- #1
- 5,848
- 552
Hey there guys! So we know that in linearized GR we work with small perturbations \gamma _{ab} of the background flat minkowski metric. In deriving the linearized field equations the quantity \bar{\gamma _{ab}} = \gamma _{ab} - \frac{1}{2}\eta _{ab}\gamma is usually defined, where \gamma = \gamma ^{a}_{a}. Under the action of an infinitesimal diffeomorphism (generator of flow), \gamma _{ab} transforms as \gamma' _{ab} = \gamma _{ab} + \partial _{b}\xi _{a} + \partial _{a}\xi _{b} (this comes out of the lie derivative of the minkowski metric with respect to the flow generated by this vector field). This implies that \bar{\gamma' _{ab}} = \bar{\gamma _{ab}} + \partial _{b}\xi _{a} + \partial _{a}\xi _{b} - \eta _{ab}\partial^{c}\xi _{c}. Since we have the freedom to then fix the gauge by choosing some \xi ^{a}, we can take one satisfying \partial ^{b}\partial _{b}\xi _{a} = -\partial ^{b}\bar{\gamma _{ab}} which, after differentiating the expression for \bar{\gamma' _{ab}}, gives \partial^{b}\bar{\gamma' _{ab}} = 0. Apparently we can then conclude from this that \partial^{b}\bar{\gamma _{ab}} = 0 but why is that? Is it because in a background flat space - time we can regard \partial^{b}\bar{\gamma' _{ab}} = 0 as a covariant equation due to being able to treat \triangledown _{a} as \partial _{a} therefore, since \bar{\gamma '_{ab}}, \bar{\gamma _{ab}} are related by a diffeomorphism, the equation must remain invariant under the transformation \bar{\gamma '_{ab}}\rightarrow \bar{\gamma _{ab}} (in the context of GR)?
Last edited:
