenomenon in Nature it seems. I’ll just conclude with quoting Feynman once again (Vol. I-52-9):

“The marvelous thing about it all is that for such a wide range of important phenomena–nuclear forces, electrical phenomena, and gravitation–over a tremendous range of physics, all the laws for these seem to be symmetrical. On the other hand, this little extra piece says, “No, the laws are not symmetrical!” How is it that Nature can be almost symmetrical, but not perfectly symmetrical? […] No one has any idea why. […] Perhaps God made the laws only nearly symmetrical so that we should not be jealous of His perfection.”

Hmm… That’s the last line of the first volume of his Lectures (there are three of them), and so that should end the story really.

However, I would personally not like to involve God in such discussions. When everything is said and done, we are talking atomic decay processes here. Now, I’ve already said that I am not a physicist (my only ambition is to understand some of what they are investigating), but I cannot accept that these decay processes are entirely random. I am not saying there are some ‘inner variables’ here. No. That would amount to challenging the Copenhagen interpretation of quantum mechanics, which I won’t.

But when it comes to the weak force, I’ve got a feeling that neutrino physics may provide the answer: the Earth is being bombarded with neutrinos, and their ‘intrinsic parity’ is all the same: all of them are left-handed. In fact, that’s why weak interactions which emit neutrinos or antineutrinos violate P-symmetry! It’s a very primitive statement – and not backed up by anything I have read so far – but I’ve got a feeling that the weak force does not only involve emission of neutrinos or antineutrinos: I think they enter the equation as well.

That’s preposterous and totally random statement, you’ll say.

Yes. […] But I feel I am onto something and I’ll explore it as good as I can–if only to find out why I am so damn wrong. I can only say that, if and when neutrino physics would allow us to tentatively confirm this random and completely uninformed hypothesis, then we would have an explanation which would be much more in line with the answers that astrophysicists give to questions related to other observable asymmetries such as, for example, the imbalance between matter and anti-matter.

However, I know that I am just babbling now, and that nobody takes this seriously anyway and, hence, I will conclude my series on CPT symmetry right here and now. 🙂

Time reversal and CPT symmetry (I)

In my previous posts, I introduced the concept of time symmetry, and parity and charge symmetry as well. However, let’s try to explore T-symmetry first. It’s not an easy concept – contrary to what one might think at first.

The arrow of time

Let me start with a very ‘common-sense’ introduction. What do we see when we play a movie backwards? […]

We reverse time. When playing some movie backwards, we look at where things are coming from. And we see phenomena that don’t make sense, such as: (i) cars racing backwards, (ii) old people becoming younger (and dead people coming back to life), (iii) shattered glass assembling itself back into some man-made shape, and (iv) falling objects defying gravity to get back to where they were. Let’s briefly say something about these unlikely or even impossible phenomena before a more formal treatment of the matter:

  1. The first phenomenon – cars racing backwards – is unlikely to happen in real life but quite possible, and some crazies actually do organize such races.
  2. The last example – objects defying gravity – is plain impossible because of Newton’s universal law of gravitation.
  3. The other examples – the old becoming young (and the dead coming back to life), and glass shards coming back together into one piece – are also plain impossible because of some other ‘law’: the law of ever increasing entropy.

However, there’s a distinct difference between the two ‘laws’ (gravity versus increasing entropy). As one entry on Physics Stack Exchange notes, the entropy law – better known as the second law of thermodynamics – “only describes what is most likely to happen in macroscopic systems, rather than what has to happen”, but then the author immediately qualifies this apparent lack of determinism, and rightly so: “It is true that a system may spontaneously decrease its entropy over some time period, with a small but non-zero probability. However, the probability of this happening over and over again tends to zero over long times, so is completely impossible in the limit of very long times.” Hence, while one will find some people wondering whether this entropy law is a ‘real law’ of Nature – in the sense that they would question that it’s always true no matter what – there is actually no room for such doubts.

