# A Coordinate Transformations in Linearized GR

#### 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 any one be kind enough to explain to me how the metric transformation is gotten from the coordinate transformation?

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#### Orodruin

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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 $x+ \epsilon \xi$ $$g_{ab} (x+ \epsilon \xi) d(x+ \epsilon \xi)^a d(x+ \epsilon \xi)^b = (g_{ab} + \epsilon (g_{ac} \xi^c_{,b} + g_{bc} \xi^c_{,a}))dx^a dx^b + \mathcal O (\epsilon^2) \approx (g_{ab} + \epsilon (\mathcal L_\xi g)_{ab}) dx^a dx^b$$

• vanhees71 and Woolyabyss

#### Orodruin

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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

#### Woolyabyss

Much appreciated guys. I understand now thanks.

• kent davidge

#### samalkhaiat

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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 any one 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 $$- \frac{1}{\epsilon} \left( \bar{g}_{ab} - g_{ab}\right) (x) \equiv - \left( \mathcal{L}_{Y}g\right)_{ab} (x) = Y^{c}\partial_{c}g_{ab} + g_{ac}\partial_{b}Y^{c} + g_{cb}\partial_{a}Y^{c} .$$

• Woolyabyss

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