Scale Invariant Classical Field Theory

  • #1
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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?
 

Answers and Replies

  • #2
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Thanks for the post! This is an automated courtesy bump. Sorry you aren't generating responses at the moment. Do you have any further information, come to any new conclusions or is it possible to reword the post?
 

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