I Snell's law from a mechanical model

[greypilgrim]

Hi.
I'm trying to derive Snell's law from a mechanical model: An axle of length $d$ with two wheels that run at speeds $v_1$ or $v_2$, depending on the medium they are in ($v_1>v_2$ for the moment). The $x$-axis is the interface, $v_1$ for $y>0$ and $v_2$ for $y<0$. The wheels come in at an angle $\alpha_1$ from the top left and the origin is where the first wheel (the right one in direction of motion) enters the slow medium. Here's what I did:

1. Calculated the turn radius (measured to the outer, i.e. left, wheel) $r$ using proportionality with $d$, $v_1$ and $v_2$.
2. Calculated the centre of the turn using trigonometry with $(r-d)$ and $\alpha_1$.
3. The turn is over when the left wheel enters the slow medium. Hence I intercepted the turn circle of radius $r$ with the $x$-axis and took the positive solution.
4. Calculated the exit angle $\alpha_2$.

Now I did this for a couple of random numerical values and got exactly the same angle as with Snell's law. However, as soon as I try this with variables, the expressions (especially the solution of the quadratic equation in step 3) get hopelessly complicated and not even Mathematica was able to simplify this to something resembling Snell's law.

Is there a better way to do this, or maybe a cleverer choice of the coordinate system?

 

[Ibix]

Seems easy enough if you just impose the condition that the turn center must be the same height above/below the interface at the time each wheel crosses it. Check you aren't doing anything silly like using $r$ for both the turn radius and the angle of refraction (consider possible lack of case sensitivity if you're using $r$ and $R$).

 

[greypilgrim]

I don't think I'm doing anything conceptually wrong. As I said, I got the correct result with different random numbers. It's that the symbolic calculation becomes intractable at step 3 because the solution of the quadratic equation is a huge formula that doesn't seem to simplify.

 

[Ibix]

I didn't get a quadratic. Post what you did get to before it became unmanageable.

 

[A.T.]

Now I did this for a couple of random numerical values and got exactly the same angle as with Snell's law.

That's surprising if you used an axle of constant length. In most wavefront diagrams the beam width seems to change.

From: https://www.telescope-optics.net/reflection.htm

 

[greypilgrim]

I got it now. I found that the $x$-position of where the left wheel enters the slow medium does not need to be calculated, and then it becomes straightforward.

However, before that, I finally found an expression for $x$, that is still ugly (and eventually also leads to the correct result), but not as unmanageable as the Mathematica solution suggested.

Nice, so this memory aid is not just that, but even works quantitatively!

What I wonder now is if the exiting beam after the curve is somehow shifted compared to the usual version with two straight lines that meet at the interface. But I'm not sure if that might even depend on the point on the axle I'm looking at...

 

[greypilgrim]

Is it just coincidence that this mechanical model leads to Snell's law as well, or is it more than just an analogy to the propagation of light?

 

[Ibix]

I lean towards coincidence. The underlying physics is very different, and (besides @A.T.'s comment about beam width) you get different behaviour at the boundary - rigid rotation versus discontinuous slope.

 

[greypilgrim]

you get different behaviour at the boundary - rigid rotation versus discontinuous slope.

Seems like the same discontinuous slope emerges in the limit of the axle length going to zero.

What about taking the approach via Fermat's principle, light taking the path of shortest time? Can the axle and wheels be written in Lagrangian or Hamiltonian formulation, and how would that look like? Seems tricky because of the discontinuity in velocity at the boundary, how to account for the kinetic energy changing?

 

[hutchphd]

I lean towards coincidence.

What about taking the approach via Fermat's principle,

I believe this leads to the principle of stationary phase which is dictated by least action (or time in this case). I don't have the time/energy to work it out in detail right now...

 

[Ibix]

Seems like the same discontinuous slope emerges in the limit of the axle length going to zero.

If you have an axle of length $L$ with one wheel having linear speed $v$ and the other $v/n$ then the wheels follow concentric circles of radius $R$ and $R+L$ which must satisfy $$\begin{equation}\frac{v}{R+L}=\frac{v}{Rn}\end{equation}$$so that they have equal angular velocity. Requiring that the circular motion starts when one wheel enters the medium and stops when the other one enters leads to $$\begin{equation}R\sin(i)=(R+L)\sin(r)\end{equation}$$You can solve both of these for $(R+L)/R$ and equate the other sides to get Snell's Law. But note that in the case $L=0$, (1) reduces to $n=1$ and (2) reduces to $\sin(i)=\sin(r)$. In other words, your model tells you that a zero-length axle doesn't "refract" at the boundary at all.

Can the axle and wheels be written in Lagrangian or Hamiltonian formulation

All physical systems can be expressed that way. But look at the diagram in post #5, specifically a wavefront crossing the boundary. Will an axle ever have that shape? If not, an axle is never going to accurately model a wavefront crossing a boundary, however complicated a mathematical approach you take.

 

[greypilgrim]

But note that in the case $L=0$, (1) reduces to $n=1$ and (2) reduces to $\sin(i)=\sin(r)$. In other words, your model tells you that a zero-length axle doesn't "refract" at the boundary at all.

That's why I said "limit", not "case $L=0$".

But look at the diagram in post #5, specifically a wavefront crossing the boundary. Will an axle ever have that shape?

No, but neither will a real wavefront be clipped off sharply on the sides as in this diagram.

It just find it hard to believe it to be utter coincidence for those systems to follow the exact same law, but I might just have to accept that.

 

[A.T.]

I lean towards coincidence. The underlying physics is very different, and (besides @A.T.'s comment about beam width) you get different behaviour at the boundary - rigid rotation versus discontinuous slope.

I guess it depends on what one means by the word "coincidence". On the math level, one could also call it a "neat feature" that those two differences (width vs. slope continuity) cancel each other, yielding the same final refraction angle.

It just find it hard to believe it to be utter coincidence for those systems to follow the exact same law, but I might just have to accept that.

On the physics level, you might have a point that it follows from least action principle. There was a nice video by Derek Mueller on this recently: