Conformal Invariance Klein Gordon Action

[FedEx]

Wald Appendix D talks on why $$g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\phi$$ is not conformally invariant when n is not equal to 2.

I want to prove that the Klein Gordon Action (V=0) is not conformally invariant.

However the term that I have in the action is just
$$g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi$$ which IS clearly invariant in any dimension since $$g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi = \nabla^{\nu}\phi\nabla_{\nu}\phi$$ where the RHS is just an inner product of two vectors which indeed remains invariant.

PS: $$\phi\rightarrow\Omega^s\phi$$ for s=0;

 

[Physics Monkey]

What happens when the conformal transformation is not just a constant? How do $\phi$ and $\partial_\mu \phi$ transform?

Also, what do mean by the PS when you say s=0?

 

[FedEx]

I think i should have mentioned the context. Wald proves that [tex]g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\phi[\tex] is not conformally invariant in n greater than 2 dimensions by making a conformal transformation (for which \phi transforms as \Omega^s \phi and that long transformation for Ricci scalar). In the final expression he gets terms like

(2s+n-2)(long expression) + s(long) + s(n+s-3)(long) + (Klein gordon term)(\Omega^s-2)

Which is not zero for n greater than 2 for any choice of s.(true for (n=2,s=0))

In our procedure, let say that \phi'(x)-->\phi(\Lamda^-1 x) where Lambda is the transformation matrix. In this case, isn't Klein Gordon conformally invariant then?

I think a better understanding of active & passive diffeos is what i need.

 

[samalkhaiat]

Wald Appendix D talks on why $$g^{\mu\nu}\nabla_{\mu}\nabla_{\nu}\phi$$ is not conformally invariant when n is not equal to 2.

I want to prove that the Klein Gordon Action (V=0) is not conformally invariant.

However the term that I have in the action is just
$$g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi$$ which IS clearly invariant in any dimension since $$g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi = \nabla^{\nu}\phi\nabla_{\nu}\phi$$ where the RHS is just an inner product of two vectors which indeed remains invariant.

PS: $$\phi\rightarrow\Omega^s\phi$$ for s=0;

Thank you for the email.

Give me a few days to make some notes. For now, I just want to tell you that the conformal weight of some field $d$ (the s in your post) is not an arbitrary number, it depends on the spin of the field and on the dimension of the space:
$$d(\phi) = \frac{n-2}{2}, \ \ d(\psi) = \frac{n-1}{2}.$$

As for the active vs pssive transformation: I just say, dot follow that garbage because it confuses you more. The important point is to understant the connection between Weyl rescalling plus diffeomorphism on one hand and the flat space conformal group SO(2,4)one the other. They are equivalent to each other, i.e. Weyl and diffeomorphism invariances of a theory in curved space is equivalent to conformal invariance of the same theory in flat space. This is the content of Zumino's theorem which I will explain to you in few days.

Regards

Sam

 

[FedEx]

Give me a few days to make some notes! Are you kidding me! My tutors back here wouldn't take so much effort to help! I have been looking at your other posts as well. Respect.

Okay now back to Physics.

I have convinced myself with the following :

1) The Klein Gordon action without the mass term, is conformally invariant for n=2. Since the action is Weyl Invariant. And since a conformal transformation can be undone by a Weyl Transform. One can state that the action is conformally invariant.

2) The equation: $$g_{\mu\nu}(x)=\Omega^2(x)g_{\mu\nu}$$ has the same argument! i.e x. One would expect to have x' in the LHS and x in the RHS or the vice versa depending on whether you are looking at it actively or passively(sorry had to use that :P). Because of this peculiarity the first thing that follows is the conformal killing equation rather than the usual killing equation. Which means that the metric changes! Which means that the interval changes! Which means that the inner product changes! Which means that the logic that i was following in the original post is wrong, (as it should have been)

$$g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi$$which IS clearly invariant in any dimension since$$g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi = \nabla^{\nu}\phi\nabla_{\nu}\phi$$ where the RHS is just an inner product of two vectors which indeed remains invariant.


