Total energy derivation: energy as a time component of a four-vector

1. Jul 10, 2013

Raze

I'm kind of new to special relativity, I mean beyond what they tell you in survey courses. In any case, I've heard there was a relationship between energy and time in special relativity (I've actually heard someone say the "time component" of energy), and I've always been fascinated with E = mc2 (who isn't?), so I've spent some time trying to derive it. My question arises from a questionable application of the definition of work that I use to get the total energy equation. I'm not sure it's a legal move, since to get it I integrate over time instead of over distance. Usually when I see it "derived" it's by showing how the special relativity kinetic energy equation is the same as the classical kinetic energy equation when c >> u. Otherwise I've never really seen it derived, hence the reason I wanted to explore this.

The thing is, I get correct results for both kinetic energy and total energy, so I'm wondering if there is something to it, or if it is merely coincidence and my own lack of understanding.

In any case, here is my methodology (using only two dimensions for simplicity):

(1) Start with a 2-dimensional space time vector: (dx, cdt) because c makes the time unit a distance unit (I've seen this done repeatedly in special relativity texts and tutorials, so I'm assuming this is allowed)

(2) Divide both components by proper time (to make it compatible with a lorentz transformation): (γu, γc)

(3) Multiply both components by mass to get momentum: (γmu, γmc)

(4) Take the derivative with respect to time of both components to get force: (γm du/dt + mu dγ/dt, mc dγ/dt)

(5) Integrate over distance... HERE IS THE QUESTIONABLE PART.

With the spatial component of this "vector," it is straight forward Calculus II. I divide it into two integrals to make it easier, then use trig substitution and voila , out pops the CORRECT term for kinetic energy in special relativity: mc2 (γ - 1) (I made it a definite integral from 0 to u). This is definitely the right answer (I am willing to show my math if needs be, but it would be kind of tedious). It sits neatly in the spatial component of my new "vector."

However, when I tried this with time, I ended up with γmcu, Since energy is conserved with respect to time, I expected to get TOTAL energy in the time component: γmc2, which means my answer is right for only one value of u, one that the object can never achieve. So then I adjusted my strategy, and did the one thing I have a question about:

Since I started off with cdt as my time component, I figured, what will happen if instead of at step (5) integrating with dx, I integrate with cdt? The result was a shockingly easy integral. After doing everything to get force, my time component of force was:

mc dγ/dt

mc dγ/dt*cdt

and of course dt/dt cancels to 1, leaving only

mc2 dγ from 0 to γ

which is just

γmc2.

That's the right answer that I expected, because it is total energy. Essentially all I did was multiply dt by a scalar of c and use that instead of dx for my integration. But I'm still not sure if it is a legal move. I feel like I cheated. Here's why:

First of all, what the heck does "work" mean in this context? Purely mathematically it seems to make sense to me, but in my experience work only applies to space, unless you're taking about dW/dt, which I am definitely NOT talking about. Did I just sort of pull that out of thin air? Or is this legal because we're talking about a four-dimensional space where time is treated similar to a spatial direction?

Second, I kind of feel like the constant c is used arbitrarily here just to make the vector. I know c is special, but it does feel kind of like an arbitrary construction.

What would you guys say about the validity of this method of deriving the total energy by treating it as sort of a "time" component of a displacement vector? Is it compatible with physics or just a waste of time?

2. Jul 10, 2013

Staff: Mentor

The first thing to realize is that $E = m c^2$ only applies to an object with nonzero rest mass that is at rest. If the object is moving, the correct equation is $E^2 = m^2 c^4 + p^2 c^2$, where $p$ is the object's momentum.

Yes, that's questionable, but the issues start even before that. See above and further comments below.

Everything looks OK up to here; the vector you have come up with is usually called the 4-momentum of the object.

Here you should be taking derivatives with respect to proper time, not coordinate time; otherwise your result won't be a 4-vector. See, for example, here:

http://en.wikipedia.org/wiki/Four-force

Even before we get to the questions you asked, there's another issue: distance is frame-dependent. The integral you are writing down is therefore not frame-invariant for two reasons: you took derivatives with respect to coordinate time, not proper time, to get the force (see above), and you're integrating over a distance that's frame-dependent.

You're right to wonder if this is justified. A good question to start with is, what would this mean, physically? You raise some good concerns with this later in your post.

I've already made some comments above, but a further one is, deriving total energy is not the same as deriving $E = m c^2$, because, as I noted above, that formula is only valid for an object at rest. That's not to say that deriving total energy by integrating work done is not a worthwhile thing to do: it is. But you need to be careful about how you do it, and what you expect the result to be.

3. Jul 10, 2013

Raze

I appreciate your response, and I'm going to look into integrating with respect to proper time instead of coordinate time (I had assumed since I divided by proper time in the beginning that I was good to go), but your post has left me with a little confusion, which I will elaborate on below.

Just one thing: while I said I was fascinated with E = mc2, what I set out to derive was total energy, not rest energy. In any case I'm still not clear with what you posted because it seems to me that $E^2 = m^2 c^4 + p^2 c^2$ is the same as γ(u) mc2 (which is what I derived), just written in a different form:

Assume E =γ(u) mc2

Then E2 = m2c4/(1 - u2/c2)

Then E2 - E2 u2/c2 = m2c4

Then we have

E2 = m2c4 + E2 u2/c2

Then using the initial assumption again,

E2 = m2c4 + γ2m2c4 u2/c2

Which would be simplified to:

E2 = m2c4 + (γmu)2c2

E2 = m2c4 + p2c2

So basically, what I derived, γ(u)mc2 seems to be exactly the same as
$E^2 = m^2 c^4 + p^2 c^2$​

except that γ(u)mc2 is the square root of E2.

