Recent content by ProfDawgstein

Thanks! Schutz GR, Carrol GR, Schutz MM, Cahill, Lovelock, Ohanian, ... should do the trick :)

Ok, done :) Just plug in, replace ab by ml, use symmetry of g_ml. Thank you both. Now I know what to prove next - or maybe I should finally start a serious book about this :rolleyes: I could do that for sure, but I haven't been that much into this, that I think about things like...

It wasn't much work, actually. But that still boils down to ##\frac{\partial g^{ml}}{\partial x^p} \frac{\partial g_{ml}}{\partial {x^i}} == \frac{\partial g^{ml}}{\partial x^i} \frac{\partial g_{ml}}{\partial {x^p}}##, doesn't it? Using your definition, I get the first term to be...

Hey, I have been doing a few proofs and stumbled across this little problem. Trying to show the symmetry of the Ricci tensor by using the Riemann tensor definition ##R^m_{\ ikp} = \partial_k \Gamma^m_{\ ip} - \partial_p \Gamma^m_{\ ki} + \Gamma^a_{\ ip} \Gamma^m_{\ ak} - \Gamma^a_{\ ik}...

Oh yes, I know. I just wanted to see what I get. Taking the trace / contracting with ##\eta## made it easier to compare both equations. If you have ##h_{ab}## or ##h^{ab}## shouldn't matter, right? It just has to be on the same levels as in the other equation. At the moment I am just...

I don't know, but maybe this helps... http://home.comcast.net/~peter.m.brown/gr/geodesic_equation.htm

In this exercise ##T_{\mu\nu} = 0##. ##\partial_\mu \partial^\mu h_{\alpha\beta} - \partial_\mu \partial_\beta h^{\mu}_{\ \ \alpha} - \partial_\mu \partial_\alpha h_{\beta}^{\ \ \mu} + \eta_{\alpha\beta} \partial_\mu \partial_\nu h^{\nu \mu} + \partial_\alpha \partial_\beta h -...

[ limits to the rescue :) ] So you emailed him? Cool. I don't think I need anything else right now, thanks.

Okay, after 1 hour of 'index porn' I got this result... ##\partial_\mu \partial^\mu h - \partial_\mu \partial_\nu h^{\mu\nu} = 0## 1) I treated ##h^{\mu\nu}## as symmetric 2) I contracted out the ##\eta##'s 3) Divided by the common factors (0.5 in mine, 4 in 3.38) If I do the same for...

Wow, the obvious again... I have seen way too many indices today, thanks alot! Kind of stressed to finish this, not my own book... I will do the whole thing again tomorrow, and will post my results here. Any tips/hints to make the calculation easier? I think my starting point should be...

Another Exercise... Ohanian Exercise 6 The Euler-Lagrange Equation given: ##\frac{\partial}{\partial x^\mu} \frac{\partial \mathcal{L}}{\partial (\partial_\mu h^{\alpha\beta})} - \frac{\partial \mathcal{L}}{\partial h^{\alpha\beta}} = 0## ##\frac{\partial \mathcal{L}}{\partial...

I was wondering about the whole thing using ##\delta## notation... Using the Lagrangian ##\mathcal{L} = -\frac{1}{4}F^{\mu\nu}F_{\mu\nu}## where ##\delta F_{\mu\nu} = \delta (\partial_\mu A_\nu - \partial_\nu A_\mu) = \delta (\partial_\mu A_\nu) - \delta (\partial_\nu A_\mu) =...

Wow, nice explanation. Thanks! I need to find a book which does this a lot more, the other books didn't spend too much time on anything :( What does Wald say?

Actually I am working on it right now [hahah], but thanks :) EDIT: done ##\partial_\mu F^{\mu\nu} = 0## looks good

Two pages and 400 indices later, yes. It was a total mess, especially that little index mistake :D NOW I will try your way. Where ##\frac{\partial F_{\rho\sigma}}{\partial (\partial_{\mu}A_{\nu})} = \frac{\partial (\partial_\rho A_\sigma - \partial_\sigma A_\rho )}{\partial...