I have been reading these notes ("Minkowski Spacetime: A Hundred Years Later", by Vesselin Petkov) 1, in which the author states (in the middle of the text on page 137) that

"The only Lorentz invariant interaction between two particles (at least those for which a notion of conserved total energy and momentum could be defined) is a contact interaction. Only if the interaction occurs only when the two particles are at a single spacetime point could the system be Lorentz invariance and conserve energy and momentum."

Why is this the case? I thought that the reason why we occur that interaction terms in the Lagrangian density are evaluated at a single point was to ensure that the interaction is local and that this notion of locality be Lorentz invariant (i.e. that the interaction is local in all inertial frames of reference)? $^{(\ast)}$

$^{(\ast)}$ I think the following is the reason why this is the case, please correct me if I'm wrong.

[In particular, the reason interaction terms must be evaluated at single spacetime points is because of the following:

  1. If the two objects are space-like separated then there will be no inertial reference frame in which they are located at the same spatial point and as such any direct interaction between them will certainly be non-local as they will be able to directly (and superluminally) affect one another regardless of their spatial separation - this is action at a distance which is physically untenable in relativity.

  2. If the two objects are time-like separated then again this implies non-locality in the interaction of two objects, as there will be multiple inertial frames, but only one in which the objects are located at the same point in space. Therefore, they will be non-local in space apart from in one frame of reference, and as such, if we demand Lorentz invariance and locality in space, then we must have locality in time.

  3. For light-like separation the argument is a similar one as for the time-like case, as the objects will still be located at distinct points in all frames, and hence they will be non-local in space (as they would be able to directly interact without being in physical contact). However, it is less straightforward to show as we can't use arclength between points to distinguish them (mathematically) as the interval is zero. Nonetheless, it is still possible to do so by choosing an appropriate affine parametrisation. (I have to admit I'm a bit shaky on this argument- improvements/clarification would be appreciated).

Hence the only case in which the interaction is local in all inertial frames of reference is when it occurs at a single point in spacetime. (Of course, an object located at a given point in spacetime can also interact with objects within its immediate neighbourhood. In the discrete case, this can only be achieved through interacting with objects at adjoining spacetime points. In the continuum case, this can only be achieved through coupling to derivative terms, with the derivatives evaluated at the point at which the object is situated).]

  • $\begingroup$ Hi Will - for future reference, please don't make large numbers of edits to your posts. $\endgroup$
    – David Z
    Jul 23, 2015 at 13:52
  • $\begingroup$ @DavidZ Sorry about that, kept on realising my mistakes too late. $\endgroup$
    – Will
    Jul 23, 2015 at 17:23

1 Answer 1


These to me sound like they're two sides of the same coin. If you lose locality of interaction, then you lose locality of energy conservation, and you therefore have, among other things, combinations of energy transfer which simply push the energy out to $\infty$ instantaneously, creating a pathological global violation.

I am not sure that I buy your statement about time-separated interactions, since timelike vectors preserve causality and so the obvious problem isn't obvious. Instead, I would point out that if you can "teleport" the energy and momentum through time nonlocally, then you can teleport it through space nonlocally: Alice and Bob both operate science labs on spaceships; they together put a set of particles in a trajectory where it will surely be at 4-position $r^\bullet$ in their mutual relativistic future: they agree in advance on that point in spacetime and the amount of energy-momentum to transfer. Then they separate and at some spacelike separated interval both do the following: Alice dumps the requested energy-momentum into the particles at $r^\bullet$; Bob calculates $r^\bullet + \delta r^\bullet$ for some time shortly thereafter and absorbs the requested energy-momentum from those particles at that point in spacetime. (This applies even if the interaction is not a bidirectional information transfer, but usually it is and you've got bigger causality problems if that's true.)

The author is certainly not intending the most literal reading of his statement: for of course, both particles have energy-momentum 4-vectors at all points in their trajectories and if you choose a "present moment" before either side of the interaction and compare it with a "present" after both sides of the interaction are complete, before other interactions have occurred, if the interaction preserves energy-momentum in the usual sense then these two total-momentums will be equal. In other words, the system has "eventual conservation", the way some non-ACID databases have "eventual consistency."

But presumably the author means that in the time "between" these two moments, total energy and momentum between the system appears to fluctuate, and you can use this fluctuation to push the "eventuality" as far away as you want, achieving semi-permanent energy-momentum.

  • $\begingroup$ Would what I've put about locality of interactions be a correct understanding at all then (I've updated it somewhat), at least in the case where they are spatially separated? $\endgroup$
    – Will
    Jul 19, 2015 at 16:18

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