# Re-visiting relativity and four-vectors: the proper time, the tensor and the four-force

Pre-script (dated 26 June 2020): Our ideas have evolved into a full-blown realistic (or classical) interpretation of all things quantum-mechanical. In addition, I note the dark force has amused himself by removing some material. So no use to read this. Read my recent papers instead. ðŸ™‚

Original post:

My previous post explained how four-vectors transformÂ from one reference frame to the other. Indeed, a four-vector isÂ notÂ just some one-dimensional array of four numbers: it represent somethingâ€”a physical vector that… Well… Transforms like a vector. ðŸ™‚ So whatÂ vectorsÂ are we talking about? Let’s see what we have:

1. We knew the position four-vector already, which we’ll write as xÎ¼Â = (ct, x, y, z) = (ct, x).
2. We also proved that AÎ¼Â = (Î¦,Â Ax, Ay, Az) = (Î¦,Â A)Â is a four-vector: it’s referred to as the four-potential.
3. We also know the momentumÂ four-vector from theÂ LecturesÂ on special relativity. We write it as pÎ¼Â = (E, px, py, pz) = (E, p), with E = Î³m0, p = Î³m0v, and Î³ = (1âˆ’v2/c2)âˆ’1/2Â or, forÂ cÂ = 1, Î³ =Â (1âˆ’v2)âˆ’1/2

To show that it’s notÂ just a matter of adding some fourth t-component to aÂ three-vector, Feynman gives the example of the four-velocity vector. We have vxÂ = dx/dt,Â vyÂ = dy/dt and vzÂ = dz/dt, but a vÎ¼Â = (d(ct)/dt, dx/dt, dy/dt, dz/dt) = (c, dx/dt, dy/dt, dz/dt) ‘vector’ is, obviously, not a four-vector. [Why obviously? The inner productÂ vÎ¼vÎ¼ Â is not invariant.] In fact,Â Feynman ‘fixes’ the problem by noting that ct, x, y and z have the ‘right behavior’, but the d/dt operator doesn’t. The d/dt operator isÂ not an invariant operator. So how does he fix it then? He tries the (1âˆ’v2/c2)âˆ’1/2Â·d/dt operator and, yes, it turns out we do get a four-vector then. In fact, we get that four-velocity vector uÎ¼Â that we were looking for:[Note we assume we’re using equivalentÂ time and distance units now, soÂ cÂ = 1 andÂ v/c reduces to a new variable v.]

Now how do we know this is four-vector? How can we prove this one? It’s simple. We can get it from ourÂ pÎ¼Â = (E, p) by dividing it by m0, which is an invariantÂ scalar in four dimensions too. Now, it is easy to see that a division by an invariantÂ scalar doesÂ notÂ change the transformation properties. So just write it all out, and you’ll see that pÎ¼/m0Â = uÎ¼Â and, hence, that uÎ¼Â is a four-vector too. ðŸ™‚

We’ve got an interesting thing here actually: division by an invariant scalar, or applying that (1âˆ’v2/c2)âˆ’1/2Â·d/dt operator, which is referred to as an invariant operator, on a four-vector will give us another four-vector. Why is that? Let’s switch to compatible time and distance units soÂ c = 1 so to simplify the analysis that follows.

#### The invariant (1âˆ’v2)âˆ’1/2Â·d/dt operatorÂ and the proper time s

Why is theÂ (1âˆ’v2)âˆ’1/2Â·d/dt operator invariant? Why does it ‘fix’ things? Well… Think about the invariant spacetime interval (Î”s)2Â = Î”t2Â âˆ’ Î”x2Â âˆ’ Î”y2Â âˆ’Â Î”z2Â going to the limit (ds)2Â = dt2Â âˆ’ dx2Â âˆ’ dy2Â âˆ’ dz2Â . Of course, we can and should relate this to an invariant quantity s = âˆ« ds. Just like Î”s, this quantity also ‘mixes’ time and distance. Now, we could try to associate some derivative d/ds with it because, as Feynman puts it, “it should be a nice four-dimensional operation because it is invariant with respect to a Lorentz transformation.” Yes. It should be. So let’s relate ds to dt and see what we get. That’s easy enough: dx = vxÂ·dt,Â dy = vyÂ·dt, dz = vzÂ·dt, so we write:

