# How can a wave travel if it is spread over all space?

The wave equation in one-dimension is

$$\nabla^2\psi = \dfrac{1}{c^2} \dfrac{\partial ^2\psi}{\partial t^2}$$

and in one dimension one possible solution to this equation is the function

$$\psi(x,t)=A \cos (kx-\omega t)$$

where $k/\omega = 1/c$. Now, if we fix $t = t_0$ this thing is spread over all space. Now, the intuition I have about waves is that a wave is a disturbance on a medium that propagates with a certain speed. So it should be a disturbance that travels.

But these solutions they are, for any instant of time spread over all space. It doesn't seem possible to think about that as a disturbance that starts somewhere and travels.

So how can these waves travel if they are already spread over all space?

• How does anything constrained to be stationary move? (Or maybe I'm just misunderstanding here?) – Kyle Kanos May 16 '15 at 16:01
• if your questions is about real waves you should know that they don't extend to infinity they are finite. if it is about theoretical plane waves then only local disturbances (eg peaks) will travel but there is no energy transport. – Azad May 16 '15 at 16:10

Every function of the form: $$\psi(x, t) = A\cos[k(x\pm ct) + \phi]$$ is a solution of the wave equation $$\frac{\partial^2\psi}{\partial x^2} = \frac{1}{c^2}\frac{\partial^2\psi}{\partial t^2}$$ The equation is linear, and this means that the general solution is in fact, any linear combination of the possible solutions, something that we can express as follows: $$\psi(x, t) = \int_{-\infty}^{+\infty}dk\ [A_+(k)\cos(k(x-ct)) + A_-(k)\cos(k(x+ct)) + B_+(k)\sin(k(x-ct)) + B_-(k)\sin(k(x+ct))\ ]$$

The above integral is just the Fourier series for $\psi$, with the wave equation enforcing $k^2 = \frac{\omega^2}{c^2}$.

Now, solutions of this type are often called wave packets, as by suitable choice of the coefficients, the $A$s and the $B$s, you can represent any pulse that travels according to the wave equation. These (pulse-like) wave packets propagate like disturbances with speed $c$.

To see this easily, let $A_-$, and $B_+$ and $B_-$ all vanish for some pulse $\psi_1(x, 0)$, so that we only have the first cosine terms. Let's assume that at $t=0$, the maximum of the pulse is at $x=0$, where $x-ct = 0$. This means that the maximum of the pulse occurs when the argument of all the cosines vanish i.e. when $x = ct$ (because the only position dependence is in the cosines). This means that the peak travels with speed $c$, and the same argument carries over (for this wave equation) to any other part with a particular value of $\psi$.

Here's a sample wave packet propagating according to the wave equation (taken from the linked article)

It's worth noting that this pulse's motion ultimately comes from the phase velocity (i.e. the change of phase $x-ct$) of each of the $\cos$ terms, which is the motion of those solutions themselves. They do, of course, travel locally - the regions of constant phase move with speed $c$; it is simply because of the fact that the waveform repeats periodically that you can't think of it as a disturbance that originates somewhere.

You're correct that a wave is a disturbance that propagates at a certain speed, but the disturbance is not a single change. Whatever is being disturbed (the local pressure, the static electric field) will be restored to its "normal" status. If there is a series of waves (a wave train) the medium will be disturbed, restored, disturbed, restored.

The disturbance isn't "already spread over all space." The equation you cite describes the amount of disturbance at any one point in space and time for a continuous sinusoidal wave train. They aren't "spread out;" they are descriptive of the medium if the wave has actually reached location x.

This is a plane wave. Its an idealization of a wave that doesn't exist. Waves that actually exist are spherical waves. A sinusoidal train of spherical wave are represented as following: $$\psi(r, t) = \frac{A}{r}\cos(kr - \omega t)$$

However, the solution of a general wave for spherical coordinates can be expressed as: $$\psi(r, t) = \frac{1}{r}F(r - vt) + \frac{1}{r}F(r + vt)$$

If we fix the time $t=t_0$, you can now see you can't say anymore the wave is spread all over the space. In particular, if it is a pulse, it is spread only over a spherical shell. If that's the case why we use plane waves in the first place? Simple. It is easier. And some times can hold as a good approximation, say, in the case of too large $r$ in an spherical wave.

• Why plane waves don't exist and spherical waves do? Both plane and spherical waves are just mathematical idealizations (for example, real waves have finite temporal duration). Pulses are idelizations, too. We only can make wave packets, with certain spread in position and spread in wavevector (wich are inversely proportional). And any wave packet can be described by a superposition of plane waves, or spherical waves. – Bosoneando May 16 '15 at 17:26
• This pretty good answer would benefit from more emphasis on the notion that some real cases (like $r \gg \lambda$) being good approximations to plane waves and less on what configurations are really real. But bringing the subject up at all is important. – dmckee --- ex-moderator kitten May 16 '15 at 19:36
• The idea that plane waves are fiction and spherical waves are real is incorrect. In narrow tubes, sound waves propagate as plane waves when the wavelength is larger than the tube radius. This is how a pipe organ works. – jcandy Nov 8 '17 at 9:04