Derivation of gauge condition in linearized GRby WannabeNewton Tags: condition, derivation, gauge, linearized 

#1
Feb1113, 11:58 AM

C. Spirit
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Hey there guys! So we know that in linearized GR we work with small perturbations [itex]\gamma _{ab}[/itex] of the background flat minkowski metric. In deriving the linearized field equations the quantity [itex]\bar{\gamma _{ab}} = \gamma _{ab}  \frac{1}{2}\eta _{ab}\gamma [/itex] is usually defined, where [itex]\gamma = \gamma ^{a}_{a}[/itex]. Under the action of an infinitesimal diffeomorphism (generator of flow), [itex]\gamma _{ab}[/itex] transforms as [itex]\gamma' _{ab} = \gamma _{ab} + \partial _{b}\xi _{a} + \partial _{a}\xi _{b}[/itex] (this comes out of the lie derivative of the minkowski metric with respect to the flow generated by this vector field). This implies that [itex]\bar{\gamma' _{ab}} = \bar{\gamma _{ab}} + \partial _{b}\xi _{a} + \partial _{a}\xi _{b}  \eta _{ab}\partial^{c}\xi _{c}[/itex]. Since we have the freedom to then fix the gauge by choosing some [itex]\xi ^{a}[/itex], we can take one satisfying [itex]\partial ^{b}\partial _{b}\xi _{a} = \partial ^{b}\bar{\gamma _{ab}}[/itex] which, after differentiating the expression for [itex]\bar{\gamma' _{ab}}[/itex], gives [itex]\partial^{b}\bar{\gamma' _{ab}} = 0[/itex]. Apparently we can then conclude from this that [itex]\partial^{b}\bar{\gamma _{ab}} = 0[/itex] but why is that? Is it because in a background flat space  time we can regard [itex]\partial^{b}\bar{\gamma' _{ab}} = 0[/itex] as a covariant equation due to being able to treat [itex]\triangledown _{a}[/itex] as [itex]\partial _{a}[/itex] therefore, since [itex]\bar{\gamma '_{ab}}, \bar{\gamma _{ab}}[/itex] are related by a diffeomorphism, the equation must remain invariant under the transformation [itex]\bar{\gamma '_{ab}}\rightarrow \bar{\gamma _{ab}}[/itex] (in the context of GR)?




#2
Feb1113, 04:45 PM

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I don't see the difference. You use the gauge transformation to put γ_{ab} into the Hilbert gauge and then just drop the prime.




#3
Feb1113, 04:52 PM

C. Spirit
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Hi Bill! Thanks for responding. My question is why are we allowed to drop the prime? Thanks again mate.




#4
Feb1113, 08:41 PM

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Derivation of gauge condition in linearized GR
Seems to be a discussion after Eq 20, 21 of http://www.tapir.caltech.edu/~chirata/ph236/lec08.pdf. Something about nonuniqueness of the gauge.




#5
Feb1213, 06:59 AM

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By the way, in gravity this gauge condition IS called the Hilbert gauge, not the "Lorentz gauge", which pertains to electromagnetism. And "Lorentz gauge" itself is incorrect. Quoting Wikipedia, 


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