Recent content by dwd40physics

I'm being honest in that your explanation does not actually explain how to do an inverse transformation. It just says do it. I've fudged something to get the answer I need now but not sure I understand how I got it to to be honest.

you asked if I could do the forward transformation so I answered that part, I don't know how to do the backward transform.

so I get ##\tilde{g}_{\mu\nu} (x + \xi) = (\delta^{\rho}{ }_{\mu} - \partial{_{\mu}}\xi^{\rho})(\delta^{\sigma}{ }_{\nu} - \partial{_{\nu}}\xi^{\sigma})g_{\rho\sigma}(x)##?

I don't get how to do an inverse transform for this - could you show me some steps.

it is a function of y

do you mean the RHS ?if so yes

yes I see how you get the forward one, I did the forward one already and you get the kronecka deltas and minus partial derivatives, still not sure how to do LHS

sorry i think my computer is bugging apologies

Still not sure would this be ##g_{\mu\nu}(x) = \frac{\partial{y_\rho}}{\partial{x_\mu}}\frac{\partial{y_\sigma}}{\partial{x_\nu}}\tilde{g}_{\rho\sigma} (y)## ?

How do I apply inverse to 2.4, I tried but couldn't do it

i don't get this part:

is ##\tilde{g}_{\mu \nu}## different from ##g_{\mu \nu}## in terms of it is the same function but expressed in different co-ordinates or are they completely different functions ?

yeah I think lots of Uk universities are more tight with uploading accessible content, yeah I wasn't sure if I could have help on the screenshots - I know other books go more abstract as you've said. how does $$\tilde{g}_{uv}$$ differ from $${g}_{\rho\sigma}$$

yes you've found correct course - lecture notes have been produced this year.