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What is an example of a Lorentz-invariant probability measure on Minkowski space other than the delta measure at the origin?

Euclidean-invariance is easy to attain: the uniform measure on a ball does the trick, and more generally one may take any probability measure on $\mathbb R_{\geq 0}$ and multiply the round measure on the sphere in polar coordinates.

But in Minkowski space, the level curves of the "norm" are not compact, so it seems not so straightforward. It's important to me that this measure have finite (nonzero) volume.

I'm not sure I've chosen the right tags for this question; any help would be appreciated.

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Probably I should have thought this through a bit more before asking, but

There are no Lorentz-invariant probability distributions on Minkowski space, at least under reasonable regularity assumptions.

Proof:

It suffices to consider 1+1 dimensions. Let $(t,x)$ be the standard (time,space) coordinates. First, if $\rho(t,x)dxdt$ is the measure, then because Lorentz transformations are linear with determinant 1, the Jacobian of a Lorentz transformation is identically 1, so Lorentz-invariance demands that the function $\rho$ is Lorentz-invariant, so that $\rho(t,x) = \sigma(\sqrt{t^2 - x^2})$ for some nonnegative function $\sigma$. If $\sigma(m)$ is bounded below by some positive number in a neighborhood of some $m$ (e.g. if $\sigma(m) \neq 0$ and $\sigma$ is continuous in a neighborhood of $m$), then it suffices to show that the Lebesgue measure of the region between two mass shells (i.e. a region of the form $\{(x,t) \mid m^2 - \epsilon \leq t^2 - x^2 \leq m^2 + \epsilon\}$) is infinite. If my calculations are correct, the area diverges logarithmically.

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It seems to me that one can smear out your example of the delta measure at the origin. Parametrize Minkowski space in terms of the coordinates $x^{\mu } $. Pick four suitable fixed 4-vectors $a_0^{\mu } $, $a_1^{\mu } $, $a_2^{\mu } $, $a_3^{\mu } $ and form, say, the measure (not bothering to normalize) $$ p(x)=\exp (- (a_0 \cdot x)^2 - (a_1 \cdot x)^2 - (a_2 \cdot x)^2 - (a_3 \cdot x)^2 ) \ . $$ By construction, this is Lorentz-invariant. Suitable, as far as I understand, are any $a_0 $, $a_1 $, $a_2 $, $a_3 $ that yield a finite result when we integrate $p$ over all of Minkowski space. In my current rest frame, I'd pick $a_{\rho }^{\mu } = \delta_{\rho }^{\mu } $.

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  • $\begingroup$ Thanks! But I don't see why this should be Lorentz-invariant. If I apply a Lorentz transformation $\gamma$, don't I get a new measure of the form $q(x) = \exp(-\sum (\gamma^{-1}(a_i) \cdot x)^2)d^4x$? I guess that means this is a finite-dimensional subspace of all measures which is closed under Lorentz transformations, but I was hoping for a single measure which is Lorentz-invariant. I'm also not sure the total measure is really finite, although perhaps that could be addressed by setting the measure to zero on space-like vectors, say... $\endgroup$
    – Tim Campion
    Commented Jun 20, 2019 at 17:43
  • $\begingroup$ Yes, it occurred to me after posting that it's really a functional form that's invariant that you want. I'm of course thinking of also transforming the $a_i $ - say I'm looking at a physical apparatus described by the $a_i $, and the motion of an object described by $x^{\mu } $, and now I decide to look at all this from a different Lorentz frame. I do think the measure as given is finite - I've "Euclideanized" it ... $\endgroup$
    – Michael Engelhardt
    Commented Jun 20, 2019 at 17:51

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