Can you bend light to go in a circle?

Question

Is it possible to bend light so that it forms a circle and goes round and round indefinitely without losing energy?

Answer 1

How could one manipulate light? It does not have mass, it does not have electric charge. For that matter, it also does not have any color or weak charge. There seems no way to change its direction of motion.

Black Hole

General relativity describes how masses can create curvature in spacetime. If you have enough mass, it will get curved significantly. Light will follow this curvature, because light will go “straight” which will become curved in curved spacetime. Right at the Schwarzschild radius of a black hole, the escape velocity is the speed of light. That means that a photon there trying to go straight away from the black hole will not get any further, although it moves with the speed of light.

That is not a closed orbit, of course. As Jerry Schirmer pointed out in the comments, a closed orbit happens at $r = 3M$ where $M$ is the mass of the black hole. The problem with this orbit is that it is unstable. Any perturbation will either send the photon away from the black hole or lets it spiral into the singularity. Either way it breaks from the closed orbit.

Since a photon has an energy, it also creates spacetime curvature. A moving photon will therefore radiate gravitational waves, although they will be minuscule. However, they are sufficient perturbation to prevent the orbit from being closed forever. This could be prevented by using a solid ring of light such that the mass density along the orbit is constant. Then no gravitational waves would be emitted.

If the Hawking temperature of the black hole does not exactly match the temperature of the ambient universe (think of the cosmic microwave background), the black hole will grow or shrink. This will change the radius of the orbit and also prevents an orbiting photon for eternity.

All in all this is very unstable and will not work out.

See also:

• https://en.wikipedia.org/wiki/Schwarzschild_radius
• https://en.wikipedia.org/wiki/Kruskal%E2%80%93Szekeres_coordinates

Wave Optics

Another possibility is to use refraction of light. If you have an optical medium with different optical densities (different index of refraction $n$), light will also bend. This is how a lens works. With the right setup of lenses one can refract light to go around a path. You could even set up three mirrors and let the light go round and round in a triangle!

The optical fiber is a bit more sophisticated, it has a gradient of the optical density and can therefore smoothly direct the light around a curve.

Quantum Electrodynamics

With quantum electrodynamics, there is the tiny interaction of light rays with other light rays. Although light has no charge in itself, it can couple to virtual charged fermions and create a closed loop that couples four photons in total. If you have enough light around in a particular configuration, one could bend light rays with that. However, I fear that this is not realizable in any experiment.

See also:

• https://en.wikipedia.org/wiki/Euler%E2%80%93Heisenberg_Lagrangian

The Point?

Another valid issue was raised in the comments: If you would have this situation successfully set up, how would you know that it is working? If you try to observe the photon, you would change it. If it radiates something to the outside (scattered light, gravitational waves), it would lose energy over time and leave the orbit.

Answer 2

Edited version, with additional information and correcting that @Jerry Schirmer was wrong. He was right on the photon sphere.

This expands on part of the answer having to do light orbits near Black Holes (BHs), and actually in other gravitational fields. You can indeed have closed light orbits near but outside BHs and it is interesting what they represent. You could also have closed light curves in cosmology, but only in certain cases and not in all.

Around a spherical (static, Schwarzschild) BH there is only one possible way in which light can orbit: it is at a distance R = 3/2 $R_s$ = 3M, with M the BH mass and $R_s$ the horizon or Schwarzschild radius of the BH. That was correctly pointed out by @Jerry Schirmer in his comments. The sphere at that radius is called the photon sphere, and a photon at that distance moving horizontally will orbit and come back around. Anything closer in or further out is not a possible closed orbit for light.

See the math and physics at Wikipedia at https://en.m.wikipedia.org/wiki/Photon_sphere

You can also see there (though it doesn't derive it mathematically) that for a Kerr BH (stationary, spinning) the only circular orbit is at the equatorial plane, and there are two possible different orbits, along the BH rotation and against it.

