Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

145 submissions , 122 unreviewed
3,930 questions , 1,398 unanswered
4,852 answers , 20,624 comments
1,470 users with positive rep
501 active unimported users
More ...

Lie derivative of Dirac Delta

+ 4 like - 0 dislike
730 views

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.

This post imported from StackExchange Physics at 2014-06-11 21:30 (UCT), posted by SE-user user30835
asked Jun 9, 2014 in Theoretical Physics by user30835 (20 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2014-06-11 21:30 (UCT), posted by SE-user Jerry Schirmer
and apparently \pounds doesn't work in the version of mathjax that SE uses.

This post imported from StackExchange Physics at 2014-06-11 21:30 (UCT), posted by SE-user Jerry Schirmer
@JerrySchirmer to do this in mathjax write \it\unicode{xA3} to produce $\it\unicode{xA3}$ --- weird, I know, but it's MathJax's quirk...

This post imported from StackExchange Physics at 2014-06-11 21:30 (UCT), posted by SE-user Alex Nelson
I've always preferred the caligraphic L: $\mathcal L_T$.

This post imported from StackExchange Physics at 2014-06-11 21:30 (UCT), posted by SE-user Robin Ekman
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...

This post imported from StackExchange Physics at 2014-06-11 21:30 (UCT), posted by SE-user user30835

I'd start by thinking about the pullback of distributions. If $\delta^3_\gamma:f\mapsto f\circ\gamma$, something like $\Phi^*\delta^3_\gamma:f\mapsto f\circ\Phi^{-1}\circ\gamma$ might make sense (you might need to throw in a factor of $|\det T\Phi^{-1}|$ as well...)

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
p$\hbar$ysic$\varnothing$Overflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
To avoid this verification in future, please log in or register.




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...