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Niles
Jan31-10, 12:45 PM
1. The problem statement, all variables and given/known data
Hi all

If we look at a harmonic wave with constant amplitude, Ψ(x,t) = Asin(kx-ωt), then a point with constant magnitude (e.g. a crest) moves such that kx-ωt is constant in time.

Now we look at a plane wave Ψ(r,t) = Aexp(i[kr-ωt]). Will a point with constant magnitude (i.e. the whole plane) also move such that i(kr-ωt) is constant in time? If yes, then doesn't this mean that the phase for a plane wave is constant for all times?

ideasrule
Jan31-10, 08:08 PM
1. The problem statement, all variables and given/known data
Now we look at a plane wave Ψ(r,t) = Aexp(i[kr-ωt]). Will a point with constant magnitude (i.e. the whole plane) also move such that i(kr-ωt) is constant in time?


Yes, because if you take the real part of that equation, you'll get an equation almost identical to the one you posted for one-dimensional waves.


If yes, then doesn't this mean that the phase for a plane wave is constant for all times?

It means that whatever phase you choose, you can always find a point in space with that phase at any time. That's logical: the wave spreads, but it's not as if one phase "disappears": it just moves at its phase velocity.

Niles
Feb1-10, 02:54 AM
It means that whatever phase you choose, you can always find a point in space with that phase at any time. That's logical: the wave spreads, but it's not as if one phase "disappears": it just moves at its phase velocity.

Hmm, I don't get that. Say we position the plane wave such that the wavevector lies along the x-axis, i.e. it propagates along the x-axis. It is obvious that (as you said) the real part of the plane wave is just what I wrote in my first example of my OP. Hence all points on that specific plane wave have the same phase, and hence they must maintain that phase as they propagate.

With this explanation I cannot see why I can choose any arbitrary phase; there should only be one?

Thanks.

vela
Feb1-10, 04:22 AM
What do you mean when you say

all points on that specific plane wave have the same phase

Niles
Feb1-10, 04:24 AM
I mean that our plane wave has the form Ψ(r,t) = Aexp(i[kx-ωt]) (we have aligned it along the x-axis), so each point on the plane wave for some x will have the same phase, i.e. kx-ωt is the same for all points on that plane.

vela
Feb1-10, 04:40 AM
OK, that's what I thought you meant, but your wording seemed kind of funny, so I wanted to make sure. I'm not sure I understand your question then.
With this explanation I cannot see why I can choose any arbitrary phase; there should only be one?
What do you mean about choosing a phase? Choosing it for what?

Niles
Feb1-10, 04:47 AM
What do you mean about choosing a phase? Choosing it for what?

I mean it with respect to this post:

It means that whatever phase you choose, you can always find a point in space with that phase at any time. That's logical: the wave spreads, but it's not as if one phase "disappears": it just moves at its phase velocity.

ideasrule's post does not make sense, if there is only one phase that stays constant.

vela
Feb1-10, 05:11 AM
I think ideasrule just meant if you arbitrarily pick a phase, you can find its corresponding plane, and that plane of constant phase, a wavefront, will propagate at the phase velocity. If you choose a different phase, you're talking about a different wavefront, but it will also propagate with the same phase velocity.

What I found confusing about your initial post was you asked if "the phase for a plane wave is constant for all times." I think you meant "wavefront," not "plane wave." The plane wave fills all of space. The phase at a particular point in space will change with time as the wave propagates, and at an instant in time, different points in space will generally have different phases. A wavefront is a plane of constant phase, and it will propagate with the phase velocity. By definition, its phase won't change over time.

Niles
Feb1-10, 05:13 AM
I have to go to school now, but when I get home, I will reply.

Niles
Feb1-10, 11:23 AM
Ok, I agree. My explanations were not that detailed, but I think I get it now. Thanks.