That being said, the character of the entropy law and the universal law of gravitation is obviously somewhat different – because they describe different realities: the entropy law is a law at the level of a system (a room full of air, for example), while the law of gravitation describes one of the four fundamental forces.

I will now be a bit more formal. What’s time symmetry in physics? The Wikipedia definition is the following: “T-symmetry is the theoretical symmetry (invariance) of physical laws under a time reversal (T) transformation.” Huh?

A ‘time reversal transformation’ amounts to inserting –t (minus t) instead of t in all of our equations describing trajectories or physical laws. Such transformation is illustrated below. The blue curve might represent a car or a rocket accelerating (in this particular example, we have a constant acceleration a = 2). The vertical axis measures the displacement (x) as a function of time (t). , and the red curve is its T-transformation. The two curves are each other’s mirror image, with the vertical axis (i.e. the axis measuring the displacement x) as the mirror axis.

Time reversal 2

This view of things is quite static and, hence, somewhat primitive I should say. However, we can make a number of remarks already. For example, we can see that the slope (of the tangent) of the red curve is negative. This slope is the velocity (v) of the particle: v = dx/dt. Hence, a T-transformation is said to negate the velocity variable (in classical physics that is), just like it negates the time variable. [The verb ‘to negate’ is used here in its mathematical sense: it means ‘to take the additive inverse of a number’ — but you’ll agree that’s too lengthy to be useful as an expression.]

Note that velocity (and mass) determines (linear and angular) momentum and, hence, a T-transformation will also negate p and l, i.e. the linear and angular momentum of a particle.

Such variables – i.e. variables that are negated by the T-transformation – are referred to as odd variables, as opposed to even variables, which are not impacted by the T-transformation: the position of the particle or object (x) is an example of an even variable, and the force acting on a particle (F) is not being negated either: it just remains what it is, i.e. an external force acting on some mass or some charge. The acceleration itself is another ‘even’ variable.

This all makes sense: why would the force or acceleration change? When we put a minus sign in front of the time variable, we are basically just changing the direction of an axis measuring an independent variable. In a way, the only thing that we are doing is introducing some non-standard way of measuring time, isn’t it? Instead of counting from 0 to T, we count from 0 to minus T.

Well… No. In this post, I want to discuss actual time reversal. Can we go back in time? Can we put a genie back into a bottle? Can we reverse all processes in Nature and, if not, why not?

Time reversal and time symmetry are two different things: doing a T-transformation is a mathematical operation; trying to reverse time is something real. Let’s take an example from kinematics to illustrate the matter.

Kinematics

Kinematics can be summed up in one equation, best known as Newton’s Second Law: F = ma = m(dv/dt) = d(mv)/dt.  In words: the time-rate-of-change of a quantity called momentum (mv) is proportional to the force on an object. In other words: the acceleration (a) of an object is proportional to the force (F), and the factor of proportionality is the mass of the object (m). Hence, the mass of an object is nothing but a measure of its inertia.

The numbering of laws (first, second, etcetera) – usually combining some name of a scientist – is often quite arbitrary but, in this case (Newton’s Laws), one can really learn something from listing and discussing them in the right order:

  1. Newton’s First Law is the principle of inertia: if there’s no (other) force acting on an object, it will just continue doing what it does–i.e. nothing or, else, move in some straight line according to the direction of its momentum (i.e. the product of its mass and its velocity)–or further engage with the force it was already engaged with.
  2. Newton’s Second Law is the law of kinematics. In kinematics, we analyze the motion of an object without caring about the origin of the force causing the motion. So we just describe how some force impacts the motion of the object on which it is acting without asking any questions about the force itself. We’ve written this law above: F = ma.
  3. Finally, Newton’s Third Law is the law of gravitation, which describes the origin, the nature and the strength of the gravitational force. That’s part of dynamics, i.e. the study of the forces themselves – as opposed to kinematics, which only looks at the motion caused by those forces.

