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 wave equation is often written in the form


involving the Laplace-Beltrami operator $\Delta$. However, the Laplace-Beltrami operator $\Delta$ is defined only in the presence of Riemannian metric $g$. Indeed, one can think of $\Delta u$ as of difference between $u$ and its mean value in the balls, and the notion of ball only makes sense in the presence of a metric.

What is the physical meaning of metric $g$ here? My guess is it's connected with the property of isotropy of space: in the vacuum, the waves should propagate uniformly in all directions, and this situation corresponds to Euclidean metric. The case of general Riemannian (= ‘perturbed Euclidean’) metric should then correspond to the propagation of waves in a medium of some sort, so that the property of ambient space isotropy is not satisfied, hence balls in it do not look like usual ‘Euclidean’ balls, and wave equation takes its general form $(\partial^2_t-\Delta)u=0$.

Is my guess true? If not, what is the correct interpretation of metric in the general wave equation?

share|improve this question
Good question but I think you essentially answered it yourself: the correct geometrical way of thinking about Laplacian is as an averaging operator over a ball given by your metric. But since this is an infinitesimal statement, you can actually regard the deformed ball simply as an ellipsoid with length and orientation of axes determined by the eigenvalue problem of the metric. –  Marek Jul 28 '11 at 19:47
I like this question/thinking. However I'm confused about the second equation. Shouldn't that one be the vector wave equation? –  whoplisp Jul 28 '11 at 21:38
@whoplisp Two equations in the original post are identical. It is an equation for time-dependent scalar field u. What made you think of equation for vector field? –  Akater Jul 28 '11 at 22:43
If you go into the Maxwell equations with anisotropic medium you don't generally end up with a scalar equation involving u but you actually have 6 equations for $E_x$, $E_y$, $E_z$, $B_x$, $B_y$ and $B_z$. Those are coupled in non-homogeneous space. –  whoplisp Jul 29 '11 at 7:28
I was talking about an abstract wave equations, not Maxwell equations. However, this remark is very interesting. I have to to think about it. Do you mean that physically meaningful scalar fields sometimes lose their invariant properties in anisotropic medium, and one is then forced to deal with vector fields? –  Akater Jul 29 '11 at 15:47
show 1 more comment

1 Answer

up vote 0 down vote accepted

The interpretation of the metric as anisotropy may or may not be physical, depending on your point of view. The main problem with this interpretation is that it requires two distinct metrics. One of the lasting influences of general relativity in other fields of physics is the idea of treating objects by thinking about their intrinsic geometry. So when you are dealing with the propagation of the wave in a medium, and you have a Riemannian metric $g$ with respect to which the wave propagates exactly like $\partial_t^2 - \triangle$, you want to use that Riemannian metric $g$ as the preferred one when you are studying the equations. You are of course free to impose a background coordinate system and attach to it a secondary metric, but that secondary metric can be argued to be unphysical in the situation, if it does not actually affect the physics of the situation, but rather only the way your present your measurements.

Example: consider the case where your metric is $4 dx^2 + dy^2 + dz^2$. If you are only given this metric, you would not say that "one unit" in the $x$ direction is the same length as "one unit" in the $y$ direction. In fact, you would be tempted to redefine your coordinate system with $x' = \frac{1}{2}x$ such that the metric "look" more symmetric. In general, given any symmetric, positive definite bilinear form on a vector space, you can always find an orthonormal basis. So if the relevant geometry for the physics is described by a single Riemannian metric, then there is automatically a notion of infinitesimal isotropy built-in to the physics. (This is related to the notion of local Lorentz covariance in relativity, as well as to the notion of frame bundles in Riemannian geometry.)

The important thing to keep in mind is that the spectral theorem for positive definite bilinear forms can be stated in the following way:

Given any finite dimensional real vector space $V$, and any two positive definite symmetric bilinear forms $A$ and $B$ on $V$. We can find a basis $\{ e_1, e_2, e_3, e_4\}$ in which $A$ and $B$ are simultaneously diagonalized.

In other words, if you have a physical problem in which there are two distinct natural Riemannian metrics used in the equations, what you have is that the two parts of the physics corresponding to the use of the two Riemannian metrics are anisotropic relative to each other. Another way of saying the same thing is that the wave propagation in a medium that is governed by $\partial_t^2 - \triangle$ is still uniform in all directions, it is just that now the correct notion of directions should be the one given by the Riemannian metric.

Another example: imagine you are driving around on a plane. Because of friction (and other dissipative forces), to maintain constant velocity you need to keep the accelerator pedal depressed. Now above you there is a pigeon. Because of drag, for it to maintain constant air velocity it also needs to keep putting in energy. Now there is a wind blowing at the height where the pigeon flies. From your point of view, the pigeon's motion is anisotropic: because to maintain the same ground velocity with wind and against wind requires different amount of energy input. But from the pigeon's point of view, your motion is anisotropic: the pigeon's "stationary" coordinate system would be one that is being blown around by the wind!

With that said, from the point of view of studying the wave equation in a medium, the notion of ambient space isotropy is not all that important. So while the interpretation is perfectly valid, it is not something that you necessarily need to keep in mind. When one studies wave propagation in a non-linear medium (keywords: crystal optics, elastic waves), the notion of anisotropy more often refers to the fact that the longitudinal components of a wave and the various polarisations of the traverse components of a wave can have anisotropic propagation behaviour relative to each other. Like Whoplisp wrote in his comments, the general case is governed by a system of equations that cannot be reduced to just the simple scalar wave equation.

share|improve this answer
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.