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Lagrange multipliers and gauge conditions

  1. May 26, 2007 #1
    Hello, I was hoping someone would be able to clarify a problem Ive got. A lagrange multiplier can be introduced into an action to impose a constraint right?

    I was wondering what relation lagrange multipliers have to gauge conditions, which are imposed by hand. Am I correct in saying that lagrange multipliers are fields, and hence must be solved for in the resulting field equations arising from an action? They are dynamical?

    My question is in regards to the Einstein aether model, in which a vector field u_{a} is introduced into the gravitational action. The term L*(u_{a}u^{a} + 1) is added to the action to make u_{a} a unit timelike vector (L is the lagrange multiplier). But what is the difference between introducing this term into the action and leaving it out, finding the field equations, and then imposing u_{a}u^{a}=-1 by hand?

    Thank you for any help,

    Ste
     
  2. jcsd
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