Hodge duality and differential forms

  • Thread starter PhyAmateur
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105
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Main Question or Discussion Point

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?
 

Answers and Replies

112
19
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"?
 
105
2
$$\nabla^2$$ is the normal laplacian of (x,y,z). t is chosen like this such that dt is a 1 form.
 
112
19
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.
 
105
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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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