Mymymy.
There are quite some people here trying to discuss advanced concepts who lack some basic mathematics of wave propagation such as the one on a cord (used to be called string but that is doomed to make a conceptual link with advanced stuff it has nothing to do with)
To try to put some things in order, let us first consider a piano cord, but of a very very big piano. Doing some elementary Newtonian mechanics, you can find the wave equation (d^2 u/dx^2) - 1/v^2 (d^2 u/dt^2) = 0
Here, u(x,t) is the sideways displacement of the cord from its rest position, and v is a constant which is calculated from the tension strength on the cord and the mass density of the cord.
If this is not clear to you, honestly, I think the whole discussion on EM waves, photons and so on is absolutely not accessible.
The above wave equation is a partial differential equation which has as a general solution: u(x,t) = f1(x-vt) + f2(x+vt), where f1 and f2 are two completely arbitrary functions. In this one-dimensional case we're lucky that we can write explicitly down the general solution ; in most of the cases this is not true, and we have to resort to some special techniques ; one of those techniques is called Fourier transforms.
If you calculate the Fourier transform of u(x,t) and write it U(k,w), then the equation becomes:
k^2 U(k,w) - 1/v^2 w^2 U(k,w) = 0
The nice thing is that we changed the partial differential equation into an algebraic one, so we have:
(k^2 - w^2/v^2) U(k,w) = 0
This essentially tells you that for a point {k0,w0} in the k-w plane where U(k,w) is defined, to have U(k0,w0) not equal 0, you have to have that:
k0^2 - w0^2/v^2 = 0 or that v * k0 = +/- w0. For other combinations of k and w, U(k,w) = 0
The curve in the k,w plane where U is allowed to be different from 0 is called a dispersion relation, in our case:
w(k) = +/- v * k
As the differential equation is linear, we do not have to work with a general solution, it is sufficient to work with a basis set of solutions which we can then superpose. A way of doing so is by considering ONE SINGLE POINT on the dispersion curve, namely a single couple (w0, k0) satisfying the dispersion relation, where we consider U(k,w) to be different from 0.
This gives us then a HARMONIC WAVE in the x-t domain:
u(x,t) = Exp(i (k0 x + w0 t)) by inverse Fourier transformation.
We usually work with these harmonic waves but it is understood that a more general solution is a SUPERPOSITION of these harmonic solutions.
In fact, we already have to combine u(x,t) with its conjugate (the point at -w0, -k0) because u(x,t) is to be a real number, not a complex number, so we have as real solutions sin(k0x + w0 t) and cos(k0 x + w0 t).
We now come to the concept of a "wave packet". This is nothing else but such a superposition of harmonic waves, that has as a peculiar property that it is "lumped" as well in the (x,t) domain as in the (u,k) domain - this is a property of the Fourier transform.
So in order to make a wave packet, you take a U(k,w) which is non-zero over a certain "lump" of k-w space (but of course only on the dispersion curve), instead of in one single point. If you choose the amplitudes and phases of U(k,w) right, the inverse transform will give you a function u(x,t) which, for a given t-value, is lumped in x.
The funny thing is that such a wave packet propagates with speed v: the center of the wave packet, x_c, at time t is a function of t as follows:
x_c(t) = x_c(0) +/- v t
This is the underlying idea of a wave propagation, but when you have done this a few times, you already see this when looking at the harmonic solution, so in most texts, people don't bother by going through this mantra again and again.
Now consider energy. The only energy we have is kinetic energy ; you could be tempted by thinking there is a potential energy associated with a displacement u(x,t), but this is not true, as can be easily seen:
Imagine that at one side of the very long chord, you apply a "step function": u(x=0, t) = h(t). Clearly, you only do a modest amount of work when moving the point x=0 from u=0 to u=1. The solution of this will be that at time t, u = 1 for x < v*t, and u = 0 for x > v*t with a propagating step front. If having u = 1 amounts to storing energy, there would clearly be a conservation of energy violation, because you only introduced one small amount when you moved the point at x=0 at time t=0, and the part of the chord where u = 1 grows with t.
So you see that the energy (kinetic energy) of a wave motion u(x,t) is concentrated in those parts of the chord that are moving:
the energy density E(x,t) is given by 1/2 rho ( d u/dt )^2.
If you calculate the energy density of a harmonic wave sin(k0 x - w0 t) you will find that the energy density is not constant but "moves" with a speed v.
There is
In EM (classical EM), we have exactly the same equation, but it is 3-dimensional, instead of 1-dimensional. A wave packet is then a "pulse of light" which has a certain spread in frequency, and has a certain localisation in space. This "pulse of light" propagates at speed c.
Now to the "mystery" of E and B fields. First of all, you could work with the electromagnetic potential, A, and then there is only ONE potential, and you wouldn't be bothered by two fields oscillating "in phase", there would only be one field. Concerning the energy density in space, yes, it is not constant for a HARMONIC solution. So what ?
If a car is moving over a road, then its "mass density" is not constant: when the car is at point A, its mass is at point A and no mass is at point B ; while a bit later when the car is at B, the mass density at A is gone, and it is now at point B. So the same happens with a classical EM wave, at least the harmonic wave, which is a special solution.
This has nothing to do with photons ! We are working purely with classical waves here.
cheers,
Patrick.
(just kidding).
).
