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Suppose that we have a standing wave on a circle. I heard that by gradually increasing the radius of the circle, the wavelength will also increase to keep the standing wave. Is it right? If yes, what's the physics behind this?

EDIT: Since this question is marked as "ON HOLD" due to ambiguity, I should make it a bit more clear. Since I've already got my answer, this explanation is not a try to ask the question again but an attempt to demonstrate my intentions for asking this question.

Suppose that we're living in a 1-sphere expanding universe (that is just a simple circle which its radius increases with time). The metric of this space is:

$$ds^2 = a^2(t)\ d\phi^2$$

As you can see, the comoving position of galaxies (just the angle $\phi$) in this universe does not vary by time, however, since the scale factor ($a(t)$) is increasing by time, the universe is blowing up too. So now consider a case in which we have two galaxies, A and B. Due to the expansion of this universe, these two galaxies have a relative velocity which could be greater than the speed of light too. (The expansion obeys the Hubble law since the distance between galaxies is $D = a(t)\ d\phi$ and then $$\dot{D} = V = \dot{a(t)}\ d\phi = \frac{\dot{a(t)}}{a(t)}\ a(t)\ d\phi = H(t)\ D.$$ Now consider the case where an observer on galaxy A is observing galaxy B. The light of galaxy B as seen by the observer is redshifted. We can explain this redshift with two similar but a bit different interpretations. First we can say this redshift happens due to the Doppler effect related to the relative velocity of two galaxies, and second we can say that the light is redshifted due to the expansion of the universe, so it's a continues effect on a photon which is traveling between two galaxies. Ok, you may say these two interpretation are exactly the same, which I believe it's right, but you can also think of the scale factor, not as a factor that expands the universe, but a factor which decrease the speed of light as time goes on. (I may be wrong, in this case please correct me.)

Well, I was watching one of the Susskind lectures and he mentioned that if we have a standing wave on a circle, by slowly increasing the radius of the circle, the wavelength will change in a way to keep the standing wave as it was. So my question was about this claim and I think this could be related to the expansion of the universe (since he mentioned it in a cosmology course). Also Anna's answer about de Broglie wavelength clears thing up for me.

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closed as unclear what you're asking by ACuriousMind, Sebastian Riese, Kyle Kanos, Norbert Schuch, gigacyan Mar 11 '16 at 13:03

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

  • $\begingroup$ I don't understand the metric. Why isn't the coordinate $r$ used? And why isn't there a time component, ${\rm d}t^2$? $\endgroup$ – Kyle Kanos Mar 12 '16 at 15:16
  • $\begingroup$ Think of it like this, $a(t)$ is actually $r(t)$. This is the metric of space, not the space-time, here's the space-time metric: $d\tau^2 = - dt^2 + ds^2$ $\endgroup$ – Saeed Mar 12 '16 at 16:23
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Suppose you have a circle of radius $r$. In order to model a standing wave on the circle, you need to keep in mind appropriate boundary conditions. In this case, the positive integer multiple of the wavelength should be equal to the circumference of the circle, a condition for constructive interference. This is because otherwise the wave will interfere destructively with itself on the entire circle, resulting in the vanishing of the wave.

Therefore we have

$$ n \lambda = 2 \pi r$$

which implies

$$ \lambda = 2 \pi r/n$$

So basically for a fixed $n$, the wavelength required for a standing wave increases with $r$.

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  • $\begingroup$ Thanks for the answer, but my question was something else. I thing I couldn't ask it properly, let me try it again. When we increase the radius of the circle, since we're not changing the properties of the environment, the propagation speed of the wave should remain unchanged. So, if the wavelength remains the same too, it means that we couldn't have a standing wave anymore. But, it seems that the nature tend to tune the wavelength to keep the standing wave. Well, my question is why does the nature tune the wavelength? $\endgroup$ – Saeed Mar 10 '16 at 2:46
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    $\begingroup$ Since you said "Nature tunes the wave", you should provide us with a specific example. I am not aware of any cases in real life where this happens. I mean, that's what the Bohr model is about, but that's just a model and all aspects of your conceptual understanding might not apply. Bruce Lee's answer is good, except maybe you wanted him to say "if the wavelength remained the same, we would no longer have a standing wave"? $\endgroup$ – levitopher Mar 10 '16 at 3:34
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The only standing waves one can find in examples on the net are the ones on membranes and water and those are not the waves you mean, since you assume the frequency on the perimeter of the circle.

You have to go to quantum mechanics of elementary particles and the de Broglie wavelength. There one can have an electron in a magnetic field in a circle and its wavelength could be considered a standing wave on the circle. The radius is given by using the relation of the magnetic induced centripetal force equal to the centrifugal,

$$Bqv=mv^2/r . $$

A bigger radius means a larger energy so a larger de Broglie frequency. This is consistent with the other answer.

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