# I Wavevector k in Helmholtz Equation

1. Dec 16, 2016

### fog37

Hello Everyone,

Helmholtz equations derives from the wave equation by using separation of variables and assuming that the solution is indeed separable $g(x,y,z,t) = f(x,y,z) T(t)$. The solutions to Helmholtz equations are functions of space, like f(x,y,z), and do not depend on time t.

the wavevector $k = \frac{2\pi}{\lambda}$ is present in the equation. Is that a constant or a variable? Solutions can span a spectrum of $k$ values so I don't believe the $k$ in the equation can be a constant otherwise the equation would be limited to function having a single and specific k value....

Thanks

2. Dec 16, 2016

### Henryk

Different k values correspond to different waves with respect to wavelength or direction

3. Dec 16, 2016

### fog37

True, but in the equation, is the k to be take as a variable?

4. Dec 17, 2016

### Henryk

Oh, that was your question. No, wavevector k is fixed.

5. Dec 17, 2016

### eys_physics

I would say that the wave vector $\mathbf{k}$ is a parameter, since you solve the equation for fixed $\mathbf{k}$ but the value depend on the specific problem you want to solve.

6. Dec 18, 2016

### fog37

so are you saying the solutions to this equation can only have a single k since the equation itself has a single k?

7. Dec 18, 2016

### eys_physics

Presumably, this is a matter of convention. But, my personal point of view is the following (maybe other people on this forum will not agree). In your first post you are saying:
We have therefore a set of solutions $f_k$ labeled by the wave number $\mathbf{k}$. As, I said before you are then solving the equation for a fixed $\mathbf{k}$ and obtain one $f_k$. You can then repeat this and solve for arbitrary number of values of $k$ depending on which solutions on the set you are interesting in.