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I'm taking a course on waves and optics using Young and Freedman's University Physics, but I'm a bit confused about a couple of things. I've also looked at Griffiths' Introduction to Electrodynamics and Taylor's Classical Mechanics, and the three of them seem to say the following:

Young and Freedman

  • A wave is (roughly) a disturbance of a system which can propagate from one region of the system to another.
  • Derives the wave function $y(x,t) = A \cos(kx - \omega t)$ for a sinusoidal wave on a string.
  • Uses that to derive the wave equation $\frac{\partial^2 y}{\partial x^2} = \frac{1}{v^2} \frac{\partial^2 y}{\partial t^2}$, where $v$ is the phase velocity.
  • Uses Newton's second law to derive the wave equation for a general wave on a string, this time on the form $\frac{\partial^2 y}{\partial x^2} = \frac{\mu}{F} \frac{\partial^2 y}{\partial t^2}$.
  • Compares the two and concludes firstly that $v = \sqrt{F/\mu}$ for a wave on a string, and secondly that the wave equation with coefficient $1/v^2$ is valid for any wave on a string.

That last step confuses me and doesn't seem to follow from the above. The wave equation with coefficient $1/v^2$ has only been shown to be valid for sinusoidal waves, so how can they conclude that the coefficient has that form for all waves on a string?

Griffiths

  • Recognises the intrinsic vagueness of the concept but provides a tentative description (definition?): "A wave is a disturbance of a continuous medium that propagates with a fixed shape at a constant velocity."

  • Argues that a general wave function has the form $f(z,t) = g(z - vt)$ for any function $g$. (I find this pretty satisfactory.)

  • Derives the wave equation $\frac{\partial^2 f}{\partial z^2} = \frac{\mu}{T} \frac{\partial^2 f}{\partial t^2}$ for a general wave on a string in a similar manner to Young and Freedman. Rewrites the coefficient as $1/v^2$ with $v = \sqrt{T/\mu}$ but doesn't simply state that $v$ is the phase velocity.

  • Starts with the wave function $f(z,t) = g(z - vt)$ and derives the wave equation on the form $\frac{\partial^2 f}{\partial z^2} = \frac{1}{v^2} \frac{\partial^2 f}{\partial t^2}$.

  • Concludes that $v$ is the phase velocity for any wave.

  • Mentions (but does not show) that the general solution is $f(z,t) = g(z - vt) + h(z + vt)$ for some function $h$.

This seems more satisfactory. But something still confuses me: Why start with a general wave on a string? Is it simply for pedagogical reasons, or is it not sufficient to start with the wave function $f(z,t) = g(z - vt)$ and derive the wave equation from that?

Taylor

  • Also starts with a general wave on a string and derives $\frac{\partial^2 u}{\partial x^2} = \frac{1}{c^2} \frac{\partial^2 u}{\partial t^2}$ with $c = \sqrt{T/\mu}$, basically the same as Griffiths I think. He claims that $c$ "is the speed with which the waves travel" without arguing for it yet. (I find him rather vague here. It's a bit unclear exactly what kind of claim this is, whether he thinks it follows from the derivation or not, whether he wants the reader to take it for granted, or if he's going to return to it or not.)

  • Shows that the general solution has the form $u(x,t) = f(x - ct) + g(x + ct)$ for any function $f$ and $g$.

  • Considers the case where $u(x,t) = f(x - ct)$, interprets this solution physically and sees that it describes a disturbance travelling in the positive $x$-direction at speed $c$. I suppose he has now shown that $c$ is the phase velocity.

Griffiths and Taylor seem to do things in opposite directions: Griffiths starts with the wave function and derives the wave equation, and Taylor does the opposite. This leaves me confused as to what a wave is mathematically, and what relation the mathematical description has to the physical phenomenon.

That is, how is a wave defined? Is it a mathematical model which, by construction, fits many physical phenomena (and thus a wave is simply defined as a disturbance that has a wave function or is a solution to the wave equation - and if so, which of the two, function or equation, or are they equivalent?)? Or is it a more or less well-defined class of physical phenomena that just happens to be described by the wave equation (maybe not always)? In that case, what authority does the wave equation have?

Griffiths later derives differential equations for the $\mathbf{E}$ and $\mathbf{B}$ fields in empty space, compares them to the three-dimensional wave equation and concludes that electromagnetic waves exist. If this similarity to the wave equation shows and not just suggests that EM waves exist, then it seems to follow that the wave equation has some authority in determining what counts as a wave. He uses the word "imply", but I don't know what that means.

