Proof Scalar action is conformally invariant

In summary, the conversation discusses the Lagrange density for a massless scalar field and its invariance under a conformal transformation. The resulting transformed Lagrange density is shown to be equivalent to the original one, but with additional terms that need to be dealt with. It is suggested to subtract a divergence and rescale the field to obtain the same form. However, after some calculations, it is still not clear how the transformed Lagrange density reduces to the original one. Further research and reference to books may be needed to fully understand the concept.
  • #1
LAHLH
409
1
Hi,

So if we have the Lagrange density for a massless scalar field: [tex] L=\sqrt{-g}\left(-\frac{1}{2}g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi-\frac{(n-2)}{4(n-1)} R\phi^2\right) [/tex]

Then under a conformal transformation [tex] g_{\mu\nu}=\omega^{-2}\tilde{g_{\mu\nu}} [/tex], then the Ricci sclar goes to [tex] R=\omega^2\tilde{R}+2(n-1)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)-n(n-1)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right) [/tex]. We also have [tex] g^{\mu\nu}=\omega^2\tilde{g}^{\mu\nu} [/tex] and [tex] \tilde{\nabla}_{\mu}\phi=\nabla_{\mu}\phi [/tex]

Using all these suggests that,

[tex] \tilde{L}=\omega^{-1}\sqrt{-\tilde{g}}\left(-\frac{1}{2}\omega^2 \tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}\phi\tilde{\nabla}_{\nu}\phi-\frac{(n-2)}{4(n-1)} \left[\omega^2\tilde{R}+2(n-1)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)-n(n-1)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\right]\phi^2\right) [/tex]

Which can be written as,


[tex] \tilde{L}=\sqrt{-\tilde{g}}\left(-\frac{1}{2}\omega \tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}\phi\tilde{\nabla}_{\nu}\phi-\frac{(n-2)}{4(n-1)} \omega\tilde{R}\phi^2-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)\phi^2+\frac{n}{4}(n-2)\omega^{-1}\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\phi^2\right) [/tex]

The statement is that this conformal transformation leaves the scalar field theory invariant, so I was expecting to obtain the same form of the Lagrange density after applying the CT as I had before, but I can't see how the above reduces to this?
 
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  • #2
Seems that I forgot that the field should transform too, but never the less I still can't seem to show that this is conformally invariant, anyone know any books that show this?
 
  • #3
You can get rid of the term containing the second derivative of ω by subtracting a divergence. Then you need to rescale φ by some power of ω, chosen to cancel the terms with first derivatives of ω.
 
  • #4
Thanks for the reply, I have tried this and seem to still not be getting the result. Firstly my determinant was off by a power of n so I correct this, and also use [tex] \phi=\omega^{(n-2)/2} \tilde{\phi}[/tex] (Birrell and Davies)

[tex]
\tilde{L}=\omega^{-n}\sqrt{-\tilde{g}}\left(-\frac{1}{2}\omega^2 \tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}(\omega^{(n-2)/2} \tilde{\phi})\tilde{\nabla}_{\nu}(\omega^{(n-2)/2} \tilde{\phi})-\frac{(n-2)}{4(n-1)} \left[\omega^2\tilde{R}+2(n-1)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right)-n(n-1)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\right](\omega^{(n-2)/2} \tilde{\phi})^2\right)
[/tex]

[tex]
\tilde{L}=\omega^{-n}\sqrt{-\tilde{g}}\Big(-\frac{1}{2}\omega^2 \tilde{g}^{\mu\nu}\left[\frac{(n-2)}{2}\omega^{(n-4)/2}\tilde{\nabla}_{\mu}(\omega)\tilde{\phi}+\omega^{(n-2)/2}\tilde{\nabla}_{\mu}( \tilde{\phi})\right]\left[\frac{(n-2)}{2}\omega^{(n-4)/2}\tilde{\nabla}_{\nu}(\omega)\tilde{\phi}+\omega^{(n-2)/2}\tilde{\nabla}_{\nu}( \tilde{\phi})\right] [/tex] [tex]-\frac{(n-2)}{4(n-1)} \left[\omega^2\tilde{R}+2(n-1)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right)-n(n-1)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\right](\omega^{(n-2)/2} \tilde{\phi})^2\right) \Big)
[/tex]

I will have to finish this shortly as I need to go somewhere, but then I essentially expand out, and use product rule to get replace the double derivative on omega...
 
