# A physical explanation for relativistic length contraction?

My last posts were all about a possible physical interpretation of the quantum-mechanical wavefunction. To be precise, we have been interpreting the wavefunction as a gravitational wave. In this interpretation, the real and imaginary component of the wavefunction get a physical dimension: force per unit mass (newton per kg). The inspiration here was the structural similarity between Coulomb’s and Newton’s force laws. They both look alike: it’s just that one gives us a force per unit charge (newton per coulomb), while the other gives us a force per unit mass.

So… Well… Many nice things came out of this – and I wrote about that at length – but last night I was thinking this interpretation may also offer an explanation of relativistic length contraction. Before we get there, let us re-visit our hypothesis.

The geometry of the wavefunction

The elementary wavefunction is written as:

ψ = a·ei(E·t − px)/ħa·cos(px/ħ – E∙t/ħ) + i·a·sin(px/ħ – E∙t/ħ)

Nature should not care about our conventions for measuring the phase angle clockwise or counterclockwise and, therefore, the ψ = a·ei[E·t − px]/ħ function may also be permitted. We know that cos(θ) = cos(θ) and sinθ = sin(θ), so we can write:

ψ = a·ei(E·t − p∙x)/ħa·cos(E∙t/ħ – px/ħ) + i·a·sin(E∙t/ħ – px/ħ)

= a·cos(px/ħ – E∙t/ħ) i·a·sin(px/ħ – E∙t/ħ)

The vectors p and x are the the momentum and position vector respectively: p = (px, py, pz) and x = (x, y, z). However, if we assume there is no uncertainty about p – not about the direction nor the magnitude – then we may choose an x-axis which reflects the direction of p. As such, x = (x, y, z) reduces to (x, 0, 0), and px/ħ reduces to p∙x/ħ. This amounts to saying our particle is traveling along the x-axis or, if p = 0, that our particle is located somewhere on the x-axis. Hence, the analysis is one-dimensional only.

The geometry of the elementary wavefunction is illustrated below. The x-axis is the direction of propagation, and the y- and z-axes represent the real and imaginary part of the wavefunction respectively.

Note that, when applying the right-hand rule for the axes, the vertical axis is the y-axis, not the z-axis. Hence, we may associate the vertical axis with the cosine component, and the horizontal axis with the sine component. You can check this as follows: if the origin is the (x, t) = (0, 0) point, then cos(θ) = cos(0) = 1 and sin(θ) = sin(0) = 0. This is reflected in both illustrations, which show a left- and a right-handed wave respectively. We speculated this should correspond to the two possible values for the quantum-mechanical spin of the wave: +ħ/2 or −ħ/2. The cosine and sine components for the left-handed wave are shown below. Needless to say, the cosine and sine function are the same, except for a phase difference of π/2: sin(θ) = cos(θ − π/2).

As for the wave velocity, and its direction of propagation, we know that the (phase) velocity of any wave F(kx – ωt) is given by vp = ω/k = (E/ħ)/(p/ħ) = E/p. Of course, the momentum might also be in the negative x-direction, in which case k would be equal to -p and, therefore, we would get a negative phase velocity: vp = ω/k = E/p.

The de Broglie relations

E/ħ = ω gives the frequency in time (expressed in radians per second), while p/ħ = k gives us the wavenumber, or the frequency in space (expressed in radians per meter). Of course, we may write: f = ω/2π  and λ = 2π/k, which gives us the two de Broglie relations:

1. E = ħ∙ω = h∙f
2. p = ħ∙k = h/λ

The frequency in time is easy to interpret. The wavefunction of a particle with more energy, or more mass, will have a higher density in time than a particle with less energy.

In contrast, the second de Broglie relation is somewhat harder to interpret. According to the p = h/λ relation, the wavelength is inversely proportional to the momentum: λ = h/p. The velocity of a photon, or a (theoretical) particle with zero rest mass (m0 = 0), is c and, therefore, we find that p = mvv = mcc = m∙c (all of the energy is kinetic). Hence, we can write: p∙c = m∙c2 = E, which we may also write as: E/p = c. Hence, for a particle with zero rest mass, the wavelength can be written as:

λ = h/p = hc/E = h/mc

However, this is a limiting situation – applicable to photons only. Real-life matter-particles should have some mass[1] and, therefore, their velocity will never be c.[2]

Hence, if p goes to zero, then the wavelength becomes infinitely long: if p → 0 then λ → ∞. How should we interpret this inverse proportionality between λ and p? To answer this question, let us first see what this wavelength λ actually represents.

If we look at the ψ = a·cos(p∙x/ħ – E∙t/ħ) – i·a·sin(p∙x/ħ – E∙t/ħ) once more, and if we write p∙x/ħ as Δ, then we can look at p∙x/ħ as a phase factor, and so we will be interested to know for what x this phase factor Δ = p∙x/ħ will be equal to 2π. So we write:

Δ =p∙x/ħ = 2π ⇔ x = 2π∙ħ/p = h/p = λ

So now we get a meaningful interpretation for that wavelength. It is the distance between the crests (or the troughs) of the wave, so to speak, as illustrated below. Of course, this two-dimensional wave has no real crests or troughs: we measure crests and troughs against the y-axis here. Hence, our definition depend on the frame of reference.

Now we know what λ actually represents for our one-dimensional elementary wavefunction. Now, the time that is needed for one cycle is equal to T = 1/f = 2π·(ħ/E). Hence, we can now calculate the wave velocity:

v = λ/T = (h/p)/[2π·(ħ/E)] = E/p

Unsurprisingly, we just get the phase velocity that we had calculated already: v = vp = E/p. The question remains: what if p is zero? What if we are looking at some particle at rest? It is an intriguing question: we get an infinitely long wavelength, and an infinite wave velocity.

Now, re-writing the v = E/p as v = m∙c2/m∙vg  = cg, in which βg is the relative classical velocity[3] of our particle βg = vg/c) tells us that the phase velocities will effectively be superluminal (βg  < 1 so 1/ βg > 1), but what if βg approaches zero? The conclusion seems unavoidable: for a particle at rest, we only have a frequency in time, as the wavefunction reduces to:

ψ = a·e−i·E·t/ħ = a·cos(E∙t/ħ) – i·a·sin(E∙t/ħ)

How should we interpret this?

A physical interpretation of relativistic length contraction?

In my previous posts, we argued that the oscillations of the wavefunction pack energy. Because the energy of our particle is finite, the wave train cannot be infinitely long. If we assume some definite number of oscillations, then the string of oscillations will be shorter as λ decreases. Hence, the physical interpretation of the wavefunction that is offered here may explain relativistic length contraction.

🙂

[1] Even neutrinos have some (rest) mass. This was first confirmed by the US-Japan Super-Kamiokande collaboration in 1998. Neutrinos oscillate between three so-called flavors: electron neutrinos, muon neutrinos and tau neutrinos. Recent data suggests that the sum of their masses is less than a millionth of the rest mass of an electron. Hence, they propagate at speeds that are very near to the speed of light.

[2] Using the Lorentz factor (γ), we can write the relativistically correct formula for the kinetic energy as KE = E − E0 = mvc2 − m0c2 = m0γc2 − m0c2 = m0c2(γ − 1). As v approaches c, γ approaches infinity and, therefore, the kinetic energy would become infinite as well.

[3] Because our particle will be represented by a wave packet, i.e. a superimposition of elementary waves with different E and p, the classical velocity of the particle becomes the group velocity of the wave, which is why we denote it by vg.

