Graduate Coord Transf. in Linearized GR: Understanding Metric Transformation

[Woolyabyss]

I was studying linearized GR where we make the following coordinate transformation $\tilde{x}^{a} = x^{a} + \epsilon y^{a}(x)$

This coordinate transformation is then meant to imply $g_{ab}(x) = \tilde{g}_{ab}(x) + \epsilon \mathcal{L}_{Y} g_{ab}$

Would anyone be kind enough to explain to me how the metric transformation is gotten from the coordinate transformation?

 

[Orodruin]

To some extent, this is how the Lie derivative of the metric is defined. The Lie derivative is given by
$$
\mathcal L_Y g = \lim_{\epsilon\to 0}\left[\frac{1}{\epsilon}(\gamma_{Y}(\epsilon)^* g - g)\right],
$$
where $\gamma_Y(\epsilon)$ is the flow of the vector field $Y$ for a parameter distance $\epsilon$. If you write this out in coordinates, you will get exactly your given relation.

(Note that $f^*g$ is the pullback of $g$ under the diffeomorphism $f$.)

 

[kent davidge]

Noticing that the Lie Derivative for the metric is $(\mathcal L_\xi g)_{ab} = g_{ac} \xi^c_{,b} + g_{bc} \xi^c_{,a}$
you can write down the metric in $x+ \epsilon \xi$ $$g_{ab} (x+ \epsilon \xi) d(x+ \epsilon \xi)^a d(x+ \epsilon \xi)^b = (g_{ab} + \epsilon (g_{ac} \xi^c_{,b} + g_{bc} \xi^c_{,a}))dx^a dx^b + \mathcal O (\epsilon^2) \approx (g_{ab} + \epsilon (\mathcal L_\xi g)_{ab}) dx^a dx^b$$

 

[Orodruin]

Noticing that the Lie Derivative for the metric is $(\mathcal L_\xi g)_{ab} = g_{ac} \xi^c_{,b} + g_{bc} \xi^c_{,a}$
you can write down the metric in $x+ \epsilon \xi$ $$g_{ab} (x+ \epsilon \xi) d(x+ \epsilon \xi)^a d(x+ \epsilon \xi)^b = (g_{ab} + \epsilon (g_{ac} \xi^c_{,b} + g_{bc} \xi^c_{,a}))dx^a dx^b + \mathcal O (\epsilon^2) \approx (g_{ab} + \epsilon (\mathcal L_\xi g)_{ab}) dx^a dx^b$$

This is a bit of a backwards argument. The definition of the Lie derivative is that given in #2. Based on that definition, you can derive $(\mathcal L_\xi T)_{ab} = T_{ac} \xi^c_{,b} + T_{cb} \xi^c_{,a} + \xi^c T_{ab,c}$ for an arbitrary type (0,2) tensor field $T$. Now, if you have the Levi-Civita connection, then $\nabla_\xi g = 0$ and you would have
$$
(\mathcal L_\xi g)_{ab} = g_{ac} \xi^c_{;b} + g_{cb} \xi^c_{;a}.
$$
Note that you get some Christoffel symbols out of $\xi^c g_{ab,c}$ that turn the derivatives of the $\xi$ into covariant derivatives. Thus, you get the expression you started with from applying the definition of the Lie derivative, not the other way around.

 

[Woolyabyss]

Much appreciated guys. I understand now thanks.

 

[samalkhaiat]

I was studying linearized GR where we make the following coordinate transformation $\tilde{x}^{a} = x^{a} + \epsilon y^{a}(x)$

This coordinate transformation is then meant to imply $g_{ab}(x) = \tilde{g}_{ab}(x) + \epsilon \mathcal{L}_{Y} g_{ab}$

Would anyone be kind enough to explain to me how the metric transformation is gotten from the coordinate transformation?

Just substitute \bar{x} = x + \epsilon Y in the transformation law of the metric tensor and expand to first order in \epsilon. It is more convenient to work with the inverted transformation g_{ab}(x) = \frac{\partial \bar{x}^{c}}{\partial x^{a}}\frac{\partial \bar{x}^{d}}{\partial x^{b}} \bar{g}_{cd}(\bar{x}) . Keep in mind that infinitesimal transformation means that \bar{g}_{ab} = g_{ab} + \mathcal{O}(\epsilon). Therefore, to first order in \epsilon, you can use \epsilon \ \bar{g}_{ab}(x) = \epsilon \ g_{ab}(x) when you Tylor-expand \bar{g}_{cd} (\bar{x}) and when you multiply \bar{g}_{cd}(x) by other \epsilon-terms:
g_{ab}(x) = \left( \delta^{c}_{a} + \epsilon \ \partial_{a}Y^{c}\right) \left( \delta^{d}_{b} + \epsilon \ \partial_{b}Y^{d} \right) \left( \bar{g}_{cd}(x) + \epsilon \ Y^{e}\partial_{e} g_{cd}(x) \right) . So, to first order, you can rewrite this as - \frac{1}{\epsilon} \left( \bar{g}_{ab} - g_{ab}\right) (x) \equiv - \left( \mathcal{L}_{Y}g\right)_{ab} (x) = Y^{c}\partial_{c}g_{ab} + g_{ac}\partial_{b}Y^{c} + g_{cb}\partial_{a}Y^{c} .