I Justification of Superposition of Waves with Different Speeds

[greypilgrim]

TL;DR Summary

Why can waves travelling at different speeds be superposed, as they are solutions to two different wave equations?

Hi.
To my understanding, the mathematical justification of various superposition principles in physics is that the sum of solutions to a linear differential equations is again a solution to the same equation.

This justification works for solutions of the wave equation
$$\left(\frac{1}{c^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}\right)u=0$$
with different initial conditions, i.e. amplitude and phase, so I understand how it explains interference of e.g. monochromatic light. It even works for waves of different frequency and wavelength, as long as their product $\lambda f=c$ is the same. They may even travel in opposite directions since $\left(-c\right)^2=c^2$ and only this shows up in the wave equation.

However, it fails for waves travelling at two different speeds, as they are solutions to two different wave equations. But waves are superposed all the time! Is it just an assumption that this can be done, or is there another justification for this? Maybe at the level of the single oscillators?

 

[Dale]

the sum of solutions to a linear differential equations is again a solution to the same equation.
…
it fails for waves travelling at two different speeds, as they are solutions to two different wave equations.

Can you think of a case where waves going at different speeds are solutions to a linear differential equation?

 

[Dale]

I guess there could be a material that is linear but anisotropic. Then waves with different polarizations would have different speeds. They would superpose just fine.

 

[greypilgrim]

I guess there could be a material that is linear but anisotropic. Then waves with different polarizations would have different speeds. They would superpose just fine.

Or just a dispersive medium in which waves with different frequencies travel at different speeds?

But my question still remains, i.e. why they just superpose linearly. To my understanding, the linearity of the wave equation cannot be used as justification here.

I guess one could justify it at the level of single oscillators, assuming them to behave according to Hooke's law $F=-D\cdot x$ and using the linearity here.

 

[Dale]

Or just a dispersive medium in which waves with different frequencies travel at different speeds?

Isn’t that non-linear?

But my question still remains, i.e. why they just superpose linearly. To my understanding, the linearity of the wave equation cannot be used as justification here.

I think the linearity can be used. Why do you think it cannot?

 

[greypilgrim]

Yes, that's my point. To my knowledge, the superposition of two solutions to a non-linear differential equation is generally not a solution to this equation.

Take the broadening of a pulse in a dispersive medium. Quantitatively, this is treated by performing Fourier analysis on the pulse, making all components propagate at their respective frequency-dependent speed and then performing an inverse Fourier transform.

But the application of Fourier transform and its inverse already assumes superposition to work in this situation, as it is based on decomposing the pulse into a sum (integral) of its sinusoidal components. How is this justified, given that there is no single linear differential equation that is solved by all those sinusoidal components?

 

[Dale]

To my knowledge, the superposition of two solutions to a non-linear differential equation is generally not a solution to this equation.

I agree. For non-linear differential equations you don't get superposition. I don't think that anyone claims otherwise.

I guess that I am confused about your point. In the OP you say

it fails for waves travelling at two different speeds, as they are solutions to two different wave equations. But waves are superposed all the time! Is it just an assumption that this can be done

I don't think that anyone actually claims this. I think that people recognize that in non-linear mediums you cannot use superposition. Do you have an example of where people are actually claiming superposition in such a non-linear scenario?

 

[greypilgrim]

I think I gave one?

An inverse Fourier transform is a sum (if the signal is periodic) or an integral over all frequency components, hence a superposition.

 

[Dale]

An inverse Fourier transform is a sum (if the signal is periodic) or an integral over all frequency components, hence a superposition

No. You can take a Fourier transform of any well-behaved signal. That does not in any way imply that the various frequency components are individually solutions to the same differential equation. Fourier transform is not at all the same as superposition.

 

[greypilgrim]

That does not in any way imply that the various frequency components are individually solutions to the same differential equation.

I'm not claiming that.

Fourier transform is not at all the same as superposition.

Hm. Maybe my usage of the word "superposition" is inadequate.

Let me try to avoid this: Two individual waves
$$
\begin{align*}
f\left(t,x\right)&=A_1 \sin\left(\omega_1 t-k_1 x\right) \\
g\left(t,x\right)&=A_2 \sin\left(\omega_2 t-k_2 x\right)
\end{align*}
$$
with $\frac{\omega_1}{k_1}\neq\frac{\omega_2}{k_2}$ travel through a medium because they each satisfy a wave equation that can be derived from the properties of the material (usually assumed to consist of many weakly coupled harmonic oscillators), that also fix the wave speed for that specific frequency.

