Are the space and time parts of a standing wave in 3D separable?

[PLAGUE]

TL;DR

How can we show that for stationary waves, we can separate the space as time part of a 3D wave equation?

I quote from my book:

In the case of the equation for the one-dimensional stationary wave (Eq. 4-51), in which the magnitude u of the phase velocity was constant and in which solutions were restricted through boundary conditions to a set of privileged constant values for v, we were able to separate the complete displacement function into the product of a space-dependent part and a time-dependent part. It may be shown that when u and v are constant, we may make a similar separation in the equation for displacement of the three-dimen¬ sional stationary wave.

But it doesn't show any proof that for stationary waves, we can separate the space as time part of a 3D wave equation. How can we show that for stationary waves, we can separate the space as time part of a 3D wave equation?

 

[anuttarasammyak]

I don't see the distinction between separation of variables,
$$\Psi(\mathbf{r},t)=R(\mathbf{r})T(t)$$
with trigonometric function T and a standing or stationary wave.

 

[Dale]

TL;DR: How can we show that for stationary waves, we can separate the space as time part of a 3D wave equation?

I quote from my book:


But it doesn't show any proof that for stationary waves, we can separate the space as time part of a 3D wave equation. How can we show that for stationary waves, we can separate the space as time part of a 3D wave equation?

What is the equation for a 3D stationary wave?

 

[Roberto Pavani]

The separation is not a consequence of the wave being stationary.
Rather, stationary modes arise from separable solutions of the 3D wave equation.

Starting from

$\nabla^2 \Psi - \frac{1}{u^2}\frac{\partial^2 \Psi}{\partial t^2}=0$,

assume

$\Psi(\mathbf r,t)=X(\mathbf r)T(t)$.

Substituting and dividing by XT gives

$\frac{\nabla^2 X}{X} = \frac{1}{u^2}\frac{T''}{T} = -k^2 .$

Since the left-hand side depends only on space and the right-hand side only on time, both must be equal to a constant.

This yields

$\nabla^2 X + k^2 X = 0$,

which is the Helmholtz equation, and

$T'' + \omega^2 T = 0$,

with

$\omega = uk$.

Hence the time dependence is

$T(t)=A\cos(\omega t)+B\sin(\omega t)$,

and the full solution is

$\Psi(\mathbf r,t)=X(\mathbf r)\cos(\omega t+\phi)$.

With appropriate boundary conditions, these separable solutions correspond to normal modes (standing waves).

In spherical coordinates the spatial Helmholtz equation can be further separated as

$X(r,\theta,\phi)=R(r)Y_{\ell m}(\theta,\phi)$,

where

$Y_{\ell m}$

are the spherical harmonics and the radial equation has solutions given by the spherical Bessel functions

$j_\ell(kr)$

(and, in general, also $y_\ell(kr)$).

Thus, the 1D equation

$X''+k^2X=0$

with sine/cosine solutions generalizes in 3D to Helmholtz solutions of the form

$X(r,\theta,\phi)=j_\ell(kr)Y_{\ell m}(\theta,\phi)$.

The spherical harmonics and Bessel functions are not alternative solutions: they are the angular and radial parts of the same separation of variables.


Of course, this is not the most general 3D standing-wave solution. I used spherical coordinates only to illustrate how the familiar 1D Helmholtz equation

$X''+k^2X=0$

generalizes to three dimensions. In spherical geometry, the separated solutions naturally split into an angular part (spherical harmonics) and a radial part (spherical Bessel functions).

 

[PLAGUE]

The separation is not a consequence of the wave being stationary.
Rather, stationary modes arise from separable solutions of the 3D wave equation.

Starting from

$\nabla^2 \Psi - \frac{1}{u^2}\frac{\partial^2 \Psi}{\partial t^2}=0$,

assume

$\Psi(\mathbf r,t)=X(\mathbf r)T(t)$.

Substituting and dividing by XT gives

$\frac{\nabla^2 X}{X} = \frac{1}{u^2}\frac{T''}{T} = -k^2 .$

Since the left-hand side depends only on space and the right-hand side only on time, both must be equal to a constant.

This yields

$\nabla^2 X + k^2 X = 0$,

which is the Helmholtz equation, and

$T'' + \omega^2 T = 0$,

with

$\omega = uk$.

Hence the time dependence is

$T(t)=A\cos(\omega t)+B\sin(\omega t)$,

and the full solution is

$\Psi(\mathbf r,t)=X(\mathbf r)\cos(\omega t+\phi)$.

With appropriate boundary conditions, these separable solutions correspond to normal modes (standing waves).

In spherical coordinates the spatial Helmholtz equation can be further separated as

$X(r,\theta,\phi)=R(r)Y_{\ell m}(\theta,\phi)$,

where

$Y_{\ell m}$

are the spherical harmonics and the radial equation has solutions given by the spherical Bessel functions

$j_\ell(kr)$

(and, in general, also $y_\ell(kr)$).

Thus, the 1D equation

$X''+k^2X=0$

with sine/cosine solutions generalizes in 3D to Helmholtz solutions of the form

$X(r,\theta,\phi)=j_\ell(kr)Y_{\ell m}(\theta,\phi)$.

The spherical harmonics and Bessel functions are not alternative solutions: they are the angular and radial parts of the same separation of variables.


Of course, this is not the most general 3D standing-wave solution. I used spherical coordinates only to illustrate how the familiar 1D Helmholtz equation

$X''+k^2X=0$

generalizes to three dimensions. In spherical geometry, the separated solutions naturally split into an angular part (spherical harmonics) and a radial part (spherical Bessel functions).

Is this a sufficient condition that "stationary modes arise from separable solutions"?

 

[Roberto Pavani]

As far as I understand it, not by itself.

Separation of variables shows that solutions of the form

$\Psi(\mathbf r,t)=X(\mathbf r)T(t)$

can exist. However, standing modes are usually obtained when appropriate boundary conditions select a discrete set of spatial eigenfunctions and frequencies.

So I would say that, with suitable boundary conditions, stationary (standing) modes arise from separable solutions of the wave equation.

 

[AlexB23]

TL;DR: How can we show that for stationary waves, we can separate the space as time part of a 3D wave equation?

I quote from my book:


But it doesn't show any proof that for stationary waves, we can separate the space as time part of a 3D wave equation. How can we show that for stationary waves, we can separate the space as time part of a 3D wave equation?

Which physics textbook are you using? Maybe sharing the name of the textbook so other people can read the chapter, assuming they legally own a copy?

 

[Andy Resnick]

But it doesn't show any proof that for stationary waves, we can separate the space as time part of a 3D wave equation. How can we show that for stationary waves, we can separate the space as time part of a 3D wave equation?

I wonder if the more interesting version of your question is, "are there non-separable solutions to the 3D wave equation?"

Standing waves typically are separable in space and time. However, there does appear to be a case (irregular boundary conditions), when the solution cannot be:

https://www.sciencedirect.com/science/article/abs/pii/S0022460X85800018