# Quantum Mechanics: The Other Introduction

About three weeks ago, I brought my most substantial posts together in one document: it’s the Deep Blue page of this site. I also published it on Amazon/Kindle. It’s nice. It crowns many years of self-study, and many nights of short and bad sleep – as I was mulling over yet another paradox haunting me in my dreams. It’s been an extraordinary climb but, frankly, the view from the top is magnificent. 🙂

The offer is there: anyone who is willing to go through it and offer constructive and/or substantial comments will be included in the book’s acknowledgements section when I go for a second edition (which it needs, I think). First person to be acknowledged here is my wife though, Maria Elena Barron, as she has given me the spacetime and, more importantly, the freedom to take this bull by its horns.

Below I just copy the foreword, just to give you a taste of it. 🙂

# Foreword

Another introduction to quantum mechanics? Yep. I am not hoping to sell many copies, but I do hope my unusual background—I graduated as an economist, not as a physicist—will encourage you to take on the challenge and grind through this.

I’ve always wanted to thoroughly understand, rather than just vaguely know, those quintessential equations: the Lorentz transformations, the wavefunction and, above all, Schrödinger’s wave equation. In my bookcase, I’ve always had what is probably the most famous physics course in the history of physics: Richard Feynman’s Lectures on Physics, which have been used for decades, not only at Caltech but at many of the best universities in the world. Plus a few dozen other books. Popular books—which I now regret I ever read, because they were an utter waste of time: the language of physics is math and, hence, one should read physics in math—not in any other language.

But Feynman’s Lectures on Physics—three volumes of about fifty chapters each—are not easy to read. However, the experimental verification of the existence of the Higgs particle in CERN’s LHC accelerator a couple of years ago, and the award of the Nobel prize to the scientists who had predicted its existence (including Peter Higgs and François Englert), convinced me it was about time I take the bull by its horns. While, I consider myself to be of average intelligence only, I do feel there’s value in the ideal of the ‘Renaissance man’ and, hence, I think stuff like this is something we all should try to understand—somehow. So I started to read, and I also started a blog (www.readingfeynman.org) to externalize my frustration as I tried to cope with the difficulties involved. The site attracted hundreds of visitors every week and, hence, it encouraged me to publish this booklet.

So what is it about? What makes it special? In essence, it is a common-sense introduction to the key concepts in quantum physics. However, while common-sense, it does not shy away from the math, which is complicated, but not impossible. So this little book is surely not a Guide to the Universe for Dummies. I do hope it will guide some Not-So-Dummies. It basically recycles what I consider to be my more interesting posts, but combines them in a comprehensive structure.

It is a bit of a philosophical analysis of quantum mechanics as well, as I will – hopefully – do a better job than others in distinguishing the mathematical concepts from what they are supposed to describe, i.e. physical reality.

Last but not least, it does offer some new didactic perspectives. For those who know the subject already, let me briefly point these out:

I. Few, if any, of the popular writers seems to have noted that the argument of the wavefunction (θ = E·t – p·t) – using natural units (hence, the numerical value of ħ and c is one), and for an object moving at constant velocity (hence, x = v·t) – can be written as the product of the proper time of the object and its rest mass:

θ = E·t – p·x = E·t − p·x = mv·t − mv·v·x = mv·(t − v·x)

⇔ θ = m0·(t − v·x)/√(1 – v2) = m0·t’

Hence, the argument of the wavefunction is just the proper time of the object with the rest mass acting as a scaling factor for the time: the internal clock of the object ticks much faster if it’s heavier. This symmetry between the argument of the wavefunction of the object as measured in its own (inertial) reference frame, and its argument as measured by us, in our own reference frame, is remarkable, and allows to understand the nature of the wavefunction in a more intuitive way.

While this approach reflects Feynman’s idea of the photon stopwatch, the presentation in this booklet generalizes the concept for all wavefunctions, first and foremost the wavefunction of the matter-particles that we’re used to (e.g. electrons).

II. Few, if any, have thought of looking at Schrödinger’s wave equation as an energy propagation mechanism. In fact, when helping my daughter out as she was trying to understand non-linear regression (logit and Poisson regressions), it suddenly realized we can analyze the wavefunction as a link function that connects two physical spaces: the physical space of our moving object, and a physical energy space.

Re-inserting Planck’s quantum of action in the argument of the wavefunction – so we write θ as θ = (E/ħ)·t – (p/ħ)·x = [E·t – p·x]/ħ – we may assign a physical dimension to it: when interpreting ħ as a scaling factor only (and, hence, when we only consider its numerical value, not its physical dimension), θ becomes a quantity expressed in newton·meter·second, i.e. the (physical) dimension of action. It is only natural, then, that we would associate the real and imaginary part of the wavefunction with some physical dimension too, and a dimensional analysis of Schrödinger’s equation tells us this dimension must be energy.

This perspective allows us to look at the wavefunction as an energy propagation mechanism, with the real and imaginary part of the probability amplitude interacting in very much the same way as the electric and magnetic field vectors E and B. This leads me to the next point, which I make rather emphatically in this booklet:  the propagation mechanism for electromagnetic energy – as described by Maxwell’s equations – is mathematically equivalent to the propagation mechanism that’s implicit in the Schrödinger equation.

I am, therefore, able to present the Schrödinger equation in a much more coherent way, describing not only how this famous equation works for electrons, or matter-particles in general (i.e. fermions or spin-1/2 particles), which is probably the only use of the Schrödinger equation you are familiar with, but also how it works for bosons, including the photon, of course, but also the theoretical zero-spin boson!

In fact, I am personally rather proud of this. Not because I am doing something that hasn’t been done before (I am sure many have come to the same conclusions before me), but because one always has to trust one’s intuition. So let me say something about that third innovation: the photon wavefunction.

III. Let me tell you the little story behind my photon wavefunction. One of my acquaintances is a retired nuclear scientist. While he knew I was delving into it all, I knew he had little time to answer any of my queries. However, when I asked him about the wavefunction for photons, he bluntly told me photons didn’t have a wavefunction. I should just study Maxwell’s equations and that’s it: there’s no wavefunction for photons: just this traveling electric and a magnetic field vector. Look at Feynman’s Lectures, or any textbook, he said. None of them talk about photon wavefunctions. That’s true, but I knew he had to be wrong. I mulled over it for several months, and then just sat down and started doing to fiddle with Maxwell’s equations, assuming the oscillations of the E and B vector could be described by regular sinusoids. And – Lo and behold! – I derived a wavefunction for the photon. It’s fully equivalent to the classical description, but the new expression solves the Schrödinger equation, if we modify it in a rather logical way: we have to double the diffusion constant, which makes sense, because E and B give you two waves for the price of one!

[…]

In any case, I am getting ahead of myself here, and so I should wrap up this rather long introduction. Let me just say that, through my rather long journey in search of understanding – rather than knowledge alone – I have learned there are so many wrong answers out there: wrong answers that hamper rather than promote a better understanding. Moreover, I was most shocked to find out that such wrong answers are not the preserve of amateurs alone! This emboldened me to write what I write here, and to publish it. Quantum mechanics is a logical and coherent framework, and it is not all that difficult to understand. One just needs good pointers, and that’s what I want to provide here.

As of now, it focuses on the mechanics in particular, i.e. the concept of the wavefunction and wave equation (better known as Schrödinger’s equation). The other aspect of quantum mechanics – i.e. the idea of uncertainty as implied by the quantum idea – will receive more attention in a later version of this document. I should also say I will limit myself to quantum electrodynamics (QED) only, so I won’t discuss quarks (i.e. quantum chromodynamics, which is an entirely different realm), nor will I delve into any of the other more recent advances of physics.

In the end, you’ll still be left with lots of unanswered questions. However, that’s quite OK, as Richard Feynman himself was of the opinion that he himself did not understand the topic the way he would like to understand it. But then that’s exactly what draws all of us to quantum physics: a common search for a deep and full understanding of reality, rather than just some superficial description of it, i.e. knowledge alone.

