Recent content by JunhoPhysics

In classical field theories, I believe I understood how to derive a Noether charge that corresponds to a symmetry of action. And there is no problem in understanding its time independence. But in quantum field theory, it looks like the two different approaches, 1) Canonical quantization...

Oh wait... This means ##\chi## is infact orthogonal to ##\partial_t,\partial_r,\partial_\theta,\partial_\phi## all of them. Haha... it looks like my original question about this Kerr BH was not a problem in the first place... But anyway I learned a lot from your help. Thanks!

?? I checked this just a few minutes ago. Also, if it does not vanish, can you explain what is wrong in the following? $$\begin{align} \chi^2=0&=g_{tt}+2g_{t\phi}\Omega_H+g_{\phi\phi}\Omega_H^2\nonumber\\ &=g_{tt}+g_{t\phi}\Omega_H+\Omega_H(g_{t\phi}+g_{\phi\phi}\Omega_H)\nonumber\\...

I thought ##\chi_\mu=(g_{tt}+g_{t\phi}\Omega_H,0,0,g_{t\phi}+g_{\phi\phi}\Omega_H)=(0,0,0,0)## is true. In post #9, I have shown ##\chi^2=0=g_{tt}+2g_{t\phi}\Omega_H+g_{\phi\phi}\Omega^2## on the horizon. But in the calculation I checked ##g_{tt}+g_{t\phi}\Omega_H## and...

Thanks a lot! And I just found that ##\chi_\mu## is in fact a zero vector (even though ##\chi^\mu## is not a zero vector) on the event horizon so it is obviously parallel to ##n_\mu##. Finally everything perfectly makes sense. Again, I really appreciate for your help!

Yes, I absolutely agree with your general argument (except a minor thing, that is ##n\neq\partial_r##). But my question was, when we try to check this general argument with a specific example, 1) On a Kerr BH's event horizon, ##\chi## and ##n~(n_\mu=(0,1,0,0))## both are null, normal(=tangent)...

1. One thing I can tell about the direction of ##\partial_r## is that it is normal to ##\Sigma:r-r_H=0## by definition. Then since ##\Sigma## is a null surface, it would also be tangent to ##\Sigma##. I am not sure what you mean by "the direction of the basis vector on that surface." 2. From...

Oh, so "Killing" was important! I mean, for any Killing vector field ##\chi## and its Killing horizon ##\Sigma:\chi^\nu\chi_\nu=0##, the normal vector of ##\Sigma## can be written as ##n_\mu=\nabla_\mu(\chi^\nu\chi_\nu)##. And then ##\chi## is orthogonal to ##n## since ##\chi^\mu...

I am really sorry, but I still don't get it. I do understand that "##\chi## is normal to ##\Sigma~\Leftrightarrow~\chi## is tangent to ##\Sigma##" when ##\Sigma## is a null hypersurface. But my question is, the only information that we have on ##\chi## is that it is a Killing vector field and...

Thanks! Now I got the answer for some part of my questions but not for all of them, so please let me summarize them. 1. Now I see why there are no two linearly indept null tangent vectors on a null hypersurface. (If ##\sigma_1## and ##\sigma_2## both are normal to a normal vector of the null...

In p.244 of Carroll's "Spacetime and Geometry," the Killing horizon ##\Sigma## of a Killing vector ##\chi## is defined by a null hypersurface on which ##\chi## is null. Then it says this ##\chi## is in fact normal to ## \Sigma## since a null surface cannot have two linearly independent null...