Recent content by bloby

Ok, yes it is possible but way too complicate here. (E is expressed in the basis I mentioned above. You are expressing things in ##\{e_x,e_y,e_z\}## here so to take the dot product you would have to express E in ##\{e_x,e_y,e_z\}##) Keep things in shperical coordinates, it is much simpler here...

The domain is right: 1<r<2 and 0<##\rho##<2 pi, but the vectors are expressed in the basis ##\{e_{r},e_{\theta},e_{\rho}\}##. The unit normal outside the surface is then very simple. Can you find dS in this basis?

From the problem statement I would rather say (r, ##\theta##, ##\rho##) are spherical coordinates.

I would say ##\Gamma (E)=\Sigma (E+\Delta) -\Sigma (E)=\Delta \frac{\partial \Sigma}{\partial E}+...##

(1-cos(u)^2)^1/2=|sin(u)|, not sin(u), then two times the integral from -pi/2, pi/2

I see no mistakes, BUT a(a^2-(r cos (u))^2)^-1/2 r is never negative in your domain, so if you get zero the mistake is in the computation of the integral which you did not give us. EDIT: sorry did not see post #6 / sorry LCKurtz did not see your post... time to wake up

My two cents for the OP: Yes, yes, yes. No. To avoid belives, you would have to set an experiment that measures the speed of light of the moving system, and you would find the astonishing fact that the result you obtain and the one obtained by your friend moving with the system agree. Yes...

Hello, for inline latex ##\#\### before and after the expression. 'You would just do' : not exactly, it is what you have to prove assuming ##2^k \geq k+1## holds.

Rather ##T^a_d## the same indices than LHS. The indices are related to basis element. They must be consistent within an equation, like ##v^i=\frac{dx^i}{dt}##, not ##v^i=\frac{dx^j}{dt}##. The 3rd and 4th line of the thumbnail are the same(after corrections) with renamed indices .

Ok, you're welcome.

Hello. Where do e comes from?? Take ##A^i=B^i_jC^j##. j is a dummy index, there is a summation over j. i can be 1, 2 or 3. This equation tells that the equality is true for all three values of i. ##A^k=B^k_jC^j## is exactly the same as ##A^i=B^i_jC^j##, you must not keep track of indices from...

Yes and a, b and c can be 1, 2 or 3 giving 27 gamma's. You have to calculate all the gamma's, but a lot of them are equals or vanish. What did you find for the metric in these coordinates?

Do you see why only one equation remains? How ##\square ## is expressed with low index nabla's only? Do you know the expression for ##A_{{\alpha}{;}{\beta}}## ? (I use latin indices for space components and greek indices for space-time components) note: if you past an url it becomes...

You are right it looks strange for me too. I would rather perform the integration by part first, then we have ##0=\delta S =\delta S^1 - \delta S^2##, with ##\delta S^1=\int d^d x \omega_a(x) \partial_{\mu} j^{\mu}##, for arbitrary small changes in ##\Phi##, in particular those given by the...

For what I understand your last equation in the OP gives the general variation of the action no matter if the transformation is a symmetry and with ##\omega_a## dependent of position.(This is the 'elegant trick': assume a general ##\omega## and put it constant at the end) Your last equation in...