So let’s get on with it. I am not saying this is going to be easy reading. In fact, I blogged about much easier stuff than this in my blog—treating only aspects of the whole theory. This is the whole thing, and it’s not easy to swallow. In fact, it may well too big to swallow as a whole. But please do give it a try. I wanted this to be an intuitive but formally correct introduction to quantum math. However, when everything is said and done, you are the only who can judge if I reached that goal.

Of course, I should not forget the acknowledgements but… Well… It was a rather lonely venture, so I am only going to acknowledge my wife here, Maria, who gave me all of the spacetime and all of the freedom I needed, as I would get up early, or work late after coming home from my regular job. I sacrificed weekends, which we could have spent together, and – when mulling over yet another paradox – the nights were often short and bad. Frankly, it’s been an extraordinary climb, but the view from the top is magnificent.

I just need to insert one caution, my site (www.readingfeynman.org) includes animations, which make it much easier to grasp some of the mathematical concepts that I will be explaining. Hence, I warmly recommend you also have a look at that site, and its Deep Blue page in particular – as that page has the same contents, more or less, but the animations make it a much easier read.

Have fun with it!

Jean Louis Van Belle, BA, MA, BPhil, Drs.

# Relativistic transformations of fields and the electromagnetic tensor

We’re going to do a very interesting piece of math here. It’s going to bring a lot of things together. The key idea is to present a mathematical construct that effectively presents the electromagnetic force as one force, as one physical reality. Indeed, we’ve been saying repeatedly that electromagnetism is one phenomenon only but we’ve been writing it always as something involving two vectors—the electric field vector E and the magnetic field vector B—and, while Lorentz’ force law F = q(E + v×B) makes it clear we’re talking one force only, there’s a way of writing it all up that is much more elegant.

I have to warn you though: this post doesn’t add anything to the physics we’ve seen so far: it’s all math, really and, to a large extent, math only. So if you read this blog because you’re interested in the physics only, then you may just as well skip this post. However, the mathematical concept we’re going to present is that of the tensor and… Well… You’ll have to get to know that animal sooner or later anyway, so you may just as well give it a try right now, and see whatever you can get out of this post.

The concept of a tensor further builds on the concept of the vector, which we liked so much because it allows us to write the laws of physics as vector equations, which do not change when going from one reference frame to another. In fact, we’ll see that a tensor can be described as a ‘special’ vector cross product (to be precise, we’ll show that a tensor is a ‘more general’ cross product, really). So the tensor and vector concepts are very closely related, but then… Well… If you think about it, the concept of a vector and the concept of a scalar are closely related, too! So we’re just moving up the value chain, so to speak: from scalar fields to vector fields to… Well… Tensor fields! And in quantum mechanics, we’ll introduce spinors, and so we also have spinor fields! Having said that, don’t worry about tensor fields. Let’s first try to understand tensors tout court. 🙂

So… Well… Here we go. Let me start with it all by reminding you of the concept of a vector, and why we like to use vectors and vector equations.

#### The invariance of physics and the use of vector equations

What’s a vector? You may think, naively, that any one-dimensional of numbers is a vector. But no! In math, we may call any one-dimensional array of numbers a ‘vector’, perhaps, but in physics, a vector does represent something real, something physical, and so a vector is only a vector if it transforms like a vector under the transformation rules that apply when going from one another frame of reference, i.e. one coordinate system, to another. Examples of vectors in three dimensions are: the velocity vector v, or the momentum vector p = m·v, or the position vector r.

[Needless to say, the same can be said of scalars: mathematicians may define a scalar as just any real number, but it’s not in physics. A scalar in physics refers to something real, i.e. a scalar field, like the temperature (T) inside of a block of material. In fact, think about your first vector equation: it may have been the one determining the heat flow (h), i.e. h = −κ·T = (−κ·∂T/∂x, −κ·∂T/∂y, −κ·∂T/∂z). It immediately shows how scalar and vector fields are intimately related.]

Now, when discussing the relativistic framework of physics, we introduced vectors in four dimensions, i.e. four-vectors. The most basic four-vector is the spacetime four-vector R = (ct, x, y, z), which is often referred to as an event, but it’s just a point in spacetime, really. So it’s a ‘point’ with a time as well as a spatial dimension, so it also has t in it, besides x, y and z. It is also known as the position four-vector but, again, you should think of a ‘position’ that includes time! Of course, we can re-write R as R = (ct, r), with r = (x, y, z), so here we sort of ‘break up’ the four-vector in a scalar and a three-dimensional vector, which is something we’ll do from time to time, indeed. 🙂

We also have a displacement four-vector, which we can write as ΔR = (c·Δt, Δr). There are other four-vectors as well, including the four-velocity, the four-momentum and the four-force four-vectors, which we’ll discuss later (in the last section of this post).

So it’s just like using three-dimensional vectors in three-dimensional physics, or ‘Newtonian’ physics, I should say: the use of four-vectors is going to allow us to write the laws of physics vector equations, but in four dimensions, rather than three, so we get the ‘Einsteinian’ physics, the real physics, so to speak—or the relativistically correct physics, I should say. And so these four-dimensional vector equations will also not change when going from one reference frame to another, and so our four-vector will be vectors indeed, i.e. they will transform like a vector under the transformation rules that apply when going from one another frame of reference, i.e. one coordinate system, to another.

What transformation? Well… In Newtonian or Galilean physics, we had translations and rotations and what have you, but what we are interested in here now, is ‘Einsteinian’ transformations of coordinate systems, so these have to ensure that all of the laws of physics that we know of, including the principle of relativity, still look the same. You’ve seen these transformation rules. We don’t call them the ‘Einsteinian’ transformation rules, but the Lorentz transformation rules, because it was a Dutch physicist (Hendrik Lorentz) who first wrote them down. So these rules are very different from the Newtonian or Galilean transformation rules which everyone assumed to be valid until the Michelson-Morley experiment unequivocally established that the speed of light did not respect the Galilean transformation rules. Very different? Well… Yes. In their mathematical structure, that is. Of course, when velocities are low, i.e. non-relativistic, then they yield the same result, approximately, that is. However, I explained that in my post on special relativity, and so I won’t dwell on that here.

Let me just jot down both sets of rules assuming 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.

The Galilean or Newtonian rules are the simple rules on the right. Going from one reference frame to another (let’s call them S and S’ respectively) is just a matter of adding or subtracting speeds: if my car goes 100 km/h, 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. We could also rotate our reference frame, and our Newtonian vector equations would still look the same. As Feynman notes, smilingly, it’s what a lot of armchair philosophers think relativity theory is all about, but so it’s got nothing to do with it. It’s plain wrong!

In any case, back to vectors and transformations. The key to the so-called invariance of the laws of physics is the use of vectors and vector operators that transform like vectors. For example, if we defined A and B as (Ax, Ay, Az) and (Bx, By, Bz), then we knew that the so-called inner product Awould look the same in all rotated coordinate systems, so we can write: AB A’•B’. So we know that if we have a product like that on both sides of an equation, we’re fine: the equation will have the same form in all rotated coordinate systems. Also, the gradient, i.e. our vector operator  = (∂/∂x, ∂/∂y, ∂/∂z), when applied to a scalar function, gave three quantities that also transform like a vector under rotation. We also defined a vector cross product, which yielded a vector (as opposed to the inner product, i.e. the vector dot product, which yields a scalar):

So how does this thing behave under a Galilean transformation? Well… You may or may not remember that we used this cross-product to define the angular momentum L, which was a cross product of the radius vector r and the momentum vector p = mv, as illustrated below. The animation also gives the torque τ, which is, loosely speaking, a measure of the turning force: it’s the cross product of r and F, i.e. the force on the lever-arm.

