Re: Waves

[David Bowman <dbowman@TIGER.GEORGETOWNCOLLEGE.EDU>]

Regarding James Wheeler's example:

> Quick calculation for a transverse wave on a string.
>
> Let T= tension on the string.
> Let mu= mass per unit length of the string.
> Let the string lie along the x axis, and
> let the displacement of the wave be in the y direction.
>
> The energy density of the wave is then:
>        0.5*mu*(dy/dt)^2 + 0.5*T*(dy/dx)^2.
>
> The current of energy (flux) is:
>        -T*(dy/dt)*(dy/dx).

So far so good.

> Dividing this by the speed of the wave gives:
>        -sqrt(mu*T)*(dy/dt)*(dx/dt)
> which would be a momentum flux for the wave.
> (sqrt is the square root function)

Sorry I don't think this last step gives the momentum flux of the wave.
It gives the flux of the *magnitude* (i.e. absolute value) of the momentum
which is merely a scaled version of the energy flux.  The E vs. p
dispersion relation for the wave eqn. is E = |p|*c (actually this is the
relation between the energy and the momentum of the quantized progagating
phonons along the string, and this is equivalent to the relation
omega = |k|*c relating frequency to wave number) where c = sqrt(T/mu).  If
the absolute value function was not taken in this relationship then you
might be nearly correct (up to a factor of c) for the momentum density.
Because the momentum is a signed quantity (or vector if you will) it is
possible to show that the total momentum for the excitations of the string
(assuming the endpoints are fixed) is zero.  All momentum concentrations
and fluxes are mere local imbalances that both total to zero and average
out to zero over multiple cycles of the waves.  This is not necessarily
true of the energy flux which can propagate unidirectionally with running
waves until they reflect off of the end points and the energy returns to
its starting place.  But this time scale is given by the length of the
string divided by the wave speed, not by the wave frequency like the
momentum reversals are.

If you thought that you could get the momentum *density* from the energy
*flux* by dividing by c^2 like you can from the Poynting flux for EM waves
then I think you you were misled in this string case.  Waves for a scalar-
valued disturbance field propagating in a 1-dimensional medium are
different than 4-vector-potential-valued waves propagating in 3 dimensions.
BTW, the momentum flux is given by the stress tensor, not c times the
momentum density anyway.  Of course, since I have not worked it all out,
I'm not quite sure at the moment just what the expression for the 1 X 1
stress tensor actually happens to be for this string system, but I kind of
doubt that it would come out to your final expression.

David Bowman
dbowman@georgetowncollege.edu