Can Continuous Wave Values Exist with Finite Boundary Conditions?

[robousy]

If I have a finite boundary, say of length L. Is it possible to demonstrate that if I were to allow all possible CONTINUOUS values of a wave to exist (with unit amplitude) then deconstrutive interference destroys all waves except those with wavelength:

k=\frac{n\pi}{L}

Where n =0,1...

Thanks!

 

[Pere Callahan]

Do you mean a superposition of ALL continuous functions on L ...? Supposing this makes any sense I would expect the sum to be the zero function, since for each function in your sum, there is also its negative ...How would you expect to get multiple possible result (n=0,1...) for a certain sum of all continuous functions..?

 

[HallsofIvy]

The difficulty is that you haven't said what boundary conditions you are applying. Are you assuming, without saying so, that you are considering only waves that are 0 at 0 and L?

 

[Pere Callahan]

Are you assuming, without saying so, that you are considering only waves that are 0 at 0 and L?

Even if so, the sum would still be zero.

 

[John Creighto]

If I have a finite boundary, say of length L. Is it possible to demonstrate that if I were to allow all possible CONTINUOUS values of a wave to exist (with unit amplitude) then deconstrutive interference destroys all waves except those with wavelength:

k=\frac{n\pi}{L}

Where n =0,1...

Thanks!

Destructive interference is not what destroys the waves as energy cannot be created or destroyed. Without losses, either waves transmitted outside the boundary, or attenuation within the boundary, the waves will persist forever, and the signal will remain the same amplitude. The typical boundary condition used for black body radiation is a box with high conductivity. In which case waves which do not have nodes at the boundary cause a current which dissipates energy.

 

[robousy]

Do you mean a superposition of ALL continuous functions on L ...? Supposing this makes any sense I would expect the sum to be the zero function, since for each function in your sum, there is also its negative ...How would you expect to get multiple possible result (n=0,1...) for a certain sum of all continuous functions..?

No, just a superposition of all wavelengths (greater than zero oviously) with no boundary conditions.

We usually impose that the wave function goes to zero at 0 and L (Dirichlet BC's), but I want to see if its possible to demonstrate this is the case physically...eg those waves that don't go to zero at the boundaries will interfere deconstructively with all the others that don't go to zero at the boundary when we take all frequencies from zero to infinity.

 

[robousy]

The difficulty is that you haven't said what boundary conditions you are applying. Are you assuming, without saying so, that you are considering only waves that are 0 at 0 and L?

Precisely not, in fact.

I'm hoping to show that the boundary conditions (wave goes to zero at 0 and L) have a physical origin.

 

[Pere Callahan]

No, just a superposition of all wavelengths (greater than zero oviously) with no boundary conditions.

I understand that, and still think that this superposition will be zero.

We usually impose that the wave function goes to zero at 0 and L (Neumann BC's).

What you are talking about are Dirichlet boundary conditions. Von neumann B.C. means prescribing the normal derivative at the boundary (in this case, just the ordinary derivative)

 

[robousy]

Destructive interference is not what destroys the waves as energy cannot be created or destroyed. Without losses, either waves transmitted outside the boundary, or attenuation within the boundary, the waves will persist forever, and the signal will remain the same amplitude. The typical boundary condition used for black body radiation is a box with high conductivity. In which case waves which do not have nodes at the boundary cause a current which dissipates energy.

No, good point but I certainly appreciate that. I'm trying to understand from a physical perspective why only standing waves form between Casimir plates.

You see, typically the quantum vacuum has the freedom to take any frequency from zero to \infty, but in the presence of boundaries, the degrees of freedom are modified and only standing waves form. I'm hoping to understand what it is about the boundaries that 'inhibits' the vacuum, and if perhaps is some consequence of interference.

 

[robousy]

I understand that, and still think that this superposition will be zero.

Me too! I want to show it mathematically.

What you are talking about are Dirichlet boundary conditions.

Yes, I 'knew' that. Slip of the brain...thanks.

:)