Help with Diffeomorphisms

  1. 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.
     
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  3. fzero

    fzero 2,696
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    Chegg
    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].
     
  4. 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?
     
  5. fzero

    fzero 2,696
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    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).
     
  6. Re: Diffeomorphisms

     
  7. fzero

    fzero 2,696
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    Re: Diffeomorphisms

     
  8. 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.
     
  9. fzero

    fzero 2,696
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    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].
     
  10. 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?
     
  11. fzero

    fzero 2,696
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    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.
     
  12. 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?
     
  13. 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.
     
  14. fzero

    fzero 2,696
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    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]?
     
  15. 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.
     
  16. fzero

    fzero 2,696
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    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.
     
  17. 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.
     
    Last edited: Apr 13, 2011
  18. fzero

    fzero 2,696
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    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].
     
  19. 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?
     
  20. fzero

    fzero 2,696
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    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?
     
  21. 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.
     
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