**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. 🙂

**Original post**:

My previous post explained how four-vectors *transform* from one reference frame to the other. Indeed, a four-vector is *not *just some one-dimensional array of four numbers: it represent something—a *physical* vector that… Well… Transforms like a vector. 🙂 So what *vectors *are we talking about? Let’s see what we have:

- We knew the
*position*four-vector already, which we’ll write as x_{μ }= (*c*t, x, y, z) = (*c*t,**x**). - We also
*proved*that A_{μ }= (Φ, A_{x}, A_{y}, A_{z}) = (Φ,**A**) is a four-vector: it’s referred to as the*four-potential*. - We also know the
*momentum*four-vector from the*Lectures*on special relativity. We write it as p_{μ }= (E, p_{x}, p_{y}, p_{z}) = (E,**p**), with E = γm_{0},**p**= γm_{0}**v**, and γ = (1−v^{2}/*c*^{2})^{−1/2}or, for*c*= 1, γ = (1−v^{2})^{−1/2}

To show that it’s *not *just a matter of adding some fourth t-component to a *three-vector*, Feynman gives the example of the four-velocity vector. We have v_{x }= dx/dt, v_{y }= dy/dt and v_{z }= dz/dt, but a v_{μ }= (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 product v_{μ}v_{μ } is not invariant.] In fact, Feynman ‘fixes’ the problem by noting that *c*t, x, y and z have the ‘right behavior’, but the d/dt operator doesn’t. The d/dt operator is *not* an *invariant *operator. So how does he fix it then? He tries the (1−v^{2}/*c*^{2})^{−1/2}·d/dt operator and, yes, it turns out we do get a four-vector then. In fact, we get that four-velocity vector u_{μ} that we were looking for:[Note we assume we’re using *equivalent *time and distance units now, so *c *= 1 and *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 our p_{μ }= (E, **p**) by dividing it by m_{0}, which is an *invariant *scalar in four dimensions too. Now, it is easy to see that a division by an invariant scalar does *not* change the transformation properties. So just write it all out, and you’ll see that p_{μ}/m_{0} = u_{μ} and, hence, that u_{μ} is a four-vector too. 🙂

We’ve got an interesting thing here actually: division by an invariant scalar, or applying that (1−v^{2}/*c*^{2})^{−1/2}·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 so *c = *1 so to simplify the analysis that follows.

**The invariant (1−v**^{2})^{−1/2}·d/dt operator and the proper time s

^{2})

^{−1/2}·d/dt operator and the proper time s

Why is the (1−v^{2})^{−1/2}·d/dt operator invariant? Why does it ‘fix’ things? Well… Think about the invariant spacetime interval (Δs)^{2 }= Δt^{2 }− Δx^{2 }− Δy^{2 }− Δz^{2} going to the limit (ds)^{2 }= dt^{2 }− dx^{2 }− dy^{2 }− dz^{2} . Of course, we can and should relate this to an invariant quantity s = ∫ ds. Just like Δs, 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 = v_{x}·dt, dy = v_{y}·dt, dz = v_{z}·dt, so we write:

(ds)^{2 }= dt^{2 }− v_{x}^{2}·dt^{2 }− v_{y}^{2}·dt^{2 }− v_{z}^{2}·dt^{2 }⇔ (ds)^{2 }= dt^{2}·(1 − v_{x}^{2 }− v_{y}^{2 }− v_{z}^{2}) = dt^{2}·(1 − v^{2})

and, therefore, ds = dt·(1−v^{2})^{1/2}. So our operator d/ds is equal to (1−v^{2})^{−1/2}·d/dt, and we can apply it to *any *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 = (1−v ^{2})^{−1/2}·d/dt operator is an invariant operator for four-vectors.**

For example, if we apply it to x_{μ }= (t, x, y, z), we get the very same four-velocity vector μ_{μ}:

dx_{μ}/ds = u_{μ} = p_{μ}/m_{0}

Now, if you’re somewhat awake, you should ask yourself: what** is **this s,

*really*, and what is this operator all about? Our new function s = ∫ ds is

*not*the distance function, as it’s got both time and distance in it. Likewise, the invariant operator d/ds = (1−v

^{2})

^{−1/2}·d/dt has both time and distance in it (the distance is implicit in the v

^{2}factor). Still, it is referred to as

**the**

*along the path of a particle. Now why is that? If it’s got distance*

**proper time***and*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, Δx, Δy^{ }and Δz must be zero, because it’s *not* moving in its own reference frame. So the (Δs)^{2 }= Δt^{2 }− Δx^{2 }− Δy^{2 }− Δz^{2} reduces to (Δs)^{2 }= Δt^{2}, and so we’re only adding time to s. Of course, this view of things implies that the 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 *intuitive *way. But let’s move on.

