Recent content by S. Leger

Yeah, that's it! Sorry but it took some time to check all your steps in the calculations and to prove it handwritten by myself, especially the indices.. But everything turns out right now und it's pretty clear. With all this information I understand what the trick (in andriens post) is. fzero...

fzero, I'm sorry if I have not yet completely understood your hint but does this mean that $$\delta^a_b$$ (Kronecker) is the same as $$\delta$$ (variation)? And isn't the result I get from this not $$\delta h =h^{bc}h_{ca}h$$ rather than $$\delta h=h^{ab}\delta h_{ab}h$$?

This is what I found in another thread in this forum: https://www.physicsforums.com/threads/variation-of-the-metric-tensor-determinant.658572/ take a look at #3 to see what I mean. What does andrien mean?

Thank you samalkhaiat, Yes this is one of those shorthand and easy ways to variate the metric determinant. But I am searching the derivation of the above equation ( with the levi cevita) behause I heard that this was an easy "trick" to get the result

the variation of the determinant should be w.r.t. the metric, i.e. $$h_{ab}$$ I thought that the second term is right because there is only a variation in "part" $$h_{\alpha_1 \beta_1}$$ of the field. But you're right, there is no relation to the right term - and that's my problem. Any hint how...

This is my attempt for a general solution, for n dimensions: $$h=det(h_{ab})=\frac{1}{n!}\epsilon^{\alpha_1...\alpha_n}\epsilon^{\beta_1...\beta_n} h_{\alpha_1\beta_1}...h_{\alpha_n\beta_n}$$ and so $$\delta h=\frac{1}{(n-1)!}\epsilon^{\alpha_2...\alpha_n}\epsilon^{\beta_2...\beta_n}...

Thanks a lot fzero! I'm working on a more general solution, which I will post as soon as I can express it. It's based on the fact that the derivation of the above determinant is a subdeterminant of the determinant itself. Multiplicated with the matrix will result in the determinant again - and...

Homework Statement I'm trying to calculate the variation of the following term for the determinant of the metric in the polyakov action: $$h = det(h_{ab}) = \frac{1}{3!}\epsilon^{abc}\epsilon^{xyz}h_{ax}h_{by}h_{cz}$$ I know that there are some other ways to derive the variation of a metric...