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.


1 Answer 1


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


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