  • #5
So expanding:

[tex] \tilde{L}=\omega^{-n}\sqrt{-\tilde{g}}\Big(-\frac{1}{2}\omega^2 \tilde{g}^{\mu\nu}\left[\frac{(n-2)^2}{4}\omega^{(n-4)}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\omega)\tilde{\phi}^2+\omega^ {(n-2)}\tilde{\nabla}_{\mu}( \tilde{\phi})\tilde{\nabla}_{\nu}( \tilde{\phi})+(n-2)\omega^{(n-3)}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi}\right]
[/tex]
[tex]
+\left[ -\frac{(n-2)}{4(n-1)}\omega^2\tilde{R}-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\omega\left(\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right)+\frac{n}{4}(n-2)\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\right]\omega^{(n-2)} \tilde{\phi}^2 \Big)

[/tex]


Finally,

[tex] \tilde{L}=\sqrt{-\tilde{g}}\Big(-\tilde{g}^{\mu\nu}\frac{(n-2)^2}{8}\omega^{-2}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\omega)\tilde{\phi}^2-\frac{1}{2}\tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}( \tilde{\phi})\tilde{\nabla}_{\nu}( \tilde{\phi})-\frac{(n-2)}{2}\tilde{g}^{\mu\nu}\omega\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi}
[/tex]
[tex]
-\frac{(n-2)}{4(n-1)}\omega^2\tilde{\phi}^2\tilde{R}-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\omega\tilde{\phi}^2\left(\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right)+\frac{n}{4}(n-2)\tilde{\phi}^2\tilde{g^{\mu\nu}}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right) \Big)

[/tex]


Now we can use product rule e.g. [tex]\tilde{\phi}^2\tilde{\nabla}_{\m u}\tilde{\nabla}_{\nu}\omega\right) =\tilde{\nabla}_{\m u}\left(\tilde{\phi}^2}\tilde{\nabla}_{\nu}\omega\right)\right)-2\tilde{\phi}\tilde{\nabla}_{\m u}\left(\tilde{\phi}\right)\left(\tilde{\nabla}_{\nu}\omega\right)[/tex]

But now what?
 
  • #6
I made a mistake the last line should be:

[tex] \tilde{L}=\sqrt{-\tilde{g}}\Big(-\tilde{g}^{\mu\nu}\frac{(n-2)^2}{8}\omega^{-2}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu} (\omega)\tilde{\phi}^2-\frac{1}{2}\tilde{g}^{\mu\nu}\tilde{\nabla}_{\mu}(\tilde{\phi})\tilde{\nabla}_{\nu}( \tilde{\phi})-\frac{(n-2)}{2}\tilde{g}^{\mu\nu}\omega^{-1}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi} [/tex]
[tex]-\frac{(n-2)}{4(n-1)}\tilde{\phi}^2\tilde{R}-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\omega^{-1}\tilde{\phi}^2\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)+\frac{n}{4}(n-2)\tilde{\phi}^2\tilde{g^{\mu\nu}}\omega^{-2}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right) \Big)
[/tex]

From which one has the conformally the Lagrangian in the conformal frame, plus a bit I want to get ride of:

[tex] \tilde{L}=L_0+\sqrt{-\tilde{g}}\Big(-\tilde{g}^{\mu\nu}\frac{(n-2)^2}{8}\omega^{-2}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu} (\omega)\tilde{\phi}^2-\frac{(n-2)}{2}\tilde{g}^{\mu\nu}\omega^{-1}\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi} [/tex]
[tex]-\frac{1}{2}(n-2)\tilde{g^{\mu\nu}}\omega^{-1}\tilde{\phi}^2\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)+\frac{n}{4}(n-2)\tilde{\phi}^2\tilde{g^{\mu\nu}}\omega^{-2}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right) \Big)
[/tex]

Which looks prettier as:

[tex] \tilde{L}=L_0+\sqrt{-\tilde{g}}\Big(\tilde{g^{\mu\nu}}\omega^{-2}\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2\frac{(n^2-4)}{8} -\frac{(n-2)}{2}\tilde{g}^{\mu\nu}\omega^{-1}\left(\tilde{\nabla}_{\mu}(\omega)\tilde{\nabla}_{\nu}(\tilde{\phi})\tilde{\phi}
+\left(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2\Big)
[/tex]
 
