### Question

In Euclidean three-space we can define paraboloidal coordinates ##\left(u,v,\phi\right)## via$$

x=uv\cos{\phi}\ \ y=uv\sin{\phi}\ \ z=\frac{1}{2}\left(u^2-v^2\right)

$$(a) Find the coordinate transformation matrix between paraboloidal and Cartesian coordinates ##{\partial x^\alpha}/{\partial x^{\beta^\prime}}## and the inverse transformation. Are there any singular points in the map?

x=uv\cos{\phi}\ \ y=uv\sin{\phi}\ \ z=\frac{1}{2}\left(u^2-v^2\right)

$$(a) Find the coordinate transformation matrix between paraboloidal and Cartesian coordinates ##{\partial x^\alpha}/{\partial x^{\beta^\prime}}## and the inverse transformation. Are there any singular points in the map?

(b) Find the basis vectors and the basis one-forms in terms of Cartesian basis vectors and forms.

(c) Find the metric and inverse metric in paraboloidal coordinates.

(d) Calculate the Christoffel symbols.

(e) Calculate the divergence ##\nabla_\mu V^\mu## and Laplacian ##\nabla_\mu\nabla^\mu f##.

### Overview

I completely screwed up calculation of the metric, getting one that was very non-diagonal and would have been hard work and I had to go back to spherical polar coordinates to see where I had gone wrong, then it all worked except for a sign error in the inverse transformation matrix. It was fairly obvious which component contained the error.The way that a covariant tensor ends up in the wrong coordinates after using the general tensor transformation law (e.g. at (81)) is very odd and needs further investigation. I think it's just inevitable and would often need transformation back into the correct coordinate system.

I learned a bit more about basis vectors, but this exercise mainly seemed to be teacher-torture with plenty of exercise on differentiation. The Laplacian ##\nabla_\mu\nabla^\mu f## is not in the index of this book. I hope it is discussed somewhere. I spent about half the time developing drawings of the basis vectors and one-forms but sadly the animation broke down. It was a surprise that the coordinate transformation matrix and its inverse not only take tensors back and forth between coordinate systems but are also inverses in the matrix sense. I suppose I should have known.

When calculating the Christoffel symbols, my first shot was 77% correct. I suppose if this was an exam question that would be quite good, except that I have taken too much time!

### Answer

See: Ex 3.04 Paraboloidal coordinates.pdf (15 pages, 147 equations)

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ReplyDeleteHi,George. When i calculated the divergence of vevtor V,i found that i use two different formulas (3.32)and(3.34) in carroll's to get different results, which confusing me. Then i found that when u calculated the connection, u first fixed the up index of Γ and next u fixed the two down index together of g,like (94) (104) and (114).However, when we choose different μυρ of g, it turns out that the index of g are not fixed, some of them are going to be vanish.Excuse my poor English！Thx u！

DeleteGlad you are still at it Chun! Is there a question here? Or have I made another mistake?

DeleteOne mistake:I find that you fixed the down index of g_μυ(like g_11 g_22 g_33) at the beginning when you calculated connection Γ.But after you fixed them, some ∂_σg_μυ which should be vanish do not be vanish.This resulted in a miscalculation of the connection Γ.Am I making myself clear? If not, I can show you my manuscript.

DeleteNot clear! I suppose you are at equations (94) and onwards. I checked a few they seem OK. Do send me your manuscript: george@general-relativity.net. I will check again later today.

DeleteI have sent my handscript to your email.I hope it won't be classified as spam:(

ReplyDeleteI think you have some errors in the calculation of the Cristoffel symbols, namely the fact that you have non-vanishing symbols even when all of the three indices are different. This should not be the case (in fact these symbols should vanish) since the metric is diagonal; this follows from the previous exercise.

ReplyDeleteQuite right! And I marked them up in the pdf, but I was too lazy to correct them. Chun first pointed this out over a year ago.

DeleteYour (43) is wrong and hence everything that follows is wrong too. It is clearer to write something like \hat{u} = \frac{\partial}{\partial u} = \frac{\partial x}{\partial u} \frac{\partial}{\partial x} = \frac{\partial x}{\partial u} \hat{x}.

ReplyDelete##\hat{u} = \frac{\partial}{\partial u} = \frac{\partial x}{\partial u} \frac{\partial}{\partial x} = \frac{\partial x}{\partial u} \hat{x}##

ReplyDelete