The components of L are:

Now, we find that these three numbers, or objects if you want, transform in exactly the same way as the components of a vector. However, as Feynman points out, that’s a matter of ‘luck’ really. It’s something ‘special’. Indeed, you may or may not remember that we distinguished axial vectors from polar vectors. L is an axial vector, while r and p are polar vectors, and so we find that, in three dimensions, the cross product of two polar vectors will always yields an axial vector. Axial vectors are sometimes referred to as pseudovectors, which suggests that they are ‘not so real’ as… Well… Polar vectors, which are sometimes referred to as ‘true’ vectors. However, it doesn’t matter when doing these Newtonian or Galilean transformations: pseudo or true, both vectors transform like vectors. 🙂

But so… Well… We’re actually getting a bit of a heads-up here: if we’d be mixing (or ‘crossing’) polar and axial vectors, or mixing axial vectors only, so if we’d define something involve and p (rather than r and p), or something involving and τ, then we may not be so lucky, and then we’d have to carefully examine our cross-product, or whatever other product we’d want to define, because its components may not behave like a vector.

Huh? Whatever other product we’d want to define? Why are you saying that? Well… We actually can think of other products. For example, if we have two vectors a = (ax, ay, az) and b = (bx, by, bz), then we’ll have nine possible combinations of their components, which we can write as Tij = aibj. So that’s like Lxy, Lyz and Lzx really. Now, you’ll say: “No. It isn’t. We don’t have nine combinations here. Just three numbers.” Well… Think about it: we actually do have nine Lij combinations too here, as we can write: Lij = ri·pj – rj·pi. It just happens that, with this definition, only three of these combinations Lij are independent. That’s because the other six numbers are either zero or the opposite. Indeed, it’s easy to verify that Lij = –Lji , and Lii  = 0. So… Well… It turns out that the three components of our L = r×p ‘vector’ are actually a subset of a set of nine Lij numbers. So… Well… Think about it. We cannot just do whatever we want with our ‘vectors’. We need to watch out.

In fact, I do not want to get too much ahead of myself, but I can already tell you that the matrix with these nine Tij = aibj combinations is what is referred to as the tensor. To be precise, it’s referred to as a tensor of the second rank in three dimensions. The ‘second rank’, aka as ‘degree’ or ‘order’ refers to the fact that we’ve got two indices, and the ‘three dimensions’ is because we’re using three-dimensional vectors. We’ll soon see that the electromagnetic tensor is also of the second rank, but it’s a tensor in four dimensions. In any case, I should not get ahead of myself. Just note what I am saying here: the tensor is like a ‘new’ product of two vectors, a new type of ‘cross’ product really (because we’re mixing the components, so to say), but it doesn’t yield a vector: it yields a matrix. For three-dimensional vectors, we get a 3×3 matrix. For four-vectors, we’ll get a 4×4 matrix. And so the full truth about our angular momentum vector L, is the following:

1. There is a thing which we call the angular momentum tensor. It’s a 3×3 matrix, so it has nine elements which are defined as: Lij = ri·pj – rj·pi. Because of this definition, it’s an antisymmetric tensor of the second order in three dimensions, so it’s got only three independent components.
2. The three independent elements are the components of our ‘vector’ L, and picking them out and calling these three components a ‘vector’ is actually a ‘trick’ that only works in three dimensions. They really just happen to transform like a vector under rotation or under whatever Galilean transformation! [By the way, do you know understand why I was saying that we can look at a tensor as a ‘more general’ cross product?]
3. In fact, in four dimensions, we’ll use a similar definition and define 16 elements Fij as Fij = ∇iAj − ∇jAi, using the two four-vectors ∇μ and Aμ (so we have 4×4 = 16 combinations indeed), out of which only six will be independent for the very same reason: we have an antisymmetric vector combination here, Fij = −Fji and Fii = 0. 🙂 However, because we cannot represent six independent things by four things, we do not get some other four-vector, and so that’s why we cannot apply the same ‘trick’ in four dimensions.

However, here I am getting way ahead of myself and so… Well… Yes. Back to the main story line. 🙂 So let’s try to move to the next level of understanding, which is… Well…

Because of guys like Maxwell and Einstein, we now know that rotations are part of the Newtonian world, in which time and space are neatly separated, and that things are not so simple in Einstein’s world, which is the real world, as far as we know, at least! 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 of x and t as well. [Yes, please scroll all the way up and have a look at the transformation on the left-hand side!]

So 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 quite appropriately, and so it’s space-time – i.e. ‘some kind of union of space and time’ as Minkowski termed it  that transforms.

So that’s why physicists use four-vectors to keep track of things. These four-vectors always have three space-like components, but they also include one so-called time-like componentIt’s the only way to ensure that the laws of physics are unchanged when moving with uniform velocityIndeed, any true law of physics we write down must be arranged so that the invariance of physics (as a “fact of Nature”, as Feynman puts it) is built in, and so that’s why we use Lorentz transformations and four-vectors.

In the mentioned post, I gave a few examples illustrating how the Lorentz rules work. Suppose we’re looking at some spaceship that is moving at half the speed of light (i.e. 0.5c) and that, inside the spaceship, some object is also moving at half the speed of light, as measured in the reference frame of the spaceship, then we get the rather remarkable result that, from our point of view (i.e. our reference frame as observer on the ground), that object is not going as fast as light, as Newton or Galileo – and most present-day armchair philosophers 🙂 – would predict (0.5+ 0.5c = c). We’d see it move at a speed equal to = 0.8c. Huh? How do we know that? Well… We can derive a velocity formula from the Lorentz rules:

So now you can just put in the numbers now: vx = (0.5c + 0.5c)/(1 + 0.5·0.5) = 0.8c. See?

Let’s do another example. 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 us? The Galilean transformation rules say its speed should be 1.5c, but that can’t be true of course, and the Lorentz rules save us once more: vx = (0.5c + c)/(1 + 0.5·1) = c, so it turns out that the speed of light does not depend 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!” 🙂

So let’s now apply relativity to electromagnetism. Indeed, that’s what this post is all about! However, before I do so, let me re-write the Lorentz transformation rules for = 1. We can equate the speed of light to one, indeed, when measure time and distance in equivalent units. It’s just a matter of ditching our seconds for meters (so our time unit becomes the time that light needs to travel a distance of one meter), or ditching our meters for seconds (so our distance unit becomes the distance that light travels in one second). You should be familiar with this procedure. If not, well… Check out my posts on relativity. So here’s the same set of rules for = 1:

They’re much easier to remember and work with, and so that’s good, because now we need to look at how these rules work with four-vectors and the various operations and operators we’ll be defining on them. Let’s look at that step by step.

#### Electrodynamics in relativistic notation

Let me copy the Universal Set of Equations and Their Solution once more:

The solution for Maxwell’s equations is given in terms of the (electric) potential Φ and the (magnetic) vector potential A. I explained that in my post on this, so I won’t repeat myself too much here either. The only point you should note is that this solution is the result of a special choice of Φ and A, which we referred to as the Lorentz gauge. We’ll touch upon this condition once more, so just make a mental note of it.

Now, E and B do not correspond to four-vectors: they depend on x, y, z and t, but they have three components only: Ex, Ey, Ez, and Bx, By, and Bz respectively. So we have six independent terms here, rather than four things that, somehow, we could combine into some four-vector. [Does this ring a bell? It should. :-)] Having said that, it turns out that we can combine Φ and A into a four-vector, which we’ll refer to as the four-potential and which we’ll will write as:

Aμ = (Φ, A) = (Φ, Ax, Ay, Az) = (At, Ax, Ay, Az) with At = Φ.

So that’s a four-vector just like R = (ct, x, y, z).

How do we know that Aμ is a four-vector? Well… Here I need to say a few things about those Lorentz transformation rules and, more importantly, about the required condition of invariance under a Lorentz transformation. So, yes, here we need to dive into the math.

