3
$\begingroup$

In my special relativity course we defined 4-momentum $P$ as $mU$ where $U$ is the 4-velocity and $m$ is restmass. Then by definition $P$ is a 4-vector. We then defined $E$ via $E/c = P^0$ and claimed that $E$ is a conserved energy quantity based on the first 2 terms of its Taylor expansion. This was presented as being a not-completely-rigorous step, with some mention of Noether's theorem (beyond the scope of the course), but to me it seems much worse than that - I don't see how it gives any indication that $E$ is conserved. As a result I don't understand why 4-momentum is conserved either. Since these conservation laws underpin all relativistic kinematics I would expect there to be a convincing justification accessible at this level, but I just can't find one. Am I missing something about how these things are defined?

| cite | improve this question | | | | |
$\endgroup$
  • $\begingroup$ To narrow down the question a bit: for a system of particles in Newtonian physics I can define total momentum as the sum of individual momenta, similar for energy. They are defined such that when particles interact with conservative forces, total energy and momentum are conserved. So in a collision between particles I can confidently say that total energy and momentum are conserved, and use this to solve a problem. In a collision between particles in a relativistic setting, why can I state the same thing about 4-momentum? $\endgroup$ – Danny Duberstein Apr 7 at 15:37
3
$\begingroup$

We then defined $E$ via $E/c=P^0$ and claimed that $E$ is a conserved energy quantity based on the first 2 terms of its Taylor expansion.

Taylor expanding $E$ around $v=0$ gives that $E = mc^2 + \frac{1}{2}mv^2 + \mathcal O(v^4)$, which means that at low velocities (ignoring the constant term, because constant shifts in energy are irrelevant in Newtonian mechanics), this definition of $E$ coincides with the definition of non-relativistic kinetic energy.

I am guessing that your instructor wanted to use this fact to argue that this makes it plausible that $E$ is the correct relativistic generalization of kinetic energy. It's not a proof, but a proof doesn't seem to be what your instructor was after.

As far as the 4-momentum being conserved, it generally isn't, just like in non-relativistic physics. If the object in question is under the influence of some kind of force, then neither its linear momentum nor its kinetic energy are going to be constant. If we introduce the concept of the four-force $\mathbf F$ and generalize Newton's 2nd Law to

$$\mathbf F = \frac{d}{d\tau}\mathbf P $$

then momentum and kinetic energy conservation follow in exactly the same way that they do in non-relativistic physics.

| cite | improve this answer | | | | |
$\endgroup$
2
$\begingroup$

You are better to take conservation of energy and momentum as a fundamental empirical principle that underpins everything else in physics. You can prove them from Newton's laws, but that only covers classical mechanics. Actually conservation of energy and momentum is a fundamental principle contained in Einstein's equation in general relativity, and they can be proven within quantum field theory, but this is much more advanced. I prefer to show Newton's laws from conservation of energy and momentum.

| cite | improve this answer | | | | |
$\endgroup$
1
$\begingroup$

When you extend Newtonian mechanics to special relativity you use the principle that the equations of SR must reduce to the Newtonian in the non-relativistic limit. By this principle you can find suitable new definitions of relativistic energy and momentum, as you describe.

So, why are relativistic enery and momentum conserved? There are two slightly different answers. It is an empirical fact that energy and momentum is conserved in Newtonian mechanics. It is reasonable to postulate that the relativistic generalizations have the same property (like Einstein did). By now energy and momentum conservation in relativistic systems is also confirmed by experiments and shows that the postulate indeed is correct.

You can also view energy and momentum in a Noether sense. In this framework we instead postulate certain symmetries and derive conserved quantities from that. The quantities energy and momentum are defined as quantities that are conserved as a consequence of certain symmetries. It is however important to keep in mind that the Noether procedure is not a proof that energy and momentum is conserved, you just derive it from other (in some sense more fundamental) principles.

So, that they are conserved is most importantly an observed fact, just as $F=ma$ is ”true” because the experiments confirm it.

Hope it helps! I was also very confused about this when I studied the subject.

| cite | improve this answer | | | | |
$\endgroup$
0
$\begingroup$

1) $E$ being a conserved quantity is a principle of physics; your theory must by construction preserve whatever it defined to be $E$. The Taylor expansion you mentioned is a plausibility argument to convince you that $mU$ is indeed the correct relativistic generalization of $E$ in newtonian mechanics.

2) You are correct about Noether's theorem. In a nutshell, given a symmetry of the Lagrangian(that captures information about your system), there exists a conserved quantity. In this way of thinking, (roughly)$E$ is the conserved current corresponding to translational symmetry of your system. Here is the key point-

Translational symmetry is essentially the homogeneity of space(time). If you recall, the derivation of Lorentz transformations assume homogeneity, isotropy and such nice things. Then, BY CONSTRUCTION, your theory will have a corresponding conserved current-the energy.

3) It is perhaps useful to point out that an inertial frame is defined to be the one where space is homogeneous and isotropic, and time homogeneous(we are looking at conservative systems, ofcourse). By construction, your nice conservation laws follow. See Landau's initial pages for a discussion.

| cite | improve this answer | | | | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.