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The four-momentum in special relativity is defined as: $$p = \bigg(\frac{E}{c},\ p_x,\ p_y,\ p_z\bigg)$$ where $$p_x = \gamma (mv_x)$$ $$p_y = \gamma (mv_y)$$ $$p_z = \gamma (mv_z)$$

When solving problems, it is assumed that the individual components of this four-vector at any time will have a constant value. In other words, we assume that this quantity is conserved. How can we prove this fact?

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Consider the action of a free relativistic particle of mass $m$, given by,

$$S = -m\int dt \, \sqrt{1-\dot x^2}$$

and we can find the Euler-Lagrange equations imply,

$$\frac{\partial}{\partial t}\frac{\partial L}{\partial \dot x} = \frac{\partial}{\partial t} \frac{m\dot x}{\sqrt{1-\dot x^2}} = 0$$

since $\partial L/\partial x = 0$. Note that we can identify $\partial L/\partial \dot x = p = \gamma m\dot x$. Thus, the Euler-Lagrange equations are the statement that the relativistic momenta are conserved. The Hamiltonian is given by a Legendre transform,

$$H = p\dot x - L = \sqrt{m^2 + p^2}$$

and one can see that,

$$\frac{\partial H}{\partial t} = \frac{p \dot p}{\sqrt{m^2+p^2}} = 0$$

since $\dot p =0$ by the Euler-Lagrange equations implying conservation of momentum. We thus have that the energy and relativistic momenta are conserved and so $p^\mu$ itself is conserved.


Both of these conservation laws are a consequence of the fact the system is invariant under translations in time, and the action does not depend on the position of the particle, since only its derivative $\dot x$ appears in the action. By Noether's theorem, this implies energy and momentum conservation.

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  • $\begingroup$ Is there any way to show the conservation law without the use of Lagrangian Mechanics? I am not yet familiar with it. $\endgroup$
    – Anonymous
    Jul 10 '17 at 16:49
  • $\begingroup$ Playing devil's advocate here: how do you know it's the right action without invoking that it gives the right answer? $\endgroup$ Jul 10 '17 at 17:08
  • $\begingroup$ @ZeroTheHero It's analogous to the Nambu-Goto action for the relativistic string that sweeps out a worldsheet, except for a particle it sweeps out a wordline. Moreover, the action written in a different form, is equivalent to the one that gives rise to the geodesic equations in general relativity, that is, $\sqrt{g_{\mu\nu}\dot x^\mu \dot x^\nu}$. $\endgroup$
    – JamalS
    Jul 10 '17 at 17:18
  • $\begingroup$ @JamalS I don't want to belabour the point too much because I know you have the right action. My observation is that the starting point of this is the experimental observation that the 4-momentum as given in the OP's question is conserved. Whatever action you come up with has to include this conservation law as a consequence. You can come up with all the elegant arguments in the world as to why this or that action should be right, if it is doesn't include conservation of 4-momentum, it's not going anywhere. $\endgroup$ Jul 10 '17 at 17:24
  • $\begingroup$ @ZeroTheHero I think an argument can be made analogous to bubbles, in that they minimise their surface area with respect to the volume they enclose. Since the action is a one-dimensional version of such an action, maybe there is some way to make a similar argument, that isn't self-referential with respect to the conservation of momentum. $\endgroup$
    – JamalS
    Jul 10 '17 at 17:38
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If a Lagrangian has the symmetry to be translationally invariant, by Noether's theorem, you can show the corresponding conserved quantities are energy and momentum.

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There is no a priory way of showing that 4-momentum as defined is conserved. It is however a reasonable guess and, more importantly, it is experimentally verified that this definition leads to conservation of 4-momentum.

I can't for the life of me immediately think of another reasonable definition - it's a 4-vector, it reduces to the usual conservation in the limit of small velocities, etc - but if this definition had failed the experimental test someone would have come up with another definition compatible with experiment.

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  • $\begingroup$ But JamalS has provided a justification for it using the techniques of Lagrangian mechanics. What is wrong with that? $\endgroup$
    – Anonymous
    Jul 10 '17 at 17:06
  • $\begingroup$ @Anonymous see my comment to Jamal's answer. (Not suggesting JamalS is wrong in anyway.) $\endgroup$ Jul 10 '17 at 17:09
  • $\begingroup$ So, it's just an experimental fact and not a consequence of the postulates? $\endgroup$
    – Anonymous
    Jul 10 '17 at 17:59
  • $\begingroup$ @Anonymous As per my last comment to JamalS, AFAIK it is primarily an experimental fact. The form of the 4-vector is dictated by the requirement of properly transforming under Lorentz - that's theoretical. Of course as mentioned above it is immediately included in any theory, and - I want to emphasize this - I don't know of any other reasonable alternative that would compete with the current form. Basically what theory shows is that, if conservation of 4- momentum as defined in your question holds in one inertial frame, it must hold in any other Lorentz-connected frame. $\endgroup$ Jul 10 '17 at 18:06

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