- #1

cristo

Staff Emeritus

Science Advisor

- 8,107

- 73

Suppose the metric tensor is form-invariant under the transformation [itex]x\rightarrow\tilde{x}[/itex], so we require [tex]g_{ij}(x)=\frac{\partial\tilde{x}^k}{\partial x^i}\frac{\partial \tilde{x}^l}{\partial x^j}g_{kl}(\tilde{x})[/tex](*)

Now, suppose we take the infinitesimal transformation [tex]\tilde{x}^i=x^i+\varepsilon \xi^i(x)[/tex]

The textbook then says that (*) takes on, in order [itex]\varepsilon[/itex] the form [tex]0=\frac{\partial \xi^i(x)}{\partial x^k}g_{il}(x)+\frac{\partial\xi^j(x)}{\partial x^l}g_{kj}(x)+\xi^i(x)\frac{\partial g_kl(x)}{\partial x^i}[/tex].

I can't see how the author arrives at this. My calculation of (*) using the transformation gives [tex] g_{ij}(x)=\frac{\partial}{\partial x^i}\left[x^k+\varepsilon\xi^k(x)\right]\frac{\partial}{\partial x^j}\left[x^l+\varepsilon\xi^l(x)\right]g_{kl}(\tilde{x})[/tex]

And, on calculating the derivatives[tex]g_{ij}(x)=\left(\frac{\partial x^k}{\partial{x^i}} +\varepsilon \frac{\partial \xi^k}{\partial x^i}\right)\left(\frac{\partial x^l}{\partial x^j}+\varepsilon\frac{\partial\xi^l}{\partial x^j}\right)g_{kl}(\tilde{x})[/tex]

Expanding, and noting that [tex]\frac{\partial x^i}{\partial x^j}=\delta^i_j[/tex] gives [tex] g_{ij}(x)=g_{ij}(\tilde{x})+\varepsilon\frac{\partial\xi^k}{\partial x^i}g_{kj}(\tilde{x})+\varepsilon\frac{\partial \xi^k}{\partial x^j}g_{il}(\tilde{x})[/tex]

From here, I don't see how to get the metric tensors on the RHS in terms of x. I wonder if anyone could help me with this, or point out an error?