PS: I would love to know the Zumino theorem. Also, if there is anything wrong in the above logic, please let me know.

Thanks a ton. :-)

 

[samalkhaiat]

$$g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi$$which IS clearly invariant in any dimension since$$g^{\mu\nu}\nabla_{\mu}\phi\nabla_{\nu}\phi = \nabla^{\nu}\phi\nabla_{\nu}\phi$$ where the RHS is just an inner product of two vectors which indeed remains invariant.PS: I would love to know the Zumino theorem. Also, if there is anything wrong in the above logic, please let me know.

Thanks a ton. :-)

$\nabla_{a}\phi \nabla^{a}\phi$ is (scalar) invariant under diffeomorphism. This does not mean that it is invariant under Wyle transformation. There are examples of massless and Lorentz invariant actions that are conformally non-invariant.
Ok, let us see what is going on in this erea.

Let $A[\phi]$ be a matter field theory that can be made generally covariant, $A[\phi , g]$, by minimal substitutions:
$$\eta_{ab}\rightarrow g_{ab}(x), \ \ \partial_{a}\phi \rightarrow \nabla_{a}\phi .$$
By “generally covariant theory” we mean a theory invariant under diffeomorphism, i.e. general coordinates transformation.
Next, we study the change in $A[\phi , g]$ brought about by infinitesimal changes in the fields,
$$g_{ab}(x) \rightarrow g_{ab}(x) + \delta g_{ab}(x),$$
$$\phi (x) \rightarrow \phi (x) + \delta \phi (x).$$
This gives us
$$\delta A = \int d^{n}x \left( \frac{\delta A}{\delta \phi} \delta \phi + \frac{\delta A}{\delta g_{ab}} \delta g_{ab}\right).$$
If we use the definition of the energy-momentum tensor
$$\frac{\delta A}{\delta g_{ab}} = \frac{1}{2} \sqrt{-g} \ T^{ab},$$
we find
$$\delta A = \int d^{n}x \left( \frac{1}{2}\sqrt{-g} \ T^{ab}\delta g_{ab} + \frac{\delta A}{\delta \phi}\delta \phi \right). \ \ (1)$$
Let us now consider the special case where the field variations are given by
$$\delta g_{ab}(x) = 2 \lambda (x) g_{ab}(x), \ \ (2a)$$
$$\delta \phi (x) = - d \lambda (x) \phi (x), \ \ \ (2b)$$
where $\lambda (x)$ is an arbitrary small positive function and $d$ is a real number which we take to be the canonical dimension of the (scalar) matter field $\phi$ ($d = (n-2)/2$).
So, what we are doing is simply looking at the change in $A[\phi , g]$ due to arbitrary redefinition of the fields:
$$g_{ab}= \Omega^{2}(x) g_{ab},$$
$$\phi = \Omega^{-d}\phi ,$$
with $\Omega (x) \approx 1 + \lambda (x)$. This is the Wyle transformations. They act on the function space of the fields not on their coordinate labels. It is like, making local gauge transformations.
Ok, inserting eq(2a,b) in eq(1) and using the fact that $\lambda (x)$ is arbitrary, we find
$$\frac{\delta A}{\delta \lambda} = \sqrt{-g}g_{ab}T^{ab} - d \frac{\delta A}{\delta \phi}\phi .$$
This (so called trace identity) shows the connection between Wyle rescaling and conformal transformations: If we use the field equation of the matter field
$$\frac{\delta A}{\delta \phi}=0,$$
