Recent content by Jokar

Hi, I gave the article and the equation number. I am trying to understand the section 5 equation 9 of this article. This is the link. https://adsabs.harvard.edu/full/1953MNRAS.113...34S According to the notation of the paper, in frame 1 the vector potential is ##(-1, 0, 0, 0)## In the...

Sorry, still can't find the difference According to the notation of the paper, in frame 1 the vector potential is $$(-1, 0, 0, 0)$$ In the rotating reference frame it is $$(-\sqrt{1+(\omega r)^2}, \omega y, -\omega x, 0)$$. Now in the first reference frame the electromagnetic tensor is...

I am trying to understand the section 5 equation 9 of this article. This is the link. https://adsabs.harvard.edu/full/1953MNRAS.113...34S

Hi, I am trying to understand the section 5 equation 9 of this article. ( I was just trying to copy the cutting of the section. However, some how I can't attach image.) [Moderator's note: attachment deleted. See post #9 for link.]

Suppose there are two reference frames. One is rotating with respect to another with rotational velocity ##\omega##. Now if in one of the reference frames the vector potential is $$(1, 0, 0, 0)$$ then in the other reference frame it will be $$(\sqrt{1-(\omega r)^2}, 0, \omega r, 0)$$. Now in...

Thanks. I did not replace it in the left side. So I was not getting the answer. Thank you very much.

Any help?

Thanks. I am also doing the same thing. But I think there are some steps in between. If you take the derivative then you will get $$ \frac{dp_\gamma}{d\tau} = -\frac{1}{2} \frac{\partial {g^{\alpha\beta}}}{\partial x^\gamma} (p_\alpha - A_\alpha)(p_\beta - A_\beta) +...

I have a Hamiltonian $$ H = \frac{1}{2} g^{\alpha \beta}\left(p_\alpha- A_\alpha\right)\left(p_\beta- A_\beta\right) $$ I want to calculate the equation of motion. How can I calculate the equation of motions $$ \frac{dx^\mu}{d\tau} = g^{\mu\nu}(p_\nu - A_\nu) $$ This one is straight...

Thank you very much. This was the coordinate that I was looking for.

Can you please elaborate. The 4D Riemann Curvature tensors are ## 0 ## as they won't change due to coordinate transform. So, at a given instant ## \tau ##, we can assume two Minkowski coordinate frames moving with respect to each other. And therefore, ## \frac{d t}{d\tau} \ne \frac{d...

OK. Let' rephrase my original question based on all the answers. The line element in the K frame ## d\tau^2 = -dt^2 + dx^2 + dy^2 + dz^2 ## According to the above proposition in K' frame ## d\tau^2 = - dt'^2 + dx'^2 + dy'^2 + dz'^2 + \omega^2 (x'^2 + y'^2) dt'^2 + 2\omega (x' \, dy' - y' \...

What kind of coordinate transform? Thats what is the question. One thing you can do is ## t = t' ## ## r = r' ## ## \theta = \theta' + w * t' ## ## z = z' ## But there is the problem. In this coordinate transform you are assuming that ## t = t' ##. How can that be true? Because in the ## K'...

Thanks for your reply. Thats what I said. In K frame all are orthogonal. But not in K' frame. So, what will be the metric in the K' frame. Do you have any idea or any reference? It will be helpful.

Sorry, I don't understand. Let's say all the coordinates are orthogonal in K. Now in K' frame the particle is moving with velocity wr. So, the particle's time will show some time dilation and t cannot be equal to t'. Can you please elaborate? Maybe I have some problem in understanding.