In the setting of general relativity, I came across a source term of the wave equation of the following form:

$$ \frac{1}{\sqrt{q}}\,\delta^{(3)}(p-\gamma(t)) $$

where $p\in M$ is a point in our 4d spacetime and $\gamma(t)$ is a trajectory that the source takes in the 4d spacetime. $\sqrt{q}$ is the 3d metric determinant of a preferred 3d slicing of $M$. $\delta^{(3)}(p-\gamma(t))$ is a 3d Delta distribution which means that we should have

$$ \int_M\,\delta^{(3)}(p-\gamma(t))\,f(p)\,d^3x=f(\gamma(t))\,. $$

Of course, this is rather the physics short hand notation that $\delta^{(3)}(p-\gamma(t))$ is a map $C^\infty_c(M)\to C^\infty(\mathbb{R})$ that maps a function $f$ to $f\circ \gamma$.

We would like to show that the Lie derivative of the source along a certain vector field $T$ vanishes if $\gamma(t)$ is a Killing trajectory, that is $\gamma$ is an integral curve of $T$.

However, we are very confused about the rigorous treatment of this expression. Our intuition states the following:

  1. The delta distribution should transform as scalar density of weight 1 under changes of the 3d frames and as a scalar under changes of the frame along the forth direction. However, we are not sure how to make this rigorous, nor how to find the Lie derivative of such a combined expression.

  2. The object $1/\sqrt{q}$ should be an inverse object, that is it is a scalar density of weight $-1$ under changes of the 3d frame and a regular scalar along the forth direction.

  3. Combining these two statements, it would make sense that the original object $$ \frac{1}{\sqrt{q}}\,\delta^{(3)}(p-\gamma(t)) $$ is a regular 4d scalar. This would also make a lot of sense because it serves as source term of a regular 4d scalar wave equation.

  4. Finally, it makes somehow sense that the Delta distribution is invariant under the Killing vector field $T$ iff the trajectory $\gamma$ is an integral curve of $T$. But we are not sure how to prove this and how to deal rigorously with the delta distribution.

  • $\begingroup$ Why can't you just prove this using the chain rule? $\pounds_{T}\delta^{3}(p-\gamma(t)) = -\pounds_{T}\gamma_{t}\pounds_{T}\delta^{3}(p-\gamma(t))$ obviously vanishes if $\gamma$ flows along an isometry. $\endgroup$ – Jerry Schirmer Jun 9 '14 at 20:30
  • $\begingroup$ and apparently \pounds doesn't work in the version of mathjax that SE uses. $\endgroup$ – Jerry Schirmer Jun 9 '14 at 20:31
  • $\begingroup$ @JerrySchirmer to do this in mathjax write \it\unicode{xA3} to produce $\it\unicode{xA3}$ --- weird, I know, but it's MathJax's quirk... $\endgroup$ – Alex Nelson Jun 9 '14 at 20:55
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    $\begingroup$ I've always preferred the caligraphic L: $\mathcal L_T$. $\endgroup$ – Robin Ekman Jun 9 '14 at 21:40
  • $\begingroup$ Thanks a lot for your suggestions so far, but up to now I'm even not convinced what the proper definition of the Lie derivative of the delta distribution is. In particular, because this delta distribution is not a regular 4 density, but this combined 3-density plus scalar. Maybe somebody of you can make it mathematically more rigorous... $\endgroup$ – LFH Jun 10 '14 at 22:33

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