Why is the frequency constant when waves pass through an interface?

[McLaren Rulez]

What the title says.. I'm looking for a general reason that will apply to everything from waves on a string to EM waves. Why is it that the wavelength is the quantity that changes while the frequency stays the same? Thank you.

 

[Meir Achuz]

The boundary conditions at the interface musst hold for all times.
This requires the two frequencies to be the same.

 

[McLaren Rulez]

Could you elaborate please? I realize that the boundary conditions must require the frequency to be constant but I want to see a proof for that. Thank you.

 

[Matisse]

Here is an intuitive argument.

Imagine that a device calculates crests in the first substance (let's say glass) and a separate device calculates crests in the second substance. Both devices must count the same number of crests(let's say 3 E 10) for if they did not crests would accumulate at the interface which is impossible.

This is equivalent to a boundary condition.

 

[sophiecentaur]

There is a sort of Null Argument about this. By how much would you expect the frequency to change - up or down? If you can't think why is should then it won't.
But the boundary conditions is the clincher argument. Right on the boundary between the two media, there would have to be a step change in the phases of the EM fields on either side, which would be continuously changing as one frequency strobed through the other. However would that work?

 

[McLaren Rulez]

Matisse, your argument is good! I can see it intuitively now. Is it possible to show that this is indeed equivalent to the boundary condition?

sophiecentaur, I'm not sure if this is the kind of Null Argument you mentioned. After all, letting it change is a weaker condition that asking it to stay constant right? Regarding the second part of your post, could you explain why there has to be a phase change in the field on either side of the boundary?

Thank you both for the help.

 

[jtbell]

If the oscillation on both sides of the boundary don't have the same frequency, then at some point in time and at some locations along the boundary, you must have situations such as: the wave immediately next to one side of the boundary is 1/8 of the way through a cycle and moving "upwards", while the immediately adjacent portion of the wave on the other side of the boundary is 5/8 of the way through a cycle and moving "downwards."

This is impossible because adjacent portions of any wave are tied together by some physical mechanism which allows the wave to exist in the first place. With water waves and sound waves, it's intermolecular forces. With electromagnetic waves, it's the interrelationship of the electric and magnetic fields via Maxwell's equations.

 

[McLaren Rulez]

Thank you very much, everyone!

 

[netheril96]

Boundary conditions, most of the time, is independent of time. And monochromatic waves evolve proportional to [tex]e^{-i\omega t}[/tex]. Thus for the boundary conditions to be met, the time evolution must be synchronous, or boundary conditions will be met at one time, and not other times.

 

[sophiecentaur]

Matisse, your argument is good! I can see it intuitively now. Is it possible to show that this is indeed equivalent to the boundary condition?

sophiecentaur, I'm not sure if this is the kind of Null Argument you mentioned. After all, letting it change is a weaker condition that asking it to stay constant right? Regarding the second part of your post, could you explain why there has to be a phase change in the field on either side of the boundary?

Thank you both for the help.

jtbell explained what I meant about the phase.
and, yes, the easiest condition would have to be the frequency staying the same or else you'd have to involve some incredible mechanism (not involving energy being added, too) to make anything else happen.