Scale Invariant Classical Field Theory

[spaghetti3451]

Homework Statement

A class of interesting theories are invariant under the scaling of all lengths by $x^{\mu} \rightarrow (x')^{\mu}=\lambda x^{\mu}$ and $\phi(x) \rightarrow \phi'(x) = \lambda^{-D}\phi(\lambda^{-1}x)$.

Here $D$ is called the scaling dimension of the field.

Consider the action for a real scalar field given by $S = \int d^{4}x\ \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}-g\phi^{p}$.

Find the scaling dimension $D$ such that the derivative terms remain invariant.

For what values of $m$ and $p$ is the scaling $x^{\mu} \rightarrow (x')^{\mu}=\lambda x^{\mu}$ and $\phi(x) \rightarrow \phi'(x) = \lambda^{-D}\phi(\lambda^{-1}x)$ a symmetry of the theory? How do these conclusions change for a scalar field living in an $(n+1)$-dimensional spacetime instead of a $3+1$-dimensional spacetime?

In $3+1$ dimensions, use Noether's theorem to construct the conserved current $D^{\mu}$ associated to scaling invariance.

Homework Equations

The Attempt at a Solution

Under the scaling of all lengths by $x^{\mu} \rightarrow (x')^{\mu}=\lambda x^{\mu}$ and $\phi(x) \rightarrow \phi'(x) = \lambda^{-D}\phi(\lambda^{-1}x)$, where $D$ is called the scaling dimension of the field, the given action for the real scalar field transforms as follows:

$S = \int d^{4}x\ \frac{1}{2}\partial_{\mu}\phi(x)\partial^{\mu}\phi(x)-\frac{1}{2}m^{2}\phi^{2}(x)-g\phi^{p}(x)$

$\rightarrow \int d^{4}x\ \Big|\frac{1}{\lambda^{4}}\Big| \Big[ \frac{1}{2}({\Lambda^{\rho}}_{\mu}\lambda^{-D}\partial_{\rho}\phi)(\lambda^{-1}x)({\Lambda_{\sigma}}^{\mu}\lambda^{-D}\partial^{\sigma}\phi)(\lambda^{-1}x)-\frac{1}{2}m^{2}(\lambda^{-D}\phi)^{2}(\lambda^{-1}x)-g(\lambda^{-D}\phi)^{p}(\lambda^{-1}x)\Big]$, where the factor of $\Big|\frac{1}{\lambda^{4}}\Big|$ is the Jacobian for the coordinate transformation

Therefore, the scaling dimension $D$ such that the derivative terms remain invariant is found as follows:

$\Big|\frac{1}{\lambda^{4}}\Big| \Big[ \frac{1}{2}({\Lambda^{\rho}}_{\mu}\lambda^{-D}\partial_{\rho}\phi)(\lambda^{-1}x)({\Lambda_{\sigma}}^{\mu}\lambda^{-D}\partial^{\sigma}\phi)(\lambda^{-1}x)\Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

$\Big(\frac{1}{\lambda^{4}}\Big)(\lambda^{-D})(\lambda^{-D}) \Big[ \frac{1}{2}({\Lambda^{\rho}}_{\mu}{\Lambda_{\sigma}}^{\mu})(\partial_{\rho}\phi)(\lambda^{-1}x)(\partial^{\sigma}\phi)(\lambda^{-1}x) \Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

$\lambda^{-4-2D} \Big[ \frac{1}{2}{\Lambda^{\rho}}_{\mu}{(\Lambda^{-1})^{\mu}}_{\sigma}(\partial_{\rho}\phi)(\lambda^{-1}x)(\partial^{\sigma}\phi)(\lambda^{-1}x) \Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

$\lambda^{-4-2D} \Big[ \frac{1}{2}{\eta^{\rho}}_{\sigma}(\partial_{\rho}\phi)(\lambda^{-1}x)(\partial^{\sigma}\phi)(\lambda^{-1}x) \Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

$\lambda^{-4-2D} \Big[ \frac{1}{2}(\partial_{\rho}\phi)(\lambda^{-1}x)(\partial^{\rho}\phi)(\lambda^{-1}x) \Big]=\frac{1}{2}(\partial_{\nu}\phi)(\lambda^{-1}x)(\partial^{\nu}\phi)(\lambda^{-1}x)$

so that

$\lambda^{-4-2D}=\lambda^{0}$
$D=-2$

Am I correct so far?

 

[member38644]

Please help.

Yes, you are correct. The scaling dimension for the field is -2 in this case. To determine the values of m and p for which the scaling is a symmetry of the theory, we can look at the terms involving the field in the action. Since the scaling of the field is given by $\phi(x) \rightarrow \phi'(x) = \lambda^{-D}\phi(\lambda^{-1}x)$, the term $m^2\phi^2$ scales as $m^2\phi^2 \rightarrow \lambda^{-2D}m^2\phi^2$. For the scaling to be a symmetry, the term must remain invariant, so we have $\lambda^{-2D}m^2 = m^2$, which implies $D=0$. Therefore, for the scaling to be a symmetry, we must have $D=0$, which means that the scaling dimension of the field must be 0. This is only possible if $m=0$, since $D=-2$ for this particular theory. So, the scaling is a symmetry of the theory if $m=0$ and $D=0$, which implies that the field has no mass and its scaling dimension is 0.

If the scalar field lives in an (n+1)-dimensional spacetime instead of a 3+1-dimensional spacetime, the result is still the same. The scaling dimension of the field must be 0 for the scaling to be a symmetry of the theory, which means that the field has no mass. The only difference is that the Jacobian for the coordinate transformation will be different, but the result will still be the same.

To construct the conserved current associated to scaling invariance using Noether's theorem, we start by considering the infinitesimal transformation of the field:

$\phi(x) \rightarrow \phi'(x) = \phi(x) + \delta \phi(x)$

where $\delta \phi(x)$ is the infinitesimal change in the field. This transformation should leave the action invariant, so we have:

$S = \int d^{4}x\ \frac{1}{2}\partial_{\mu}\phi\partial^{\mu}\phi-\frac{1}{2}m^{2}\phi^{2}-g\phi^{p}$

##\rightarrow S' = \int d