Understanding the wave equation in 3dimensions

Question

I am learning about waves and the wave equation in lectures, and there was something interesting my lecturer said which I have not been able to find about online or in a book.

With regards to the three dimensional wave equation $$ \frac{\partial^2 u}{\partial t^2} = c^2 \nabla^2 u $$

he said that you add the second derivatives because what drives the motion of the wave is the curvature which gives rise to a force. As these forces are orthogonal, we can add them. Thus the sum of the curvatures in each direction determines the motion.

Could anyone explain this or elaborate?

Thank you.

Answer 1

This may not be a direct answer to your question but it may give you an insight about Laplacian operator.

Now assume that we have a field $A(x,y,z)$. Let us expand this, for simplicity, around the origin. $$A(x,y,z)=A(0)+(x\partial_x+y\partial_y+z\partial_z)A(0)+(xy\partial_x\partial_y+xz\partial_x\partial_z+yz\partial_y\partial_z)A(0)+\frac{1}{2}(x^2\partial_x^2+y^2\partial_y^2+z^2\partial^2_z)A(0)+\text{Higher Order Terms}$$ where $A(0):=A(0,0,0)$. If we now take a volume integral over a cube around the origin, say $-\epsilon<x,y,z<\epsilon$, we get

$$\int A(x,y,z)dV=A(0)V+0+0+ \frac{4}{3}\epsilon^5(\partial_x^2+\partial_y^2+\partial_z^2)A(0)+\text{Higher Order Terms}$$ where $V:=(2\epsilon)^3$. We can define the average value of $A$ around the origin as $$A_{ave}(0):=\frac{1}{V}\int A(x,y,z) dV$$hence our equation becomes $$\Delta A(0)=\frac{6}{\epsilon^2}\left(A_{ave}(0)-A(0)-\text{Higher Order Terms} \right)$$

We can kill the Higher Order Terms by taking $\epsilon\rightarrow 0$, hence our equation becomes $$\Delta A(0)=\lim_{\epsilon\rightarrow0}\frac{6}{\epsilon^2}\left(A_{ave}(0)-A(0)\right)$$

Therefore, Laplacian of a function at one point gives the difference between the functions value at that point and the average of the functions values in the infinitesimal neighborhood. Above, we used Cartesian coordinates for simplicity, hence the ugly $\frac{6}{\epsilon^2}$ in the front, but the result should be much more elegant in the spherical coordinates!

Back to your question: Since the difference of the average of surrounding and the point itself is actually related to the curvature (every point is average of its surroundings in a flat space), the RHS of the wave equation is indeed curvature induced force, and wave equation simply relates the change of the field due to the curvature (or more simply, it relates how a field changes in time as its values are not properly distributed in space, meaning that values at points being not equal to averages of their surroundings) .

Answer 2

Laplace operator often shows up in equations that describe the flow of something: Fourier equation for the heat, Schrodinger equation for probability density of quantum particles, Lapalce's equation for incompressible liquids, d'Alambert equation for electromagnetic waves, etc.

The best example to understand this concept is hydrodynamics. It is possible to define the velocity field of the fluid we are dealing with. Let's call it $\vec{v}$. Obviously, this is the flow of our fluid. Now let's notice that this velocity field can be written as the gradient of a scalar field, which is called potential flow: $$ \vec{v}=\nabla \varphi $$ So, it's very easy to observe that the Laplacian of the scalar potential field is nothing more than the divergence of the flow: $$ \nabla\cdot\vec{v} =\mathrm{div}\vec{v}= \Delta\varphi $$ In other words: the laplacian of a field can be seen as the divergence (i.e. the source or the waste) of a flow. Concerning our hydrodynamic example, if a fluid is incompressible, the velocity field must be divergence-free in all points, so one naturally finds Laplace equation: $$ \Delta\varphi=0 $$
In a compressible fluid, the divergence of the velocity field can be non-zero, and one can write the famous wave eqution: $$ \Delta \varphi= \frac{1}{c^2}\frac{\partial^2 \varphi}{\partial t^2} $$ In other words, if the Laplacian of the potential field (which is equivalent to the divergence of the associated flow) is non-zero in a certain point, then there exist an "acceleration" of the potential field. It's somehow similar to Newton's second equation: $$ \vec{F}=m\ddot{\vec{x}} $$ In this sense the Laplacian can be seen as a "driving force" and the propagation speed can be seen as an inverse mass. Notice that in classical mechanics the force is given by the (opposite of) gradient of a potential, while here the "force" is given by the curvature of the potential.

Mathematically a Laplacian measures the curvature of a 3-variables function: $$ \Delta = \frac{\partial^2}{\partial x^2}+\frac{\partial^2}{\partial y^2}+\frac{\partial^2}{\partial z^2} $$ exactly as $\frac{\partial^2}{\partial x^2}$ measures the curvature of a one-variable function. Basically you have 3 independent curvatures, one for each direction, but this is not a surprise: the divergence of the associated flow, $\vec{v}$ is made up of 3 addends, one for each direction: $$ \mathrm{div}\vec{v} = \frac{\partial \vec{v}}{\partial x}+\frac{\partial \vec{v}}{\partial y }+\frac{\partial \vec{v}}{\partial z} $$