(ds)2Â = dt2Â âˆ’ vx2Â·dt2Â âˆ’ vy2Â·dt2Â âˆ’Â vz2Â·dt2Â â‡”Â (ds)2Â = dt2Â·(1 âˆ’ vx2Â âˆ’ vy2Â âˆ’Â vz2) = dt2Â·(1 âˆ’ v2)

and, therefore, ds = dtÂ·(1âˆ’v2)1/2. So our operator d/ds is equal to (1âˆ’v2)âˆ’1/2Â·d/dt, and we can apply it toÂ anyÂ four-vector, as we are sure that, as an invariant operator, it’s going to give us another four-vector. I’ll highlight the result, because it’s important:

The d/ds = (1âˆ’v2)âˆ’1/2Â·d/dt operator is an invariant operator for four-vectors.

For example, if we apply it to xÎ¼Â = (t, x, y, z), we get the very same four-velocity vector Î¼Î¼:

dxÎ¼/ds = uÎ¼Â = pÎ¼/m0

Now, if you’re somewhat awake, you should ask yourself: what is this s, really,Â and what is this operator all about? Our new function s = âˆ« ds is notÂ the distance function, as it’s got both time and distance in it. Likewise, the invariant operatorÂ d/ds = (1âˆ’v2)âˆ’1/2Â·d/dt has both time and distance in it (the distance is implicit in the v2Â factor). Still, it is referred to as the proper timeÂ along the path of a particle. Now why is that? If it’s got distance andÂ time in it, why don’t we call it the ‘proper distance-time’ or something?

Well… The invariant quantity s actually is the time that would be measured by a clock that’s moving along, in spacetime, with the particle. Just think of it: in the reference frame of the moving particle itself, Î”x, Î”yÂ and Î”zÂ must be zero, because it’s not moving in its own reference frame. So theÂ (Î”s)2Â = Î”t2Â âˆ’ Î”x2Â âˆ’ Î”y2Â âˆ’Â Î”z2Â reduces to (Î”s)2Â = Î”t2, and so we’re only adding time to s. Of course, this view of things implies that theÂ proper time itself is fixed only up to some arbitrary additive constant, namely the setting of the clock at some event along the ‘world line’ of our particle, which is its path in four-dimensional spacetime. But… Well… In a way, s is the ‘genuine’ or ‘proper’ time coming with the particle’s reference frame, and so that’s why Einstein called it like that. You’ll see (later) that it plays a very important role in general relativity theory (which is a topic we haven’t discussed yet: we’ve only touched special relativity, so no gravity effects).

OK. I know this is simple and complicated at the same time: the math is (fairly) easy but, yes, it may be difficult to ‘understand’ this in some kind of intuitiveÂ way. But let’s move on.

#### The four-force vector fÎ¼

We know the relativistically correct equation for the motionÂ of some charge q. It’s just Newton’s Law F = dp/dt = d(mv)/dt. The only difference is that we areÂ not assuming that m is some constant. Instead, we use the pÂ = Î³m0vÂ formula to get:

How can we get a four-vector for the force? It turns out that we get it when applying our new invariant operator to the momentum four-vector pÎ¼Â = (E, p), so we write:Â fÎ¼Â =Â dpÎ¼/ds. But pÎ¼Â = m0uÎ¼Â =Â m0dxÎ¼/ds, so we can re-write this asÂ fÎ¼Â =Â d(m0Â·dxÎ¼/ds)/ds, which gives us a formula which is reminiscent of the Newtonian F = ma equation:

WhatÂ isÂ this thing? Well… It’s not so difficult to verify that the x, y and z-components are just our old-fashioned Fx,Â FyÂ andÂ Fz, so these are the components of F. The t-component is (1âˆ’v2)âˆ’1/2Â·dE/dt. Now, dE/dt is the time rate of change of energy and, hence, it’s equal to the rate of doing work on our charge, which is equal to Fâ€¢v. So we can write fÎ¼ as:

#### The force and the tensor

We will now derive that formula which we ended the previous postÂ with. We start with calculating the spacelike components of fÎ¼ from the Lorentz formula F = q(E + vÃ—B). [The terminology is nice, isn’t it? The spacelike components of the four-force vector!Â Now thatÂ sounds impressive, doesn’t it? But so… Well… It’s really just the old stuff we know already.]Â So we start with fxÂ =Â Fx, and write it all out:

