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QF, action, EoM, mass of particle, on-shell

  1. Jul 18, 2017 #1
    1. The problem statement, all variables and given/known data

    Question attached.

    avtiommas.png
    I am stuck on part d, but give my workings to all parts of the question below.

    2. Relevant equations


    3. 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.
     
  2. jcsd
  3. Jul 23, 2017 #2
    Thanks for the thread! 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? The more details the better.
     
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