How to compute the wave operator with respect to a metric?

Question

I am reading a paper where the author is working with the following metric $$g = \frac{-dt^2 + dx^2}{t^2}$$ on the manifold $M=[0,\infty)\times \mathbb{R}^3.$ He then goes on to state that $$\Box_g u = (t\partial_t)^2 u - 3 t\partial_t u -t^2 \Delta u.$$ I am not very well versed in geometry, so I am not sure how to derive the above expression. I know that the usual wave operator is $u_{tt}-\Delta u$ but I am guessing that because we are working with the metric $g$ we have some modifications. Any comments on how to derive the above expression will be much appreciated.

Answer 1

Take a look at Definition of Tensor Laplacian; more specifically, the end of the answer where I write the coordinate formula for the Laplacian of a smooth function: \begin{align} \Delta_gu&=\frac{1}{\sqrt{|g|}}\frac{\partial}{\partial x^a}\left(\sqrt{|g|}g^{ab}\frac{\partial u}{\partial x^b}\right)\tag{$*$} \end{align} In that post, I merely stated that the above coordinate formula follows from the abstract definition $\Delta_gu= \text{div}(\text{grad}_gu)$. Recall that the gradient is the vector field associated to $du$ via the metric (below $g^{\sharp}:T^*M\to TM$ is the inverse musical isomorphism): \begin{align} \text{grad}_g(u):=g^{\sharp}(du)=g^{\sharp}\left(\frac{\partial u}{\partial x^a}\,dx^a\right)=\frac{\partial u}{\partial x^a}g^{\sharp}(dx^a)= \frac{\partial u}{\partial x^a}g^{ab}\frac{\partial}{\partial x^b} \end{align} i.e the gradient has components $(\text{grad}_g(u))^b=g^{ab}\frac{\partial u}{\partial x^a}$. Once you have the formula for the components of the gradient, you need to calculate the divergence. For this, you need to know the Voss-Weyl formula, which I prove here. So, this establishes the formula $(*)$.

Now, for your specific case, I assume of course that you mean $dx^2+dy^2+dz^2$ in your definition of the metric. In this case, the components written out as a matrix are: \begin{align} [g_{ab}]= \begin{pmatrix} -\frac{1}{t^2}&0&0&0\\ 0&\frac{1}{t^2}& 0&0\\ 0&0&\frac{1}{t^2}&0\\ 0&0&0&\frac{1}{t^2} \end{pmatrix}\quad\text{and} \quad [g^{ab}]= \begin{pmatrix} -t^2&0&0&0\\ 0&t^2&0&0\\ 0&0&t^2&0\\ 0&0&0&t^2 \end{pmatrix} \end{align} and $\sqrt{|g|}:=\sqrt{|\det [g_{ab}]|}=\frac{1}{t^4}$ (the upstairs indices mean take the inverse matrix). So, carrying out the Einstein summation in $(*)$, we get \begin{align} \Delta_gu&=t^4\frac{\partial}{\partial t}\left(\frac{1}{t^4}\cdot (-t^2)\frac{\partial u}{\partial t}\right)+ \sum_{i=1}^3t^4\frac{\partial}{\partial x^i}\left(\frac{1}{t^4}\cdot t^2\frac{\partial u}{\partial x^i}\right)\\ &=-t^4\left[\frac{1}{t^2}\frac{\partial ^2u}{\partial t^2}+\left(-\frac{2}{t^3}\right)\frac{\partial u}{\partial t}\right]+t^2\sum_{i=1}^3\frac{\partial^2u}{(\partial x^i)^2}\\ &=-t^2\frac{\partial^2u}{\partial t^2}+2t\frac{\partial u}{\partial t}+t^2\sum_{i=1}^3\frac{\partial^2u}{(\partial x^i)^2} \end{align} So, I guess the $3\partial_tu$ in your post should actually be $2\partial_tu$.

What I found is the overall minus of what you wrote, so perhaps the author is defining $\Box_g$ as $-\Delta_g$ (which is weird, because I've always used the box operator to mean the Laplacian given by the metric), or perhaps the overall signature of the metric should have an extra minus sign. Anyway, I leave this minus sign issue to you to figure out based on the conventions.