But bodies with mass, and enough momentum, can go inside the photosphere and still come out, in an elliptical orbit. Also, an accelerated observer (i.e., not free falling, say one with rocket motors blasting away), can be inside the photon sphere and hold its radial distance or fly out.

But any photon (or light) sent inwards, at the photon sphere will fall into the BH, and any sent outwards from inside the photon sphere, but outside the horizon, escapes permanently.

Those light orbits are not stable, a slight kick in will make the light go into the horizon, and a slight kick outwards will have it escape. The orbits won't last long.

Note that to orbit the photon sphere distance has to be outside the body, if it is not a BH. So you can have those orbits around BHs, but it could also happen outside a small and dense enough neutron star. It's unlikely, I've read that there is a small chance around a neutron star, with that photon sphere outside the surface of the neutron star, and obviously no horizon.

As for other gravitational conditions, it is possible to have a cosmological solution where the spatial hypersurfaces are closed 3 spheres, i.e., the closed positive curvature Robertson Walker solution to the Einstein equations for the universe. That solution is not favored by the data which indicates a most likely flat universe, but the uncertainties don't totally rule it out. A light ray will go around the universe and come back behind you - if you waited long enough for that trip you'd see yourself. EDITED HERE FROM DVORAK's COMMENT BELOW As he points out the universe is expanding too fast for the light to go around, even a closed universe. Probably then the only way would be a topological non trivial universe with some region or boundary connected to another, such as in a flat PacMan topology. END EDIT. But there is some astronomical search still for possible multiple images of a galaxy or cluster, which could indicate that geometry is responsible. Of course there's been no such findings.

In the movie interstellar there are physically semi-accurate simulated images of the BH. It is a different story, we see light around it but it is NOT the photon sphere. See below what it looks like. It's from the PSE question and answers at What does this depiction of a black hole in the movie Interstellar mean? The disk through the center of the BH is the accretion disk of matter orbiting around and getting pulled in - it's highly energetic, lots of collisions and very hot. The circle is the image of light sources behind the BH, they bend around it; we see those in images from other real astronomical heavy objects, but usually not as well defined and sometimes just multiple images of the same few stars behind it.

Also see here http://hubblesite.org/explore_astronomy/black_holes/encyc_mod3_q11.html how BHs can bend light, similarly to the more idealized/cinemantic effects in the movie, right below

Answer 3

The spatially-closed, lightlike path arising in non-Minkowskian spacetime has already been addressed well in detail by Bob Bee's Answer and Martin Ueding's Answer, so I'm going to focus on an answer grounded wholly on Maxwell's equations for lossless dielectric mediums in flat, Minkowskian spacetime.

In this case, the answer is most definitely yes - it is basically the idea of an optical fiber loop, and, moreover, it can theoretically be done in a perfectly lossless way. It is not as weird or wonderful as the idea seems at first; indeed, it is simply a particular case of a resonant cavity mode called a whispering gallery mode. I have sketched two two-dimensional (i.e. of infinite extent in the $z$-direction out of the page, and with $z$-translational invariance symmetry) dielectric structures below and we'll analyze them in cylindrical polar co-ordinates below; analogous discussions hold for a circular cross section optical fiber bent into a torus and analyzed with toroidal co-ordinates but the much more doable problem below illustrates the physical principles well.

The structure on the left is a high refractive index ring of finite radius surrounded by low refractive index regions. That on the right is a dielectric region surrounded by a perfect conductor. I should think that a reasonable interpretation of your question is "can we set up a field with the Poynting vector $\mathbf{S}$ tangent to the ring, or substantially in the direction of increasing polar angle, as shown below?".

The answer (I sketch how to show this further down) is definitely yes. What you end up with are whispering gallery modes of the structures, i.e. in the left hand structure, the Poynting vector points tangent to the ring channel (in the large structure limit) and in both structures the the field's phase everywhere varies as $e^{i\,\nu\,\varphi}$, where $\nu$ is an integer - a very large big one if the ring is many wavelengths wide for the correct phase velocities.