With these definitions and clarifications, we are now well armed to tackle the subject of T-symmetry in kinematics (we’ll discuss dynamics later). Suppose some object – perhaps an elementary particle but it could also be a car or a rocket indeed – is moving through space with some constant acceleration a (so we can write a(t) = a). This means that v(t) – the velocity as a function of time – will not be constant: v(t) = at. [Note that we make abstraction of the direction here and, hence, our notation does not use any bold letters (which would denote vector quantities): v(t) and a(t) are just simple scalar quantities in this example.]

Of course, when we – i.e. you and me right here and right now – are talking time reversal, we obviously do it from some kind of vantage point. That vantage point will usually be the “now” (and quite often also the “here”), and so let’s use that as our reference frame indeed and we will refer to it as the zero time point: t = 0. So it’s not the origin of time: it’s just ‘now’–the start of our analysis.

Now, the idea of going back in time also implies the idea of looking forward – and vice versa. Let’s first do what we’re used to do and so that’s to look forward.

At some point in the future, let’s call it t = T, the velocity of our object will be equal to v(T) = v(0) + aT. Why the v(0)? Well… We defined the zero time point (t = 0) in a totally random way and, hence, our object is unlikely to stop for that. On the contrary: it is likely to already have some velocity and so that’s why we’re adding this v(0) here. As for the space coordinate, our object may also not be at the exact same spot as we are (we don’t want to be to close to a departing rocket I would assume), so we can also not assume that x(0) = 0 and so we will also incorporate that term somehow. It’s not essential to the analysis though.

OK. Now we are ready to calculate the distance that our object will have traveled at point T. Indeed, you’ll remember that the distance traveled is an infinite sum of infinitesimally small products vΔt: the velocity at each point of time multiplied by an infinitesimally small interval of time. You’ll remember that we write such infinite sum as an integral:

Eq 1

[In case you wonder why we use the letter ‘s’ for distance traveled: it’s because the ‘d’ symbol is already used to denote a differential and, hence, ‘s’ is supposed to stand for ‘spatium’, which is the Latin word for distance or space. As for the integral sign, you know that’s an elongated S really, don’t you? So its stands for an infinite sum indeed. But lets go back to the main story.]

We have a functional form for v(t), namely v(t) = v(0) + at, and so we can easily work out this integral to find s as a function of time. We get the following equation:

Eq 2

When we re-arrange this equation, we get the position of our object as a function of time:

Eq 3

Let us now reverse time by inserting –T everywhere:

Eq 4

Does that still make sense? Yes, of course, because we get the same result when doing our integral:

Eq 5

So that ‘makes sense’. However, I am not talking mathematical consistency when I am asking if it still ‘makes sense’. Let us interpret all of this by looking at what’s happening with the velocity. At t = 0, the velocity of the object is v(0), but T seconds ago, i.e. at point t = -T, the velocity of the object was v(-T) = v(0) – aT. This velocity is less than v(0) and, depending on the value of -T, it might actually be negative. Hence, when we’re looking back in time, we see the object decelerating (and we should immediately add that the deceleration is – just like the acceleration – a constant). In fact, it’s the very same constant a which determines when the velocity becomes zero and then, when going even further back in time, when it becomes negative.

Huh? Negative velocity? Here’s the difference with the movie: in that movie that we are playing backwards, our car, our rocket, or the glass falling from a table or a pedestal would come to rest at some point back in time. We can calculate that point from our velocity equation v(t) = v(0) + at. In the example below, our object started accelerating 2.5 seconds ago, at point t = –2.5. But, unlike what we would see happening in our backwards-playing movie, we see that object not only stopping but also reversing its direction, to go in the same direction as we saw it going when we’re watching the movie before we hit the ‘Play Backwards’ button. So, yes, the velocity of our object changes sign as it starts following the trajectory on the left side of the graph.

time reversal

What’s going on here? Well… Rest assured: it’s actually quite simple: because the car or that rocket in our movie are real-life objects which were actually at rest before t = –2.5, the left side of the graph above is – quite simply – not relevant: it’s just a mathematical thing. So it does not depict the real-life trajectory of an accelerating car or rocket. The real-life trajectory of that car or rocket is depicted below.

real-life car

So we also have a ‘left side’ here: a horizontal line representing no movement at all. Our movie may or may not have included this status quo. If it did, you should note that we would not be able to distinguish whether or not it would be playing forward or backwards. In fact, we wouldn’t be able to tell whether the movie was playing at all: we might just as well have hit the ‘pause’ button and stare at a frozen screenshot.

