Unraveling Dirac's General Relativity Equation

dEdt
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There's an equation from Dirac's book on General Relativity that I don't get at all. It isn't derived; instead it's treated as almost self-evident, when it isn't.

Dirac begins by defining ##p^\mu## to be the "mass 4-current" of a continuous distribution of matter (matter in this case really being matter, ie not including the E&M field). This is defined such that ##p^\mu=\rho_0 v^\mu##, where ##\rho_0## is the mass density of a particular element of matter in that element's rest frame, and ##v^\mu## is just the four-velocity of the matter at the same point. Basically, ##p^\mu## is just the mass flux, in the same way as ##J^\mu## is the charge flux.

Dirac is trying to derive the equations of motion for a distribution of matter using the action principle. He does this by first making "arbitrary variations in the position of an element of matter" and seeing how it affects ##p^\mu##. He then asserts that if each element of matter is displaced from ##x^\mu## to ##x^\mu +b^\mu##, where ##b^\mu## is small, then ##\delta p^\mu=(p^\nu b^\mu - p^\mu b^\nu),_{\nu}##.

As I stated at the beginning, he doesn't derive this equation. I was hoping that someone else could describe where this equation comes from. Thanks.
 
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Well he does give an argument supporting this expression for δpμ but doesn't really derive it. However I believe the expression he winds up with is just the Lie derivative of pμ with respect to b.

(The Lie derivative of a vector quantity is more familiar, but here pμ is a vector density and you get an additional term in that case involving bμ.)
 
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Yes, he did give an argument for his expression, but when reading it I was reminded of Nathaniel Bowditch's quote about Laplace: "I never came across one of Laplace's 'Thus it plainly appears' without feeling sure that I have hours of hard work before me to fill up the chasm and find out how it plainly appears." I felt the same way about Dirac's "the generalization is evidently".

Could you explain why the Lie derivative of ##p^\mu## with respect to ##b^\mu## would give ##\delta p^\mu##? It isn't obvious to me, though granted I am still a novice to the mathematics of GR.
 
dEdt said:
Could you explain why the Lie derivative of ##p^\mu## with respect to ##b^\mu## would give ##\delta p^\mu##?
Dirac says, "Let us suppose each element of matter is displaced from zμ to zμ + bμ with bμ small. We must determine the change in pμ at a given point."

While according to Wikipedia, the Lie derivative is "the change of a tensor field along the flow of another vector field." Sounds like the same thing!

It can get a lot more formal from then on. Roughly, the idea is this. You do a point transformation, that is you actually pick up and move the points a small amount. (Dirac's "displacing the matter") You construct a new coordinate system x' in which each point now has the same coordinate values that it did in the original coordinate system x. We say the coordinate system has been dragged along. Likewise you drag along every scalar, vector and tensor quantity by saying when displaced it has the same components in x' that it originally had in x. Then these quantities have been dragged along also, and the Lie derivative is how much they have changed if you just look at them in x.

Using Dirac's notation bμ for the small displacement, the Lie derivative of a scalar s is bμμs. That is, the directional derivative of s along b. For each tensor index you get a correction term.

For a contravariant vector vμ the Lie derivative is bσσvμ - vσσbμ.

A covariant index has a similar correction term but with a plus sign.

For a density w there's a correction term: bμμw + w∂μbμ.

This is all explained in more detail in Wikipedia!
 
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