RG flow of quadrupole coupling in 6+1 dimension electrostatic problem

[DaniV]

Homework Statement

Given a classical action which describes electrostatics in 6+1 dimensions
$$\frac{S_\text{eff}}{T}=\frac{1}{2} \int d^6x(\nabla\phi)^2+\frac{S_{p.p}}{T}$$

where the one particle action, that represents the energy of the monopole dipole and quadrupole induced by charge distribution around the origin is:

$$\frac{S_{p.p}}{T}=-q\phi(0)-D\nabla_i\phi(0)\nabla^i\phi(0)-Q\nabla_i\nabla_j\phi(0)\nabla^i\nabla^j\phi(0)+...$$

Now we add to the theory a scalar field $\sigma$ that couples to the potential $\phi$ so the action becomes:

$$\frac{S_\text{eff}}{T}=\int d^6x\left[(\nabla \phi)^2+(\nabla \sigma)^2+\lambda\sigma (\nabla \phi)^2\right]+\frac{S_{p.p}}{T}$$

and the system is subjected to an external quadrupole background $$\bar \phi=\frac{1}{2}\rho_{ij}x^ix^j$$ when $Tr\rho_{ij}=0$
i.e. $\phi \to \phi+ \bar \phi$ where $\phi$ describes the classical response of the charge distribution at the origin to the external perturbation $\bar \phi$, and we assume that $\frac{q^2\lambda^2}{Q^2}\ll1$

I want to calculate the RG flow of the coupling $Q$,
using Minimal subtraction renormalization scheme and dimensional regularization.

Relevant Equations

Euler Lagrange for fields:
$$\frac{\partial \mathcal L}{\partial \phi}=\partial_{\mu}(\frac{\partial \mathcal L}{\partial (\partial _{\mu}\phi)})$$

dimensional regularization:
we put $$2\epsilon=d-4$$ when d is the dimension , solving divergent integrals by expanding to small $\epsilon$ and then as we approach to final result we take $\epsilon \to 0$

Minimal subtraction renormalization scheme:
defining new couplings using the old ones in way that they cancel divergent part of the solution.

RG flow:
the flow of the coupling constant in the dependence of energy scale $\mu$ given by the beta function $\beta(Q)=\mu \frac{\partial Q}{\partial \mu}$

I tried to do a Euler Lagrange equation to our Lagrangian:
$$\frac{S_\text{eff}}{T}=\int d^6x\left[(\nabla \phi)^2+(\nabla \sigma)^2+\lambda\sigma (\nabla \phi)^2\right]+\frac{S_{p.p}}{T}$$
and then I would like to solve the equation using perturbation theory when $Q$ or somehow $$\frac{q^2\lambda^2}{Q^2}\ll1$$ but when itried to do the equation I got that is not depend on Q at all because the relevant term in $$S_{p.p}$$ is just derivatives of the field $\phi$ in the point zero (so its acting like constant) or am I wrong?

 

[member 553083]

A:To do the Euler-Lagrange equation, you need to take derivative of the action with respect to the respective fields. The only term containing $\lambda$ is the second one, so we will concentrate there. $$\delta S = \int d^6x \left[2\lambda \sigma \nabla \cdot \phi \delta \nabla \phi + \lambda \sigma (\nabla \phi)^2 \delta \sigma \right]$$We can use integration by parts on the first term and use the fact that the field is periodic to obtain$$\delta S = \int d^6x \left[-2\lambda \phi \nabla \cdot \sigma \delta \nabla \phi + \lambda \sigma (\nabla \phi)^2 \delta \sigma \right]$$Integrating the first term by parts again, and using periodicity of the field again, we obtain$$\delta S = \int d^6x \left[2\lambda \phi \nabla^2 \sigma \delta \phi + \lambda \sigma (\nabla \phi)^2 \delta \sigma \right]$$The Euler-Lagrange equation then is$$\nabla^2 \sigma = 0$$$$\nabla^2 \phi - \lambda \sigma (\nabla \phi)^2 = 0$$These equations are independent of $Q$.