Recent content by Bartolius

Didn't you leave a factor in the second part of the variation? shouldn't it be $$ \delta S = \int d^{10} x \delta \sqrt{-g}\left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]^2 +2 \int d^{10} x \sqrt{-g} \left[ f(x_1, \ldots, x_{10}) +\delta(y)A\right]\delta\left[ f(x_1, \ldots, x_{10})...

ok, maybe I was doing the calculations too fast, I'll check later, thank you! as for the smeared out function: yes it was the starting point of the calculation, the delta distribution was smeared out on a circle (the directions y and z combine to give a subspace with a circle as a boundary) but...

Yes because I have to differentiate on ## A ##, so there is actually no reason for the ## \delta (y) ## to disappear. Also, I would like to use the ## \delta ## whenever I have them to integrate away the y coordinate, for the other terms of the action are 9 dimensional (the field ## A ## lives...

Homework Statement I have to write equations of motion for a field, namely ## A ##. The full action has actually three terms, but my problem is with a part of the action reading: $$ S =\int d^{10}x \sqrt{-g} [ f(x_1, ... , x_{10}) + \delta (y) A ]^2 $$ In the 10 x's there is of course the...

No, because R is defined to be R^\alpha _\alpha = g^{\nu \gamma} R^\alpha _{\nu \alpha \gamma} and you can obtain that form by exchange of the last two indices, hence the minus sign

The real error you made in the normalization resides in the use you made of the orthonormality relations of the eigenfunctions: it is not the simple product of the two eigenfunctions that must result 1 or 0.

Hi everyone, it is a real pleasure to have found such a qualified forum to talk about my passion and hopefully future work. I am an Italian student in Theoretical Physics at the University of Florence, my main interests are QFT and Relativity. I joined this forum to seek help and deeper insights...