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Fermat's principle says that the path taken between two points by a ray of light is the path that can be traversed in the least time.

It occurred to me today that maybe the path is actually the one that covers the smallest distance through spacetime, and that in flattish places the distance would be measured using the Minkowski metric. It seems to me that under typical circumstances this ought to be slightly different from the path of least time, but the difference should be small because the $c$ factor in the Minkowski distance formula is very big. And perhaps this difference is too small to be detected unless someone was specifically looking for it.

Is any of this correct? I know hardly anything about physics, and I can't tell if this is correct, slightly wrong, absurdly wrong, or totally incoherent. If the latter, let me know and I will try to elaborate on what I meant.

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  • $\begingroup$ Not going to make this an answer because, after thinking on it, I have a question myself. But I think that, yes, if you take the principle to be expressed as "least time" you are right. However, IIRC a more modern expression of the principle is that the light takes a path that minimizes the number of oscillations it undergoes between source and destination. This is a number that should be independent of reference frame. Anyone know if I am remembering this correctly? $\endgroup$ – bob.sacamento Aug 4 '17 at 15:09
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In general relativity, it's not entirely clear what "least time" means, since you have to ask "whose time are you talking about"? Are you talking about the time as measured by the emitter? The receiver? Someone else "far away" from both of them? All three of these quantities might be different.

Rather, it's more productive to say that light rays travel on null geodesics in spacetime. A geodesic is a path that is "locally straight"; what this means is that there aren't any "nearby" paths between point A and point B that have a significantly larger or smaller integrated spacetime interval. So you're not wrong to say that the path of the light ray "travels the smallest distance through spacetime", if you replace "travels the smallest distance" with "is a stationary point of the spacetime interval". There's a similarity between these two notions, but the latter is more mathematically precise.

If you're familiar with the calculus of variations, you might want to take a quick look at the wikipedia page on geodesics in general relativity, particularly the bit above "curves of stationary interval". More detailed lecture notes can also be found here, or in many other places by searching for calculus of variations geodesic spacetime or some variant thereof.

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    $\begingroup$ Okay, suppose the experiment and the observer are in the same reference frame (That's something that can happen, right?) in otherwise empty space. Then the observer sets up some simple refraction experiment and very carefully measures the path of the light ray. Will they find that it is the path of least time in their frame of reference, or that it deviates slightly from that path? $\endgroup$ – MJD Aug 4 '17 at 16:00
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    $\begingroup$ This is a nicely-written, accessible answer. It's always refreshing when someone takes the time to do this. $\endgroup$ – uhoh Aug 4 '17 at 16:08
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    $\begingroup$ The spacetime interval traveled by light is zero, is it not? Is the distance light travels through spacetime not zero? $\endgroup$ – safesphere Aug 4 '17 at 23:04
  • $\begingroup$ @safesphere: It is, if travelling through a vacuum, since $dx^2 + dy^2 + dz^2 = (c\; dt)^2$. If travelling through matter at speed $\vec{v}=v \hat{x}$, it travels slower, so $ds^2 = dx^2 - (c\; dt)^2 = (v\; dt)^2 - (c\; dt)^2 < 0$, since $v<c$. $\endgroup$ – jvriesem Aug 5 '17 at 20:46
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    $\begingroup$ @MJD: I think what you're asking is whether the null geodesic principle I've postulated is equivalent to the principle of least time so long as you're talking about a single reference frame. I'm pretty sure the answer is yes; I'll try to work through a full proof and post it in the next couple of days. $\endgroup$ – Michael Seifert Aug 6 '17 at 14:24
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It is the least-time path.

Fermat's principle was itself built on two earlier observations: on the one hand the Greeks had noticed that in reflections, light traveled along the least-distance path; on the other hand Snell had discovered and Descartes had popularized that in refraction light obeyed a "law of sines" which said that the ratio of the speed of light in the two mediums was equal to the ratio of the sines of the angles that light made with the interface's normal-direction.

Now if light always took the least-distance path, light would not bend when it entered other mediums where its speed is different, like prisms or water. They are exactly equivalent if we do not include this refraction. What Fermat derived was that you could get Snell's law by assuming that instead of taking the least-distance path, light takes the least-time path, which is equivalent for the purposes of travelling in straight lines or reflections, but is different for refractions.

Now when light transitions from one medium to another, conservation of energy needs to hold at the boundary, which by $E=hf$ means that the photons coming in cannot change their frequency but must respond entirely by changing their wavelength. You can also understand this purely in a wave context without any quanta: the wave is continuous across the boundary which means that these peaks of wave-fronts must be consistent across the interface, but this means that the time rate of peaks entering the new medium must be the exact same as the time rate of peaks exiting the old one, which means the frequency is the same in both. In a deep sense this tells us that "time" is what we want to look at. Let me give you a bit of a taste for this.

