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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.

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    $\begingroup$ I think his idea is to see the second derivative as potential droping (or some other phrases, similar to which in fluid dynamics, can't remenber the exact term). So that gives the wave the motivation to propagate. This kind of make sense. Normally the curvature is the boundary condition of the defferential equation. But eventually this is the Laplace equation. So it's like the baisc equation of plane wave. Thinking of that, what's the connection between this and the geodesic equation in general relativity? $\endgroup$ – Turgon Jan 28 '17 at 16:28
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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) .

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  • $\begingroup$ Nice. Are you sure about the constant 4 in the computation of the first integral? Isn't $1/3$ missing (from integrating $x^2$, etc.)? $\endgroup$ – user130529 Feb 22 '17 at 19:13
  • $\begingroup$ For each term, we have $$\int_{-\epsilon}^{\epsilon} \frac{x^2}{2} dxdydz=2\epsilon^2\int_{-\epsilon}^{\epsilon} x^2dx=\frac{4\epsilon^5}{3}$$ Since we have 3 such terms: $4\epsilon^5$ $\endgroup$ – Soner Feb 23 '17 at 16:59
  • $\begingroup$ That's exactly what I thought, but the thing is that you do not need to add them, they appear separately in $(\partial^2x+\partial^2y+\partial^2z)A(0)$. $\endgroup$ – user130529 Feb 23 '17 at 17:25
  • $\begingroup$ You're welcome, I just wanted to make sure your nice answer is correct down to this small detail. $\endgroup$ – user130529 Feb 23 '17 at 21:21
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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: $$ \mathrm{div}\vec{v} = \frac{\partial \vec{v}}{\partial x}+\frac{\partial \vec{v}}{\partial y }+\frac{\partial \vec{v}}{\partial z} $$

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  • $\begingroup$ Thank you for your reply! However I am slightly confused here as I thought that the divergence was a scalar. you cannot simple add scalar forces in different directions. Or should the 'u' in my equation above actually be a vector? Generally, I thought that wavefunctions were just functions and not vectors... $\endgroup$ – Meep Feb 22 '17 at 16:10
  • $\begingroup$ The divergence of a vector field is indeed a scalar. And yes, wavefunctions are scalar functions. Don't confuse the scalar field $u$ with vector field $\vec{v}$. As far as quantum mechanics, you can look at $\psi(\vec{X})$ as a scalar potential field. In fact its gradient, $\nabla \psi$, which is a vector field, is proportional to quantum-mechanic velocity. $\endgroup$ – AndreaPaco Feb 22 '17 at 16:29
  • $\begingroup$ Thank you for your reply. That is one of my difficulties though. If these curvatures, which are proportional to forces in the orthogonal directions, are scalars, then we cannot simply add them? Perhaps my lecturer was meaning something else. Or I suppose that even if the sum of the curvatures doesn't really have any particular meaning, it is true that the greater the curvatures the greater the force will be... $\endgroup$ – Meep Feb 22 '17 at 17:14
  • $\begingroup$ Curvatures are scalar quantities and, as such, you can simply add them (Look at the structure of the Laplace operator). Yes, the idea is that the bigger the (total) curvature, the bigger is this sort of generalized force. $\endgroup$ – AndreaPaco Feb 22 '17 at 18:01
  • $\begingroup$ @21joanna12, is everything clear? $\endgroup$ – AndreaPaco Mar 2 '17 at 8:24

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