Laplace-Beltrami operator for a vector field

Question

For a scalar field $\varphi$, the "wave" operator is defined as follows:

$$\Box \varphi \equiv g^{ab}\nabla_a\nabla_b~\varphi = \frac{1}{\sqrt{|g|}}\partial_a\left\{\sqrt{|g|}~g^{ab}~\partial_b~\varphi\right\}~,$$

where $g_{ab}$ represents the (spacetime) metric, $g$ the metric determinant and the indices $a,b = 0, 1, 2, ...., n-1$ in $n$ dimensions. This expression is a way to write down the wave operator without using covariant derivatives.

I want to know how to obtain a similar expression in the case of a vector field $A^a$. For example in Maxwell's equation in curved spacetime, one has the equation $\Box A^a - \nabla^a(\nabla_bA^b) - R^a_bA^b= 0$.

In such a case, is a similar expansion valid? I hesitate to think so, because one then obtains terms like $\partial_mA^a$ which are not tensorial.

Answer 1

The (diffeomorphism) covariant derivative is designed precisely to avoid the non-tensorial terms you are worried about. For a vector field, \begin{align} \nabla_a A^b = \partial_a A^b + A^c \Gamma^a_{cb}. \end{align} Assuming you want a connection with no torsion, there is a unique expression for the Christoffel symbol in terms of the metric. Namely \begin{align} \Gamma^a_{cb} = \frac{1}{2} g^{ad} \left ( \partial_b g_{dc} + \partial_c g_{db} - \partial_d g_{bc} \right ). \end{align} One more formula is needed to compute $\Box A^b \equiv \nabla^a \nabla_a A^b$ because this is now a covariant derivative of a two-index tensor. This is \begin{align} \nabla_c T_a^{\,\,b} = \partial_c T_a^{\,\,b} + \Gamma^b_{cd} T_a^{\,\,d} - \Gamma^d_{ac} T_d^{\,\,b} \end{align} which can be guessed by taking the special case of a tensor formed by multiplying two vectors and applying the Leibniz rule.

With these steps, it is clearly possible to expand Maxwell's equation in terms of partial derivatives and the metric (suitable for entry into a computer) but the form might not be very compact. In the scalar case, most of the simplification comes from the fact that the inner covariant derivative is just an ordinary partial derivative. We therefore have only one Christoffel symbol and the contraction of its indices is what produces $\sqrt{|g|}$ through \begin{equation} \Gamma^a_{ba} = \partial_b \log \sqrt{|g|}. \end{equation} Maybe there's some massaging of this form that also applies in the Maxwell case but I haven't thought about it.