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Derivation of gauge condition in linearized GR

by WannabeNewton
Tags: condition, derivation, gauge, linearized
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WannabeNewton
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Feb11-13, 11:58 AM
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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)?
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Bill_K
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Feb11-13, 04:45 PM
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I don't see the difference. You use the gauge transformation to put γab into the Hilbert gauge and then just drop the prime.
WannabeNewton
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Feb11-13, 04:52 PM
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Hi Bill! Thanks for responding. My question is why are we allowed to drop the prime? Thanks again mate.

atyy
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Feb11-13, 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 non-uniqueness of the gauge.
Bill_K
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Feb12-13, 06:59 AM
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My question is why are we allowed to drop the prime?
It's just notation. Whether you call it γab or γ'ab or something else entirely, it represents the gravitational perturbation in the Hilbert gauge.

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,
The Lorenz condition is named after Ludvig Lorenz. It is a Lorentz invariant condition, and is frequently called the "Lorentz condition" because of confusion with Hendrik Lorentz, after whom Lorentz covariance is named.
WannabeNewton
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Feb12-13, 08:05 AM
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Thanks Bill and atyy!


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