Does that make sense? Yes. There are no forces acting on this object here and, hence, there is no arrow of time.

Dynamics

The numerical example above is confusing because our mind is not only thinking about the trajectory as such but also about the force causing the particle—or the car or the rocket in the example above—to move in this or that direction. When it’s a rocket, we know it ignited its boosters 2.5 seconds ago (because that’s what we saw – in reality or in a movie of the event) and, hence, seeing that same rocket move backwards – both in time as well as in space – while its boosters operate at full thrust does not make sense to us. Likewise, an obstacle escaping gravity with no other forces acting on it does not make sense either.

That being said, reversing the trajectory and, hence, actually reversing the effects of time, should not be a problem—from a purely theoretical point at least: we should just apply twice the force produced by the boosters to give that rocket the same acceleration in the reverse direction. That would obviously means we would force it to crash back into the Earth. Because that would be rather complicated (we’d need twice as many boosters but mounted in the opposite direction), and because it would also be somewhat evil from a moral point of view, let us consider some less destructive examples.

Let’s take gravity, or electrostatic attraction or repulsion. These two forces also cause uniform acceleration or deceleration on objects. Indeed, one can describe the force field of a large mass (e.g. the Earth)—or, in electrostatics, some positive or negative charge in space— using field vectors. The field vectors for the electric field are denoted by E, and, in his famous Lectures on Physics, Feynman uses a C for the gravitational field. The forces on some other mass m and on some other charge q can then be written as F = mC and F = qE respectively. The similarity with the F = ma equation – Newton’s Second Law in other words – is obvious, except that F = mC and F = qE are an expression of the origin, the nature and the strength of the force:

  1. In the case of the electrostatic force (remember that likes repel and opposites attract), the magnitude of E is equal to E = qc/4πε0r2. In this equation, εis the electric constant, which we’ve encountered before, and r is the distance between the charge q and the charge qcausing the field).
  2. For the gravitational field we have something similar, except that there’s only attraction between masses, no repulsion. The magnitude of C will be equal to C = –GmE/r2, with mE the mass causing the gravitational field (e.g. the mass of the Earth) and G the universal gravitational constant. [Note that the minus sign makes the direction of the force come out alright taking the existing conventions: indeed, it’s repulsion that gets the positive sign – but that should be of no concern to us here.]

So now we’ve explained the dynamics behind that x(t) = x(0) + v(0)·t + (a/2)·tcurve above, and it’s these dynamics that explain why looking back in time does not make sense—not in a mathematical way but in philosophical way. Indeed, it’s the nature of the force that gives time (or the direction of motion, which is the very same ‘arrow of time’) one–and only one–logical direction.

OK… But so what is time reversibility then – or time symmetry as it’s referred to? Let me defer an answer to this question by first introducing another topic.

Even and odd functions

I already introduced the concept of even and odd variables above. It’s obviously linked to some symmetry/asymmetry. The x(t) curve above is symmetric. It is obvious that, if we would change our coordinate system to let x(0) equal x(0) = 0, and also choose the origin of time such that v(0) = 0, then we’d have a nice symmetry with respect to the vertical axis. The graph of the quadratic function below illustrates such symmetry.

Even functionFunctions with a graph such as the one above are called even functions. A (real-valued) function f(t) of a (real) variable t is defined as even if, for all t and –t in the domain of f, we find that f(t) = f(–t).

We also have odd functions, such as the one depicted below. An odd function is a function for which f(-t) = –f(t).

Odd function

The function below gives the velocity as a function of time, and it’s clear that this would be an odd function if we would choose the zero time point such that v(0) = 0. In that case, we’d have a line through the origin and the graph would show an odd function. So that’s why we refer to v as an odd variable under time reversal.

