Recent content by Psi-String

I noticed that. But I can't derive it out if I use different indices... I don't know how to turn g^{\mu \sigma}g_{\lambda \rho}(\nabla_\sigma \nabla^\rho \nabla _\mu K^\lambda) into \nabla _\lambda \nabla^\mu \nabla_\mu K^\lambda

I think I figure it out! :redface: \nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma = (g^{\mu \sigma} \nabla_\sigma) (g_{\sigma \mu}\nabla^\mu)(\nabla_\mu K^\sigma) = g^{\mu \sigma}g_{\sigma \mu} (\nabla _\sigma \nabla^\mu \nabla_\mu K^\sigma) + g^{\mu \sigma}(\nabla^\mu \nabla_\mu...

Hi~ R_{\mu \lambda} \nabla ^\mu K^\lambda =0 is due to the Killing Equation and the symmetry of Ricci tensor R_{\mu \lambda} = R_{\lambda \mu} But it seems like \nabla ^\mu \nabla _\sigma \nabla _\mu K^\sigma \neq \nabla _\sigma \nabla ^\mu \nabla _\mu K^\sigma ?? Thanks for Reply!

Homework Statement I'm currently self-studying Carroll's GR book and get stuck by proving the following identity: K^\lambda \nabla _\lambda R = 0 where K is Killing vector and R is the Ricci ScalarHomework Equations Mr.Carroll said that it is suffice to show this by knowing: \nabla _\mu...

well, this is an integration I need to carry out when solving Griffiths' problem. The solution manual only give the result as above, it didn't take out step by step, and I don't think it is that obvious that we can solve it just by one equation XDDDD

Hi! Could someone give me an idea how the following integration be carried out?? \int \frac{r dr}{ \left( r^2 + a^2 - 2rau \right)^{3/2}} = \frac{ru - a}{a(1-u^2)\sqrt{r^2 + a^2 - 2rau}} where u and a are constant. I have encounter such integration several times in electrodynamics. I can...

Thanks! It does provide plenty of help!

Could you sketch the idea/method how to get the diagonal tensor? This consequence is due to Maxwell equations, in fact, we use maxwell equations and some vector calculus to derive it. But this consequence are so "simple" that I think it might have some simple way that can comes to the same...

Why do the non-diagonal vanishes:confused: Despite the Kronecker delta term which will vanishes, there are still have terms that will not vanish.( \epsilon_0 E_i E_j + \frac{1}{\mu _0} B_i B_j ). I think I could understand what the stress tensor means. But what I'm curious about is why...

Maxwell stress tensor: T_{ij} = \epsilon_0 \left( E_i E_j - \frac{1}{2} \delta_{ij} E^2 \right) + \frac{1}{\mu_0} \left( B_i B_j - \frac{1}{2} \delta_{ij} E^2 \right) We can interpret T as the force per unit area acting on the surface. But what surprises me is, T_{ij} = T_{ji}, i.e. the...

How about Partial Differential Equations: An introduction, by W. A. Strauss?? Is this a good book for self-study after learning one semester of ODE??

I know I can't find the direction of the produced photon by only using conservation of energy, I also need conservation of momentum. But I don't see why the magnitude won't be the same by two different methods. I don't understand your last sentence. What do you mean of "outgoing" wave and...

Problem:An electron flies toward +x direction with velocity of 0.9c, while a positron flies toward -y direction with the same velocity. Assuming their speed is so fast that they collide and annihilate at the origin, what will be the magnitude and the direction of the wave vector of the generated...

I see. Thanks a lot!

J77, could you explain more details about your example to me?? Is this kind of couple relates to interference :confused: Thanks for help!