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What are the laplacian operators from the three following two dimensional metrics of one variable dependence :

\begin{align} (A) && d\mathcal{l}^2 = e^{2w(x_2)} \left( dx_1^2 + dx_2^2 \right) && \nabla_1\nabla^1 + \nabla_2\nabla^2 = ? \\ (B) && d\mathcal{l}^2 = \left( e^{4w(x_2)} dx_1^2 + dx_2^2 \right) && \nabla_1\nabla^1 + \nabla_2\nabla^2 = ? \\ (C) && d\mathcal{l}^2 = \left( dx_1^2 + e^{4w(x_2)} dx_2^2 \right) && \nabla_1\nabla^1 + \nabla_2\nabla^2 = ? \end{align}

My answer for all three was \begin{align} \nabla_1\nabla^1 + \nabla_2\nabla^2 = \dfrac{\partial^2}{\partial x_1^2} + e^{-w(x_2)} \dfrac{\partial}{\partial x_2} \left( e^{w(x_2)} \dfrac{\partial}{\partial x_2} \right) , \end{align} however I do not believe my answer is correct.

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I'll do (A). Recall, that in coordinates, we have that $$\Delta_g u=\frac{1}{\sqrt{|g|}}\partial_a\left[g^{ab}\sqrt{|g|}\partial_b[u]\right],$$ where $|g|=\det(g_{ab})$. Note for this case $g_{ab}=e^{2w(x_2)}\delta_{ab}$, where $\delta$ is our usual Euclidean metric (i.e., it's conformally equivalent). Then $g^{ab}=e^{-2w(x_2)}\delta^{ab}$ and $\sqrt{|g|}=e^{2w(x_2)}$ (since $|\delta|=1$, and we're in dimension $2$). Computing, we see that $$\Delta_g u =e^{-2w(x_0)}\partial_a\left[e^{-2w(x_2)}\delta^{ab}e^{2w(x_2)}\partial_b[u]\right]=e^{-2w(x_2)}\delta^{ab}\partial_a\partial_bu=e^{-2w(x_2)}\Delta_\delta u,$$ where $\Delta_\delta$ is our usual Euclidean Laplacian.

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  • $\begingroup$ Is this different if I use the Christoffel symbols to calculate the Laplacian? I used $\Gamma_{a a b} = \partial_b g_{aa}$/2? $\endgroup$ – linuxfreebird Jul 3 '14 at 20:12
  • $\begingroup$ It would turn out to be the same if you use the Christoffel symbols, though it would be a longer calculation. I avoid computing Christoffel symbols at all costs normally, but in this case, they are quite simple. $\endgroup$ – Matt Jul 4 '14 at 9:16

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