- #1

cristo

Staff Emeritus

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## Main Question or Discussion Point

I know this isn't technically special or general relativity, but I'm posting this here since, hopefully, people in this forum will be familiar with the question!

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?

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?