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Help with Diffeomorphismsby latentcorpse
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#19
Apr1311, 12:40 PM

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#20
Apr1311, 01:24 PM

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Leave [tex]f[/tex] arbitrary, but note that you have [tex](\partial/\partial y^\alpha )f(y^\alpha)[/tex] on the LHS and [tex](\partial/\partial x^\mu) f(y^\alpha)[/tex] on the RHS. You want to use the chain rule on the RHS and equate coefficients.
[/QUOTE] What happened to your [itex]\phi^*(f)[/itex] on the RHS  you have just made it [itex]f[/itex], no? Anyway, let's say I try with the metric in EF coords.... we have [itex]g_{\mu \nu} X^\mu X^\nu[/itex] with [itex]X^\mu=(0,1,0,0)[/itex] so we would have [itex]g_{\mu \nu} X^\mu X^\nu=g_{rr}X^rX^r=g_{rr}=0[/itex]? I'm sure I've done this before. I can't understand what I am doing wrong! I assume that checking it's timelike is the only way? Presumably then t would be spacelike in this region? How would we prove this since there is no t coordinate in the EF coord metric? Thanks again. 


#21
Apr1311, 02:04 PM

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The simplest way I'd work this out is by using the Schwarzschild coordinates for the region [tex]r<2M[/tex]. Then it's almost trivial to show that the radial and time coordinates get switched. One can complain that the coordinates are not valid past [tex]r=2M[/tex], but I don't see anything wrong with using EF to extend through [tex]r=2M[/tex] and then transforming back. One can obviously show the same thing in EF coordinates, it just seems like it's going to be a lot harder. One can see how it works qualitatively by looking at the light cones in the Finklestein diagram on pg 18 of Townsend's BH notes. It's actually a little bit complicated to work those lightcones out, but is sort of a useful exercise if your morals prevent you from believing my suggestion above. 


#22
Apr1311, 02:56 PM

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#23
Apr1311, 04:26 PM

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Secondly, that the metric is the "same" does not mean that [itex]X^\mu X_\mu[/itex] has the same value in both regions. Just try to compute it and see for yourself. 


#24
Apr1311, 04:50 PM

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[itex]( \phi_*(X))^\alpha \left( \frac{\partial}{\partial y^\alpha} \right)_{\phi(p)} f(y^\beta) = X^\mu \left( \frac{\partial}{\partial x^\mu} \right)_p (\phi^* f)(x^\nu)[/itex] [itex]( \phi_*(X))^\alpha \left( \frac{\partial}{\partial y^\alpha} \right)_{\phi(p)} f(y^\beta) = X^\mu \left( \frac{\partial}{\partial x^\mu} \right)_p f(y^\beta)[/itex] using [itex] (\phi^*f)(x^\nu)=f(\phi(x^\nu))=f(y^\beta)[/itex] so [itex]( \phi_*(X))^\alpha = X^\mu \left( \frac{\partial y^\alpha}{\partial x^\mu} \right)_p [/itex] One little problem though is that in the 2nd last line I had the d/dy evaluated at phi(p) but then when I cross multiplied I now miraculously have it evaluated at p. How does that work? Taking t=0 and r=1 I find [itex]v=1+2M \log{  \frac{12M}{2M}  }[/itex] So [itex]X^\mu=(1+2M \log{  \frac{12M}{2M}  },1,0,0)[/itex] Surely those components are wrong? 


#25
Apr1311, 05:27 PM

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[tex] \left( \frac{\partial}{\partial x^\mu} \right)_p = \left( \frac{\partial y^\alpha}{\partial x^\mu} \right)_p \left( \frac{\partial}{\partial y^\alpha} \right)_{\phi(p)} [/tex] 


#26
Apr1311, 06:48 PM

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where [itex]A^\mu{}_\nu = \frac{\partial x'^\mu}{\partial x^\nu}[/itex] So what is this [itex]A[/itex] matrix? I find it has non zero cpts [itex]A=\text{diag} ( \frac{\partial v}{\partial t} , 1 , 1 , 1)[/itex] but then [itex]v=t+r+2M \log{  \frac{r}{2M}1 } \Rightarrow \frac{\partial v}{ \partial t}=1[/itex] but then we just have a identity matrix which doesn't seem very useful? 


#27
Apr1311, 08:16 PM

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#28
Apr1411, 04:43 AM

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But things don't really simplify much when I work out [itex]g_{\mu \nu} X'^\mu X'^\nu[/itex]? Also, in Chapter 34, he says that outgoing radial null geodesics inside the region r<2M will have decreasing r and therefore reach the r=0 singularity in finite affine parameter. This makes sense to me physically but I don't see how the maths backs it up. We have shown that [itex]\frac{dr}{d \tau}=1[/itex] for these radial null geodesics and so wouldn't they have increasing r i.e. be able to escape from the black hole? 


#29
Apr1411, 11:40 AM

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#30
Apr1411, 11:54 AM

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What do you mean by fix it? The only way I can think of to see what the radial coordinate is doing along the geodesic is to find its rate of change wrt the affine parameter and that is given as [itex]\frac{dr}{d \tau}=1[/itex] so why does that not mean the geodesic will have increasing r? And do you have any advice about my attempt to do that calculation in the last post? Am I going about it on the right lines? Thanks again! 


