# Killing's equation from infinitesimal transformation

1. Feb 26, 2007

### 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?

2. Feb 26, 2007

### dextercioby

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

You'll get the remaining term easily.

3. Feb 26, 2007

### cristo

Staff Emeritus
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?

4. Feb 26, 2007

### AlphaNumeric

Standard Taylor expansion,

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

5. Feb 26, 2007

### cristo

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

6. Feb 27, 2007

### cristo

Staff Emeritus
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!

7. Feb 27, 2007

### AlphaNumeric

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

8. Feb 27, 2007

### cristo

Staff Emeritus
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)

9. Feb 27, 2007

### coalquay404

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.

Last edited: Feb 27, 2007
10. Feb 27, 2007

### cristo

Staff Emeritus
Yup, I got it, thanks.