# Expanding a metric tensor in a Taylor series around another metric tensor

I am doing problem 5.11 in Guidry, which asks the following:

using $$g_{\mu \nu}(x) = \frac{\partial x'^{\alpha}}{\partial x^{\mu}} \frac{\partial x'^{\beta}}{\partial x^{\nu}} g_{\alpha \beta}(x')$$ and the transformation $$x^{\mu} \rightarrow x'^{\mu} = x^{\mu} + \epsilon K^{\mu}$$ to show that to first order, $$g_{\mu \nu}(x) = (\delta^{\alpha}_{\mu}+\epsilon\partial _{\mu}K^{\alpha})(\delta^{\beta}_{\nu}+\epsilon\partial _{\nu}K^{\beta})g_{\alpha \beta}(x').$$ Then, by expanding $$g_{\alpha \beta}(x')$$ in a Taylor series around $$g_{\alpha \beta}(x)$$ show that to order $$\epsilon$$ the above equation implies that $$\partial _{\nu}K_{\mu} + \partial _{\mu}K_{\nu} + K^{\gamma}\partial_{\gamma}g_{\mu \nu} = 0.$$ The part that I am stuck at is expanding $$g_{\alpha \beta}(x')$$ in a Taylor series. The solution set says it should be $$g_{\alpha \beta}(x) + \frac{\partial g_{\alpha \beta}}{\partial x^{\gamma}} \vert_x\Delta x^{\gamma} = g_{\alpha \beta}(x')$$ and that $$\Delta x^{\gamma} = \epsilon K^{\gamma}.$$ However, I do not understand why $$\Delta x^{\gamma} = \epsilon K^{\gamma}$$ and how to interpret $$x^{\gamma}$$ here.

• Commented Jun 19, 2022 at 8:02
• thanks so much, I tried looking this up before but I guess I didn't use the proper key words! @Eletie Commented Jun 19, 2022 at 16:14

$$\Delta x^\gamma = x'^\gamma - x^\gamma$$. Here $$\gamma$$ is just an index.