Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

The relativistic wave equation is $$\square\varphi=\rho$$ where $\varphi$ is the field, $\rho$ is the source, and $\square$ is the D'Alembert operator, defined by $$\square=\nabla_\mu\nabla^\mu=g^{\mu\nu}\nabla_\mu\nabla_\nu$$ where $g^{\mu\nu}$ is the inverse metric tensor and $\nabla$ is the covariant derivative operator. In Minkowski spacetime, this can be reduced to $$\square=\partial_\mu\partial^\mu=\eta^{\mu\nu}\partial_\mu\partial_\nu$$ where $\partial$ is the partial derivative operator. Using Cartesian coordinates $(t,x,y,z)$, this can be further reduced to $$\square=\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}-\frac{\partial^2}{\partial y^2}-\frac{\partial^2}{\partial z^2}$$ Is this characterization of the wave equation correct? In particular, what is the correct distinction between using $\partial$ and using $\nabla$, and the correct distinction between curved spacetime and curvilinear coordinates, when manipulating these expressions?

share|improve this question
Just a small caveat to add to the existing answers: beware that some authors may write $\square$ in general relativity, when they really mean the Lichnerowicz operator, which does depend on the curvature, not just the connection coefficients. Usually you should be able to tell the difference based on the context. –  JamalS May 17 at 8:02
add comment

2 Answers 2

There isn't any difference here between using curved spacetime and curvilinear coordinates. While one could tell "true" curvature in spacetime by, say, calculating scalars like $R^{\alpha\beta\gamma\delta} R_{\alpha\beta\gamma\delta}$ or $R^\mu{}_\mu$ and seeing if they vanish, the fact is your differential operator doesn't care. Put another way, all you care about is whether or not the connection coefficients $\Gamma^\sigma_{\mu\nu}$ vanish, not whether some special coordinate-independent quantity does.

Because of this, Minkowski spacetime often but not always implicitly involves Cartesian coordinates, otherwise one could probably just call it "flat." If your spacetime is flat but your coordinates are not Cartesian, there will be terms you omitted from your third equation. In any spacetime and for any scalar1 $\varphi$, \begin{align} \square\varphi & = g^{\mu\nu} \nabla_\mu \nabla_\nu \varphi \\ & = g^{\mu\nu} \nabla_\mu (\partial_\nu \varphi) \\ & = g^{\mu\nu} (\partial_\mu \partial_\nu \varphi - \Gamma^\sigma_{\mu\nu} \partial_\sigma \varphi) \\ & = (\partial_\mu \partial^\mu - g^{\mu\nu} \Gamma^\sigma_{\mu\nu} \partial_\sigma) \varphi. \tag{1} \end{align}

So if in coordinates $(t,x,y,z)$ we have $g_{\mu\nu} = \mathrm{diag}(-1,1,1,1)$, then sure, $\Gamma^\sigma_{\mu\nu} = 0$ and $\square\varphi = \partial_\mu \partial^\mu \varphi$. The same flat spacetime in spherical coordinates $(t,r,\theta,\phi)$ -- $g_{\mu\nu} = \mathrm{diag}(-1,1,r^2,r^2\sin^2\theta)$ -- will however have some nonzero connection coefficients. In fact, you should recognize the second term in (1) as being necessary, given that there are single derivatives in the spherical Laplacian: $$ \nabla^2\varphi = \left(\partial_r^2 + \frac{2}{r} \partial_r + \frac{1}{r^2} \partial_\theta^2 + \frac{1}{r^2\tan\theta} \partial_\theta + \frac{1}{r^2\sin^2\theta} \partial_\phi^2\right) \varphi. $$

The thing to remember is covariant derivatives reduce to partial derivatives generally only in the following two circumstances:

  1. $\nabla_\mu \varphi = \partial_\mu \varphi$ in any spacetime in any coordinates for a scalar $\varphi$.
  2. $\nabla_\mu T^\alpha{}_\beta = \partial_\mu T^\alpha{}_\beta$ in flat spacetime in Cartesian coordinates, where $g_{\mu\nu} = \eta_{\mu\nu}$, for any tensor $T$ (with as many indices as you want).

Other than those two times, you have to resort to the general rule $$ \nabla_\mu T^\alpha{}_\beta = \partial_\mu T^\alpha{}_\beta + \Gamma^\alpha_{\mu\sigma} T^\sigma{}_\beta - \Gamma^\rho_{\mu\beta} T^\alpha{}_\rho, $$ where extra indices follow the same pattern as $\alpha$ or $\beta$ depending on if they are up or down.

1 Note that if $\varphi$ is a higher-rank tensor, the simplification $\nabla\varphi \to \partial\varphi$ no longer holds. Also, in the context of wave equations, often the differential operator you want is not the d'Alambertian but rather the de Rham operator, which adds a coupling between the Ricci tensor and your tensor being differentiated (a "curvature coupling"). This reduces to the d'Alambertian on scalars.

share|improve this answer
add comment

You can only promote to partials in Minkowski for the special case of Cartesian coordinates. As far as these simple equations are concerned, there is no direct distinction between curvilinear coordinates and curved spacetime. All that matters is that the $\Gamma_{ab}{}^{c}$ are nonzero.

Note, however, that there are other generalizations of the D'Alembertian operator for curved spacetime that involve the Ricci scalar. These do explicitly tell a difference between curved spacetime and curvilinear coordinates.

share|improve this answer
So the partial differential operator can be properly used in place of the covariant differential operator only in Minkowski space under Cartesian coordinates, correct? Also, what are some examples of these generalizations of the D'Alembertian operator? –  user1667423 May 17 at 5:19
@user1667423 check out en.wikipedia.org/wiki/… –  Danu May 17 at 10:37
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.