Velocity curve

A very particular and very interesting example of an even function is the cosine function – as illustrated below.

Cosine functionNow, we said that the left side of the graph of the trajectory of our car or our rocket (i.e. the side with a negative slope and, hence, negative velocity) did not make much sense, because – as we play our movie backwards – it would depict a car or a rocket accelerating in the absence of a force. But let’s look at another situation here: a cosine function like the one above could actually represent the trajectory of a mass oscillating on a spring, as illustrated below.

oscillating springIn the case of a spring, the force causing the oscillation pulls back when the spring is stretched, and it pushes back when it’s compressed, so the mechanism is such that the direction of the force is being reversed continually. According to Hooke’s Law, this force is proportional to the amount of stretch. If x is the displacement of the mass m, and k that factor of proportionality, then the following equality must hold at all times:

F = ma = m(d2x/dt2) = –kx ⇔ d2x/dt= –(k/m)x

Is there also a logical arrow of time here? Look at the illustration below. If we follow the green arrow, we can readily imagine what’s happening: the spring gets stretched and, hence, the mass on the spring (at maximum speed as it passes the equilibrium position) encounters resistance: the spring pulls it back and, hence, it slows down and then reverses direction. In the reverse direction – i.e. the direction of the red arrow – we have the reverse logic: the spring gets compressed (x is negative), the mass slows down (as evidence by the curvature of the graph), and – at some point – it also reverses its direction of movement. [I could note that the force equation above is actually a second-order linear differential equation, and that the cosine function is its solution, but that’s a rather pedantic and, hence, totally superfluous remark here.]

temp

What’s important is that, in this case, the ‘arrow of time’ could point either way, and both make sense. In other words, when we would make a movie of this oscillating movement, we could play it backwards and it would still make sense. 

Huh? Yes. Just in case you would wonder whether this conclusion depends on our starting point, it doesn’t. Just look at the illustration below, in which I assume we are starting to watch that movie (which is being played backwards without us knowing it is being played backwards) of the oscillating spring when the mass is not in the equilibrium position. It makes perfect sense: the spring is stretched, and we see the mass accelerating to the equilibrium position, as it should.

temp2

What’s going on here? Why can we reverse the arrow of time in the case of the spring, and why can’t we do that in the case of that particle being attracted or repelled by another? Are there two realities here? No. There’s only. I’ve been playing a trick on you. Just think about what is actually happening and then think about that so-called ‘time reversal’:

  1. At point A, the spring is still being stretched further, in reality that is, and so the mass is moving away from the equilibrium position. Hence, in reality, it will not move to point B but further away from the equilibrium position.
  2. However, we could imagine it moving from point A to B if we would reverse the direction of the force. Indeed, the force is equal to –kx and reversing its direction is equivalent to flipping our graph around the horizontal axis (i.e. the time axis), or to shifting the time axis left or right with an amount equal to π (note that the ‘time’ axis is actually represented by the phase, but that’s a minor technical detail and it does not change the analysis: we just measure time in radians here instead of seconds).

It’s a visual trick. There is no ‘real’ symmetry. The flipped graph corresponds to another situation (i.e. some other spring that started oscillating a bit earlier or later than ours here). Hence, our conclusion that it is the force that gives time direction, still holds.

Hmm… Let’s think about this. What makes our ‘trick’ work is that the force is allowed to change direction. Well… If we go back to our previous example of an object falling towards the center of some gravitational field, or a charge being attracted by some other (opposite) charge, then you’ll note that we can make sense of the ‘left side’ of the graph if we would change the sign of the force.