we find
$$\frac{\delta A}{\delta \lambda} = \sqrt{-g} \ g_{ab} \ T^{ab}. \ \ (3)$$
Thus, the invariance of the covariant action under Wyle rescaling implies (and is implied by) the trace-less-ness of the symmetric energy-momentum tensor. i.e., if the covariant action is, at the same time, invariant under Wyle rescaling, then that action, restricted to flat space, is invariant under the conformal group $SO(2,n)$. Bellow, I will demonstrate this on the transformations level. In real life though, covariant actions obtained by the minimal substitutions are not invariant under Wyle rescaling (as we will see bellow, the action of massless scalar field provides an example). In this case, eq(3) relates the trace of the energy-momentum tensor to the symmetry breaking part of the action.
The minimal action for massless scalar field is given by
$$A[\phi , g] = \frac{1}{2}\int d^{n}x \sqrt{-g} \ g^{ab} \ \partial_{a}\phi \ \partial_{b}\phi .$$
This action which is invariant under diffeomorphism is not invariant under Wyle transformations. To see this, we use eq(2a,b) and
$$\delta \sqrt{-g} = n \lambda (x) \ \sqrt{-g},$$
to calculate the change in the action. You can easily verify that
$$\delta A = ( \frac{n}{2} - d - 1 ) \int d^{n}x \sqrt{-g} \ \lambda (x) g^{ab}\partial_{a} \phi \ \partial_{b}\phi - \frac{d}{2}\int d^{n}x \sqrt{-g} \ g^{ab}\ \partial_{a} \lambda \ \partial_{b}\phi^{2}.$$
The first term on the right hand side vanishes because, in n-dimension, the scalar field has
$$d = \frac{n}{2} - 1 .$$
So, after integrations by parts, we end up with
$$\delta A = \frac{n-2}{4}\int d^{n}x \sqrt{-g} \ \phi^{2}\left( \frac{1}{\sqrt{-g}}\ \partial_{b}( \sqrt{-g}\ g^{ab}\partial_{a}\lambda ) \right). \ \ (4)$$
Thus, for $n > 2$ and $\lambda \neq \mbox{constant}$, the action is not invariant under Wyle transformations. Notice that the integral in eq(4) is symmetric under interchanging $\lambda$ with $\phi^{2}$, so if we introduce the box operator
$$\Box f(x) = \frac{1}{\sqrt{-g}}\partial_{b}\left( \sqrt{-g} \ g^{ab}\partial_{a}f(x)\right),$$
we can write eq(4) as
$$\delta A = \frac{n-2}{4}\int d^{n}x \sqrt{-g}\ \phi^{2}(x)\Box \lambda = \frac{n-2}{4} \int d^{n}x \sqrt{-g}\ \lambda (x) \Box \phi^{2}.$$
Form this we calculate the symmetry breaking part of the action as given in eq(3)
$$\frac{1}{\sqrt{-g}}\frac{\delta A}{\delta \lambda} = g_{ab} \ T^{ab} = \frac{n-2}{4}\Box \phi^{2}.$$
However, if we add to the minimal action $A$ a term
$$B = -\frac{1}{12}\int d^{4}x \sqrt{-g}\ \phi^{2}R,$$
then the total action $A + B$ is Wyle invariant. This follows because, under Wyle rescaling, (putting $n = 4$ for simplicity) the scalar curvature transforms as
$$\delta R = -2 \lambda R + 6 \Box \lambda .$$
If we now define a new energy-momentum tensor by
$$\theta^{ab} = \frac{2}{\sqrt{-g}}\frac{\delta}{\delta g_{ab}}(A + B),$$
we find that it is traceless
$$g_{ab}\theta^{ab}= 0.$$