  • #7
I think I have it now:

Take [tex]\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2= \tilde{\nabla}_{\mu}\left[\omega(\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2\right]-\omega(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2-\omega(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})[/tex]

If we include [tex] \tilde{g}^{\mu\nu} [/tex] the first term is a total divergence so vanishes. So we are left with:

[tex]\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2= -\omega(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2-\omega(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi}) (1)[/tex]

Now using the product rule again on the first term:

[tex] (\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2=\tilde{\nabla}_{\mu}(\tilde{\phi}^2(\tilde{\nabla}_{\nu}\omega)) -(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})[/tex]

again if we bring in the external metric the first term here is a total divergence so vanishes leaving:

[tex] (\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\omega)\tilde{\phi}^2= -(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})[/tex]

Plugging this into (1)

[tex]\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\omega\right)\tilde{\phi}^2= +\omega(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})-\omega(\tilde{\nabla}_{\nu}\omega)2\tilde{\phi}(\tilde{\nabla}_{\mu}\tilde{\phi})=0 [/tex]

So that takes care of the first term in the undesired Lagrangian density. We also have:

[tex] \tilde{\nabla}_{\mu}(\omega)\tilde{\nabla} _{\nu}(\tilde{\phi}) \tilde{\phi}[/tex]

[tex]
\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\tilde{\phi}\right)\tilde{\phi}= -\omega(\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\tilde{\phi})\tilde{\phi}-\omega(\tilde{\nabla}_{\nu}\tilde{\phi})(\tilde{\nabla}_{\mu}\tilde{\phi})
[/tex]
Use product rule again on the first term:

[tex](\tilde{\nabla}_{\mu}\tilde{\nabla}_{\nu}\tilde{\phi})\tilde{\phi}=\tilde{\nabla}_{\mu}\left[\tilde{\phi}(\tilde{\nabla}_{\nu}\tilde{\phi})\right]-(\tilde{\nabla}_{\mu}\tilde{\phi})(\tilde{\nabla}_{\nu} \tilde{\phi}) =-(\tilde{\nabla}_{\mu}\tilde{\phi})(\tilde{\nabla}_{\nu} \tilde{\phi})[/tex]

So we get:


[tex]
\left(\tilde{\nabla}_{\mu}\omega\right)\left(\tilde{\nabla}_{\nu}\tilde{\phi}\right)\tilde{\phi}= -\omega\left[ -(\tilde{\nabla}_{\mu}\tilde{\phi})(\tilde{\nabla}_{\nu} \tilde{\phi})\right]-\omega(\tilde{\nabla}_{\nu}\tilde{\phi})(\tilde{\nabla}_{\mu}\tilde{\phi})=0
[/tex]

The only other term not yet considered is:

This everything undesired vanishes.

I have used the symmetry of the metric and the divergence theorem to convert total divergences into surface terms that one can set to zero.
 

1. What is "scalar action" in physics?

Scalar action refers to a mathematical quantity that describes the dynamics of a system in a particular physical theory. In simple terms, it is a measure of the total energy or "action" of a system, which is a crucial concept in understanding the behavior of physical systems.

2. What does it mean for an action to be "conformally invariant"?

An action is said to be conformally invariant if it remains unchanged under conformal transformations. Conformal transformations are mathematical operations that preserve angles but can change distances between points. In physics, conformal invariance is a property that some physical theories possess, and it allows for a deeper understanding of certain physical systems.

3. How is it proven that scalar action is conformally invariant?

The proof of conformal invariance for scalar action involves using mathematical techniques such as group theory and differential geometry. The key step is to show that the action remains unchanged under conformal transformations, which requires careful mathematical analysis and manipulation of equations.

4. What are some examples of physical theories that exhibit conformal invariance?

One of the most well-known examples is the theory of general relativity, which describes the behavior of gravity. Other examples include certain field theories in particle physics and condensed matter physics, such as the Ising model and the Kondo model.

5. Why is conformal invariance important in physics?

Conformal invariance is important because it allows us to gain a deeper understanding of physical theories and systems. It also has practical applications in various fields such as cosmology, where conformal invariance is used to study the behavior of the universe on a large scale. Additionally, it has connections to other fundamental concepts in physics such as symmetry and conservation laws.

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