# Recent content by grey_earl

1. ### A problem with Wald's General Relativity

But yes, you can, since μ and ν are not Lorentz indices, and the first theorem (that v^a = ∂_b s^{ab}) only makes a statement about Lorentz indices. U is antisymmetric in the first and second index, and separately in the third and fourth, but not in the first and third. So only in that case...
2. ### Coordinate matters

True, but that's not a point on the manifold (sure you meant φ=2π, no?). My φ coordinate is in the range [0, 2π), so it includes 0 but excludes 2π. So everything ok there. (if you want strictest mathematical rigor: φ ∊ ℝ/{2πℤ})
3. ### Coordinate matters

With mathematical theorems you must be careful in intepretation, they assure only and exactly only what they state. For each manifold of dimension n (so in my toy example n=2) there is an atlas containing n+1 charts, so you don't need more. Of course you can take more if you wish, and there are...
4. ### A problem with Wald's General Relativity

Yes, with your notation T^{ab} comes out zero. With the notation I proposed also, I see it now. But note that in the original problem the derivatives should act on the first and third index of U, so that T doesn't vanish. So you should take W^c_{μν} = ∂_d U^c_μ^d_ν, sorry for the confusion before.
5. ### Coordinate matters

Which point is not covered by the coordinate system I gave you?
6. ### Coordinate matters

So now you know that there are intrinsically curved manifolds where you only need one chart. The topology doesn't fully specify the metric; there are infinitely many possibilities to bestow a given topological manifold with a metric. We didn't understand each other saying "trivial" :) There...
7. ### Pell's Equation

Yes, I see. I'm sorry, don't know more. From n = 3/2 +- 1/2 \sqrt{2/3 ( m^2 - 5/2 )}, you see that 2/3 ( m^2 - 5/2 ) must be a perfect odd square for n to be integer, so you can equally solve 2/3 ( m^2 - 5/2 ) = (2k+1)^2, with k = 0,1,2,... This can be reexpressed as m = +- \sqrt{6 k^2 + 6 k...
8. ### A problem with Wald's General Relativity

The original theorem was that for any v^a with d_a v^a = 0, there is an antisymmetric tensor s^{ab} such that v^a = d_b s^{ab} = - d_b s^{ba}. So you should have T^a_μ = ∂_c W^{ac}_μ, which is minus the result you gave. Of course you can always redefine W to be -W, but let's stick with mine for...
9. ### Pell's Equation

Ok, integer values and Pell's equation, I refer you to https://www.physicsforums.com/showthread.php?t=78855 and http://en.wikipedia.org/wiki/Chakravala_method
10. ### Pell's Equation

For which variable should you solve? I assume for n, then from 3/2 (2n-3)^2 + 5/2 = m^2 I get (2n-3)^2 = 2/3 ( m^2 - 5/2 ) and then 2n-3 = +- \sqrt{2/3 ( m^2 - 5/2 )}, so n = 3/2 +- 1/2 \sqrt{2/3 ( m^2 - 5/2 )}. If you should solve for m, m = +- \sqrt{3/2 (2n-3)^2 + 5/2}. But since that's...
11. ### A problem with Wald's General Relativity

Could you show your calculation in a bit more detail, please? Otherwise it is difficult seeing where you went wrong, as I haven't done the calculation.
12. ### Coordinate matters

I don't want to offend you in any way, if you felt I have done this. But it seems to me that you don't understand some (subtle) points, and I want to explain them to you. I may have misinterpreted your statement you gave in your post before the last, but to me you were stating that H is a...
13. ### Coordinate matters

The parameter H is intrinsically defined in de Sitter space: it's R/12 in the four-dimensional case, where R is the curvature scalar. I don't need any embedding for it, period. If you want to embed de Sitter space, then H is the radius as seen from the embedding space, but H is not a...
14. ### Coordinate matters

True; technically this is a coordinate singularity is you use polar coordinates for the sphere. However, you can for example obtain its area by integrating dA = sin θ dθ dφ over the whole sphere, and so it's an integrable singularity. Those typically don't cause any problems, so let's include...
15. ### Coordinate matters

(To avoid confusion: space in the following is short for spacetime.) Global coordinates for me are coordinates that cover the whole manifold, p.ex. cartesian coordinates for Minkowski space. Every point of Minkowski space is described by one coordinate tuple (x,y,z,t), and to every coordinate...