#### Four-vectors and invariance under Lorentz transformations

When you were in high-school, you learned how to rotate your coordinate frame. You also learned that the distance of a point from the origin does not change under a rotation, so you’d write r’= x’+ y’+ z’= r= x+ y+ z2, and you’d say that r2 is an invariant quantity under a rotation. Indeed, transformations leave certain things unchanged. From the Lorentz transformation rules itself, it is easy to see that

c·t’– x’– y’–z ‘2 = c·t–x– y – z2, or,

if = 1, that t’– x’– y’– z’2 = t– x– y – z2,

is an invariant under a Lorentz transformation. We found the same for the so-called spacetime interval Δs = ΔrcΔt2, which we write as Δs = Δr– Δt2 as we chose our time or distance units such that = 1. [Note that, from now on, we’ll assume that’s the case, so = 1 everywhere. We can always change back to our old units when we’re done with the analysis.] Indeed, such invariance allowed us to define spaceliketimelike and lightlike intervals using the so-called light cone emanating from a single event and traveling in all directions.

You should note that, for four-vectors, we do not have a simple sum of three terms. Indeed, we don’t write x+ y+ z2 but t– x– y – z2. So we’ve got a +−−− thing here or, it’s just another convention, we could also work with a −+++ sum of terms. The convention is referred to as the signature, and we will use the so-called metric signature here, which is +−−−. Let’s continue the story. Now, all four-vectors aμ = (at, ax, ay, az) have this property that:

at– ax– ay– az2 = at– ax– ay – az2.

So. Yes. 🙂 But… Well… We can say that our four-potential vector is a four-vector, but so we still have to prove that. So we need to prove that Φ’– Ax– Ay– Az2 = Φ– Ax– Ay – Az2 for our four-potential vector Aμ = (Φ, A). So… Yes… How can we do that? The proof is not so easy, but you need to go through it as it will introduce some more concepts and ideas you need to understand.

In my post on the Lorentz gauge, I mentioned that Maxwell’s equations can be re-written in terms of Φ and A, rather than in terms of E and B. The equations are:

The expression look rather formidable, but don’t panic: just look at it. Of course, you need to be familiar with the operators that are being used here, so that’s the Laplacian ∇2 and the divergence operator • that’s being applied to the scalar Φ and the vector A. I can’t re-explain this. I am sorry. Just check my posts on vector analysis. You should also look at the third equation: that’s just the Lorentz gauge condition, which we introduced when deriving these equations from Maxwell’s equations. Having said that, it’s the first and second equation which describe Φ and A as a function of the charges and currents in space, and so that’s what matters here. So let’s unfold the first equation. It says the following:

In fact, if we’d be talking free or empty space, i.e. regions where there are no charges and currents, then the right-hand side would be zero and this equation would then represent a wave equation, so some potential Φ that is changing in time and moving out at the speed c. Here again, I am sorry I can’t write about this here: you’ll need to check one of my posts on wave equations. If you don’t want to do that, you should believe me when I say that, if you see an equation like this:

then the function Ψ(x, t) must be some function

Now, that’s a function representing a wave traveling at speed c, i.e. the phase velocity. Always? Yes. Always! It’s got to do with the x − ct and/or x + ct  argument in the function. But, sorry, I need to move on here.

The unfolding of the equation with Φ makes it clear that we have four equations really. Indeed, the second equation is three equations: one for Ax, one for Ay, and one for Az respectively. The four quantities on the right-hand side of these equations are ρ, jx, jy and jz respectively, divided by ε0, which is a universal constant which does not change when going from one coordinate system to another. Now, the quantities ρ, jx, jy and jz transform like a four-vector. How do we know that? It’s just the charge conservation law. We used it when solving the problem of the fields around a moving wire, when we demonstrated the relativity of the electric and magnetic field. Indeed, the relevant equations were:

You can check that against the Lorentz transformation rules for = 1. They’re exactly the same, but so we chose t = 0, so the rules are even simpler. Hence, the (ρ, jx, jy, jz) vector is, effectively, a four-vector, and we’ll denote it by jμ = (ρ, j). I now need to explain something else. [And, yes, I know this is becoming a very long story but… Well… That’s how it is.]

It’s about our operators , ∇•, × and ∇, so that’s the gradient, the divergence, curl and Laplacian operator respectively: they all have a four-dimensional equivalent. Of course, that won’t surprise you. 😦 Let me just jot all of them down, so we’re done with that, and then I’ll focus on the four-dimensional equivalent of the Laplacian  ∇•∇ = ∇, which is referred to as the D’Alembertian, and which is denoted by 2, because that’s the one we need to prove that our four-potential vector is a real four-vector. [I know: is a tiny symbol for a pretty monstrous thing, but I can’t help it: my editor tool is pretty limited.]

Now, we’re almost there. Just hang in for a little longer. It should be obvious that we can re-write those two equations with Φ, A, ρ and j, as:

Just to make sure, let me remind you that Aμ = (Φ, A) and that jμ = (ρ, j). Now, our new D’Alembertian operator is just an operator—a pretty formidable operator but, still, it’s an operator, and so it doesn’t change when the coordinate system changes, so the conclusion is that, IF jμ = (ρ, j) is a four-vector – which it is – and, therefore, transforms like a four-vector, THEN the quantities Φ, Ax, Ay, and Az must also transform like a four-vector, which means they are (the components of) a four-vector.

So… Well… Think about it, but not too long, because it’s just an intermediate result we had to prove. So that’s done. But we’re not done here. It’s just the beginning, actually. Let me repeat our intermediate result:

Aμ = (Φ, A) is a four-vector. We call it the four-potential vector.

OK. Let’s continue. Let me first draw your attention to that expression with the D’Alembertian above. Which expression? This one:

What about it? Well… You should note that the physics of that equation is just the same as Maxwell’s equations. So it’s one equation only, but it’s got it all.

It’s quite a pleasure to re-write it in such elegant form. Why? Think about it: it’s a four-vector equation: we’ve got a four-vector on the left-hand side, and a four-vector on the right-hand side. Therefore, this equation is invariant under a transformation. So, therefore, it directly shows the invariance of electrodynamics under the Lorentz transformation.

To wrap this up, I should also note that we can also express the gauge condition using our new four-vector notation. Indeed, we can write it as:

It’s referred to as the Lorentz condition and it is, effectively, a condition for invariance, i.e. it ensures that the four-vector equation above does stay in the form it is in for all reference frames. Note that we’re re-writing it using the four-dimensional equivalent of the divergence operator •, but so we don’t have a dot between ∇μ and Aμ. In fact, the notation is pretty confusing, and it’s easy to think we’re talking some gradient, rather than the divergence. So let me therefore highlight the meaning of both once again. It looks the same, but it’s two very different things: the gradient operates on a scalar, while the divergence operates on a (four-)vector. Also note the +−−− signature is only there for the gradient, not for the divergence!

You’ll wonder why they didn’t use some • or ∗ symbol, and the answer: I don’t know. I know it’s hard to keep inventing symbols for all these different ‘products’ – the ⊗ symbol, for example, is reserved for tensor products, which we won’t get into – but… Well… I think they could have done something here. 😦

In any case… Let’s move on. Before we do, please note that we can also re-write our conservation law for electric charge using our new four-vector notation. Indeed, you’ll remember that we wrote that conservation law as:

Using our new four-vector operator ∇μ, we can re-write that as ∇μjμ = 0. So all of electrodynamics can be summarized in the two equations only—Maxwell’s law and the charge conservation law:

OK. We’re now ready to discuss the electromagnetic tensor. [I know… This is becoming an incredibly long and incredibly complicated piece but, if you get through it, you’ll admit it’s really worth it.]

