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Hodge duality and differential forms

  1. Mar 26, 2015 #1
    If we have,$$A=d[(\bar{\alpha}-\alpha)(dt+\lambda)]$$
    where $$\alpha$$ is a complex function and $$\lambda$$ is a 1-form. t here represents the time coordinate.

    If we want to calculate $$d\star A=0$$ where $$\star$$ is hodge star, we get if I did my calculations correctly $$\nabla^2[(\bar{\alpha}-\alpha)(dt+\lambda)]=0$$

    From the product rule, we get $$(\bar{\alpha}-\alpha)\nabla^2(dt+\lambda)+ (dt+\lambda)\nabla^2 (\bar{\alpha}-\alpha) +2 \nabla^2(\bar{\alpha}-\alpha)\nabla^2(dt+\lambda)=0$$
    Does it mean here that each term must equal zero and thus since $$(\bar{\alpha}-\alpha)$$ can not be zero in the first term so $$\nabla^2(dt+\lambda)=0$$ thus killing with it the third term in the equation and only leaving the second term?
    That is simplifying the above equation, we get, $$(dt+\lambda)\nabla^2 (\bar{\alpha}-\alpha)=0$$ thus leaving us with $$\nabla^2 (\bar{\alpha}-\alpha)=0$$

    Can this step be done?
     
  2. jcsd
  3. Mar 26, 2015 #2
    I suppose you're working on a Lorentzian manifold, Minkowski space?
    What is ##\nabla^2##, the Hodge-Laplacian?

    If you're doing physics, why choose a particular "time coordinate"?
     
  4. Mar 26, 2015 #3
    $$\nabla^2$$ is the normal laplacian of (x,y,z). t is chosen like this such that dt is a 1 form.
     
  5. Mar 29, 2015 #4
    Right, so you're working in a Newtonian spacetime. Maybe using the fact that
    $$\Delta = \left( d + d^* \right)^2 = d d^* + d^* d $$
    with ##d^*## being the codifferential will help. Btw, if you're doing Cartan calculus ##\nabla^2## is a bad notation for the Hodge-Laplace / Laplace-Beltrami operator, because it suggests that you're applying the connection twice.
     
  6. Mar 29, 2015 #5
    So yes, considering that I used this notation istead of $\nabla^2$ is my final answer correct or at least is the procedure correct that I considered all terms must be set equal to zero?
     
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