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Contact interaction in Lagrangians

  1. Jun 6, 2013 #1
    Hi all,

    If I take an action involving two point particles coupled together by a delta function contact interaction is it possible to carry out the variation with respect to the fields? For e.g.

    S = \int dt \frac{1}{2} \dot{x}^{2} + \int \int dt \dot{x}(t) \delta^{D}\left(x(t) - y(t')\right) \dot{y}(t')dt' + \int dt' \frac{1}{2}\dot{y}^{2}


    working in, say, D dimensions. I think it will be possible to integrate the delta function out in D=1, but not in higher target space dim. The problem I have is in how to do the variation of the delta function term. Physically it is producing an interaction every time the worldlines of the particles intersect and I've tried writing this as a sum over such points - where [itex]x(t_{0})=y(t')[/itex] - of [itex]\frac{\delta(t - t_{0})}{\dot{x}(t_{0})}[/itex] but this is valid only in D = 1 and I still can't get the variation correct.

    Any help would be appreciated.
    Last edited: Jun 6, 2013
  2. jcsd
  3. Jun 7, 2013 #2
    where are your space integral parts.If you are doing a one dimensional problem then you will have D=1,if you are doing a D dimensional one your eqn will be modified and the effect of delta function is to just make x(t)=y(t') in second integral after space integration but nevertheless what you have written for first integral and third (not to mention second one) just does not qualify.
  4. Jun 8, 2013 #3
    There are no space integrals - it is a (gauge fixed) action for point particles, proportional to the length of the worldline in D dimensional target space plus some interaction terms. The domain is one dimensional, as for the ordinary relativistic point particle.

    Sorry for the confusion
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