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Conformal inv of scalar wave equation

  1. May 24, 2012 #1
    I'm trying to prove the conformal invariance (under [itex]g_{\mu\nu}\to\omega^2 g_{\mu\nu}[/itex]) of


    I've found that this equation is invariant upto a quantity proportional to

    g^{\mu\nu}\left[-\omega(\nabla_{\mu}\nabla_{\nu}\omega)\phi+(\omega^2-\omega)(\nabla_{\mu}\omega)(\nabla_{\nu} \phi)-\frac{n-4}{4}(\nabla_{\mu}\omega)(\nabla_{\nu} \phi)\right]

    Here [itex]\bar{\phi}=\omega^{(2-n)/2}\phi[/itex], and the conformal transformations of other quantitities like box and the Ricci scalar are those found in Carroll Appendix.

    How can I get rid of this extra unwanted quantity? (or have I simply made an algebraic error of some kind)

  2. jcsd
  3. May 24, 2012 #2


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    Did you use this identity for the wave operator? ◻φ = (-g)-1/2μ[(-g)1/2 gμννφ]
  4. May 25, 2012 #3
    I didn't, I just used



    \bar{R}=\omega^{-2}R-2(n-1)g^{\alpha\beta}\omega^{-3}(\nabla_{\alpha}\nabla_{\beta}\omega)-(n-1)(n-4)g^{\alpha\beta}\omega^{-4}(\nabla_{\alpha}\omega)(\nabla_{\beta}\omega) [/tex]

    as well as [itex]\bar{\phi}=\omega^{(2-n)/2}\phi[/itex].

    I plugged these into


    The full result I obtain is

    [tex] \omega^{-(n+2)/2}\left(\Box\phi+\frac{1}{4}\frac{n-2}{n-1}R\phi\right)+\omega^{-(n+6)/2}(n-2)g^{\mu\nu}\left[-\omega(\nabla_{\mu}\nabla_{\nu}\omega)\phi+(\omega ^2-\omega)(\nabla_{\mu}\omega)(\nabla_{\nu} \phi)-\frac{n-4}{4}(\nabla_{\mu}\omega)(\nabla_{\nu} \phi)\right] [/tex]

    The first terms are obviously what I want, but the second square bracket is unwanted.

    At which point are you thinking I should use the identity you posted?

    thanks for the help.
  5. May 25, 2012 #4

    Physics Monkey

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    You might try checking your algebra for the [itex] \Box \bar{\phi} [/itex] part again, because it looks to me like it should work out although I don't see your error. Did you properly compute the derivatives of [itex] \bar{\phi} [/itex] and not just [itex] \phi [/itex]?

    Alternatively, you could try the differential form [itex] \omega = 1 + \delta [/itex] to simply things.

    Also, do you know that there are mathematica packages that do this kind of algebra for you? I only mention this because I never felt that I learned much from such exercises.
  6. May 25, 2012 #5
    [tex]\bar{\Box}\bar{\phi}=\omega^{-2}\Box\bar{\phi}+(n-2)g^{\alpha\beta}\omega^{-3}(\nabla_{\alpha}\omega)(\nabla_{\beta}\bar{\phi} ) [/tex]

    What I did was, for example in the first term of this on RHS :[itex]\omega^{-2}\Box\bar{\phi}[/itex], I wrote

    [tex]\omega^{-2}\Box\bar{\phi}=\omega^{-2}g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\left(\omega^{(2-n)/2}\phi\right) =\omega^{-2}g^{\mu\nu}\nabla_{\mu}\left[\frac{2-n}{2}\omega^{-n/2}(\nabla_{\nu}\omega)\phi+\omega^{(2-n)/2}\nabla_{\nu}\phi\right][/tex]

    Ultimately I ended up with

    [tex]\omega^{-(n+2)/2}\Box\phi+\omega^{-(n+4)/2}(n-2)g^{\mu\nu}\left[\frac{n}{4}\omega^{-1}(\nabla_{\mu}\omega)(\nabla_{\nu}\omega)\phi-\frac{1}{2}(\nabla_{\mu}\nabla_{\nu}\omega)\phi -(\nabla_{\mu}\omega)(\nabla_{\nu}\phi)\right] [/tex]

    The first piece being what I want, the second square bracketted bit again going to the unwanted piece when combined with the rest of box and the Ricci term.
  7. May 25, 2012 #6


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    This is one mistake. You're trying to treat covariant derivatives like partial derivatives. ∇μφ is a vector, and when you take the divergence of a vector you need to include a Christoffel symbol. The identity I mentioned above uses instead factors of √g to avoid the Christofel symbol. Try using the identity and see what you get.

    EDIT: I think I see at least two more mistakes, but let's do one at a time!
  8. May 25, 2012 #7
    where did I make this mistake? I was tring to be very conscious of not doing that. omega and phi are scalars so the [itex]\nabla_{\nu}[/itex] cov deriv acts as a partial deriv and this is how I generate the second equality in my second line of equations in my post prev to yours. Or are you thinking of somewhere else? My expansion of the rightmost side of my second line of equations would continue by using the product rule (valid I think for the CovDeriv) to take [itex] \nabla_{\mu}[/itex] of the product of scalars and 1-forms in the square brackets...
    Last edited: May 25, 2012
  9. May 25, 2012 #8


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    Do you agree that the identity is right?
    Do you agree that if you do use it, you will get a different answer?