#### The electromagnetic tensor

The whole analysis above was done in terms of the Φ and A potentials. It’s time to get back to our field vectors E and B. We know we can easily get them from Φ and A, using the rules we mentioned as solutions:

These two equations should not look as yet another formula. They are essential, and you should be able to jot them down anytime anywhere. They should be on your kitchen door, in your toilet and above your bed. 🙂 For example, the second equation gives us the components of the magnetic field vector B:

Now, look at these equations. The x-component is equal to a couple of terms that involve only y– and z-components. The y-component is equal to something involving only x and z. Finally, the z-component only involves x and y. Interesting. Let’s define a ‘thing’ we’ll denote by Fzy and define as:

So now we can write: Bx = Fzy, By = Fxz, and Bz = Fxy. Now look at our equation for E. It turns out the components of E are equal to things like Fxt, Fyt and Fzt! Indeed, Fxt = ∂Ax/∂t − ∂At/∂x = Ex!

But… Well… No. 😦 The sign is wrong! Ex = −∂Ax/∂t−∂At/∂x, so we need to modify our definition of Fxt. When the t-component is involved, we’ll define our ‘F-things’ as:

So we’ve got a plus instead of a minus. It looks quite arbitrary but, frankly, you’ll have to admit it’s sort of consistent with our +−−− signature for our four-vectors and, in just a minute, you’ll see it’s fully consistent with our definition of the four-dimensional vector operator ∇μ = (∂/∂t, −∂/∂x, −∂/∂y, −∂/∂z). So… Well… Let’s go along with it.

What about the Fxx, Fyy, Fzz and Ftt terms? Well… Fxx = ∂Ax/∂x − ∂Ax/∂x = 0, and it’s easy to see that Fyy and Fzz are zero too. But Ftt? Well… It’s a bit tricky but, applying our definitions carefully, we see that Ftt must be zero too. In any case, the Ftt = 0 will become obvious as we will be arranging these ‘F-things’ in a matrix, which is what we’ll do now. [Again: does this ring a bell? If not, it should. :-)]

Indeed, we’ve got sixteen possible combinations here, which Feynman denotes as Fμν, which is somewhat confusing, because Fμν usually denotes the 4×4 matrix representing all of these combinations. So let me use the subscripts i and j instead, and define Fij as:

Fij = ∇iAj − ∇jAi

with ∇i being the t-, x-, y- or z-component of ∇μ = (∂/∂t, −∂/∂x, −∂/∂y, −∂/∂z) and, likewise, Ai being the t-, x-, y- or z-component of Aμ = (Φ, Ax, Ay, Az). Just check it: Fzy = −∂Ay/∂z + ∂Az/∂y = ∂Az/∂y − ∂Ay/∂z = Bx, for example, and Fxt = −∂Φ/∂x − ∂Ax/∂t = Ex. So the +−−− convention works. [Also note that it’s easier now to see that Ftt = ∂Φ/∂t − ∂Φ/∂t = 0.]

We can now arrange the Fij in a matrix. This matrix is antisymmetric, because Fij = – Fji, and its diagonal elements are zero. [For those of you who love math: note that the diagonal elements of an antisymmetric matrix are always zero because of the Fij = – Fji constraint: just use k = i = j in the constraint.]

Now that matrix is referred to as the electromagnetic tensor and it’s depicted below (we plugged back in, remember that B’s magnitude is 1/c times E’s magnitude).

So… Well… Great ! We’re done! Well… Not quite. 🙂

We can get this matrix in a number of ways. The least complicated way is, of course, just to calculate all Fij components and them put them in a [Fij] matrix using the as the row number and the as the column number. You need to watch out with the conventions though, and so i and j start on t and end on z. 🙂

The other way to do it is to write the ∇μ = (∂/∂t, −∂/∂x, −∂/∂y, −∂/∂z) operator as a 4×1 column vector, which you then multiply with the four-vector Aμ written as a 4×1 row vector. So ∇μAμ is then a 4×4 matrix, which we combine with its transpose, i.e. (∇μAμ)T, as shown below. So what’s written below is (∇μAμ) − (∇μAμ)T.

If you google, you’ll see there’s more than one way to go about it, so I’d recommend you just go through the motions and double-check the whole thing yourself—and please do let me know if you find any mistake! In fact, the Wikipedia article on the electromagnetic tensor denotes the matrix above as Fμν, rather than as Fμν, which is the same tensor but in its so-called covariant form, but so I’ll refer you to that article as I don’t want to make things even more complicated here! As said, there’s different conventions around here, and so you need to double-check what is what really. 🙂

Where are we heading with all of this? The next thing is to look at the Lorentz transformation of these Fij = ∇iAj − ∇jAcomponents, because then we know how our E and B fields transform. Before we do so, however, we should note the more general results and definitions which we obtained here:

1. The Fμν matrix (a matrix is just a multi-dimensional array, of course) is a so-called tensor. It’s a tensor of the second rank, because it has two indices in it. We think of it as a very special ‘product’ of two vectors, not unlike the vector cross product a × b, whose components were also defined by a similar combination of the components of a and b. Indeed, we wrote:

So one should think of a tensor as “another kind of cross product” or, preferably, and as Feynman puts it, as a “generalization of the cross product”.

2. In this case, the four-vectors are ∇μ = (∂/∂t, −∂/∂x, −∂/∂y, −∂/∂z) and Aμ = (Φ, Ax, Ay, Az). Now, you will probably say that ∇μ is an operator, not a vector, and you are right. However, we know that ∇μ behaves like a vector, and so this is just a special case. The point is: because the tensor is based on four-vectors, the Fμν tensor is referred to as a tensor of the second rank in four dimensions. In addition, because of the Fij = – Fji result, Fμν is an asymmetric tensor of the second rank in four dimensions.

3. Now, the whole point is to examine how tensors transform. We know that the vector dot product, aka the inner product, remains invariant under a Lorentz transformation, both in three as well as in four dimensions, but what about the vector cross product, and what about the tensor? That’s what we’ll be looking at now.

#### The Lorentz transformation of the electric and magnetic fields

Cross products are complicated, and tensors will be complicated too. Let’s recall our example in three dimensions, i.e. the angular momentum vector L, which was a cross product of the radius vector r and the momentum vector p = mv, as illustrated below (the animation also gives the torque τ, which is, loosely speaking, a measure of the turning force).

The components of L are:

Now, this particular definition ensures that Lij turns out to be an antisymmetric object:

So it’s a similar situation here. We have nine possible combinations, but only three independent numbers. So it’s a bit like our tensor in four dimensions: 16 combinations, but only 6 independent numbers.

Now, it so happens that that these three numbers, or objects if you want, transform in exactly the same way as the components of a vector. However, as Feynman points out, that’s a matter of ‘luck’ really. In fact, Feynman points out that, when we have two vectors a = (ax, ay, az) and b = (bx, by, bz), we’ll have nine products Tij = aibj which will also form a tensor of the second rank (cf. the two indices) but which, in general, will not obey the transformation rules we got for the angular momentum tensor, which happened to be an antisymmetric tensor of the second rank in three dimensions.

To make a long story short, it’s not simple in general, and surely not here: with E and B, we’ve got six independent terms, and so we cannot represent six things by four things, so the transformation rules for E and B will differ from those for a four-vector. So what are they then?

Well… Feynman first works out the rules for the general antisymmetric vector combination Gij = aib− ajbi, with aand bj the t-, x-, y- or z-component of the four-vectors aμ = (at, ax, ay, az) and bμ = (bt, bx, by, bz) respectively. The idea is to first get some general rules, and then replace Gij = aib− ajbi by Fij = ∇iAj − ∇jAi, of course! So let’s apply the Lorentz rules, which – let me remind you – are the following ones:

So we get:

The rest is all very tedious: you just need to plug these things into the various Gij = aib− ajbi formulas. For example, for G’tx, we get:

Hey! That’s just G’tx, so we find that G’tx = Gtx! What about the rest? Well… That yields something different. Let me shorten the story by simply copying Feynman here:

So… Done!