Quantum theory gives us a radical reinterpretation of the least-time principle, saying that light maybe takes all paths from one point to another, but we only see the paths which matter most for its constructive interference. So, you may know that two waves can interfere constructively, $\cos(2\pi~f~t) + \cos(2\pi~f~t) = 2\cos(2\pi~f~t),$ or destructively, $\cos(2\pi~f~t) + \cos(2\pi~f~t +\pi) = 0,$ or perhaps by some combination of the two. The idea here is that we have a sum of a lot of different paths which each take some time $T$ to get there, so the final intensity of the wave is going to be a big sum of terms looking like $\cos(2\pi~f~t + 2\pi~f~T)$.

Now we take some path with some time $T$ and a bunch of the paths nearby it, what happens? Well usually those nearby paths will have a range of times, some longer and some shorter. Call the time difference $\delta T$. Since $f$ is a very large number any path that's more than about a micrometer or so longer is going to see a phase difference $f~\delta T > 1/2$ and we will get a bunch of constructive and destructive interferences which it turns out all cancel each other out in the big picture. So we derive that if light takes all paths, light cannot propagate at all! Great, right?

Well, okay, there's a catch. The catch is that if you look at any local extremum path -- could be longest-time or shortest-time -- then all of the nearby paths take approximately the same time and $\delta T = 0.$ It's sort of like, if we're talking about a big hilly landscape with rounded tops and valleys: if a few people are standing next to each other then it's very likely that some of them are standing above or below the other ones, because they're probably on a slope and then the slope puts some of them above or below the other ones. But there's a couple of places where this doesn't happen: at the very tops of hills or the bottoms of valleys, where in order to be "top" or "bottom" in fact locally the surface needs to be "flat". So then all of these people standing next to each other near the top or the bottom of the valley have to be at the same height. This is the same argument but the "people" are "paths light can take," and the "height" is the "time it takes" which turns out to be a wave phase term $2\pi~f~T.$

We can also play this game with distances but again one needs to alter the wavelengths when we consider refraction, because wavelength is not the same across interfaces. But since frequency is, we can just keep the same "minimum time" principle and make it work for refraction as well, in a way that we can't for minimum-distance.

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    $\begingroup$ So is Fermat's principle non-relativistic, then? $\endgroup$ – jpmc26 Aug 5 '17 at 17:48
  • $\begingroup$ I mean, it has to be special-relativistic because it comes from classical electromagnetism which is the whole raison d'etre of special relativity. However it is hard for me to phrase in a "co-moving" way, because the obvious thing would be to say "use the proper time!" but the proper time for light's paths are always 0. In addition I'd need to think about frequency shifts due to the relativistic doppler effect and such before I'd be totally comfortable saying "even though photons don't have a reference frame, Fermat's principle is Lorentz-invariant and applies in all reference frames." $\endgroup$ – CR Drost Aug 6 '17 at 4:30
  • $\begingroup$ And of course we know the result for general relativity which is that you look at the null geodesics and those are light's paths; the same "quantum" argument doesn't necessarily work because there is no satisfactory quantum gravity theory, but arguably classical waves on a manifold have to have a similar effect. $\endgroup$ – CR Drost Aug 6 '17 at 4:33
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  1. Well, the Fermat's principle is usually discussed in the context of a non-moving optical system fixed relative to a laboratory frame. The time $t$ that is (locally) extremized is the laboratory time $t$, not e.g. the proper time $\tau$. (The latter may not make much sense for massless particles!)

  2. Hence the system has a preferred reference frame, namely the laboratory frame. Therefore it is less interesting (but certainly possible and straightforward) to go to other inertial frames by applying Lorentz transformations to the laboratory time $t$.

  3. The Fermat's principle is certainly an approximation to full QED. See e.g. this & this Phys.SE posts. But it is not an approximation of SR once one realizes that the extremizing time $t$ is the laboratory time.

  4. Finally let us mention that the Fermat's principle looks superficially like the stationary action principle for the geodesic equation. See e.g. this & this Phys.SE posts. But the devil is in the details: In that stationary action principle, it's the proper time $\tau$ that's extremized (for massive point particles). For massless point particles, see this & this Phys.SE posts.

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Two comments. The first is about terminology. Space-time mixes time and space; it is the 4-dimensional "dimension" that we live in. Speaking about time or spatial distances separately requires us to agree on a reference frame (at the simplest an inertial one, but of course as we orbit the Earth, Sun, & Milky Way, our frame isn't inertial). Anyway, you need to be careful in talking about "time" or "distance" in space-time. And if you don't understand Relativity (either flavor, depending on the discussion), then you probably should avoid using either term. The other comment is about the least time path. Light (in a vacuum) travels at c. Nothing can travel faster. Often SR space time diagrams are represented as graphs with one spatial dimension (often the x-direction, but sometimes as a projection of all three space-like distances from the origin) and time as the vertical dimension. The units are almost always chosen so that light travels at a 45° angle. So, for any point A on this graph, if I mark a point B so that the angle between A and B, measured by parallels to the space axis, say, is LESS than 45°, and then draw a path between the points, then that path is "space-like" and hypothetical travel along it would be faster than light. These paths are not allowed. Your scenario would require such a disallowed path.

