Laplace operator and tensor calculus:

Question

I'm studying Tensor calculus and I found this interesting problem:

Show that: $$ \Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)$$

Here's some attempts, hope it helps, even I find them useless!

Well, we know that: $$\Delta F=\nabla\cdot \nabla F $$ And : $$\nabla \cdot \mathbf{V}=\nabla_iv ^i$$Using it : $$\Delta F=\nabla_i (g^{ik}\partial_kF)$$

That's the only advance I've made till now, I'm thinking about a property but I'm not that much certain about its validity here.

$$\Delta F=g^{ik}\nabla_i(\partial_k F)$$

Being true or false I think it's not useful to derive this formula.

Answer 1

Well, the Laplace operator is a composite operator:

$$ \Delta F = div\ grad\ F = \nabla\cdot\nabla F $$

and as you wrote

$$ (grad\ F)^r = (\nabla F)^r = \frac{\partial F}{\partial x^k}\,g^{rk} = V^r $$

You obtain the divergence by contraction of the derivation operator $\nabla$ and we emphasize that the contraction has to be performed on the covariant derivative:

$$ div\ \boldsymbol{V} = \nabla_iV^i = V^i_{\phantom{i};\,i}= \frac{\partial V^i}{\partial x^i} + V^r\; \Gamma^i_{ir} $$

By use of a property of the levi-Civita connection coefficients

$$ \Gamma^i_{ki} = \frac{1}{2} g^{ij} \frac{\partial g_{ij}}{\partial x^k} = \frac{1}{2g} \frac{\partial g}{\partial x^k} = \frac{\partial \,log \sqrt{|g|}}{\partial x^k} $$

you can write further

$$ div\ \boldsymbol{V} = \nabla_iV^i = V^i_{\phantom{i};\,i}= \frac{\partial V^i}{\partial x^i} + V^r\; \Gamma^i_{ir} = \frac{\partial V^r}{\partial x^r} + V^r\; \frac{\partial \,log \sqrt{|g|}}{\partial x^r} = \frac{1}{\sqrt{|g|}}\; \frac{\partial}{\partial x^r} (\sqrt{|g|}\; V^r) $$

Finally, substituting $V^r$ gives the desired result:

$$ \Delta F = div\ grad\ F = \frac{1}{\sqrt{|g|}}\; \frac{\partial}{\partial x^r} (\sqrt{|g|}\; \frac{\partial F}{\partial x^k}\,g^{rk} ) $$

Answer 2

Not from the first principles, but based on physical intuition proof looks as follows. Consider the action for scalar field: $$ S = \int d^D x \ \sqrt{g} g^{\mu \nu} \partial_\mu \phi \partial_\nu \phi $$ This is the only viable GR covariant expression for the action of scalar field without free indices, and $d^D x \sqrt{g}$ is an invariant volume element. Integrating this expression by parts, one gets: $$ S = -\int d^D x \ \phi \partial_\mu (\sqrt{g} g^{\mu \nu} \partial_\nu \phi) = - \int d^D x \ \sqrt{g} \phi \frac{1}{\sqrt{g}}\partial_\mu (\sqrt{g} g^{\mu \nu} \partial_\nu \phi) = -\int d^D x \ \phi \Delta \phi $$ Where we have assumed that boundary terms vanis, and recovered in the last equality the invariant volume element.

Answer 3

Here's a quick derivation of this problem:

As you said: $$\Delta F= \nabla\ .\nabla F$$ And always using your steps: $$\nabla \ .\ F=\nabla_iv^i$$ And for those who don't know why he involved the "($g^{ik}\partial_k F$)" well that's the contravariant components of the gradient operator. \begin{align} \Delta F&= \nabla_iv^i\\ &=\nabla_i\left(g^{ik}\partial_kF\right)\\ &=g^{ik}\nabla_i\left(\partial_k F\right) \end{align}

Recall: $$\nabla_i(\partial_k F)=\partial_{ik}F-\Gamma_{ik}^l\partial_lF$$

Hence: $$\Delta F=g^{ik}(\partial_{ik}F-\Gamma_{ik}^l\partial_lF)$$

Another recall: :) $$\nabla\ .\ \mathbf{V}=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(v^i \sqrt{\vert g\vert}\right)\quad{(1)}$$

Involving the contravariant components of $\mathbf{grad}F$ in $(1)$ we got the following: $$\bbox[silver,5px,border:2px solid teal] {\Delta F=\frac{1}{\sqrt{\vert g\vert}}\partial_i\left(\sqrt{\vert g\vert} g^{ik}\partial_kF\right)}$$ and that's true, because when $g^{ik}=\delta^{ik}$ we get the classic expression of the Laplacien operator: $$\Delta \ F=\partial_{kk}F.$$