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    $\begingroup$ Waves are disturbances that propagate. But only in ideal cases without dispersion, as a function of only (x-vt), with the same velocity for all frequencies. And then there are solitons, those would also count as waves in my opinion. $\endgroup$ – Pieter Apr 2 '17 at 9:47
  • $\begingroup$ See physics.stackexchange.com/a/395212/45664 $\endgroup$ – user45664 Mar 26 '18 at 20:06
  • $\begingroup$ Return to Young and Freedman for a moment. Take a segment of the sinusoidal string and use Newton's laws to derive the wave equation. What do you arrive at? Use dimensional analysis to compare the coefficients $v$ and $\sqrt{\frac{F}{\mu}}$. Are they the same physical quantities? $\endgroup$ – R004 Apr 22 '18 at 1:36
  • $\begingroup$ See: What is a wave? geo.umass.edu/faculty/wclement/nature_wave.pdf $\endgroup$ – JohnS May 14 '18 at 15:23
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Mathematically

The waves are solutions of the wave equation: $$\Delta f - \frac{1}{v^2}\frac{\partial^2 f}{\partial t^2} = 0$$

This equation can be solved by many tools. The most elegant method is probably Fourier transformation; it allows us to separate the solution in coordinates (that's useful in some physics applications). The solution you mentioned $f = g(z-v/t) + h(z+v/t)$ is only one solution, but it doesn't cover the whole picture (understand the whole space of solutions).

Physically

We can find wave equation in few basics examples. The first one could be the string vibrating for small initial deviations. The other wave equation can be found in Maxwell equation for field $\vec E, \vec B$ or for scalar and vector potentials $\phi, \vec A$. It only means that these waves are physical, but these waves must still satisfy Maxwell equations.

In physics, this is only a special case of what we call waves. We have Klein-Gordon waves, Dirac waves (you will learn this in quantum field theory) or very simple equation from electrodynamics: waves in conductors. All these waves don't satisfy the wave equation in the classical meaning of \eqref{A}. But we can still call it waves.

So mathematically waves are strictly solutions of the wave equation \eqref{A}, physically we call disturbance in the space that are time-dependent or physical entities carrying information that is propagating from one place to another.

The word "wave" has its origin. Mathematicians in history (in post-renessaince era mostly) were working on the description of the musical instruments. So the first origin of waves was from mathematicians studying physical nature. I can recommend you this article [*].

Edit:

When Maxwell equations are reduced into wave equations, it really means that electromagnetic waves exists. That is because of existence of $\vec E, \vec B$. These fields exists, they carry momentum and energy. So if they are solution of the wave equation, they must be waves.

This led many of physicists to think that there is an aether, which is a medium where electromagnetic waves (as light) can be disturbance. And this is the problem known in special relativity when Michelson and Moorley proved that aether (even if it exists) is not important for electromagnetic waves. This experiment was proof for Einstein special relativity and it started the modern era of the physics.

[*] https://www.jstor.org/stable/41134001?seq=1#page_scan_tab_contents

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  • $\begingroup$ +1 Please explain to me why the solution for E and B are not shifted by 90° or solutions of sin and cousin. If necessary and after I get it how to express this question right I could ask this question of course. $\endgroup$ – HolgerFiedler Apr 3 '17 at 3:19
  • $\begingroup$ So to be completely explicit: Mathematically, waves are solutions to the wave equation. I suppose e.g. solutions to the Klein-Gordon and Dirac equations qualify as well? Physically, waves is something more vague, but we can derive mathematical descriptions that may vary across different types of waves. However, if a function satisfies any of the wave equations, it follows that the function describes a wave phenomenon. Is this correctly understood? $\endgroup$ – Danny Hansen Apr 3 '17 at 19:42
  • $\begingroup$ @HolgerFiedler I am not sure if I understand. Vectors $\vec E$ and $\vec B$ have to satisfy Maxwell equation. For plane waves it means that $\vec E$ is perpendicular to $\vec B$. When you say shift I think of phase shift, it means $\sin(\cdots + \pi/2$, what I don't see a sense in. $\endgroup$ – Jimmy Found Apr 4 '17 at 5:44
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    $\begingroup$ @DannyHansen I think this problems is connected with another one: "What is the difference between mathematics and physics?" There are many mathematical models that are made by mathematicians and then physicists do only these models that are in a good relationship with nature. It means waves are of the models motivated by waving objects (strings, membranes) and it is one of the model that is mutual. Only physicist can find another examples that could be also called waves, but mathematicians still see only the first waves that they described in post-renaissance. I hope it's helpful now. $\endgroup$ – Jimmy Found Apr 4 '17 at 6:03
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A wave is a disturbance in space and time. As a consequence, it must be described (in 1d) by a function if the type $f(kx-\omega t)$ or $g(kx+\omega t)$ where the argument $kx-\omega t$ is dimensionless; $f$ describes a wave propagating in the positive $x$-direction since, to keep the phase $kx-\omega t$ with increasing $t$ implies increasing $x$. $g$ describes a wave propagating in the negative $x$-direction.