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Firstly, in connexion with optics, Fermat's principle is always and approximation: it defines an approximation, namely the first term in the WKB Approximation to the solution of either the quasi-time-harmonic Maxwell's equations or, in a more general setting, the Helmholtz equation. Here the WKB scale parameter is the wavelength, which is taken to be small compared to the features of a medium in question. As explained beautifully in CR Drost's Answer, the intuition behind this is that Fermat's principle defines paths of stationary phase around which diffraction effects are added to get the complete solution to the problem and is extremely widely applicable without approximation, i.e. Fermat's principle will give, without approximation, the first term in the relevant WKB expansion in all the situations discussed below. These include all special relativistic (flat spacetime) problems and static general relativistic ones too.

The extremized quantity, or the optical Lagrangian, is the optical path length expressed as a phase difference between the ends of the ray path - in waves or radians, for example. A careful reading and mulling over CR Drost's answer clearly shows this fact. So it is not probably not helpful to think of it as least time principle since, as in Michael Siefert's answer this is can be problematic in relativity. In contrast, the phase field of a steady state optical excitation in a medium is a scalar field i.e. it transforms as such.

In Special Relativity, one can see how the principle plays out in two different, relatively boosted inertial frames by looking at the steady-state optical field in question from the two different frames at the instant when the origins of their co-ordinate systems co-incide. Let's put one of the frames at rest relative to the material mediums that the field is established in. Fermat's principle plays out in the wonted way in this frame.

The relatively moving observer sees the medium in Lorentz-transformed co-ordinates. Maxwell's equations are still Lorentz covariant with the medium present, but the medium properties and the constitutive relationships transform radically. Intuitively you can see this is so; the Lorentz Fitzgerald contraction changes the medium's optical density anisotropically. In fact, if we have a simple, anisotropic medium in the rest frame with electric and magnetic constants $p_e$ and $p_m$ (One shuns epsilons and mus in these kinds of calculations to avoid confusion with Greek indices on tensors), the relatively moving observer sees an anisotropic magnetoelectric constant such that:

$$\vec{D} = p_e\,\vec{E} + c^{-1} \vec{v}\times \vec{H}$$ $$\vec{B} = p_m\,\vec{H} - c^{-1} \vec{v}\times \vec{E}$$

where, naturally, $\vec{v}$ is the relative velocity.

The upshot of all of this is that both observers calculate the same scalar phase field from their version of Maxwell equations and so a ray path is an extremal optical path length path in one frame if and only if it is an extremal path in the other. So we see that the Fermat principle gives us the same rays in both cases.

A quirk here is that, in the anisotropic medium as seen by the relatively moving observer, Snell’s law does not apply to rays at interfaces, although it does apply to wave vectors. Phase fronts are not needfully normal to the Poynting vectors. This is the same situation as in an anisotropic crystal. But Fermat’s principle still applies.

In General Relativity we must be a little careful. The optical Fermat principle applies time-invariant mediums. Therefore, it cannot be applied (at least I am not aware of any extension) to nonstatic spacetime - or at least one without a timelike Killing field - with or without material mediums. This is because, in the first instance, Fermat's principle applies to time-harmonic electromagnetic fields, with pulses and the like being described by Fourier superpositions of time-harmonic solutions;

But for a static, curved spacetime, the situation is similar to the special relativistic one. Different observers see a medium’s material properties and constitutive differently, but such that they would all agree on the scalar phase field for a given steady state optical excitation of the medium, and they would all calculate the same ray paths from the Fermat principle.

In fact, an empty, medium-less curved spacetime has the constitutive relationships (Plebanski's constitutive equations, see J. Plebanski Phys. Rev. 118 (1960), p1396:

$$D^i = -\epsilon_0\,\frac{\sqrt{-g}}{g_{0 0}}\,g^{i j}\,E_j + c^{-1} \varepsilon^{ij}_{\,\,k}\,\frac{g_{0j}}{g_{00}}H^k$$ $$B^i = -\mu_0\,\frac{\sqrt{-g}}{g_{0 0}}\,g^{i j}\,H_j + c^{-1} \varepsilon^{ij}_{\,\,k}\,\frac{g_{0j}}{g_{00}}E^k$$

where we have summed over spatial indices $1,\,2,\,3$ only (note the Roman, rather than Greek indices) and the $\varepsilon$ is the three dimensional, spatial Levi-Civita symbol. This observation is the starting point for the field of transformational optics: the use of metamaterial mediums to mimic propagation in the spatially curved part of static curved spacetime. These ideas show great promise for the realization of optical cloaking devices, for example: material mediums whose electric, magnetic and magnetooptic constants match the above propagate light, as does of course empty curved space, without scattering. Light can be bent around objects by such mediums without scattering and it's not too hard to see that the object to be cloaked can be hidden inside regions not reachable from an observer. See, for example:

Ulf Leonhardt and Thomas G. Philbin, "Transformation Optics and the Geometry of Light", Prog. Opt. 53, pp69-152 (2009

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protected by Qmechanic Aug 4 '17 at 21:33

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