# What are the laplacian operators for the three two dimensional metrics of one variable dependence?

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.

## 1 Answer

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.

• Is this different if I use the Christoffel symbols to calculate the Laplacian? I used $\Gamma_{a a b} = \partial_b g_{aa}$/2? – linuxfreebird Jul 3 '14 at 20:12
• 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. – Matt Jul 4 '14 at 9:16