Now assume we freeze time at $t=0$ and place all the oscillators at positions given by
$$f\left(0,x\right)+g\left(0,x\right)=A_1 \sin\left(-k_1 x\right)+A_2 \sin\left(-k_2 x\right)\enspace.$$
Now this is what I don't understand: To predict a later time $t>0$, what we would do is calculate $f\left(t,x\right)$ and $g\left(t,x\right)$ and sum them up. Since dispersion is nonlinear, why does it still behave so nicely that we can linearly combine the individual behaviour of $f\left(t,x\right)$ and $g\left(t,x\right)$?
Why don't we have to cumbersomely formulate and solve a new, probably very complicated, differential equation for this situation?
I just fail to see where and why the linearity comes in.

Also, isn't the whole field of nonlinear optics just about this not being true for high intensities? So why is this linear combination possible at low intensities if superposition does not apply due to not being solutions to the same LDE?

 

[Dale]

I think what you are describing is orthogonality rather than linearity. But I am not certain about that. I will have to think a bit more on it.

 

[Dale]

I just fail to see where and why the linearity comes in.

I think that the linearity you are describing comes in because you are specifically considering a medium that exhibits frequency dispersion.

That is the usual meaning of "dispersion" but there are also waves that exhibit amplitude dispersion. In a medium with amplitude dispersion you wouldn't get the linearity you are describing.

A medium with frequency dispersion is specifically one in which the summation you are describing works.

 

[greypilgrim]

I think that the linearity you are describing comes in because you are specifically considering a medium that exhibits frequency dispersion.

But can it be derived, or is it rather an assumption or part of the definition of "frequency dispersion"?

I think there is also a problem in the microscopic point of view. The usual derivation of the wave equation models the medium as a collection of mass points that interact with their neighbours via a linear force law with some coupling constant. The wave speed then drops out as function of the mass (or density) and that coupling constant. Since the mass of the oscillators is the same for both frequencies, the difference in wave speed must be due to different coupling constants.
Now the linearity says that both the displacements $f\left(t,x\right)+g\left(t,x\right)$ and accelerations $\partial_t^2 f\left(t,x\right)+\partial_t^2 g\left(t,x\right)$ of the oscillator at any position $x$ should add up. What kind of force law could possibly satisfy this?
Granted, this might be for mechanical waves, but I believe they exhibit dispersion as well.

 

[snorkack]

I think that the linearity you are describing comes in because you are specifically considering a medium that exhibits frequency dispersion.

That is the usual meaning of "dispersion" but there are also waves that exhibit amplitude dispersion. In a medium with amplitude dispersion you wouldn't get the linearity you are describing.

A medium with frequency dispersion is specifically one in which the summation you are describing works.

A simple example of frequency vs. amplitude dispersion is water waves.
Deep water waves (water is deep compared to wavelength) have frequency dispersion.
Waves have amplitude dispersion when amplitude becomes appreciable relative to depth.

 

[greypilgrim]

So as far as I understand, frequency dispersion simply means that the wave speed is a function of frequency, $c\left(f\right)$ with no other constraints, so just as "dispersion" has been used from the start of this thread.

So is the summation just assumed to work? But what about the problems in the microscopic picture in #13?

 

[hutchphd]

May I suggest study of the Korteweg-deVries equation... https://en.wikipedia.org/wiki/Inverse_scattering_transform
This is really good stuff but far from easy IMHO

 

[hutchphd]

TL;DR Summary: Why can waves travelling at different speeds be superposed, as they are solutions to two different wave equations?

To my understanding, the mathematical justification of various superposition principles in physics is that the sum of solutions to a linear differential equations is again a solution to the same equation.

So is the summation just assumed to work? But what about the problems in the microscopic picture in #13?

Why would it not work? You are tacitly assuming that the "dispersion" is associated with the medium. The "dispersion", for much of physics (i.e the Schrodinger Equation separation constants) is required to ensure the proposed separation of t and x yields a linear solution. The Fourier transform is agnostic on this question and remains valid.
Only after we fashion particular solutions do we assign phase and group velocity notions to $\frac {\omega} k$ and $\frac {d\omega} {dk}$
If you think these are not solutions, the way to test is by substitution (the one thing I learned in high school calculus.....)

