So I'm trying to prove eqn (223) in the notes attached in this thread: https://www.physicsforums.com/showthread.php?t=457123 I took the equation [itex]( \phi^* ( \eta) ) ( X) = \eta ( \phi_* (X))[/itex] and expanded in a coordinate basis as follows [itex]( \phi^* ( \eta) )_\mu dx^\mu X = \eta_\alpha dy^\alpha \phi_*(X)[/itex] So to get the result it seems like it should be a simple case of cross multiplying but unfortunately the [itex]X[/itex] doesn't cancel the [itex]\phi_*(X)[/itex] Thanks for any help.
Re: Diffeomorphisms The notation [itex]( \phi^* ( \eta) ) ( X) = \eta ( \phi_* (X))[/itex] always means that indices are contracted in the coordinate basis, so [tex]( \phi^* ( \eta) ) ( X) = ( \phi^* ( \eta) )_\mu X^\mu, [/tex] [tex] \eta ( \phi_* (X)) = \eta_\alpha ( \phi_* (X))^\alpha. [/tex] Just think of the usual expression for a 1-form acting on a vector: [tex]\omega(X) = \omega_\mu X^\mu[/tex].
Re: Diffeomorphisms Ok. So then we get the equality: [itex]( \phi^*( \eta ) )_\mu X^\mu = \eta_\alpha ( \phi_* ( X ))^\alpha[/itex] I don't see how we can get this into the form of (223)? Additionally, I am trying to prove (187). Given the hint, I am able to multiply by those vectors and get [itex]\nabla_X \nabla_Y Z^a- \nabla_Y \nabla_X Z^a = R^a{}_{bcd}Z^bX^cY^d = (R(X,Y)Z)^a[/itex]. But according to teh defn of the Riemann tensor (see (178)), I am missing a [itex]\nabla_{[X,Y]}[/itex] piece. Any ideas?
Re: Diffeomorphisms You can use (221). You have to use the product rule correctly on the terms [tex]\nabla_b\nabla_c(Z^a X^bY^c)[/tex] and (135).
Re: Diffeomorphisms I get [itex]( \phi_*(X))^\alpha (\frac{\partial}{\partial y^\alpha})_{\phi(p)} (f) = X^\nu (\frac{\partial}{\partial x^\nu})_p \phi^*(f)[/itex] Again, I'm having trouble getting rid of the [itex]\phi^*[/itex] on the right Ok. You appear to have multiplied from the left by [itex]Y^dX^c[/itex] on the first term but [itex]X^cY^d[/itex] on the second term. Now we are using abstract index notation so these both represent vectors rather than the components of vectors. How do you know those vectors commute? Thanks.
Re: Diffeomorphisms You need to use that [tex]\phi[/tex] is the map [tex]x^\mu\rightarrow y^\alpha[/tex] and the chain rule. They represent the components of vectors and commute. Look at the derivation of (130), the 2nd line follows from the first by commuting the components of [tex]X[/tex] and [tex]Y[/tex].
Re: Diffeomorphisms well [itex]\phi[/itex] has components [itex]y^\alpha(x^\mu)[/itex] I can see we want to introduce something like dy/dx d/dy on the RHS but I just dont see how? Right. But there he is using basis indices (see chapter 7). Although he does say in Chapter 7 that we are free to change abstract indices to basis indices to get an equation that is true in an arbitrary basis so I guess everything is ok providing we work this out in greek indices? Anyway I tried working out the following using the product rule: [itex]\nabla_Y (X^\mu \nabla_\mu Z^\nu) = Y^\rho \nabla_\rho (X^\mu \nabla_\mu Z^\nu) = Y^\rho X^\mu{}_{; \rho} \nabla_\mu Z^\nu + Y^\rho X^\mu \nabla_\rho \nabla_\mu Z^\nu[/itex] in the hope that one of those terms we get would be useful but I don't see it?
Re: Diffeomorphisms Use [tex]\phi^*(f) = f\circ \phi[/tex] applied to the coordinates [tex]x^\mu[/tex]. I'm not sure why you chose that expression. We had [tex] Y^d \nabla_X \nabla_d Z^a - X^c \nabla_Y \nabla_c Z^a, [/tex] so you want to re-express [tex] Y^d \nabla_X \nabla_d Z^a = \nabla_X(Y^d \nabla_d Z^a) - ( \nabla_X Y^d )(\nabla_d Z^a)[/tex] and similarly the other term.
Re: Diffeomorphisms Sorry I still don't get. The [itex]\phi[/itex]'s are charts so they have components [itex]x^\mu[/itex] so how would this work?