The point is that these modes are exact solutions of Maxwell's equations, so how does this square with the well known fact that when you bend an optical fiber, it will loose light, particularly in the case of the device on the left above?

Firstly, these are not a practical devices to use: there is no way to get light in or out of them. Secondly, losses do indeed arise from bends, but in these idealized structures there are resonant conditions (which manifest themselves as the eigenvalue equations I sketch below) where radiation is coupled back into the waveguiding structure near where it leaves, and with the nett result of zero loss and zero power transfer in the radial direction, owing to the device's precise shape and the tuning of this shape to the resonant frequnecy. It is well known that a constant curvature bend has modes such as described, but if one tried to exploit these for zero loss bends, you have to have transition regions along the fiber where the curvature changes so that you can access the bend, and radiation is shed at these points where the curvature changes. See:

William L Kath & G. A Kriegsmann, "Optical Tunnelling: Radiation Losses in Bent Fibre-Optic Waveguides", IMA J. App. Math. 41(2):85-103 · January 1988

The device on the right is less mysterious, since a perfectly conducting barrier clearly leaves no way for light to leave this structure. The light can bounce around indefinitely off the perfect conductor, and, if the radius of the device is large compared with the wavelength, the Poynting vector is everywhere almost exactly in the direction of increasing polar angle.

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Sketch of Solutions

I shall use the Riemann-Silberstein notation for the electromagnetic field (basically because I can lift all the equations I need from former work!); in this notation, the field variables are the positive frequency parts of the entities $\mathbf{F}_\pm = \mathbf{E} \pm i\,c\,\mathbf{B}$. The Maxwell curl equations then become the two decoupled equations:

$$i\, \partial_t \mathbf{F}_\pm = \pm c\,\nabla \times \mathbf{F}_\pm\tag{1}$$

With some grunt work, you can solve these with a solution of the form $\mathbf{F} = e^{i\,\nu\,\varphi-i\,\omega\,t}\,(F_r(r),\,F_\varphi(r),\,F_z(r))$ where we use cylindrical polar co-ordinates, $\nu$ must be an integer to make the field single valued and:

$$F_r(r) = \frac{1}{r}\left(a\,H_\nu^+(k\,r) + b\,H_\nu^-(k\,r)\right)\tag{2a}$$ $$F_\varphi(r) = \frac{i}{\nu} \frac{\mathrm{d}}{\mathrm{d}\,r}(r\,F_r(r))\tag{2b}$$ $$F_z(r) = -\frac{i\,k}{\nu} r\,F_r(r)\tag{2c}$$

where $H_\nu^\pm(k\,r)=J_\nu(k\,r)\pm i\,Y_\nu(k\,r)$ are the Hankel functions (I like to call them "inwards" and "outwards" Hankel functions owing to their asymptotic behavior of $e^{\pm i\,k\,r}$, i.e. their approach to inwards and outwards propagating waves). We also exclude the case $\nu=0$ because in this case the phase does not vary with $\phi$, i.e. this isn't a solution where the wave runs around the ring. We get a left / right circularly polarized solution $\mathbf{F}_+$ / $\mathbf{F}_-$ to the $+$ / $-$ alternative in (1) by making $k$ positive or negative, respectively, in (2).

For the device on the left, we proceed as follows.

The continuity of tangential field components at interfaces is equivalent to the continuity of the functions $G(r)=r\,F_r(r)$ and $\mathrm{d}_r (G(r))$ across the interfaces.

In the central region within the ring, the integration constants $a$ and $b$ are equal to cancel the Neumann (second kind Bessel function) logarithmic branch point at the origin so that our solution is physically reasonable. We therefore assume a solution of the form $J_\nu(k\,r)$ in the inner region. In the high refractive index region and outer region, we assume solutions of the form $G(r)=a_{co}\,H_\nu^+(k_{co}\,r) + b_{co}\,H_\nu^-(k_{co}\,r)$ in the high refractive index ("core") region and $G(r)=a_{cl}\,H_\nu^+(k_{cl}\,r) + b_{cl}\,H_\nu^-(k_{cl}\,r)$ in the "cladding" region.

