Lorentz transformation of the Dirac function

Question

Consider a Lorentz transformation that takes the $(x, y, z, t)$ coordinates of a point in Minkowski space into $(x', y', z', t')$. An electrically charged object at rest in the first reference frame has an electric charge density $rho$, while in the second reference frame the electric charge density is $rho' = gamma(v) rho$. When the object collapses into a material point at the origin, at $t = t'= 0$, $rho(x, y, z) = e delta^3(x, y, z)$ and $rho'(x', y', z') = e delta^3(x', y', z')$, where $e$ is the charge of the electron. It seems that $delta^3(x', y', z') = gamma(v) delta^3(x, y, z)$. For a boost in the $x$ direction, $delta(x') = gamma(v) delta(x)$. This equation doesn't seem to come out of the rule for $delta(f(x))$. From $x'= gamma(v) (x - v t)$ I get $delta(x'(x)) = delta(x - vt)/gamma(v) => delta(x)/gamma(v)$ when t = 0. What am I doing wrong?

J. Murray · Answer

You have to be careful here, because $delta (x')$ is not merely a function of $x$.  When you write down the expression
$$e = int d^3 r  e delta(x)delta(y)delta(z)$$
you are taking a slice of spacetime (e.g. $t=t_0$) and then integrating $rho$ over that slice.  If you boost to a different reference frame, then the slice over which you are integrating will not be characterized by constant $t'$. Upon transforming coordinates
$$x' = gamma(x-vt) implies x = gamma(x'+vt')$$
$$y'=y$$
$$z'=z$$
$$t'=gammaleft(t - frac{vx}{c^2}right)implies t = gammaleft(t' +frac{vx'}{c^2}right)$$
we see that the slice $t=0$ corresponds to the slice $t' =- gammafrac{vx'}{c^2}$.  Correspondingly,
$$x = gamma(x'+vt') = gammaleft(1-frac{v^2}{c^2}right)x'= frac{x'}{gamma}$$
which means that $delta (x) = deltaleft(frac{x'}{gamma}right) = gamma delta(x')$.

Lorentz transformation of the Dirac function

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