# I Intuitive reasoning for frequency remaining constant during refraction

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1. Aug 18, 2016

### Frank Castle

What is the intuition for why the frequency of light does not change as it passes from a less dense medium to a denser one (or vice versa)?

Classically, if we treat light in terms of waves, then intuitively, is the reason why the frequency does not change because it is determined by the emitting source, i.e. the source emit waves of light at a particular frequency. As such, there will be a certain number of wave crests crossing the boundary between the two media per unit of time, and since the boundary between the two media doesn't destroy wave crests, the number of wave crests arriving at one side of the boundary per unit time must equal the number of crests leaving the boundary on the other side, per unit time. Hence the frequency is unchanged in the transmission from one medium to another.

Could one also argue that the frequency must remain constant on the grounds of energy conservation at the boundary. Since the refraction of light, as it propagates from one medium into another, is not dissipative this requires that the energy of a light wave must be the same on either side of the boundary, and since its energy is given by $E=h\nu$, this requires that the frequency, $\nu$ of the wave must remain constant.

2. Aug 19, 2016

### blue_leaf77

I am not sure if it's intuitive enough or not, but the real thing is that the dominating response of most media to the incoming light is linear with respect to the E field. That's why the frequency is unchanged. This linear response, however, is not the only way in which matter interact with EM wave. There are higher order responses which leads to non-linear effects such as frequency doubling etc but these effects are negligible for light with natural intensity.
At a first glance, this argument sounds safe. But if you give it more thought, it doesn't explain why the wave velocity and wavelength must change such that the frequency is conserved.
In the single photon view of the process, energy conservation need not be satisfied by one-input-one-output process, although this is the most dominant one. As an illustration of the non-linear process, take two 1.5 eV photons to excite the media then if this process turns out to be non-linear the emitted photon can be a single 3 eV photon. The input and output have different colors but energy conservation is satisfied.

3. Aug 19, 2016

### nasu

Why not ask why would the frequency would change?
Classically, when light enters a medium, two things may happen. Either it does not interact with the particles of the medium and then will be no reason to change anything. Or it interacts and makes these particles vibrate. But forced vibrations have the same frequency (in linear regime) as the force inducing them. So these interactions will emit light with the same frequency as the incoming light.

Of course, this is a very basic argument and does not include nonlinear effects, fluorescence, etc.
But it applies to mechanical waves as well, and here the classical description is even closer to reality.
Think about sound waves. Do tehy change frequency when they enter water?

4. Aug 19, 2016

### Frank Castle

How does linearity ensure that the frequency is unchanged? (sorry if I'm being a bit dense)

5. Aug 19, 2016

### blue_leaf77

The quantity that is responsible for the propagation of EM wave inside a medium is the polarization $P$. If the response is linear, it means the polarization is proportional to the E field, $P \propto E$, with the constant of proportionality being independent of time. Therefore $P$ will oscillate in time in accord with $E$. If the response is not linear, e.g. quadratic, then $P\propto E^2$ and $P$ oscillates twice as fast as $E$ because $E^2 \propto (e^{i\omega t})^2 = e^{i2\omega t}$.

6. Aug 20, 2016

### Frank Castle

Ah ok. So is the point that as the electric field propagates into the medium the induced polarisation in the dipoles oscillates at the same frequency as the original electric field and so as the polarisation propagates through the medium it maintains the same frequency as the original electric field?

7. Aug 20, 2016

### blue_leaf77

Yes.

8. Aug 20, 2016

### mpresic

Think the incoming light with frequency nu is exciting the medium The medium is a collection of harmonic oscillators. The harmonic oscillators (in steady state)
responds to the driving force at frequency nu by steady state oscillation, also at frequency nu.

9. Aug 22, 2016

### Frank Castle

Is this assuming a differential equation of the form $$\ddot{x}(t)+\omega^{2}x(t)=A\cos(\Omega t)$$ and that the "transient" contribution to the solution is negligible, such that $$x(t)\simeq -\frac{A}{\Omega^{2}}\cos(\Omega t)$$

Ok, cool.
So, from this heuristic argument, would it be correct to show that this must be true, mathematically speaking, by noting that the incoming, reflected, and transmitted are of the form $$\mathbf{E}_{i}=\mathbf{E}_{i0}e^{i(\mathbf{k}_{i}\cdot\mathbf{r}-\omega_{i}t)}\\ \mathbf{E}_{r}=\mathbf{E}_{r0}e^{i(\mathbf{k}_{r}\cdot\mathbf{r}-\omega_{r}t)}\\ \mathbf{E}_{t}=\mathbf{E}_{t0}e^{i(\mathbf{k}_{t}\cdot\mathbf{r}-\omega_{t}t)}$$ respectively, where we have assumed (for simplicity) that the incoming electric field is linearly polarised such that it is orthogonal to the boundary (between the two media). At the boundary, where the three electric fields "meet", we require that $$\mathbf{E}_{i}=\mathbf{E}_{i0}e^{i(\mathbf{k}_{i}\cdot\mathbf{r}-\omega_{i}t)}=\mathbf{E}_{r}=\mathbf{E}_{r0}e^{i(\mathbf{k}_{r}\cdot\mathbf{r}-\omega_{r}t)}=\mathbf{E}_{t}=\mathbf{E}_{t0}e^{i(\mathbf{k}_{t}\cdot\mathbf{r}-\omega_{t}t)}$$ since we require continuity such that they are one and the same electric field at this point. This relationship should hold for all $t$ and for all $\mathbf{r}$ (since it shouldn't matter on the point in space at which the boundary is positioned or the time at which the electric field arrives at the boundary). As such, $$\mathbf{k}_{i}\cdot\mathbf{r}-\omega_{i}t=\mathbf{k}_{r}\cdot\mathbf{r}-\omega_{r}t=\mathbf{k}_{t}\cdot\mathbf{r}-\omega_{t}t$$ In particular, since this is valid $\forall\;\mathbf{r}$, we are free to choose $\mathbf{r}=\mathbf{0}$, and in doing so we find that $$\omega_{i}=\omega_{r}=\omega_{t}$$ i.e. the frequency of the wave is unchanged as it propagates across the boundary.

Last edited: Aug 22, 2016
10. Aug 22, 2016

### blue_leaf77

First, I think you want the fields to be propagating (not polarized) orthogonal to the interface.
Second, in the first medium the fields $\mathbf E_i$ and $\mathbf E_s$ should add up, i.e. $\mathbf E_i+\mathbf E_i = \mathbf E_t$. However, I doubt this analysis is sufficient to prove the conservation of frequency (in linear response) because in the second medium the non-linear component might also be excited in which case $\mathbf E_t$ will contain waves with original frequency and waves with, e.g. double or triple, frequencies.

11. Aug 22, 2016

### Frank Castle

Yes, you're right. Thanks for pointing that out.

Ah ok, fair enough. Couldn't one simply state that the dominant response of the medium is linear as in the the case of the polarisation vector, $\mathbf{P}$? Is the point that the electric field acts like a driving force and as such, when it propagates into the medium it polarises the atoms within the medium causing them to oscillate, analogously to harmonic oscillators, at the same frequency as the driving force (the original electric field). This polarisation then propagates through the medium, with the same frequency as the original electric field?! Is there any way one could argue it mathematically?

Last edited: Aug 22, 2016