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Re: Waves

Thank you! It was indeed a typo.

The derivation for the Energy Flux is straightforward. The extension to
momentum was pure, oversimplified, and conjectural. It and brought forth
comments of suitable complexity from David Bowman.

It is interesting to think about a travelling triangular wave:

x
a a
a a
a a
aaaaaa aaaaaa (Moving left to right.)

During the rise of the string (wave arriving) dy/dt is positive, but dy/dx
is negative, so the flux is positive. During the fall of the wave (after
the wave passes) dy/dt is negative, but dy/dx is positive, so the flux is
still positive. This makes good sense for the flux of energy, but makes
much less sense for a flow of momentum. The (canonical) momentum density
of any point in the wave is mu*dy/dt and is perpendicular to the direction
of propagation of the wave.

The Stress Energy Tensor is probably the right way to approach the
momentum density- I'm getting T0x = mu*(dy/dt)*(dy/dx) for the momentum
per unit length. I still have questions about the interpretation of this,
so I will do a little looking.

On Fri, 12 Mar 1999, James McLean wrote:

"James W. Wheeler" wrote:
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).

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)

Is this a typo? Did you mean (using P for the second quantity)
P/v = P/ sqrt(T/mu) = -sqrt(mu*T)*(dydt)*(dy/dx) ?

Why did you choose to divide by the velocity of the wave, rather than
the velocity of a point on the rope?

--
--James McLean
jmclean@chem.ucsd.edu
post doc
UC San Diego, Chemistry