# Trying proving that energy and momentum are conserved in every inertial frame

I'm trying to show, using Lorentz transformations, that if relativistic energy and momentum are conserved in $$S$$, then they are conserved in $$S'$$ too.

## A wrong proof

Let's suppose I know that in system $$S$$ energy and momentum are conserved (in isolated systems). So, taken two time $$t_1$$ and $$t_2$$ we know that $$\begin{equation} \sum_i \gamma_{u_i} m_i = \sum_j \gamma_{u_j} m_j \end{equation}$$ $$\begin{equation} \sum_i \gamma_{u_i} m_i \mathbf{u}_i = \sum_j \gamma_{u_j} m_j \mathbf{u}_j \end{equation}$$ Where sum in $$i$$ and $$j$$ are done respectively before and after (I mean at time $$t_1$$ and at time $$t_2$$). Well, exploiting Lorentz transformations is easy to show that $$\begin{equation} \gamma_{u_i} = \gamma \gamma_{u_i'} \left( 1 + \frac{u_{xi}' v}{c^2} \right) \end{equation}$$ $$\begin{equation} u_{xi} = \frac{u_{xi}' + v }{1 + \frac{u_{xi}' v}{c^2}} \end{equation}$$ So for energy and $$x$$ component of momentum, the sums in $$S'$$ became $$\begin{equation} \sum_i \gamma_{u_i'} m_i + \frac{v}{c^2 } \sum_i \gamma_{u_i'} m_i u_{xi}' = \sum_j \gamma_{u_j'} m_j + \frac{v}{c^2 } \sum_j \gamma_{u_j'} m_j u_{xj}' \end{equation}$$ $$\begin{equation} \sum_i \gamma_{u_i'} m_i u_{xi}' + v \sum_i \gamma_{u_i'} m_i = \sum_j \gamma_{u_j'} m_j u_{xj}' + v \sum_j \gamma_{u_j'} m_j \end{equation}$$ Multiplying the first by $$-\frac{c^2}{v}$$ and then sum to the second we find conservation of energy in $$S'$$ too. If instead we multiply by $$-v$$ and we sum we get the conservation of $$x$$ component of momentum. Transverse components of momentum are even simpler, we only have to exploit $$\begin{equation} u_y = \frac{u_{yi}'}{\gamma \left( 1+\frac{u_{xi}' v}{c^2} \right)} \end{equation}$$ At first glance one can say wow, great, this looks simple and powerful, like best proofs do. Unfortunately after thinking about it I realized that this beautiful proof was... completely wrong! I hadn't noticed an important detail: in $$S'$$ I do a sum with particles that are in different time (because of different positions), this is meaningless.

## Question

I fear there are no way to patch up this proof and I have to throw it in the trash. So how can I reach the goal to proof that if energy and momentum are conserved in $$S$$ so are conserved in $$S'$$ too? I fear the only way to bypass problem related to time transformations is writing locally an energy conservation equation, and then show that this equation is Lorentz-invariant. But this road doesn't seem easy, and anyway I beg you not to start writing hieroglyphics in tensor notation, I can't understand them. Thank you.

## Edit: a simple proof, but it works only for perfect gas

Let's suppose that in $$S$$ we have a box containing perfect gas in equilibrium. We have $$E_{tot} \equiv \sum_i E_i$$ where $$i$$ is the energy of the particle $$i$$. If $$S'$$ use the same definition of energy and momentum it is easy to show that for each particle $$E_i'=\gamma (E_i - v p_{xi})$$. Each particle has this energy at its own transformed time so we shouldn't sum, but if we are considering a gas, for statistical reason, we don't care if we take all the particles at the same time or we measure their energy and momentum at different time, even in this last case we will have $$\sum_i p_{xi} \approx 0$$ and $$\sum_i E_i$$ approximately always the same, no matter the time in which we do measurement for each particle (if we are not choosing different time deliberately with the intention of obtaining certain values for $$\sum_i p_{xi}$$ and $$\sum_i E_i$$ but this is not our case: different time are simply due to different $$x$$ coordinates and particles of gas are uniformly distributed in the space, this distribution is not related to their dynamical properties). But $$E_{tot}' \equiv \sum_i E_i'$$. Conclusion: $$E_{tot}' = \gamma E_{tot}$$, so for $$S'$$ system too energy is conserved. Lorentz transformations implies that if energy of the perfect gas in a box is conserved in $$S$$, then it is conserved in $$S'$$ too.

• I think the local-conservation approach is the best because it avoids this simultaneity problem. But if you can’t tolerate tensors, that’s not really an option for you, because it is the energy-momentum four-tensor that is locally conserved in all inertial frames. If you are going to learn much physics, sooner or later you are going to have to get used to tensors. – G. Smith Jul 8 '19 at 0:50

If all the energy and momenta are brought together at a single event then it is easy: just define the total 4-momentum before and after the event: $$P = \sum_i P_i, \;\;\;\; Q = \sum_i Q_i$$ In some other frame you have $$P'_i = \Lambda P_i$$, $$Q'_i = \Lambda Q_i$$ where $$\Lambda$$ is the Lorentz transformation, and therefore $$P' = \sum_i \Lambda P_i = \Lambda \sum_i P_i = \Lambda P, \\ Q' = \sum_i \Lambda Q_i = \Lambda \sum_i Q_i = \Lambda Q.$$ Hence if $$Q=P$$ then $$Q' = P'$$. QED
When the different 4-momenta are located at different events you have to think it through more carefully. In this case the sum $$\sum_i P_i$$ is a shorthand for a sum at events simultaneous in the first frame, whereas $$\sum_i P'_i$$ would be a sum at events simultaneous in the second frame, and this is a different set of events. (In more technical language, we say they lie on a different spacelike surface). As a result, you cannot simply assume that each $$P'_i$$ in the second sum is equal to $$\Lambda P_i$$ for the $$i$$'th term in the first sum, because you are talking about two different events. This is the subtle point which you probably need a textbook or other resource to help you with. What happens is that the colliding parties at any given event preserve their total energy-momentum and this allows you to gather their terms in the sum so that you can sum them over different spacelike surfaces and expect to get the same answer. Having done this, you can then revert to the method of proof as given above.