linford86 said:
Thanks for your reply! Can you give an example of either of these methods?
I can give the rough idea :) Let's pick a tensor T
<br />
T = T_{\mu\nu} dx^{\mu} \otimes dx^{\nu}<br />
and perform the coordinate transformation
<br />
x^{\lambda} \rightarrow x^{\lambda'} = x^{\lambda} - \xi^{\lambda}<br />
We know that for our T we have
<br />
T_{\mu' \nu '}(x') = \frac{\partial x^{\mu}}{\partial x^{\mu '}}\frac{\partial x^{\nu }}{\partial x^{\nu '}}T_{\mu\nu}(x) \ \ \ \ \ \ \ \ (1)<br />
For our coordinate transformation we have that
<br />
\frac{\partial x^{\mu}}{\partial x^{\mu '}} = \delta^{\mu}_{\mu '} + \frac{\partial }{\partial x^{\mu '}}\xi^{\mu}<br />
You can plug this into (1). But the Lie derivative of our tensor T with respect to \xi is given by (in components)
<br />
T_{\mu ' \nu '}(x) - T_{\mu\nu}(x)<br />
You can calculate this by performing a Taylor expansion,
<br />
T(x) = T(x' + \xi) = T(x') + \xi^{\mu}\partial_{\mu}T(x')<br />
and recognizing that, to first order in xi,
<br />
\xi^{\mu}\partial_{\mu}T(x') = \xi^{\mu}\partial_{\mu}T(x)<br />
It's been a while that I did these kind of calculations explicitly, but I think you should be able to do it. It maybe helps to first do it for a scalar field, then for a vector or covector, and so on. Notice that the signs depend on how you define your coordinate transformation with either a plus or minus sign! For a general tensor T there always will be a leading term
<br />
\xi^{\mu}\partial_{\mu}T(x)<br />
The other terms will be due to the "pulling back of the coordinate", but that will become clear if you read something about pushforwards and pullbacks; the GR text by Carroll or Nakahara's geometry book are very useful. Hope this helps :)