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I came across a proof that says if a 4-vector $P$ is conserved in one inertial frame $$P_{before}=P_{after} \text{ (sum)}$$ then Lorentz transforming to another frame gives $$P'_{before}=\Lambda P_{before} = \Lambda P_{after} = P'_{after}$$ However, I am worried that the terms in the sums $\Lambda P_{before}$ and $\Lambda P_{after}$ will not be all at the same instant of time. But in order for $\Lambda$ P to be deemed conserved the terms have to be simultaneous. How does one resolve this problem? Thanks.

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Isn't it much simpler to just write: $\partial_\mu V^\mu=0$? This automatically holds in any inertial frame, for it is manifestly Lorentz invariant. –  Danu Jan 12 at 3:35

1 Answer 1

I assume that the question is referred to special relativity.

The fact that you use sums seems to mean that you are dealing with a finite number of interacting points of matter. I expect that you are assuming that the interactions are local, i.e. they happens in single events, the interaction vertices. In this case there is no 4-momentum associated with extended interactions (like the energy momentum of the EM field for interacting charges). In this picture the points evolve along time-like or light-like geodesics, namely causal segments, until they interact with other points in single events. The number of points may change after interactions, but along each segment, the $4$-momentum is constant.

Considering a fixed interaction vertex, the 4-momentum conservation requires that the sum of ingoing $4$-momenta is equal to that of the outgoing ones. Ingoing ("before the local interaction") and outgoing ("after the local intereaction") is an absolute notion, referring to the light-cone centered on the vertex. It is clear that the total ingoing (outgoing) 4-momentum is a $4$-vector as it is sum of $4$-vectors. No reference frame has been used up to now. However if you fix a reference frame, it is obvious, in this picture, that conservation of $4$-momentum in an inertial reference implies the same for any another inertial reference frame, because "before" and "after" do not depend form the reference since are absolutely defined by the light-cone of the interaction event.

If you eventually consider a flat 3-surface defining the rest space of an inertial observer at some time $t$, the 4-momentum of the system at that time is defined as the sum of all momenta of the segments intersecting the surface (also considering many interaction vertices simultaneously). You can sum vectors applied to different events since the spacetime of special relativity is an affine space. You can easily prove that, if in each interaction vertex the 4-momentum is conserved as said above, and taking into account that along each segment describing the story of a particle the $4$-momentum is constant, then the total four momentum of the whole system does not depend on the used reference frame and the time labeling the rest 3-space of the reference frame used to sum the momenta.

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