#31
Apr1411, 12:11 PM

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#32
Apr1411, 02:05 PM

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Surely [itex]\tau[/itex], being the affine parameter, can only increase? where [itex]A^\mu _\nu = \frac{\partial x^\mu}{\partial x^\nu}[/itex] Now our sets of coords are (v,r,theta,phi) and (t,r,theta,phi) So [itex]X'^1=X'^2=X'^3=1[/itex] (I'm hoping my [itex]X'^1[/itex] component is ok there. I don't think r in EF varies with t, theta or phi so I think it's ok. So we need to work out [itex]\frac{\partial v}{\partial x^\nu}[/itex] but [itex]v=t+r_*=t+r+2M \log{ \frac{r}{2M}1}[/itex] so [itex]\frac{\partial v}{\partial t}=1[/itex] and [itex]\frac{\partial v}{\partial r} = 1 +\frac{2M}{\frac{r}{2M}1} \frac{1}{2M}=1+\frac{2M}{r2M}[/itex] so presumably these add to give [itex]\frac{\partial v}{\partial x^\nu}=2 + \frac{2M}{r2M}[/itex]? One small problem I encountered whilst reading is why in the 1st line of (35) do we only vary the derivative terms and not the metric outside the brackets? Cheers. 


#33
Apr1411, 04:27 PM

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when [tex]r<2M[/tex]. 


#34
Apr1411, 06:49 PM

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I find [itex]X'^\mu=(1\frac{2M}{2Mr},1,0,0)[/itex] and then [itex]g(X,X)= \frac{4M^2}{(2Mr)r}  \frac{6M^3}{(2Mr)r^2} + 1  \frac{2M}{2Mr}[/itex] I find this can be rewritten as [itex]g(X,X)=1\frac{2M}{2Mr} \left( \left( \sqrt{3} \frac{M}{r}  \frac{1}{\sqrt{3}} \right)^2 + \frac{8}{9} \right)[/itex] Now everything in the brackets is positive and [itex]\frac{2M}{2Mr}>1[/itex] for [itex]r<2M[/itex] so we get [itex]g(X,X)<0[/itex] Does this look ok? How about doing the same thing in Schwarzschild coordinates? What did I do wrong the 1st time I tried this and ended up finding [itex]r[/itex] to be spacelike. I'm curious as to how they agree! Cheers. 


#35
Apr1411, 09:10 PM

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[tex]  \frac{2M}{2Mr}[/tex] so there's probably something wrong with your derivation above. [tex] \delta g^{\mu\sigma} \left( g_{\sigma\nu,\rho} +\cdots \right), ~~(*)[/tex] but these all vanish because he takes [tex] g_{\sigma\nu,\rho} =0[/tex]. If you used general coordinates, the effect of the terms (*) would be to generate the covariant derivatives in the 2nd line of (355). 


#36
Apr1511, 03:44 AM

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[itex]v=t+r+2M \log{1\frac{r}{2M}}[/itex] [itex]\frac{\partial v}{\partial r} = 1 + \frac{2M}{1\frac{r}{2M}} (  \frac{1}{2M}) = 1  \frac{1}{1\frac{r}{2M}} = 1  \frac{2M}{2Mr}[/itex] so [itex]X'^\mu=(1  \frac{2M}{2Mr},1,0,0)[/itex] so [itex]g(X,X)=\begin{pmatrix} 1\frac{2M}{2Mr} & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix}  (1\frac{2M}{r}) & 1 & 0 & 0 \\ 1 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix} \begin{pmatrix} 1  \frac{2M}{2Mr} \\ 1 \\ 0 \\ 0 \end{pmatrix}[/itex] [itex]=\begin{pmatrix} 1\frac{2M}{2Mr} & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix}  ( 1  \frac{2M}{r} )( 1  \frac{2M}{2Mr} ) + 1 \\ 1 \\ 0 \\ 0 \end{pmatrix}[/itex] [itex]=\begin{pmatrix} 1\frac{2M}{2Mr} & 1 & 0 & 0 \end{pmatrix} \begin{pmatrix} \frac{2M}{2Mr}+\frac{2M}{r}  \frac{4M^2}{(2Mr)r} \\ 1 \\ 0 \\ 0 \end{pmatrix}[/itex] [itex] = (1\frac{2M}{2Mr} ) ( \frac{2M}{2Mr}+\frac{2M}{r}  \frac{4M^2}{(2Mr)r})+1[/itex] [itex]=\frac{2M}{2Mr} + \frac{2M}{r}  \frac{4M^2}{(2Mr)r}  \frac{4M^2}{(2Mr)^2}  \frac{4M^2}{(2Mr)r} + \frac{8M^3}{(2Mr)^2r}+1[/itex] This is slightly different from what I got last time (I realised I made a mistake) but it still isn't right..... Also, I tried to show (399): [itex]g_{ab}u^au^b =  \left( \frac{dt}{d \tau} \right)^2 + L^2 \cosh^2{\frac{t}{L}} \left( \frac{\partial \chi}{\partial t} \right)^2[/itex] [itex]=\left( \frac{\partial t}{\partial \tau} \right)^2 + L^2 \cosh^2{\frac{t}{L}} \left( \frac{1}{1e^{2t}} e^t \left( \frac{dt}{d \tau} \right)^2[/itex] [itex]=\left( \frac{dt}{d \tau} \right)^2 + \frac{L^2}{2} \left( e^{t/L} + e^{t/L} \right) \frac{2 e^{t}e^t}{e^{t}e^t} \left( \frac{dt}{d \tau} \right)^2[/itex] but that second term gives a cosh/sinh and also the arguments are different: one is t/L and the other is just t so I don't see how I'm going to get them to cancel? 


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