Can Continuous Wave Values Exist with Finite Boundary Conditions?

In summary, the conversation discusses the possibility of demonstrating that destructive interference destroys all waves except for those with wavelength k=\frac{n\pi}{L}, where n=0,1... The speaker also mentions the difficulty in obtaining multiple possible results for a certain sum of all continuous functions and the importance of boundary conditions in determining the behavior of waves. They also speculate about the physical origins of these boundary conditions and their effect on the vacuum.
  • #1
robousy
334
1
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:

[tex]k=\frac{n\pi}{L}[/tex]

Where n =0,1...

Thanks!
 
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  • #2
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..?
 
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  • #3
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?
 
  • #4
HallsofIvy said:
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.
 
  • #5
robousy said:
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:

[tex]k=\frac{n\pi}{L}[/tex]

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.
 
  • #6
Pere Callahan said:
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.
 
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  • #7
HallsofIvy said:
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.
 
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  • #8
robousy said:
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.

robousy said:
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)
 
  • #9
John Creighto said:
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 [tex]\infty[/tex], 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.
 
  • #10
Pere Callahan said:
I understand that, and still think that this superposition will be zero.

Me too! I want to show it mathematically.

Pere Callahan said:
What you are talking about are Dirichlet boundary conditions.

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

:)
 
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Related to Can Continuous Wave Values Exist with Finite Boundary Conditions?

1. What are modes with boundary conditions?

Modes with boundary conditions refer to the behavior of waves or oscillations in a system that is confined to a specific region with defined boundaries. The presence of these boundaries affects the way the waves or oscillations propagate and interact with each other.

2. How do boundary conditions affect the modes of a system?

Boundary conditions can affect the modes of a system in various ways, such as altering the frequency, amplitude, or shape of the waves or oscillations. They can also create interference patterns or standing waves within the system.

3. What are some examples of systems with boundary conditions?

Some common examples of systems with boundary conditions include vibrating strings, acoustic resonators, and electromagnetic waveguides. Other examples include systems with fixed or reflective boundaries, such as a guitar string or a closed pipe.

4. How do we determine the modes of a system with boundary conditions?

The modes of a system with boundary conditions can be determined by solving the equations that describe the behavior of the system. These equations take into account the boundary conditions and can be solved using mathematical techniques such as separation of variables or Fourier series.

5. Why are modes with boundary conditions important in science?

Modes with boundary conditions play a crucial role in understanding the behavior of waves and oscillations in various physical systems. They can help us understand how energy is transferred and distributed within a system, and how different factors, such as boundary conditions, can affect this behavior. This knowledge is essential in fields such as acoustics, optics, and electromagnetics.

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