. . . 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. . . .
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Assuming that this snippet can be regarded as applying to the classic
vibrating wave on a one-dimensional string; I view the parenthetical comment
as crucial. Its not surprising that if both endpoints are fixed that we
conclude that there is zero momentum transport (flux) as we may describe
such a situation as two traveling harmonic waves of equal amplitude (a
standing wave) which would be equal momentum transport in opposite
directions, summing to a net zero *since momentum is a "signed" quantity*
On the other hand what happens if the string is very long (infinitely long
shall we say) and we attach an oscillator at one end, which we turn on at t
= 0, and the other end is not fixed (or we look at times less than the
propragation time for the leading edge of the disturbance to reach the far
end. What then?