The condition of continuity of both $G(r)$ and $\mathrm{d}_r G(r)$ at each of the two interfaces yields four equations for the integration constants $a_{co}$ and $b_{co}$ in the core $a_{cl}$ and $b_{cl}$ in the cladding.

These equations are straightforward, if messy, to solve.

Now it can be shown that if $|a_{cl}| = |b_{cl}|$, the radial component of the Poynting vector (which is $\mathrm{Re}\left(-i \sqrt{\frac{\epsilon}{\mu}} (\mathbf{F}_+ \times \mathbf{F}_+^*-\mathbf{F}_- \times \mathbf{F}_-^*) \right)$ in Riemann-Silberstein notation) then the radial component of the power vanishes and we have a whispering gallery mode of the structure: no power is being transferred into or out of the structure from far off. This condition, imposed on the expressions for $a_{cl}$ and $b_{cl}$ above defines an eigenvalue equation for $k$: there are only certain frequencies where these whispering gallery modes exist. At these frequencies, the Poynting vector is tangent to the high refractive index channel. At these frequencies too, the Poynting vector integrated over the transverse plane is also nought.

Moreover, there are only a finite number of such resonances.

There are always solutions for the integration constants, and the Maxwell equations solution in this case represents the case where there is power transfer continuously across the structure from afar: the waveguide is simply immersed in a field whose power source is far off.

The device on the right is easier to analyze. Here the radial electric field components must vanish at the conductor, which gives the eigenvalue equation for $k$ as $J_\nu(k\,R)=0$, where $R$ is the waveguide's radius. If we choose a very large value of $\nu$, the field is concentrated near the outer conductor, and the Poynting vector is indeed almost perfectly tangential to the conductor in the region of the high field. It is easy to solve this eigenvalue equation numerically in something like Mathematica. For example, The eigenvalue equation $J_{500}(k\,R)=0$ has the solution $k\,R=514.859311690494$; the reader is invited to make plots of the Poynting vector of the mode defined by $F_r(r) = \frac{1}{r} J_{500}\left(k\,\frac{r}{R}\right)$ in (2).

Answer 4

The "Sagnac Effect" (and related effects) means that it is useful in any optical gyroscopes to send light around and around in a loop.

So, a Ring Laser Gyro is typically set up as a three-mirror triangle in which light goes around and around. It loses energy because e.g. the mirrors are not perfect, but it gains energy (to compensate) because it's a laser.

In an IFOG, light goes around in a fiber-optic loop for maybe 1km or so. They don't actually close the loop for obvious practical reasons: They want to put light in and take it out. It does get gradually absorbed in the fiber--nothing is perfect. As before, in principle, you could imagine putting amplification (e.g. EDFA) into the loop and closing the loop to keep the light going around forever. (But this is not a useful thing to do in practice.)

You can also look up whispering gallery microresonators. Again, light goes around and around, though not forever. After spinning around for a few km, it has mostly been absorbed.

(Whenever light interacts with matter, there will be some absorption, however slight. Nothing is perfect.)

Answer 5

In photonics, ring resonators (RR) and microtoroids are two possible examples of this. However, in the case of the RR, light which couples to the the ring will re-couple to the photonic wire which was used to couple the light in the first place and in both of these examples you have the problem of loss which is inherent to all materials which would also deplete the energy from the cavity eventually.

This is a picture of a microtoroid which was fabricated by Caltech:

And this is a picture of how the light is coupled onto the microtoroid with a laser:

Mircotoroids and Ring Resonators are very useful in sensors and detectors of molecules. If you want to learn more about such devices, it would be good to read about and understand coupled mode theory first. Some good references can be found in the works of D. Marcuse and A. Yariv.

What determines the lifetime of the photon in such a cavity is something called the Q-Factor. Researchers who study such ring resonators and other such devices work hard to get the Q-Factor as high as possible to increase the lifetime of the photons in these devices. There is a good calculator for this on RF photonics encyclopedia