**The four-force vector f**_{μ}

_{μ}

We know the relativistically correct equation for the *motion *of some charge q. It’s just Newton’s Law **F** = d**p**/dt = d(m**v**)/dt. The only difference is that we are *not *assuming that m is some constant. Instead, we use the **p** = γm_{0}**v** 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 p_{μ }= (E, **p**), so we write: f_{μ }= dp_{μ}/ds. But p_{μ }= m_{0}u_{μ} = m_{0}dx_{μ}/ds, so we can re-write this as f_{μ }= d(m_{0}·dx_{μ}/ds)/ds, which gives us a formula which is reminiscent of the Newtonian **F** = m**a **equation:

What *is *this thing? Well… It’s not so difficult to verify that the x, y and z-components are just our old-fashioned F_{x}, F_{y} and F_{z}, so these are the components of **F**. The t-component is (1−v^{2})^{−1/2}·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 **F**•**v**. So we can write f_{μ }as:

**The force and the tensor**

We will now derive that formula which we ended the previous post with. We start with calculating the spacelike components of f_{μ }from the Lorentz formula **F** = q(**E** + **v**×**B**). [The terminology is nice, isn’t it? *The spacelike components of the four-force vector!* Now *that* sounds impressive, doesn’t it? But so… Well… It’s really just the old stuff we know already.] So we start with f_{x} = F_{x}, and write it all out:

What a monster! But, ** hey! **We can ‘simplify’ this by substituting stuff by (1) the t-, x-, y- and z-components of the four-velocity vector u

_{μ }and (2) the components of our tensor F

_{μν}= [F

_{ij}] = [∇

_{i}A

_{j}− ∇

_{j}A

_{i}] with i, j = t, x, y, z. We’ll also pop in the diagonal F

_{xx}= 0 element, just to make sure it’s all there. We get:

Looks better, doesn’t it? 🙂 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 F_{μν} matrix actually is:

If you read my previous post, this matrix – or the concept of a tensor – 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 in *four*-dimensional space. We take two vectors: a_{μ} = (a_{t}, a_{x}, a_{y}, a_{z}) and b_{μ} = (b_{t}, b_{x}, b_{y}, b_{z}) and then we take *cross*-products of their components just like we did in three-dimensional space, so we write T_{ij} = a_{i}b_{j} − a_{j}b_{i}. Now, it’s easy to see that this combination implies that T_{ij} = − T_{ji} and that T_{ii }= 0, which is why we only have *six *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 T_{ij} 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 we need to use this 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. :-)]

[…] OK. Back to our four-force. It’s easy to get a similar one-liner for f_{y} and f_{z} too, of course, *as well as for *f_{t}. But… Yes, f_{t}… Is it the same thing really? Let me quickly copy Feynman’s calculation for f_{t}:

It does: remember that **v**×**B** and **v **are orthogonal, and so their dot product is zero indeed. So, to make a long story short, the four equations – one for each component of the four-force vector f_{μ }– can be summarized in the following elegant equation:

Writing this all requires a few conventions, however. For example, F_{μν} is a 4×4 matrix and so u_{ν} has to be written as a 1×4 vector. And the formula for the f_{x} and f_{t} component also make it clear that we also want to use the +−−− signature here, so the convention for the signs in the u_{ν}F_{μν} product is the same as that for the scalar product a_{μ}b_{μ}. 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. It’s usually more convenient to solve for particle motions by using the **F** = q(**E** + **v**×**B**) = (1−v^{2})^{−1/2}·d(m_{0}**v**)/dt equations, and that’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, some mathematical construct, that effectively presents the electromagnetic force as *one *force, as one physical reality. So… Well… **Here it is! 🙂**

Well… That’s it for today. Tomorrow we’ll talk about energy and about a *very *mysterious concept—the electromagnetic mass. That should be fun! So I’ll *c* *u* tomorrow! 🙂

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