Recent content by birulami

Yes, that is what I assume, with ##r_R## being just very slightly larger than the Schwarzschild radius (assuming a situation where the event horizon is at the Schwarzschild radius). So it is true that the energy of the photon is on its way to infinity at ##r_R## then?

I found http://physicspages.com/2013/05/05/schwarzschild-metric-gravitational-redshift/: \frac{\lambda_R}{\lambda_E} = \sqrt{\frac{1-2GM/r_R}{1-2GM/r_E}} where the indexes R and E are for receiver and emitter respectively, and the speed of light is normalized to 1. Most other sources on the...

Do you have an example? Not sure whether this really is an example. Just because r gets smaller than the Schwarzschild radius does not change the Schwarzschild metric, so ##g_{00}## still does not depend on ##t##.

In SRT, the line element is ##c^2ds^2 = c^2dt^2 - dx^2 -dy^2-dz^2## and ##g_{00} = 1## (or ##-1## depending on sign conventions). In the Schwarzschild metric we have g_{00}=(c^2-\frac{2 GM}{r}) . So in the first example, ##g_{00}## is constant, in the second it depends on another coordinate...

In general relativity we have c^2ds^2 = c^2dt^2 - dx^2 -dy^2-dz^2. From this we can derive the not commonly used equation: (c\frac{ds}{dt})^2 + (\frac{dx}{dt})^2 + (\frac{dy}{dt})^2 + (\frac{dz}{dt})^2 =c^2 \qquad\qquad\qquad (1) I think this is the "velocity" squared of an object relative to...

Thanks for your answers, except I don't get it where you all are heading. I did not mention the term "black hole", I did not say that the photon has a certain size and I did not pronounce any appropriateness of the Schwarzschild solution for light. All i did was toy with some physical formulas...

Photons with smaller and smaller wave lengths have a higher and higher energy and these engeries have an increasing Schwarzschild radius r_s. Consequently i can ask when half the wave length \lambda/2 is equal to r_s, such that one wave length fits into the sphere of the Schwarzschild radius. I...

Given an electromagnetic field by its components E and B. How is this related to probability amplitudes of a Schrödinger wave function for the same field. Trying the same or at least a similar question from a different angle: given the E and B field, can we derive from it, in principle, not...

And you are telling me now that this glitch in the typography of the slightly out of place indexes is the crucial point?:cry: Oh my, I thought I understood why we have upper and lower indexes? But what is the meaning of an index slighly shifted to the right?

Zee writes in Einstein Gravity in a nutshell page 186 "let us define the transpose by ##(\Lambda^T)_\sigma^\mu = \Lambda_\sigma^\mu##" and even emphasizes the position of the indexes. Yes, they are not exchanged! This must be a typo, right?

My hunch is you want to understand a wave in general first, whether it is electromagnetic or probability amplitude. For me it was quite helpful to read about the wave equation (not wave function!) as well as the Airy disk. To me the latter was an eye opener, because it seems a single hole...

A scalar potential ##\phi: \mathbb{R}^4\to\mathbb{R}## has the physical unit of energy per particle property, which can be charge or mass. Take the positional derivative and multiply by the particle property to get the force on the particle. So far gravitational and electric potential are the...

Suppose I have a parameterized line ##\phi:\mathbb{R}\to\mathbb{R}^n## given by ##\phi(t) = (x^\mu(t))|_{\mu=1}^n##. How can I tell that the line is straight. My best answer so far is that at every time ##t## the acceleration (2nd derivative) is parallel to the velocity (1st derivative), i.e...

Yeah, MathJax! This is why I was asking, because MathJax does it. But I could not figure out how to do it without the autonumber option of MathJax, which is not set here. Now I found out in the GitHub tickets of MathJax how to do it. Here is how it looks like in the blog entry. As always, right...

No hurry. I just wanted to be sure that I need not take some further action to see it published. Thanks, Harald.