Huh? Yes, I know. This is getting complicated. But think about it. The graph below might represent a charged particle being repelled by another (stationary) particle: that’s the green arrow. We can then go back in time (i.e. we reverse the green arrow) if we reverse the direction of the force from repulsion to attraction. Now, that would usually lead to a dramatic event—the end of the story to be precise. Indeed, once the two particles get together, they’re glued together and so we’d have to draw another horizontal line going in the minus t direction (i.e. to the left side of our time axis) representing the status quo. Indeed, if the two particles sit right on top of each other, or if they would literally fuse or annihilate each other (like a particle and an anti-particle), then there’s no force or anything left at all… except ifwe would alter the direction of the force once again, in which case the two particles would fly apart again (OK. OK. You’re right in noting that’s not true in the annihilation case – but that’s a minor detail).

arrow of time

Is this story getting too complicated? It shouldn’t. The point to note is that reversibility – i.e. time reversal in the philosophical meaning of the word (not that mathematical business of inserting negative variables instead of positive ones) – is all about changing the direction of the force: going back in time implies that we reverse the effects of time, and reversing the effects of time, requires forces acting in the opposite direction.

Now, when it’s only kinetic energy that is involved, then it should be easy but when charges are involved, which is the case for all fundamental forces, then it’s not so easy. That’s when charge (C) and parity (P) symmetry come into the picture.

CP symmetry

Hooke’s ‘Law’ – i.e. the law describing the force on a mass on a stretched or compressed spring – is not a fundamental law: eventually the spring will stop. Yes. It will stop even if when it’s in a horizontal position and with the mass moving on a frictionless surface, as assumed above: the forces between the atoms and/or molecules in the spring give the spring the elasticity which causes the mass to oscillate around some equilibrium position, but some of the energy of that continuous movement gets lost in heat energy (yes, an oscillating spring does actually get warmer!) and, hence, eventually the movement will peter out and stop.

Nevertheless, the lesson we learned above is a valuable one: when it comes to the fundamental forces, we can reverse the arrow of time and still make sense of it all if we also reverse the ‘charges’. The term ‘charges’ encompasses anything measuring a propensity to interact through one of the four fundamental forces here. That’s where CPT symmetry comes in: if we reverse time, we should also reverse the charges.

But how can we change the ‘sign’ of mass: mass is always positive, isn’t it? And what about the P-symmetry – this thing about left-handed and right-handed neutrinos?

Well… I don’t know. That’s the kind of stuff I am currently exploring in my quest. I’ll just note the following:

1. Gravity might be a so-called pseudo force – because it’s proportional to mass. I won’t go into the details of that – if only because I don’t master them as yet – but Einstein’s gut instinct that gravity is not a ‘real’ fundamental force (we just have to adjust our reference frame and work with curved spacetime) – and, hence, that ‘mass’ is not like the other force ‘charges’ – is something I want to further explore. [Apart from being a measure for inertia, you’ll remember that (rest) mass can also be looked at as equivalent to a very dense chunk of energy, as evidenced by Einstein’s energy-mass equivalence formula: E = mc2.]

As for now, I can only note that the particles in an ‘anti-world’ would have the same mass. In that sense, anti-matter is not ‘anti’-matter: it just carries opposite electromagnetic, strong and weak charges. Hence, our C-world (so the world we get when applying a charge transformation) would have all ‘charges’ reversed, but mass would still be mass.

2. As for parity symmetry (i.e. left- and right-handedness, aka as mirror symmetry), I note that it’s raised primarily in relation to the so-called weak force and, hence, it’s also a ‘charge’ of sorts—in my primitive view of the world at least. The illustration below shows what P symmetry is all about really and may or may not help you to appreciate the point.

muon decay

OK. What is this? Let’s just go step by step here.

The ‘cylinder’ (both in (a), the upper part of the illustration, and in (b), the lower part) represents a muon—or a bunch of muons actually. A muon is an unstable particle in the lepton family. Think of it as a very heavy electron for all practical purposes: it’s about 200 times the mass of an electron indeed. Its lifetime is fairly short from our (human) point of view–only 2.2 microseconds on average–but that’s actually an eternity when compared to other unstable particles.

In any case, the point to note is that it usually decays into (i) two neutrinos (one muon-neutrino and one electron-antineutrino to be precise) and – importantly – (ii) one electron, so electric charge is preserved (indeed, neutrinos got the name they have because they carry no electric charge).