Mathematically the functions $f$ and $g$ will be solutions of a partial differential equation involving $t$ and $x$. The PDE need not be linear (v.g. the Korteweg-de Vries equation) but, in the simplest case, the PDE will be linear: $$ v^2\frac{\partial^2f}{\partial x^2}=\frac{\partial f^2}{\partial t^2}\, , \tag{1} $$ with $v=\omega/k$. The solution to the above is called the harmonic wave because the solution is in terms of sine and cosine.

Because the PDE of (1) is linear, the sum of two solutions is also a solution. Because (1) involves second order differentials so that $v^2$ occurs, if a solution of the form $kx-\omega t$ is found then the same function but with argument $kx+\omega t$ will also be a solution and can be added.

In the case of the non-linear KdV equation: $$ \frac{\partial f}{\partial t}+\frac{\partial^3 f}{\partial x^3}-6\,f\,\frac{\partial f}{\partial x}=0\, , $$ the function $f$ is still a function of $kx-\omega t$ but the sum of two solutions is not necessarily a solution because the corresponding PDE is non-linear.

Note that the PDE for the wave on a string that you give is approximate, i.e. it should contain small non-linear terms to account for the finite flexibility of the string; as with many problems in physics, this one is only approximately linear hence the linear PDE, which works well under many circumstances.

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  • $\begingroup$ Your last point is very good, I completely overlooked that fact. So if I understand you correctly, a wave must be described by a function of the type $f(kx−ωt)$ or $g(kx+ωt)$ to be a wave, and this is true for any wave function? Also e.g. the Klein-Gordon and Dirac equations mentioned in a different answer? $\endgroup$ – Danny Hansen Apr 3 '17 at 19:50
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    $\begingroup$ @DannyHansen Yes. The form of the solution as $f(kx \pm \omega t)$ is what matters. In the case of KG (en.wikipedia.org/wiki/Klein%E2%80%93Gordon_equation#Statement) the solution is precisely $e^{i(\omega t-kx)}$. The variables $k$ and $\omega$ will satisfy $\omega^2-k^2=m^2$ rather than $\omega/k=v$ but that's secondary. See also here: physics.stackexchange.com/questions/322490/… $\endgroup$ – ZeroTheHero Apr 3 '17 at 22:01
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Thank you for asking this question, it has made me reconsider everything I thought I knew about waves. I would like to share the perspective I have settled upon, and I welcome feedback.

As see it, waves are both physical phenomena (propagating disturbances) and mathematical abstractions (those functions which satisfy the differential wave equation). I am by no means an expert, but I think that distinguishing between physical waves and mathematical waves is a first step in truly understanding the matter. I'm sure most popular texts are quite unclear about this distinction, and perhaps it would benefit us to think in these terms.

Mathematical waves may be used to model physical waves. As with any mathematical model intended to represent physical reality, there are a number of assumptions that are made. For instance, to reach the ideal linear wave equation mentioned by OP and other answers, it must be assumed the amplitude of oscillation is small. This video sequence follows a derivation of the wave equation which I'm rather fond of, and the presenter is explicitly clear about what assumptions are made in the process. If less confining assumptions are made during the derivation, one may obtain nonlinear wave equations like the one mentioned by @ZeroTheHero. I'm guessing the linear DE we call the wave equation has earned the title by being an optimal compromise between physical applicability and mathematical convenience.

Not every wave in nature needs to satisfy "the wave equation" to be considered a wave, indeed no empirical data will fit it exactly. Which is why, in some qualitative sense, there must be such a thing as a "physical wave"; and some of these physical waves may perhaps deviate significantly from our linear model (those with large amplitudes, for instance). A "mathematical wave", being the solution of "the wave equation" (or perhaps a higher-order variant of it), might agree intimately with the data obtained from a physical wave, but it is an inherently different object from the physical wave which it describes.

So, my humble perspective is that if we want to define waves in some absolute philosophical sense, we should distinguish between the "messy" physical phenomena we call waves and the ideal mathematical structures we use to model them.

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I think the wave equation can by derived from geometry alone, without using physics.

See https://physics.stackexchange.com/a/110842/45664

What is a wave? In 1D this only requires envisioning a pulse of a given shape (e.g. a Gaussian pulse to make it simple) propagating along the x axis with speed c.

Where does the wave equation come from? Envision the above. Then the 1D wave equation can derived from the envisioned geometry as shown in the above link. So I think at the most basic level a wave and the wave equation are geometric ideas.

Where actually did it come from? Apparently D'Alembert originated the wave equation. The wave equation operator is called the D'Alembertian. D'Alembert was studying the vibrating string. See https://www.macalester.edu/aratra/chapt2/chapt2_6b.html for the 'vibrating string controversy' or search for that term.

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