    Second mistake.. Notice that in each equation you started with, every term has the same number of ω's. This is ignoring derivatives, so for example ω-2, ω-3∇ω and ω-4∇ω∇ω all have the same number.

    There's undoubtedly a deep reason for this, like conformal transformations have the group property or something. More formally, the terms are homogeneous under ω → Cω where C is a constant.

    Whatever the reason, this property must preserve itself throughout your calculation. Try it and see. In particular where you have ended up with (ω2 - ω) clearly it has been violated. So if you trace back where these two terms came from, one of them must have lost an ω.
  10. May 25, 2012 #9
    The identity you posted for box phi you mean? If so yes I agree that that's correct. I started off doing the calculation using this identity too but haven't finished the algebra yet. I can't see however why doing it the way I initially set out too would not be an equivalent way however, unless I had made an error along the way of course.

    OK, I will check into this.

    Do you still believe I am treating cov derivs incorrectly?
  11. May 25, 2012 #10


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    I think that is Mistake #3! :smile: I think that this problem can't be worked correctly using covariant derivatives - you need to use the identity to write everything in terms of partial derivatives and insure that all the appearances of gμν are explicitly written out. Why? Because what you want is the new wave operator in the conformally transformed space. Using the covariant derivatives will give you the old wave operator instead.
  12. May 25, 2012 #11

    I'm not sure I agree with that. I do accept that you could do the problem the way you posted with that identity, and will try to verify that when I get chance next, but I don't buy that you can't do it using the covariant derivs as long as you are careful. I'm starting in the conformal frame and working backwards and the first term I get is

    [tex]\omega^{-(n+2)/2}\Box\phi [/tex]

    this is not [itex] \bar{\Box} [/itex] it's genuinely the box of the non-conformal frame, not the box of the conformal frame that I started with. Secondly if one needs to convert my method to your method, it's possible to just write out the Christoffel symbols in terms of the metric and I believe the two should then agree. It might even be possible at some stage to use [itex] \Gamma^{\mu}_{\mu\lambda}=(-g)^{-1/2}\partial_{\lambda}(-g)^{1/2} [/itex].

    Anyway, I will go through both methods in some details when I get chance next and see explicitly if they are equivent or not, and also recheck my algebra bearing in my the number of omegas per term feature you highlighted, thanks.
  13. May 26, 2012 #12

    Physics Monkey

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    No offense, Bill, but I strongly disagree with this statement. I can certainly imagine that some people find the partial derivatives simpler to work with, but the problem can definitely be worked out with covariant derivatives. There is not really anything subtle about these objects in the sense that they obey all the usual rules of derivatives in calculus in addition to producing valid tensors.

    The simplest way to proceed is to replace all [itex] \Box f [/itex] with [itex] g^{ab} \nabla_a \nabla_b f [/itex] for all scalars f and then work on with covariant derivatives i.e. don't reintroduce box operators until the very end.
  14. May 26, 2012 #13

    Physics Monkey

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    I just checked it explicitly and it works out, but only for the equation [itex] \Box \phi - \frac{1}{4}\frac{n-2}{n-1} R \phi [/itex]. Is it possible you have the wrong sign in your equation?

    Are you using mostly plus or mostly minus? If it's mostly plus then box acts like [itex] E^2 - p^2 [/itex] and the minus sign would be appropriate (so that positive R acts like a normal mass). This is important for quantizing the theory on a sphere.
  15. May 26, 2012 #14
    I'm working from the book Birrell and Davies and they have the plus sign [itex] \Box \phi +\frac{1}{4}\frac{n-2}{n-1} R \phi [/itex], they also use (+----) convention however.

    Actually I now realise that I've been using the transformations for the Ricci etc from Carroll who uses the opposite sign convention (d'oh!) and this is probably what has messed me up! Also the omega^2 I had above should have been an omega
    (as bill noted) and this kills that (omega^2-omega) term..
  16. May 26, 2012 #15
    So I should have really been using

    [tex]\bar{R}=\omega^{-2}R+2(n-1)g^{\alpha\beta}\omega^{-3}(\nabla_{\alpha}\nabla_{\beta}\omega)+(n-1)(n-4)g^{\alpha\beta}\omega^{-4}(\nabla_{\alpha}\omega)(\nabla_{\beta}\omega) [/tex]

    for the (+----..) conv. Is my conformal transformation for the box still valid? In the other convention Carroll has

    [tex]\bar{\Box}\bar{\phi}=\omega^{-2}\Box\bar{\phi}+(n-2)g^{\alpha\beta}\omega^{-3}(\nabla_{\alpha}\omega)(\nabla_{\beta}\bar{\phi} ) [/tex]

    Would this change?

    EDIT: It all seems to work out now! phew. thanks for all the help..
    Last edited: May 26, 2012
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