Re: Diffeomorphisms Sorry, I have been working on some other stuff for the last few days and didn't have time to look at it again until now. I still do not get how either of these work? The second one I keep getting bogged down with what to do with algebra. The first one I don't really follow what you want me to do. Thanks again.
Re: Diffeomorphisms There really isn't any algebra to get bogged down with, [tex] \nabla_X(Y^d \nabla_d Z^a) = \nabla_X \nabla_Y Z^a, [/tex] while [tex] ( \nabla_X Y^d )[/tex] is one term in the expression (130) for [tex][X,Y]^d[/tex]. As for the other part, if [tex]\phi(x^\mu) = y^\alpha[/tex], then what is [tex](f\circ \phi)(x^\mu)[/tex]?
Re: Diffeomorphisms Ok. I get us down to [itex]\nabla_X \nabla_Y Z^a - \nabla_Y \nabla_X Z^a - ( \nabla_X Y^d)(\nabla_d Z^a) + ( \nabla_Y X^c)( \nabla_c Z^a)[/itex] Now from our Riemann tensor we want those last terms to give us a [itex]-\nabla_{[X,Y]}Z^a[/itex] So I tried working backwards: [itex]\nabla_{[X,Y]}Z^a=\nabla_{XY-YX}Z^a = \nabla_{XY}Z^a-\nabla_{YX}Z^a[/itex] Now I don't know how to get to the next line there? Could you explain please? As for the other bit I would find that [itex](f \circ \phi)(x^\mu) = f(y^\alpha)[/itex] Thanks.
Re: Diffeomorphisms Just write [tex] - ( \nabla_X Y^d)(\nabla_d Z^a) + ( \nabla_Y X^c)( \nabla_c Z^a) = - ( \nabla_X Y^d-\nabla_Y X^d)(\nabla_d Z^a)[/tex] and use (130). The last part makes no sense at all, since [tex]XY[/tex] and [tex]YX[/tex] are not tangent vectors. So you need to go back a few posts and figure out what that's good for.
Re: Diffeomorphisms Ok. So I have got the RIcci identity to work out. As for the transformation, I assume I am meant to go back to [itex]( \phi_*(X))^\alpha (\frac{\partial}{\partial y^\alpha})_{\phi(p)} (f) = X^\nu (\frac{\partial}{\partial x^\nu})_p \phi^*(f)[/itex] and get rid of the [itex]\phi^*(f)[/itex] on the RHS. Should I act both sides on the coordinates [itex]x^\mu[/itex]? That's the only way I can see of being able to use the [itex]f(x^\mu)=y^\alpha[/itex] property. Also, on p90, he says that we should think of the singularity as a time rather than a place inside the black hole because r plays the role of the time coordinate. How does this work? Thanks.
Re: Diffeomorphisms Then try doing that. It's much better to try to work it out before asking the question. Hopefully you mean [itex]\phi(x^\mu)=y^\alpha[/itex] You should be able to use the metric to show that the radial vector is timelike for [tex]r<2M[/tex].
Re: Diffeomorphisms Ok. I tried taking [itex]f=y^\alpha[/itex] but then on the LHS, this cancelled the d/dy i had and similarly on the RHS it gets pulled back to an [itex]x^\mu[/itex] which cancels the d/dx we have? This may be a stupid question but what is the form of the radial vector? Would it just be [itex]X=\frac{\partial}{\partial r} \Rightarrow X^\mu(0,1,0,0)[/itex] in Schwarzschild coordinates? If so, when I calculate [itex]g_{rr}X^rX^r[/itex] I find it to be spacelike so I'm guessing that's wrong. Moreover, how does showing a vector is timelike tell us it behaves like a time coordinate/that t doesn't behave like a time coordinate?
Re: Diffeomorphisms 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. Yes. You might want to show your work. You should find that it's spacelike for [tex]r>2M[/tex] and timelike for [tex]r<2M[/tex]. How else would you decide that something is or isn't a time-coordinate?
Re: Diffeomorphisms 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? Well in the notes it says to work with the Schwarzschild coordinates so I used the Schwarzschild metric. Now I know that the metric that covers the r<2M is the one defined using ingoing Eddington Finkelstein coordinates but, we know that if we're in the region r<2M and we change back from EF to Schwarz coordinates we should get the Schwarz metric back again (but now applied to the r<2M region)> Isn't this guaranteed by Birkhoff's theorem which tells us that Schwarzschild is unique? So I guess my question is, why does it not work if you use the Schwarzschild metric? 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! [/QUOTE] 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.