So what?

Well… Now we just substitute. In fact, there are two alternative formulations of the Lorentz transformations of E and B. They are given below (note the units are such that c = 1):

In addition, there is a third equivalent formulation which is more practical, and also simpler, even if it puts the c‘s back in. It re-defines the field components, distinguishing only two:

1. The ‘parallel’ components E|| and B|| along the x-direction ( because they are parallel to the relative velocity of the S and S’ reference frames), and
2. The ‘perpendicular’ or ‘total transverse’ components E and B, which are the vector sums of the y- and z-components.

So that gives us four equations only:

And, yes, we are done now. This is the Lorentz transformation of the fields. I am sure it has left you totally exhausted. Well… If not… […] It sure left me totally exhausted. 🙂

To lighten things up, let me insert an image of how the transformed field E actually looks like. The first image is the reference frame of a charge itself: we have a simple Coulomb field. The second image shows the charge flying by. Its electric field is ‘squashed up’. To be precise, it’s just like the scale of is squashed up by a factor ((1−v2/c2)1/2. Let me refer you to Feynman for the detail of the calculations here.

OK. So that’s it. You may wonder: what about that promise I made? Indeed, when I started this post, I said I’d present a mathematical construct that presents the electromagnetic force as one force only, as one physical reality, but so we’re back writing all of it in terms of two vectors—the electric field vector E and the magnetic field vector B. Well… What can I say? I did present the mathematical construct: it’s the electromagnetic tensor. So it’s that antisymmetric matrix really, which one can combine with a transformation matrix embodying the Lorentz transformation rules. So, I did what I promised to do. But you’re right: I am re-presenting stuff in the old style once again.

The second objection that you may have—in fact, that you should have, is that all of this has been rather tedious. And you’re right. The whole thing just re-emphasizes the value of using the four-potential vector. It’s obviously much easier to take that vector from one reference frame to another – so we just apply the Lorentz transformation rules to Aμ = (Φ, A) and get Aμ‘ = (Φ’, A’) from it – and then calculate E’ and B’ from it, rather than trying to remember those equations above. However, that’s not the point, or…

Well… It is and it isn’t. We wanted to get away from those two vectors E and B, and show that electromagnetism is really one phenomenon only, and so that’s where the concept of the electromagnetic tensor came in. There were two objectives here: the first objective was to introduce you to the concept of tensors, which we’ll need in the future. The second objective was to show you that, while Lorentz’ force law – F = q(E + v×B) makes it clear we’re talking one force only, there is a way of writing it all up that is much more elegant.

I’ve introduced the concept of tensors here, so the first objective should have been achieved. As for the second objective, I’ll discuss that in my next post, in which I’ll introduce the four-velocity vector μμ as well as the four-force vector fμ. It will explain the following beautiful equation of motion:

Now that looks very elegant and unified, doesn’t it? 🙂

[…] Hmm… No reaction. I know… You’re tired now, and you’re thinking: yet another way of representing the same thing? Well… Yes! So…

OK… Enough for today. Let’s follow up tomorrow.

My son, who’s fifteen, said he liked my post on lasers. That’s good, because I effectively wrote it thinking of him as part of the audience. He also said it stimulated him to considering taking on studies in engineering later. That’s great. I hope he does, so he doesn’t have to go through what I am going through right now. Indeed, when everything is said and done, you do want your kids to take on as much math and science they can handle when they’re young because, afterwards, it’s tough to catch up.

Now, I struggled quite a bit with bringing relativity into the picture while pondering the ‘essence’ of a photon in my previous post. Hence, I’d thought it would be good to return to the topic of (special) relativity and write another post to (1) refresh my knowledge on the topic and (2) try to stimulate him even more. Indeed, regardless of whether one does or doesn’t understand any of what I write below here, relativity theory sounds fascinating, doesn’t it? 🙂 So, this post intends to present, in a nutshell, what (special) relativity theory is all about.

What relativity does

The thing that’s best known about Einstein’s (special) theory of relativity is the following: the mass of an object, as measured by the (inertial) observer, increases with its speed. The formula for this is m = γm0, and the γ factor here is the so-called Lorentz factor: γ = (1–u2/c2)–1/2. Let me give you that diagram of the Lorentz factor once again, which shows that very considerable speeds are required before relativity effects kick in. However, when they do, they kick in with a vengeance, it seems, which makes c the limit !

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 stick of one meter 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 relativistic speeds (i.e. a velocity sufficiently close to to make the relativistic effect significant), he will come back younger than his brother. The formula here is equally simple: t = γt0. 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 – again: to the external (inertial) observer that is.

These simple rules, which comes out of Einsteins’ special relativity theory, give rise to all kinds of paradoxes. You know what a paradox is: a paradox (in physics) is something that, at first sight, does not make sense but that, when the issue is examined more in detail, does get resolved and actually helps us to better understand what’s going on.

You know the twin paradox already: only of the two twins can be the younger (or the older) when they meet again. However, because one can also say it’s the guy staying on Earth that’s moving (and, hence, is ‘traveling’ at relativistic speed) – so then the reference frame of the guy in the spaceship is the so-called inertial frame, one can say the guy who stayed behind (on Earth) should be the youngest when they meet after the journey. I am not ashamed to say that this actually is a paradox that is difficult to understand. So let me first start with another.

The paradox pushes us to consider all kinds of important questions which are usually just glossed over. How does we decide if the ladder fits? Well… By closing both the front and back door of course, you’ll say. But then you mean closing them simultaneously, and absolute simultaneity does not exist: two events that appear to happen at the same time in one reference frame may not happen at the same time in another. Only the space-time interval between two events is absolute, in the sense that it’s the same in whatever reference frame we’re measuring it, not the individual space and individual time intervals. Hence, if you’re in the garage shutting those doors at the same time, then that’s your time, but if I am moving with the ladder, I will not see those two doors shutting as something that’s simultaneous. More formally, and using the definition of space-time intervals (and assuming only one space dimension x), we have:

cΔt– Δx= cΔt’– Δx’2.

In this equation, we’ll take the x and t coordinates to be those of the inertial frame (so that’s the garage on the left-hand side), while the the primed coordinates (x’ and t’) are the coordinates as measured in the other reference frame, i.e. the reference frame that moves from the perspective of the inertial frame. Indeed, note that we cannot say that one reference frame moves while the other stands still as we we’re talking relative speeds here: one reference frame moves in respect to the other, and vice versa. In any case, the equation with the space-time intervals above implies that:

c(ΔtΔt’2) – (Δx– Δx’2) = 0

However, that does not imply that the two terms on the left-hand side of the above equation are zero individually. In fact, they aren’t. Hence, while it must be true that c(ΔtΔt’2) = Δx– Δx’2, we have:

ΔtΔt’≠ 0 and Δx– Δx’2 ≠ 0 or ΔtΔt’and Δx≠ Δx’2

To put it simply, if you’re in the garage, and I am moving with the ladder (we’re talking the left-hand side situation) now, you’ll claim that you were able to shut both doors momentarily, so that Δt= 0. I’ll say: bollocks! Which is rude. I should say: my Δt’is not equal to zero. Hence, from my point of view, I always saw one of the two doors open and, hence, I don’t think the ladder fits. Hence, what I am seeing, effectively, is the situation on the right-hand side: your garage looks too short for my ladder.

You’ll say: what is this? The ladder fits or it doesn’t, does it? The answer is: no. It is ambiguous. It does depend on your reference frame. It fits in your reference frame but it does not fit in mine. In order to get a non-ambiguous answer you have to stop moving, or I have to stop moving– whatever: the point is that we need to merge our reference frames.

