In $k$-space the allowed values for standing waves in a cube of side length $L$ are given by
$$\vec{k} = \left(\frac{\pi}{L}\right) (n_1, n_2, n_3)$$
where the $n_i$ are nonnegative integers. Why are these numbers non-negative only?
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$\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$– Community BotCommented Mar 7 at 15:11
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Very good question, indeed.
At first you've made small error about wavenumber (original, not edited) definition. There can fit $n$ half-waves in a box of size $L$, so :
$$\tag 1 n = \frac{L}{\lambda /2} = \frac {2\pi L}{\pi \lambda}=\frac {kL}{\pi}$$
From (1) follows, that $$ \tag 2 k = n \frac \pi L$$
Now back to your question, why $n=0,1,2,3,\ldots$ ? Sure, mathematically you could do $-k = -n \pi/L$ to fill other spectrum part of "negative wavenumbers", BUT ... it doesn't make sense because :
- Wavenumber by definition is magnitude of wave vector $|\mathbf {k} |$, so it's an absolute value without a sign.
- Wave vector itself, if you would reverse it's direction by putting a negative sign to it like $\mathbf {k} = -\mathbf {k}$,- would mean that waves would propagate backwards now. But since we are describing standing wave modes in a box, then any form of wave propagation features, like $\mathbf {k}$ or Poyinting vector $\mathbf {S}$ or wave phase velocity doesn't make any sense here.
answered Mar 7 at 15:42
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$\begingroup$ So, in essence, since standing waves are 'static' in nature using wave propagating features only make sense if they are positive definite? $\endgroup$ Commented Mar 8 at 3:07
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$\begingroup$ Not exactly. Since standing waves are static in nature, it only makes sense to define "static features" of waves, which would not have any "feel of motion" or propagation direction. Wave vector has direction, so it's not suitable. Wavenumber is suitable, because it's just spatial frequency of wave. By putting non-negative $n$ factors to $\pi / L$,- you are defining n-th harmonics. But making it negative,- makes no sense at all, because there's no "negative frequencies". So technically as in all Physics, when you invert sign of some parameter $x \to -x$,- you need to think does it make sense. $\endgroup$ Commented Mar 8 at 7:42