# Killing's equation from infinitesimal transformation

• cristo
In summary, the author tries to derive the metric tensor on the RHS in terms of x using the infinitesimal transformation \tilde{x}^i=x^i+\varepsilon \xi^i(x), but can't seem to get it to work. He then uses the standard Taylor expansion to get g_{ij}(x).f

#### cristo

Staff Emeritus
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 $x\rightarrow\tilde{x}$, so we require $$g_{ij}(x)=\frac{\partial\tilde{x}^k}{\partial x^i}\frac{\partial \tilde{x}^l}{\partial x^j}g_{kl}(\tilde{x})$$(*)

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

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

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

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

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

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?

You need to expand the $g_{kl}(\tilde{x})$ in terms of $g_{kl}(x)$

You'll get the remaining term easily.

I'm not too sure how to do that though. Would I write $g_{kl}(\tilde{x})=g_{kl}(x+\varepsilon\xi)$? If so, how would I expand that?

Standard Taylor expansion,

$$g_{kl}(x+\epsilon \xi) = g_{lk}(x) + \epsilon \xi^{\nu} \frac{\partial (g_{kl}(x))}{\partial x^{\nu}} + \ldots$$

Standard Taylor expansion,

$$g_{kl}(x+\epsilon \xi) = g_{lk}(x) + \epsilon \xi^{\nu} \frac{\partial (g_{kl}(x))}{\partial x^{\nu}} + \ldots$$

Ahh, thanks. I don't know why I didn't think of doing that!

Ok, so I got the line I required (thanks to both of you for helping). So, I now have $$0=\frac{\partial \xi^l}{\partial x^j}g_{il}(x)+\frac{\partial \xi^k}{\partial x^i}g_{kj}(x)+\xi^m\frac{\partial g_{ij}(x)}{\partial x^m}$$. The author now transforms this into the next line $$0=\frac{\partial \xi_i}{\partial x^j}+\frac{\partial \xi_j}{\partial x^i}+\xi^m\left(\frac{\partial g_{ij}}{\partial x^m}-\frac{\partial g_{mj}}{\partial x^i}-\frac{\partial g_{im}}{\partial x^j}\right)$$.

How does he do this? I get that the first two terms are just contracted with the metric tensor, but I don't get how the last term turns into the terms in the brackets. Is it something to do with the fact that in the first line we are differentiating the metric which depends on x, but in the second line we are differentiating only the components of the metric tensor?

Any help, as always, will be much appreciated!

Is it something to do with the fact that in the first line we are differentiating the metric which depends on x, but in the second line we are differentiating only the components of the metric tensor?
No, you're still differentiating the same things, just the notation that each component of the metric is a function of position has been dropped since that's sort of obvious given you're differentiating with respect to x as it is.

The factor $$\left(\frac{\partial g_{ij}}{\partial x^m}-\frac{\partial g_{mj}}{\partial x^i}-\frac{\partial g_{im}}{\partial x^j}\right)$$ looks suspiciously like that of the Christoffel symbol (up to a metric contraction and a factor of -2)

$$\Gamma^{a}_{bc} = \frac{1}{2}g^{ad}(g_{db,c}+g_{dc,b}-g_{bc,d})$$ where $$_{,a} = \frac{\partial}{\partial x^{a}}$$

I would hazard a guess (without putting pen to paper) that you're to use the fact $$\xi$$ is Killing, so it obeys $$\xi_{a;b}+\xi_{b;a}=0$$ where $$_{;a} = \nabla_{a}$$. If you expand that out, you will find something extremely similar to the expression you're trying to derive, if not the expression itself. It's been a while since I did anything about Killing vectors and I can't remember if that is the component form of the Killing equation or not, but it's too similar to the connection $$\Gamma$$ to be unconnected (no pun intended).

I hope that helps You're right that the term is suspiciously like the Christoffel symbol (good spot there!), but I can't use the fact that $\xi$ is a Killing vector, since this exercise is supposed to be deriving that!

Ok, so I get that fact that (x) can be dropped; I imagine it was only left in earlier since the metric depended on $\tilde{x}[/tex] before. I think I may have seen where the extra terms come from. We have $$0=\frac{\partial \xi^l}{\partial x^j}g_{il}+\frac{\partial \xi^k}{\partial x^i}g_{kj}+\xi^m\frac{\partial g_{ij}}{\partial x^m}$$. Now, in order to get to the next line we want to contract [itex]\xi^l$ in the partial derivative of the first term. But, what I neglected before, was that the metric tensor and the partial derivative operator do not commute! So, $$\frac{\partial \xi_i}{\partial x^j}=\frac{\partial}{\partial x^j}(\xi^lg_{il})=\frac{\partial \xi^l}{\partial x^j}g_{il}+\xi^l\frac{\partial g_{il}}{\partial x^j}$$. Thus the extra two subtracted terms come about from this.

Does that sound right?

(after that we go on to put in $\Gamma$ and thus give the form of Killing's equation you stated above)

Ok, so I got the line I required (thanks to both of you for helping). So, I now have $$0=\frac{\partial \xi^l}{\partial x^j}g_{il}(x)+\frac{\partial \xi^k}{\partial x^i}g_{kj}(x)+\xi^m\frac{\partial g_{ij}(x)}{\partial x^m}$$. The author now transforms this into the next line $$0=\frac{\partial \xi_i}{\partial x^j}+\frac{\partial \xi_j}{\partial x^i}+\xi^m\left(\frac{\partial g_{ij}}{\partial x^m}-\frac{\partial g_{mj}}{\partial x^i}-\frac{\partial g_{im}}{\partial x^j}\right)$$

How does he do this?

This is simply a result of the Leibniz property of the partial derivative. More explicitly,

$$0=\frac{\partial \xi^l}{\partial x^j}g_{il}(x)+\frac{\partial \xi^k}{\partial x^i}g_{kj}(x)+\xi^m\frac{\partial g_{ij}(x)}{\partial x^m}$$
$$0 = \partial_j(\xi^lg_{il}) + \partial_i(\xi^kg_{kj}) + \xi^m\partial_mg_{ij} - \xi^l\partial_jg_{ij} - \xi^k\partial_ig_{kj}$$
$$0 = 2\partial_{(i}\xi_{j)} + \xi^m(\partial_m g_{ij} - \partial_jg_{im} - \partial_i g_{mj}$$

I've used $2\partial_{(i}\xi_{j)}\equiv\partial_i\xi_j + \partial_j\xi_i$ here.

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This is simply a result of the Leibniz property of the partial derivative. More explicitly,

$$0=\frac{\partial \xi^l}{\partial x^j}g_{il}(x)+\frac{\partial \xi^k}{\partial x^i}g_{kj}(x)+\xi^m\frac{\partial g_{ij}(x)}{\partial x^m}$$
$$0 = \partial_j(\xi^lg_{il}) + \partial_i(\xi^kg_{kj}) + \xi^m\partial_mg_{ij} - \xi^l\partial_jg_{ij} - \xi^k\partial_ig_{kj}$$

You should now be able to see how to obtain the desired result.

Yup, I got it, thanks.