Infinitesimal coordinate change vs. the metric

Question

I need some clarification with the problem in this post (unfortunately I cannot comment there, so I'm asking it as a separate question).

The problem in short:

We have an infinitesimal coordinate change $ x'^{\mu} \rightarrow x^{\mu}+\alpha k^{\mu} $.

Then, the metric will also change (taken from the answer of @Prahar): $$ g_{\mu \nu}^{\prime}\left(x^{\prime}\right)=\frac{\partial x^{\rho}}{\partial x^{\prime \mu}} \frac{\partial x^{\sigma}}{\partial x^{\prime \nu}} g_{\rho \sigma}(x). $$

The goal is to derive the following expression for the new metric:

$$ g_{\mu \nu}^{\prime}(x)=g_{\mu \nu}(x)-\alpha\left(g_{\mu \rho} \partial_{\nu} k^{\rho}+g_{\nu \rho} \partial_{\mu} k^{\rho}+k^{\rho} \partial_{\rho} g_{\mu \nu}\right)+\mathcal{O}\left(\alpha^{2}\right). $$

The part where I'm stuck: how exactly do we substitute $x'^\mu$? I started out with $x^\mu = x'^\mu - \alpha k^\mu $, which means $$\frac{\partial x^{\rho}}{\partial x^{\prime \mu}} = \delta^\rho_\mu - \frac{\partial \epsilon k^{\alpha}}{\partial x^{\prime \mu}}$$ but I have no idea what to do after this step.

Answer 1

For $y=x+\alpha k$, $g_{ij}'(y)=\frac{\partial x_i}{\partial y_k} \frac{\partial x_j}{\partial y_l} g_{kl}(x(y))$.

We can Taylor expand the left hand side $g'_{ij}(y)=g'_{ij}(x(y)+\alpha k)= g'_{ij}(x(y))+\alpha k^a \partial_a g_{ij}(x(y))+O(\alpha^2)$.

We can expand the right hand side: $$\left(\delta^i_k \delta^j_l -\alpha \delta^i_k \frac{\partial k_j}{\partial y_l} -\alpha \frac{\partial k_i}{\partial y_k} \delta^j_l+O(\alpha^2)\right) g_{kl}(x(y))$$

We can use the chain rule to write $\frac{\partial k_i}{\partial y_k} =\frac{\partial k_i}{\partial x_j} \frac{\partial x_j}{\partial y_k}= \frac{\partial k_i}{\partial x_k}+O(\alpha)$.

Putting these three things together, we get:

$$g'_{ij}(x(y))+\alpha k^a \partial_a g_{ij}(x(y))+O(\alpha^2)=\left(\delta^i_k \delta^j_l -\alpha \delta^i_k \frac{\partial k_j}{\partial x_l} -\alpha \frac{\partial k_i}{\partial x_k} \delta^j_l+O(\alpha^2)\right) g_{kl}(x(y))$$

Rearranging and simplifying, we get the desired result:

$$g'_{ij}(x(y))=g_{ij}(x(y))-\alpha(k^a\partial_a g_{ij}+\frac{\partial k_j}{\partial x^l} g_{il} +\frac{\partial k_i}{\partial x_k} g_{kj})+O(\alpha^2)$$