In the derivation of Snell's Law from momentum conservation, the tangential component (parallel to the plane which separates the two media) of a photon's momentum is not changed, but what is the physical explanation or insight to conclude it?
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3$\begingroup$ It says so in the paper itself: translational invariance parallel to the boundary. $\endgroup$– AegonCommented Aug 6, 2016 at 15:37
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$\begingroup$ Yes, but I don't know what it exactly means in sense of if it's a mathematical or physical property. $\endgroup$– thiagoCommented Aug 6, 2016 at 18:09
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2 Answers
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It is a manifestation of Noether's theorem.
In short, if you have translational invariance in particular direction you have a conservation of momentum in this direction.
Use of Hamilton equations lets you prove it very easily. Suppose there is a translational invariance in $i$-th direction. Hamiltonian ${\mathcal H}$ does not depend on this coordinate and partial derivative with respect to it is zero. Then:
$$ \dot{p}_i=-\frac{\partial {\mathcal H}}{\partial q_i}=0 $$
Thus $p_i=$const.
In your case Hamiltonian has translational invariance parallel to the boundary (it doesn't matter where to refract along the boundary, since medium is uniform).
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$\begingroup$ Thank you for your answer. The momentum is conserved because "there is translational invariance", but in the case of refraction what is allowing one say that there is such direction which translation is invariant? $\endgroup$– thiagoCommented Aug 7, 2016 at 2:22
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Momentum is always considered as only the perpendicular component having an effect. Same when billiard balls collide.
The reason is understood from simple classical mechanics:
During a collision, a normal force is created, always being perpendicular to the surface (to the contact point). So only perpendicular acceleration can happen, meaning only the perpendicular component of $v$ can be altered. And thus only the perpendicular component of momentum, $p_y=mv_y$, is affected.
No parallel forces means no change in parallel motion.
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1$\begingroup$ Yes, just as occurs in reflection. It's quite clear now. Thank you both for answering. Since I haven't studied CM, Steeven's answer seems more clear to me. $\endgroup$– thiagoCommented Aug 7, 2016 at 11:22
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2$\begingroup$ Sure, Steven's answer is more physical. Since the paper you was referring was using Hamiltonians I thought it would be useful to explain it in that context. $\endgroup$ Commented Aug 8, 2016 at 21:42