So, the assumption that E = γmc2 yields the same equation you gave for total energy. All you have to do is square it and manipulate it a bit. And this leaves me some what perplexed with most of your post. But again I appreciate it.

4. Jul 10, 2013

Staff: Mentor

And that's fine, as long as you realize that total energy is not $mc^2$. I only brought it up because early in your post you said you were trying to derive $E = m c^2$, so I wasn't sure if you were aware of the difference between that and total energy. It appears that you are aware of that, so there's no problem, and you can disregard my remarks about it.

5. Jul 10, 2013

Raze

Okay, well in that case, would you revisit my original post in light of that? Because I'm still not so sure that anything I did with the time coordinate was legitimate. In fact, since you mentioned integrating with respect to proper time rather than coordinate time, I'm not even sure about the kinetic energy part (even though in both instances I got the correct answer).

I figured that using proper time in the velocity vector would have taken care of that, but now I'm not so sure, so if you get any more spare time more input would be appreciated.

Thanks!

*Edited to add: I'm hoping this won't spill over into tensor math, because I barely know what they are and how to read the way their are indexed, truth to tell. But if it must, I will be forced to expand my horizons a little earlier than planned.

6. Jul 10, 2013

Staff: Mentor

The rest of the comments in my original post are still relevant; they don't have anything to do with the proper equation for $E$ in terms of $m$.

And as I commented before, you are right to be unsure.

You don't need to derive a formula for kinetic energy in order to derive a formula for total energy; in fact, in relativity it's often better not to think about kinetic energy at all, since it just complicates matters.

As I said before, that takes care of making the 4-momentum a 4-vector, but to get the 4-force you have to take another derivative, and if you don't take that derivative with respect to proper time as well, the result won't be a 4-vector.

Another question you should consider, though, is: do you actually *want* to use the 4-force vector? Or do you want the ordinary 3-force in some specific frame? I commented in my original post that distance isn't frame-invariant, so integrating the ordinary 3-force over a distance won't be frame-invariant either. But as far as that goes, energy is also not frame-invariant; it's the "t" component of 4-momentum in a specific frame. So if you're trying to figure out how energy changes as work is done, it seems like you're interested in quantities in a particular frame, not frame-invariant quantities. That's something you will need to make up your mind about to do this derivation.

7. Jul 10, 2013

Raze

Ah ha! Now that actually adds some clarity to what I'm thinking about. I suspect then that I was looking for energy in a specific frame. However, now I'm interested in whether or not I could do the entire thing again but from a 4-vector perspective. Would the same basic physics ideas still apply? (by that I mean, force is the time derivative of momentum, work is the integral of force over distance) Except with using proper time instead of coordinate time all the way through, and I suppose instead of integrating with respect to distance I would use proper length for work (is that right?). Or would I only need to worry about proper time here?

8. Jul 10, 2013

Staff: Mentor

If you're doing things this way, you would only worry about proper time, yes. The 4-force is indeed the time derivative of the 4-momentum (where "time" here of course means proper time). The Wiki page on 4-force that I linked to discusses this some.

Integrating force over a distance to get work done is different, because the concept of "work" as we're used to thinking of it is really frame-dependent. Step back to Newtonian physics for a moment: when you do work on an object, you don't change its mass, you just change its kinetic energy (and its momentum). The relativistic version of this is a 4-force that doesn't change the object's rest mass, just its 4-momentum.

But the rest mass is just the length of the 4-momentum vector (that's really what the formula $E^2 = m^2 c^4 + p^2 c^2$ is saying--just rearrange the terms and choose units such that $c = 1$, to get $E^2 - p^2 = m^2$, which is how it's usually written in relativity textbooks). So a 4-force that doesn't change the object's rest mass doesn't change the length of the 4-momentum vector, only its direction. ("Direction" in spacetime is really a way of saying which particular inertial frame is the object's rest frame.)

And *that* means the 4-force must be orthogonal to the 4-momentum (just as in Newtonian physics, a force that changes an object's direction but not its speed must be orthogonal to the velocity vector). In the object's rest frame (more precisely, in its instantaneous rest frame at the instant the force is applied), this means the 4-force is purely spatial, i.e., it has no time component (because the 4-momentum in this frame has only a time component, which is just $m$, or $m c^2$ in conventional units).

So in the object's instantaneous rest frame, we can indeed treat the 4-force just like an ordinary 3-force, and integrate it over a small proper distance to obtain work done, again in the instantaneous rest frame. But what does this work actually do? As I said above, it doesn't change the object's rest mass; it just changes the direction in spacetime of its 4-momentum, i.e., it changes its rest frame (by a small amount, because we're talking about an infinitesimal force applied for an infinitesimal amount of proper time through an infinitesimal proper distance).

If you use an integral, based on the above, to express the effect of a 4-force applied for a finite amount of proper time, you will thus find two things different from what you're used to in Newtonian physics. First, again, the object's rest mass won't change, so there will be no 4-dimensional scalar quantity that changes the way energy changes in Newtonian physics when you do work. Second, trying to integrate the force through a distance won't be possible using the 4-dimensional method, because there will be no 4-dimensional quantity you can use to express the "distance" through which the 4-force acts.

These two things are related, of course; what they're really saying is that there is no 4-dimensional analogue of "work". My comment above about forces in Newtonian physics that don't change speed but only direction was actually a clue to this: as you may remember, such forces (for example, the force exerted by a string that is swinging an object in a circular path) do no work.