Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Derivation of gauge condition in linearized GR

  1. Feb 11, 2013 #1

    WannabeNewton

    User Avatar
    Science Advisor

    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)?
     
    Last edited: Feb 11, 2013
  2. jcsd
  3. Feb 11, 2013 #2

    Bill_K

    User Avatar
    Science Advisor

    I don't see the difference. You use the gauge transformation to put γab into the Hilbert gauge and then just drop the prime.
     
  4. Feb 11, 2013 #3

    WannabeNewton

    User Avatar
    Science Advisor

    Hi Bill! Thanks for responding. My question is why are we allowed to drop the prime? Thanks again mate.
     
  5. Feb 11, 2013 #4

    atyy

    User Avatar
    Science Advisor

  6. Feb 12, 2013 #5

    Bill_K

    User Avatar
    Science Advisor

    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,
     
  7. Feb 12, 2013 #6

    WannabeNewton

    User Avatar
    Science Advisor

    Thanks Bill and atyy!
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Derivation of gauge condition in linearized GR
  1. GR as a Gauge theory ? (Replies: 1)

  2. GR as a gauge theory (Replies: 54)

Loading...