 

[greypilgrim]

You are tacitly assuming that the "dispersion" is associated with the medium. The "dispersion", for much of physics (i.e the Schrodinger Equation separation constants) is required to ensure the proposed separation of t and x yields a linear solution.

I don't know much about how "dispersion" is used elsewhere, I used as it is in optics:
Dispersion is the phenomenon in which the phase velocity of a wave depends on its frequency. (Wikipedia)
But you might be right that I think about those things the wrong way around, I'll have to think about that.

What I still can't get my head around though is that microscopic picture. Let's assume mechanical waves and a medium for which the wave equations
$$\left(\frac{1}{c_1^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}\right)f_1=0$$
with solution $f_1\left(t,x\right)=A_1 \sin\left(\omega_1 t-k_1 x+\varphi_1\right)$, and
$$\left(\frac{1}{c_2^2}\frac{\partial^2}{\partial t^2}-\frac{\partial^2}{\partial x^2}\right)f_2=0$$
with solution $f_2\left(t,x\right)=A_2 \sin\left(\omega_2 t-k_2 x+\varphi_2\right)$ hold and $c_1\neq c_2$ .

The wave equations are derived assuming a linear force law between the oscillators at position $x$ and its nearest neighbours at $x\pm\Delta x$:
$$
\begin{align*}
F_1\left(t,x\right)&=-D_1\cdot\left[f_1\left(t,x\right)-f_1\left(t,x-\Delta x\right)\right]+D_1\cdot\left[f_1\left(t,x+\Delta x\right)-f_1\left(t,x\right)\right]\approx D_1\cdot\Delta x^2\cdot\partial_x^2 f_1\left(t,x\right) \\
F_2\left(t,x\right)&=-D_2\cdot\left[f_2\left(t,x\right)-f_2\left(t,x-\Delta x\right)\right]+D_2\cdot\left[f_2\left(t,x+\Delta x\right)-f_2\left(t,x\right)\right]\approx D_2\cdot\Delta x^2\cdot\partial_x^2 f_2\left(t,x\right)
\end{align*}
$$
Since both the mass of the oscillators (or density) and the constants $D_i$ go into $c_i$ and the mass is the same in both cases, we must have $D_1\neq D_2$ .

But $g\left(t,x\right)=f_1\left(t,x\right)+f_2\left(t,x\right)$ implies $\partial_t^2 g\left(t,x\right)=\partial_t^2 f_1\left(t,x\right)+\partial_t^2 f_2\left(t,x\right)$ and by 2^nd Newton that $$F_\text{tot}\left(t,x\right)=F_1\left(t,x\right)+F_2\left(t,x\right)=D_1\cdot\Delta x^2\cdot\partial_x^2 f_1\left(t,x\right)+D_2\cdot\Delta x^2\cdot\partial_x^2 f_2\left(t,x\right)$$
which cannot be factorised and written in terms of $\partial_x^2 g\left(t,x\right)=\partial_x^2 f_1\left(t,x\right)+\partial_x^2 f_2\left(t,x\right)$ . Hence the force law governing $g\left(t,x\right)$ cannot be formulated in terms of $g$ or its derivatives.

Now let's start from the other end: Take any random initial conditions of the positions of the oscillators $h\left(0,x\right)$ and their velocities $\partial_t h\left(0,x\right)$ except sine-like. It is reasonable to assume that the force law $F\left(t,x\right)$ only depends on $h$ and its derivatives (note that I'm not imposing linearity or anything else here).
But above argument then proves that there cannot be a set of sine-like waves and constants $\left\{\left(f_i,D_i\right)\right\}_i$ such that both
$$
\begin{align*}
h\left(t,x\right)&=\sum_{i=1} ^\infty f_i \left(t,x\right) \\
F\left(t,x\right)&=\sum_{i=1} ^\infty D_i\cdot\Delta x^2\cdot\partial_x^2 f_i\left(t,x\right)
\end{align*}
$$
Hence the decomposition into sine-like functions that each obey a wave equation cannot work!

 

[hutchphd]

Hence the decomposition into sine-like functions that each obey a wave equation cannot work!

This exposition badly and cavalierly interchanges differential (continuous) and finite difference (discrete) operations. Rather than pointing out each wrong step (a tedious process at best) I suggest that you need to either

1. choose to look at only the continuous medium or
2. Look at a solid state book (maybe Ashcroft and Mermin) for phonon dynamics and Bloch waves in periodic solids

They are not interchangeable and your confusion is not trivial. You need to learn the details without incorrect assumptions.