What a monster! But,Â hey! We can ‘simplify’ this by substituting stuff by (1) the t-, x-, y- and z-components of the four-velocity vector uÎ¼Â and (2) the components of our tensor FÎ¼Î½Â = [Fij] = [âˆ‡iAjÂ âˆ’Â âˆ‡jAi] with i, j = t, x, y, z. We’ll also pop in the diagonal FxxÂ = 0 element, just to make sure it’s all there. We get:

Looks better, doesn’t it? ðŸ™‚ Of course, it’s just the same, really. This is just an exercise in symbolism. Let me insert the electromagnetic tensor we defined in our previous post, just as a reminder of what that FÎ¼Î½Â matrix actually is:

If you read my previous post, this matrixÂ â€“ or the concept of a tensorÂ â€“ has no secrets for you. Let me briefly summarize it, because it’s an important result as well. The tensor is (a generalization of) the cross-product inÂ four-dimensional space. We take two vectors:Â aÎ¼Â =Â (at, ax, ay, az) and bÎ¼Â =Â (bt, bx, by, bz) and then we takeÂ cross-products of their components just like we did in three-dimensional space, so we write TijÂ = aibjÂ âˆ’Â ajbi. Now, it’s easy to see that this combination implies that TijÂ = âˆ’ TjiÂ and that TiiÂ = 0, which is why we only have sixÂ independent numbers out of the 16 possible combinations, and which is why we’ll get a so-called anti-symmetric matrix when we organize them in a matrix. In three dimensions, the very same definition of the cross-product TijÂ gives us 9 combinations, and only 3 independent numbers, which is why we represented our ‘tensor’ as a vector too! In four-dimensional space we can’t do that: six things cannot be represented by a four-vector, so weÂ need to use thisÂ matrix, which is referred to as a tensor of the second rank in four dimensions. [When you start using words like that, you’ve come a long way, really. :-)]

[…]Â OK. Back to our four-force. It’s easy to get a similar one-liner forÂ fyÂ and fzÂ too, of course, as well as for ft. But… Yes, ft… Is it the same thing really? Let me quickly copy Feynman’s calculation for ft:

ItÂ does: remember that vÃ—B and v are orthogonal, and so their dot product is zero indeed. So, to make a long story short, the four equations â€“ one for each component of the four-force vector fÎ¼Â â€“ can be summarized in the following elegant equation:

Writing this all requires a few conventions, however. For example,Â FÎ¼Î½Â is a 4Ã—4 matrix and so uÎ½Â has to be written as a 1Ã—4 vector. And the formula for theÂ fxÂ and ftÂ component also make it clear that we also want to use the +âˆ’âˆ’âˆ’ signature here, so the convention for the signs in the uÎ½FÎ¼Î½Â product is the same as that for the scalar product aÎ¼bÎ¼. So, in short, you really need to interpret what’s being written here.

A more important question, perhaps, is: what can we do with it? Well… Feynman’s evaluation of the usefulness of this formula is rather succinct: “Although it is nice to see that the equations can be written that way, this form is not particularly useful. Itâ€™s usually more convenient to solve for particle motions by using the F = q(E + vÃ—B) = (1âˆ’v2)âˆ’1/2Â·d(m0v)/dtÂ equations, and thatâ€™s what we will usually do.”

Having said that, this formula really makes good on the promise I started my previous post with: we wanted a formula, someÂ mathematical construct, that effectively presents the electromagnetic force as oneÂ force, as one physical reality. So… Well… Here it is! ðŸ™‚

Well… That’s it for today. Tomorrow we’ll talk about energy and about a veryÂ mysterious conceptâ€”the electromagnetic mass. That should be fun! So I’llÂ c u tomorrow! ðŸ™‚

Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:

https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:

https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:

https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:

https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:

https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 16, 2020 as a result of a DMCA takedown notice from The California Institute of Technology. You can learn more about the DMCA here:

https://en.support.wordpress.com/copyright-and-the-dmca/
Some content on this page was disabled on June 20, 2020 as a result of a DMCA takedown notice from Michael A. Gottlieb, Rudolf Pfeiffer, and The California Institute of Technology. You can learn more about the DMCA here:

https://en.support.wordpress.com/copyright-and-the-dmca/