Now, we have left- and right-handed muons, and we can actually line them up in one of these two directions. I would need to check how that’s done, but muons do have a magnetic moment (just like electrons) and so I must assume it’s done in the same way as in Wu’s cobalt-60 experiment: through a uniform magnetic field. In other words, we know their spin directions in an experiment like this.

Now, if the weak force would respect mirror symmetry (but we already know it doesn’t), we would not be able to distinguish (i) the muon decay process in the ‘mirror world’ (i.e. the reflection of what’s going on in the (imaginary) mirror in the illustration above) from (ii) the decay process in ‘our’ (real) world. So that would be situation (a): the number of decay electrons being emitted in an upward direction would be the same (more or less) as the amount of decay electrons being emitted in a downward direction.

However, the actual laboratory experiments show that situation (b) is actually the case: most of the electrons are being emitted in only one direction (i.e. the upward direction in the illustration above) and, hence, the weak force does not respect mirror symmetry.

So what? Is that a problem?

For eminent physicists such as Feynman, it is. As he writes in his concluding Lecture on mechanics, radiation and heat (Vol. I, Chapter 52: Symmetry in Physical Laws): “It’s like seeing small hairs growing on the north pole of a magnet but not on its south pole.” [He means it allows us to distinguish the north and the south pole of a magnet in some absolute sense. Indeed, if we’re not able to tell right from left, we’re also not able to tell north from south – in any absolute sense that is. But so the experiment shows we actually can distinguish the two in some kind of absolute sense.]

I should also note that Wolfgang Pauli, one of the pioneers of quantum mechanics, said that it was “total nonsense” when he was informed about Wu’s experimental results, and that repeated experiments were needed to actually convince him that we cannot just create a mirror world out of ours. 

For me, it is not a problem.I like to think of left- and right-handedness as some charge itself, and of the combined CPT symmetry as the only symmetry that matters really. That should be evident from my rather intuitive introduction on time symmetry above.

Consider it and decide for yourself how logical or illogical it is. We could define what Feynman refers to as an axial vector: watching that muon ‘from below’, we see that its spin is clockwise, and let’s use that fact to define an axial vector pointing in the same direction as the thick black arrow (it’s the so-called ‘right-hand screw rule’ really), as shown below.

Axial vector

Now, let’s suppose that mirror world actually exists, in some corner in the universe, and that a guy living in that ‘mirror world’ would use that very same ‘right-hand-screw rule’: his axial vector when doing this experiment would point in the opposite direction (see the thick black arrow in the mirror, which points in the opposite direction indeed). So what’s wrong with that?

Nothing – in my modest view at least. Left- and right-handedness can just be looked at as any other ‘charge’ – I think – and, hence, if we would be able to communicate with that guy in the ‘mirror world’, the two experiments would come out the same. So the other guy would also notice that the weak force does not respect mirror symmetry but so there’s nothing wrong with that: he and I should just get over it and continue to do business as usual, wouldn’t you agree?

After all, there could be a zillion reasons for the experiment giving the results it does: perhaps the ‘right-handed’ spin of the muon is sort of transferred to the electron as the muon decays, thereby giving it the same type of magnetic moment as the one that made the muon line up in the first place. Or – in a much wilder hypothesis which no serious physicist would accept – perhaps we actually do not yet understand everything of the weak decay process: perhaps we’ve got all these solar neutrinos (which all share the same spin direction) interfering in the process.

Whatever it is: Nature knows the difference between left and right, and I think there’s nothing wrong with that. Full stop.

But then what is ‘left’ and ‘right’ really? As the experiment pointed out, we can actually distinguish between the two in some kind of absolute sense. It’s not just a convention. As Feynman notes, we could decide to label ‘right’ as ‘left’, and ‘left’ as ‘right’ right here and right now – and impose the new convention everywhere – but then these physics experiments will always yield the same physical results, regardless of our conventions. So, while we’d put different stickers on the results, the laws of physics would continue to distinguish between left and right in the same absolute sense as Wu’s cobalt-60 decay experiment did back in 1956.