 

[jasonRF]

I have been away for awhile so missed this thread. The OP is missing the fact that linear dispersive waves have 'wave equations' that look different than the one they are used to. For example, the electric field of a simple high-frequency transverse electromagnetic wave in a plasma (the ordinary mode if it is a magnetized plasma) satisfies the equation
$c^2 \, \frac{\partial^2}{\partial z^2} E(z,t) = \frac{\partial^2}{\partial t^2} E(z,t) + \omega_p^2 \, E(z,t)$
where $c$ is the speed of light and $\omega_p$ is the plasma frequency (proportional to the square root of the electron density). Here we are assuming that the frequency is high enough that the ion dynamics can be ignored, and that there are no collisions. (Edit: we are also assuming thermal effects are irrelevant, and that the wave amplitude is small so that we can linearize the fluid equations. )


You can take Fourier transforms in both time and space
$\hat{E}(k,\omega) = \int dz\, e^{i k z} \int dt\, e^{-i\omega t} E(z,t)$

yielding

$\left[ \omega^2 - \omega_p^2- c^2 k^2\right] \hat{E}(k,\omega) = 0$

Setting the expression in the brackets equal to zero yields the dispersion relation,
$c^2 k^2 = \omega^2 - \omega_p^2$
The phave velocity is then
$v_{phase} \equiv \frac{\omega}{k} = \pm c \, \sqrt{\frac{\omega^2}{\omega^2-\omega_p^2}}$
which is of course frequency dependent. The fact that it is imaginary for $\omega<\omega_p$ tells us that the waves don't propagate below the plasma frequency. Likewise, the group velocity is found to be
$v_{group} \equiv \frac{\partial \omega}{\partial k} =\frac{c^2}{v_{phase}}$
which is also frequency dependent.

This is an example where you can have linear superposition of waves traveling different speeds. You can either use a few discrete sine waves, or you can use full-up Fourier analysis. In the latter case, we start again with
$\left[ \omega^2 - \omega_p^2- c^2 k^2\right] \hat{E}(k,\omega) = 0$
or
$\left[\omega - \sqrt{\omega_p^2+c^2 k^2}\right] \left[\omega + \sqrt{\omega_p^2+c^2 k^2}\right] \hat{E}(k,\omega) = 0$
This has solutions
$\hat{E}(k,\omega) = A_1(k)\, \delta\left(\omega - \sqrt{\omega_p^2+c^2 k^2}\right) + A_2(k)\, \delta\left(\omega + \sqrt{\omega_p^2+c^2 k^2} \right)$
Mathematically, the delta functions arise because if $f$ is a generalized function that satisfies $x f(x) = 0$ then it must be true that $f = A \, \delta(x)$. Physically, they arise because waves must satisfy the dispersion relation. To simplify things, let $\Omega_1(k) = \sqrt{\omega_p^2+c^2 k^2}$ and $\Omega_2(k) = -\sqrt{\omega_p^2+c^2 k^2}$. Then we have
$\hat{E}(k,\omega) = \sum_{\alpha=1}^2 A_\alpha(k)\, \delta\left(\omega - \Omega_\alpha(k) \right)$
In the subsequent equations we will leave out the limits on teh summation for simplicity.

Now we do the inverse transform
$\begin{eqnarray*} E(z,t) & = & \frac{1}{4\pi^2} \int dk\, e^{-i k z} \int d\omega \, e^{i\omega t} \, \hat{E}(k,\omega) \\ & = & \frac{1}{4\pi^2} \int dk\, e^{-i k z} \int d\omega \, e^{i\omega t} \, \sum_\alpha \, A_\alpha(k)\, \delta\left(\omega - \Omega_\alpha(k) \right) \\ & = & \frac{1}{4\pi^2} \, \sum_\alpha \, \int dk \, A_\alpha(k) \, e^{-i\left[ k z - \Omega_\alpha(k) t\right]} \end{eqnarray*}$
This form makes it clear how the general solutions of these kinds of problems can look. You end up with a sum of wave modes, each of which has a Fourier decomposition that satisfies the dispersion relation.

Edit: the $A_\alpha (k)$ are chosen to match the initial conditions $E(z, t=0)$ and $\frac{\partial}{\partial t} E(z,t=0)$.

There are similar kinds of equations (often systems of equations) describing more complex waves in plasma, different kinds of dispersive waves in the atmosphere (eg internal waves), etc.