Domenico Fiorenza Massive vectorial field

Massive vectorial field

When approaching the Higgs model, naturally, it comes out from the theory an electro-magnetic field with mass. Recall that $A=A^{\mu}$ is the four dimensional potential vector for the E-M field, and $F^{\mu \nu} = (\partial^{\nu} A^{\mu} - \partial^{\mu} A^{\nu})$ is the E-M tensor. The Higgs model furnish us the following lagrangian density $\mathcal{L}$ :

\mathcal{L}_A = -\frac{1}{4} F^{\mu \nu} F_{\mu \nu} + \frac{1}{2} M^2 A^{\mu}A_{\mu}.

The Lagrange equations associated are:

\Box A_{\mu} + \partial_{\mu}(\partial^{\nu} A_{\nu}) - M^2 A_{\mu}.

According to the general theory we have to check that this equation is covariant, through the action of the little group. For doing this, first, we have to identify the mass and the spin of the particle. In Fourier coordinates the operator becomes:

K = \widehat{ ( \Box - M^2) \delta_{\mu}^{\rho} + \partial_{\mu} ( \partial^{\nu} ) } = \left( (k^2 - M^2) \delta_{\mu}^{\rho} - k^{\rho} k_{\mu} \right) .

We exhibit an inverse for $K$ that is:

K^{-1} = \frac{- g^{\mu \nu} + \frac{k^{\mu}k^{\nu}}{M^2}}{k^2 - M^2}.

todo: write the inverse per series

It is explicit now that the mass of the particle $M$ : we have to look at the singularities of the symbol. The kernel of the operator is easy calculated in a rest frame, i.e. for $k=(M,0,0,0)$ :

Dim(Ker(K^{-1})) = Dim(Ker( \frac{k^{\mu}k^{\nu}}{M^2})) = 3 = 2s+1

Then the massive vectorial field represents a particle of spin $s=1$ and mass $M$ . The final step is finding an action that conserves the form of the equation, and study its irreducibility. Call $U(\Lambda)$ the action over $A$ that brings $A$ from a frame $O$ to a new one, say $O'$ . Then we have

0 = \left( (k^2 - M^2) \delta_{\mu}^{\rho} - k^{\rho} k_{\mu} \right) U^{-1}(\Lambda)A^{'}_{\rho} =

= \left( (\Lambda k^{'})^{2} - M^2) \delta_{\mu}^{\rho} - \Lambda k^{'\rho} \Lambda k^{'}_{\mu} \right) U^{-1}(\Lambda)A^{'}_{\rho}

By multiplying for $U(\Lambda)$ we have

\left( (\Lambda k^{'})^{2} - M^2) \delta_{\mu}^{\rho} - U(\Lambda) \Lambda k^{'\rho} \Lambda k^{'}_{\mu} U^{-1}(\Lambda) \right) U^{-1}(\Lambda)A^{'}_{\rho}

Then forcing the covariance of the equation we have the following condition:

U(\Lambda) \Lambda k^{'\rho} \Lambda k^{'}_{\mu} U^{-1}(\Lambda) = k^{' \rho} k^{'}_{\mu}.

to finish

Revised on November 11, 2010 at 18:19:40 by giuseppe_malavolta