Finally, let us demonstrate that combining diffeomorphsims with Wyle rescaling reproduce the correct conformal transformation on the fields in flat space. Under diffeomorphism,
$$x^{a}\rightarrow \bar{x}^{a} = F^{a}(x),$$
the metric tensor and matter fields transform like
$$\bar{g}^{ab}( \bar{x}) = \frac{\partial \bar{x}^{a}}{\partial x^{c}}\frac{\partial \bar{x}^{b}}{\partial x^{d}}g^{cd}(x), \ \ (5a)$$
$$\bar{\phi}( \bar{x}) = \Sigma ( F ) \phi (x), \ \ \ \ (5b)$$
where $\Sigma (F)$ is a matrix representation denoting the spinorial nature of the matter field. For simplicity, we will work with scalar field only, i.e. we set $\Sigma = 1$. For infinitesimal diffeomorphism,
$$\bar{x}^{a} = x^{a} + f^{a}(x),$$
with $f^{a}(x)$ are arbitrary real and infinitesimal functions, eq(5a,b) give
$$\delta_{f}g^{ab}(x) = g^{cb} \partial_{c} f^{a} + g^{ac} \partial_{c} f^{b} - f^{c} \partial_{c} g^{ab}, \ \ (6a)$$
$$\delta_{f}\phi (x) = - f^{c}\partial_{c}\phi (x). \ \ \ \ \ (6b)$$
On $(M^{n}, g^{ab})$, General relativity requires that the action be rigorously invariant under eq(6a,b). Recall that, under infinitesimal Wyle’s rescaling, we had
$$\delta_{W}g^{ab}(x) = -2 \lambda (x) g^{ab}(x), \ \ \ (2a)$$
$$\delta_{W}\phi (x) = -d \lambda (x) \phi (x). \ \ \ \ (2b)$$
We will now show that, if an action is invariant under eq(2a,b) and eq(6a,b), then the flat space limit of that action is invariant under $SO(2,n)$ transformations on the matter fields.
For diffeomorphism, we choose
$$f^{a}(x) = - \epsilon x^{a}, \ \epsilon \ll 1.$$
The variations eq(6a,b) become
$$\delta_{f}g^{ab}= \epsilon \left( (x \cdot \partial ) - 2 \right) g^{ab}(x), \ \ (7a)$$
$$\delta_{f}\phi (x) = \epsilon (x \cdot \partial) \phi (x). \ \ \ \ \ (7b)$$
For Wyle’s parameter, we choose
$$\lambda (x) = - \epsilon .$$
So eq(2a,b) become
$$\delta_{W}g^{ab} = 2 \epsilon g^{ab}, \ \ \ (8a)$$
$$\delta_{W}\phi = d \epsilon \phi . \ \ \ \ \ \ (8b)$$
Combining these transformations, we see that the action must be invariant under
$$\delta_{f + W}g^{ab}(x) = \epsilon ( x \cdot \partial ) g^{ab}(x),$$
$$\delta_{f + W}\phi (x) = \epsilon \left( ( x \cdot \partial ) + d \right) \phi (x).$$
Now, if $g^{ab}= \eta^{ab}$, then $\delta g^{ab} = 0$. Thus the action in flat space is invariant under the scale transformation on $\phi (x)$.
Next, if we choose
$$f^{a}(x) = 2 x^{a}(C \cdot x) - C^{a}x^{2},$$
$$\lambda (x) = 2 (C \cdot x), \ \ C^{a}\ll 1,$$
we find
$$\delta_{f}g^{ab} = C_{c}\left( x^{2}\partial^{c} - 2 (C \cdot x) x^{c} + 4 x^{c}\right)g^{ab},$$
$$\delta_{f}\phi (x) = C_{c}\left( x^{2}\partial^{c} - 2 (C \cdot x) x^{c}\right) \phi (x),$$
and
$$\delta_{W}g^{ab} = - 4 (C \cdot x) g^{ab},$$
$$\delta_{W}\phi = - 2 d (C \cdot x ).$$
Thus
$$\delta_{f + W}g^{ab}= C_{c}\left( x^{2}\partial^{c} - 2 x^{c} ( x \cdot \partial ) \right) g^{ab},$$
$$\delta_{f + W} \phi (x) = C_{c}\left( x^{2}\partial^{c} - 2 x^{c} (x \cdot \partial ) - 2 d x^{c}\right) \phi (x).$$
Again, we see that the action in $(M^{n}, \eta^{ab})$ is invariant under the special conformal transformation on $\phi (x)$. The proof for Poincare transformations is trivial and corresponds to choosing $\lambda (x) = 0$ and $$f^{c}(x) = \omega^{c}{}_{b}x^{b} + a^{c}.$$Good Luck

Sam

 

[FedEx]

Reading. :)