The really interesting thing in this rather lengthy discussion–in my humble opinion at least–is that imaginary ‘guy in the mirror world’. Could such mirror world exist? Why not? Let’s suppose it does really exist and that we can establish some conversation with that guy (or whatever other intelligent life form inhabiting that world).

We could then use these beta decay processes to make sure his ‘left’ and ‘right’ definitions are equal to our ‘left’ and ‘right’ definitions. Indeed, we would tell him that the muons can be left- or right-handed, and we would ask him to check his definition of ‘right-handed’ by asking him to repeat Wu’s experiment. And, then, when finally inviting him over and preparing to physically meet with him, we should tell him he should use his “right” hand to greet us. Yes. We should really do that.

Why? Well… As Feynman notes, he (or she or whatever) might actually be living in an anti-matter world, i.e. a world in which all charges are reversed, i.e. a world in which protons carry negative charge and electrons carry positive charge, and in which the quarks have opposite color charge. In that case, we would have been updating each other on all kinds of things in a zillion exchanges, and we would have been trying hard to assure each other that our worlds are not all that different (including that crucial experiment to make sure his left and right are the same as ours), but – if he would happen to live in an anti-matter world – then he would put out his left hand – not his right – when getting out of his spaceship. Touching it would not be wise. 🙂

[Let me be much more pedantic than Feynman is and just point out that his spaceship would obviously have been annihilated by ‘our’ matter long before he would have gotten to the meeting place. As soon as he’d get out of his ‘anti-matter’ world, we’d see a big flash of light and that would be it.]

Symmetries and conservation laws

A final remark should be made on the relation between all those symmetries and conservation laws. When everything is said and done, all that we’ve got is some nice graphs and then some axis or plane of symmetry (in two and three dimensions respectively). Is there anything more to it? There is.

There’s a “deep connection”, it seems, between all these symmetries and the various ‘laws of conservation’. In our examples of ‘time symmetry’, we basically illustrated the law of energy conservation:

  1. When describing a particle traveling through an electrostatic or gravitation field, we basically just made the case that potential energy is converted into kinetic energy, or vice versa.
  2. When describing an oscillating mass on a spring, we basically looked at the spring as a reservoir of energy – releasing and absorbing kinetic energy as the mass oscillates around its zero energy point – but, once again, all we described was a system in which the total amount of energy – kinetic and elastic – remained the same.

In fact, the whole discussion on CPT symmetry above has been quite simplistic and can be summarized as follows:

Energy is being conserved. Therefore, if you want to reverse time, you’ll need to reverse the forces as well. And reversing the forces implies a change of sign of the charges causing those forces.

In short, one should not be fascinated by T-symmetry alone. Combined CPT symmetry is much more intuitive as a concept and, hence, much more interesting. So, what’s left?

Quite a lot. I know you have many more questions at this point. At least I do:

  1. What does it mean in quantum mechanics? How does the Uncertainty Principle come into play?
  2. How does it work exactly for the strong force, or for the weak force? [I guess I’d need to find out more about neutrino physics here…]
  3. What about the other ‘conservation laws’ (such as the conservation of linear or angular momentum, for example)? How are they related to these ‘symmetries’.

Well… That’s complicated business it seems, and even Feynman doesn’t explore these topics in the above-mentioned final Lecture on (classical) mechanics. In any case, this post has become much too long already so I’ll just say goodbye for the moment. I promise I’ll get back to you on all of this.

Post scriptum:

If you have read my previous post (The Weird Force), you’ll wonder why – in the example of how a mirror world would relate to ours – I assume that the combined CP symmetry holds. Indeed, when discussing the ‘weird force’ (i.e. the weak force), I mentioned that it does not respect any of the symmetries, except for the combined CPT symmetry. So it does not respect (i) C symmetry, (ii) P symmetry and – importantly – it also does not respect the combined CP symmetry. This is a deep philosophical point which I’ll talk about in my next post. However, I needed this post as an ‘introduction’ to the next one.