Hence, paradox solved. In fact, now that I think of it, it’s kinda funny that we don’t have such paradoxes for the relativistic mass formula. No one seems to wonder about the apparent contradiction that, if you’re moving away from me, you look heavier than me but that, vice versa, I also look heavier to you. So we both look heavier as seen from our own respective reference frames. So who’s heavier then? Perhaps no one developed a paradox because it is kinda impolite to compare personal weights? 🙂

Of course, I am joking, but think of it: it has to do with our preconceived notions of time and space. Things like inertia (mass is a measure for inertia) don’t grab our attention as much. In any case, now it’s time to discuss time dilation.

Oh ! And do think about that photo-finish picture ! It’s related to the problem of defining what constitutes a length really. 🙂

I find the twin paradox much more difficult to analyze, and I guess many people do because it’s the one that usually receives all of the attention. [Frankly, I hadn’t heard of this ladder paradox before I started studying physics.] Feynman hardly takes the time to look at it. He basically notes that the situation is not unlike an unstable particle traveling at relativistic speeds: when it does, it lasts (much) longer that its lifetime (measured in the inertial reference frame) suggests. Let me actually just quote Feynman’s account of it:

“Peter and Paul are supposed to be twins, born at the same time. When they are old enough to drive a space ship, Paul flies away at very high speed. Because Peter, who is left on the ground, sees Paul going so fast, all of Paul’s clocks appear to go slower, his heart beats go slower, his thoughts go slower, everything goes slower, from Peter’s point of view. Of course, Paul notices nothing unusual, but if he travels around and about for a while and then comes back, he will be younger than Peter, the man on the ground! That is actually right; it is one of the consequences of the theory of relativity which has been clearly demonstrated. Just as the mu-mesons last longer when they are moving, so also will Paul last longer when he is moving. This is called a “paradox” only by the people who believe that the principle of relativity means that all motion is relative; they say, “Heh, heh, heh, from the point of view of Paul, can’t we say that Peter was moving and should therefore appear to age more slowly? By symmetry, the only possible result is that both should be the same age when they meet.” But in order for them to come back together and make the comparison, Paul must either stop at the end of the trip and make a comparison of clocks or, more simply, he has to come back, and the one who comes back must be the man who was moving, and he knows this, because he had to turn around. When he turned around, all kinds of unusual things happened in his space ship—the rockets went off, things jammed up against one wall, and so on—while Peter felt nothing.

So the way to state the rule is to say that the man who has felt the accelerations, who has seen things fall against the walls, and so on, is the one who would be the younger; that is the difference between them in an “absolute” sense, and it is certainly correct. When we discussed the fact that moving mu-mesons live longer, we used as an example their straight-line motion in the atmosphere. But we can also make mu-mesons in a laboratory and cause them to go in a curve with a magnet, and even under this accelerated motion, they last exactly as much longer as they do when they are moving in a straight line. Although no one has arranged an experiment explicitly so that we can get rid of the paradox, one could compare a mu-meson which is left standing with one that had gone around a complete circle, and it would surely be found that the one that went around the circle lasted longer. Although we have not actually carried out an experiment using a complete circle, it is really not necessary, of course, because everything fits together all right. This may not satisfy those who insist that every single fact be demonstrated directly, but we confidently predict the result of the experiment in which Paul goes in a complete circle.”

[…] Well… I am not sure I am “among those who insist that every single fact be demonstrated directly”, but you’ll admit that Feynman is quite terse here (or more terse than usual, I should say). That being said, I understand why: the calculations involved in demonstrating that the paradox is what it is, i.e. an apparent contradiction only, are not straightforward. I’ve googled a bit but it’s all quite confusing. Good explanations usually involve the so-called Minkowski diagram, also known as the spacetime diagram. You’ve surely seen it before–when the light cone was being discussed and what it implies for the concepts of past, present and future. It’s a way to represent those spacetime intervals. The Minkowski diagram–from the perspective of the twin brother on Earth (hence, we only have unprimed coordinates x and (c)t)– is shown below. Don’t worry about those simultaneity planes as for now. Just try to understand the diagram. The twin brother that stays just moves along the vertical axis: x = 0. His space-traveling brother travels out to some point and then turns back, so he first travels northeast on this diagram and then takes a turn northwest, to meet up again with his brother on Earth.

The point to note is that the twin brother is not traveling along one straight line, but along two. Hence, the argument that we can just as well say his frame of reference is inertial and that of his brother is the moving one is not correct. As Wikipedia notes (from which I got this diagram): “The trajectory of the ship is equally divided between two different inertial frames, while the Earth-based twin stays in the same inertial frame.”

Still, the situation is essentially symmetric and so we could draw a similar-looking spacetime diagram for the primed coordinates, i.e. x’ and ct’, and wonder what’s the difference. That’s where these planes of simultaneity come in. Look at the wonderful animation below: A, B, C are simultaneous events when I am standing still (v = 0). However, when I move at considerable speed (v = 0.3c), that’s no longer the case: it takes more time for news to reach me from ‘point’ A and, hence, assuming news travels at the speed of light, event A appears to happen later. Conversely, event C (in spacetime) appears to have happened before event B. Now that explains these blue so-called simultaneity planes on the diagram above: they’re the white lines traveling from the past to the future on the animation below, but for the trip out only (> 0). For the trip back, we have the red lines, which correspond to the v = –0.5c situation below. So that’s the return trip (< 0).

What you see is that, “during the U-turn, the plane of simultaneity jumps from blue to red and very quickly sweeps over a large segment of the world line of the Earth-based twin.” Hence, “when one transfers from the outgoing frame to the incoming frame there is a jump discontinuity in the age of the Earth-based twin.” [I took the quotes taken from Wikipedia here, where you can find the original references.] Now, you will say, that is also symmetric if we switch the reference frames. Yes… Except for the sign. So, yes, it is the traveling brother who effectively skips some time. Paradox solved.

Now… For some real fun…

Now, for some real fun, I’d like to ask you how the world would look like when you were traveling through it riding a photon. So… Think about it. Think hard. I didn’t google at first and I must admit the question really started wracking my brain. There are some many effects to take into account. One basic property, of course, must be that time stands still around you. You see the world as it was when you reached v = c. Well… Yes and no. The fact of the matter is that, because of all the relativistic effects (e.g. aberration, Doppler shift, intensity shifts,…), you actually don’t see a whole lot. One visualization of it (visual effects of relativistic speeds) seems to indicate that (most) science fiction movies actually present the correct picture (if the animation shows the correct visualization, that is): we’re staring into one bright flash of light ahead of us as we’re getting close to v = c. Interesting…

Finally, you should also try to find out what actually happens to the clocks during the deceleration and acceleration as the space ship of that twin brother turns. You’re going to find it fascinating. At the same time, the math behind is, quite simply, daunting and, hence, I won’t even try go into the math of this thing. 🙂

Conclusion

So… Well… That’s it really. I now realize why I never quite got this as a kid. These paradoxes do require some deep thinking and imagination and, most of all, some tools that one just couldn’t find as easily as today.

The Web definitely does make it easier to study without the guidance of professors and the material environment of a university, although I don’t think it can be a substitute for discipline. When everything is said and done, it’s still hard work. Very hard work. But I hope you get there, Vincent ! 🙂 And please do look at that Youtube video by clicking the link above. 🙂

Post scriptum: Because the resolution of the video above is quite low, I looked for others, for example one that describes the journey from the Sun to the Earth, which–as expected–takes about 8 minutes. While it has higher resolution, it is far less informative. I’ll let you google some more. Please tell me if you found something nice. 🙂

# 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.

But, 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).

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.]

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:

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. 🙂