- The problem statement, all variables and given/known data
I am stuck on part d, but give my workings to all parts of the question below.
Relevant equations
The attempt at a solution
a) EoM given by E-L equations: Gives $-\partial_u\partial^u \phi + m^2 \phi - \frac{\lambda \phi^3}{3!} =0 $ b) $ L=T- V $, $ T$ the kinetic energy, $V$ the potential energy
So , assuming a signature of $(-,+,+,..) $ , $L=\frac{1}{2} (\partial_t \phi)^2-\frac{1}{2} (\partial_i \phi )^2+\frac{1}{2}m^2 \phi^2 - \frac{\lambda}{4!}\phi^4-\frac{3m^4}{2\lambda} $
So $-V= -\frac{1}{2} (\partial_i \phi )^2+\frac{1}{2}m^2 \phi^2 - \frac{\lambda}{4!}\phi^4-\frac{3m^4}{2\lambda} $
c) for a constant field the first deriviative term of the EoM vanishes. So we have :
$m^2 \phi - \lambda \frac{\phi^3}{3!} = 0 $ $ \phi (m^2-\lambda \frac{\phi^2}{3!}) =0 $ So $\phi_c = \pm m \sqrt{\frac{3!}{\lambda}} $
To check which have the lowest energy evaluate $ T+ V$
d)
I get: (I can't do tilda in LaTex so I've changed it to $\phi=\phi_c+\phi'(x) $ )
$ S= \int d^2 x (\frac{-1}{2} \partial_u \phi' \partial^u \phi' + \frac{1}{2}m^2 (\phi_c + \phi')^2 - \frac{\lambda}{4!} (\phi_c + \phi')^4 -\frac{3m^4}{2\lambda}) $
Should I expand only to first order in $\phi'(x)$?
Since $ \phi_c $ satisfies the EoM , the EoM becomes:
$ -\partial_u\partial^u \phi' + m^2 \phi' -\frac{\lambda}{3!}\phi^3 =0 $
I have no idea how get the corresponding mass. I know that if the EoM is obeyed, the particle is on-shell and $p^up_u = -m^2 $ ? I think my answer lies in this but I'm not sure what to do
Many thanks in advance.