Friday 20 September 2019

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?

(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!

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

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1. Hi,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！

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

3. One 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.

4. Not 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.

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

3. I 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.

1. Quite 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.

4. Your (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}.

5. $\hat{u} = \frac{\partial}{\partial u} = \frac{\partial x}{\partial u} \frac{\partial}{\partial x} = \frac{\partial x}{\partial u} \hat{x}$

6. Hi George, thanks so much for all of these wonderful solutions and for the work you've put in. As a GR novice these have been incredibly helpful to me in getting a feel for how to solve these kinds of problems.

I think the confusion about your (42) and (43) comes down to this: vectors themselves are generally covariant, so when we say that a vector V is *really* the components V^\mu summed over the basis vectors e_\mu, that means that if the vector components transform as V^\mu' = (\partial x^\mu' / \partial x^\mu)V^\mu, then in order that the vector itself be invariant the basis vectors have to transform in exactly the opposite way from the vector components; i.e. the fact that V^\mu e_\mu = V^\mu' e_\mu' demands that e_\mu' = (\partial x^\mu / \partial x^\mu') so that when you take the sum, the two transformation matrices – one applied to the components and one applied to the basis vectors – will cancel out and leave the sum (the real thing!) unchanged. This would resolve the tension you noticed in footnote 1.

Does this seem right? (Apologies for my ugly attempt at inline math; I'm not sure how you do it on this website!)

1. Thanks Jake! You smarten up formulas surrounding the Latex code with double # or double \$. This is what you wrote.
I think the confusion about your (42) and (43) comes down to this: vectors themselves are generally covariant, so when we say that a vector V is *really* the components $V^\mu$ summed over the basis vectors $e_\mu$, that means that if the vector components transform as $$V^{\mu'} = (\partial x^{\mu'} / \partial x^\mu)V^\mu$$, then in order that the vector itself be invariant the basis vectors have to transform in exactly the opposite way from the vector components; i.e. the fact that $$V^\mu e_\mu = V^{\mu'} e_{\mu'}$$ demands that $$e_{\mu'} = (\partial x^\mu / \partial x^{\mu'})$$ so that when you take the sum, the two transformation matrices – one applied to the components and one applied to the basis vectors – will cancel out and leave the sum (the real thing!) unchanged. This would resolve the tension you noticed in footnote 1.