# Is it possible to create a mirror that redshifts light?

Mirrors are able to reflect light but are not perfect and after a number of reflections, light loses intensity.

However I wonder, during the reflection by a different type of mirror, could the light photons lose some energy and thus be red shifted instead of just losing intensity?

I am not talking about light becoming more red, as a simple color filter on top of a mirror could do, but rather e.g. blue becoming green and green becomes red, etc.

Does such a mirror exist?

Also I am not looking for a digital system with a camera and screen. I am looking for physical processes that would shift light frequency.

It would seem there is no natural pitch shifter for acoustics too (swallowing helium gas is not pitch shifting an existing sound).

Addendum: I am looking for a static device, not a moving mirror.

Simulation:

Very simplistic simulation of how you would look like in such a mirror, assuming you are a brunette wearing a red top.

Left picture is the original. Middle is shifted: blue becomes green, green becomes red, red is invisible (black). Right: one bigger shift, blue is now red and the rest is gone to infrared.

This is just an illustration (or even dramatization, given the amount of shift), don't take it too literally in terms of redshifts I am interested in. Also this assumes there is no UV or higher energy photons in the original scene that would become visible after being shifted.

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– Chris
Mar 27 '21 at 22:30

Yes, this is possible using nonlinear optics.

This kind of frequency shift can be done using acousto-optic modulators and electro-optic modulators, and it is normally done in a transmission geometry. The basic idea is that you have a block of material whose refractive index depends on the acoustic pressure or on the local electric field, and then you make that pressure (resp. electric field) oscillate by driving it with an acoustic or radio wave. This induces an oscillation in the refractive index of the material, which then induces a frequency shift in the light that's transmitted.

If you want to work in a reflection geometry, then it's probably possible to re-work the EOM principle of operation so that the frequency shift happens to a reflected component. The simplest idea there would be to set up the driving field to create a grating in the direction of propagation, so that the incoming light suffers a Bragg reflection from it $$-$$ and then you adjust your driver so that this Bragg grating propagates over time. In essence, you create an effective "moving mirror" without any matter getting transported.

(... or, of course, you can just use a transmission modulator and put a mirror behind it.)

Now, in the real world, this won't impart a huge frequency shift. With realistic devices, the largest shift you can get from an EOM is a few GHz (whereas, for comparison, visible light is in the hundreds of THz). But the principle is there, and there is no physical law that forbids this from happening at higher frequency shifts. (Though then again, if you push too hard in this direction, then it will just start looking like a difference-frequency-generation configuration, and those are also viable candidates for the functionality you're asking about.)

Oh, and also $$-$$ note that these principles are equally applicable to acoustic waves, and it should also be possible to build equivalent devices there. (Just saying.)

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– Chris
Mar 27 '21 at 22:33

As Jon Custer said in a comment, the light returning from a receding mirror will be redshifted. This is easiest to see if you consider the problem in the rest frame of the mirror.

A symmetry argument shows that light returning from a stationary mirror can't be Doppler shifted. If the experimenter emits two light wavefronts a time $$δt$$ apart, the worldlines of the wavefronts will be time-translated copies of each other, so they will also arrive back at the experimenter separated by $$δt$$. This argument works in any GR spacetime in which the metric and the positions of the experimenter and mirror are independent of $$t$$ in some coordinate system. For example, you can't get a redshift by putting the mirror higher or lower in a gravitational field.

I can't prove the nonexistence of an exotic material that applies some sort of pitch-scaling algorithm to the waveform, similar to what is supposed to happen in tired light cosmology, but I don't see how it could work, and I don't see any workable suggestions in that article.

Emilio Pisanty's answer says that a redshifting stationary mirror is possible with nonlinear optics. Note that the symmetry argument above does apply to that device, regardless of the details. Over time scales at which the device is stationary (i.e. steady-state), the average frequency shift must be $$1$$. A simple example of a device that redshifts light some of the time is a mirror that repeatedly recedes from the light source and returns much more quickly to its starting position. This redshifts most of the light by a small factor and blueshifts a small fraction of it by a large factor (which approaches infinity as the return speed of the mirror approaches $$c$$). You can cover the mirror while it's returning so that the sender doesn't see the blueshifted light, but you can't avoid the gap in the redshifted signal. A stationary mirror that continually, uniformly redshifts light is impossible.

However, all of this also applies to any redshift mechanism that would work in tired-light cosmology. Emilio Pisanty's device may well qualify as a "pitch-scaling" device for the purposes of the last paragraph of my original answer, so I'm retracting it.

• "For example, you can't get a redshift by putting the mirror higher or lower in a gravitational field" Aren't photons redshifted in earth's gravity field? Mar 25 '21 at 19:17
• @DescheleSchilder One-way yes, round-trip no. This argument applies only to light that returns to the sender. Mar 25 '21 at 19:53

Apart from the obvious relativitic mirror, mentioned in the other answers, in linear optics a frequency shift is not possible, because linearity implies that a wave of a specific frequency/wave number is only changed in amplitude, not in frequency by a transfer element.

Therefore, what you might look for is nonlinear optics. Might be that there are some materials where the degree of nonlinearity can be controlled by an external field or something, but I am no expert in that. Plus, I think this will not influence the frequency of the harmonics/subharmonics generated by the nonlinear medium, but only their amplitude.

PS: if there is a way to modulate the amplitude of the light fast enough, you will automatically get side bands, which correspond to "shifted frequencies". That is basically the principle of AM radio transferred to light. The faster you modulate, the bigger the shift. But again, no clue about the experimental realization. And you would only shift a fraction of the incoming intensity.

• Actually Emilio Pisanty's method is light modulation by phonons, no? Mar 28 '21 at 6:45

In addition to a receding mirror that @joncuster mentioned and @benrg elaborated on, you can coat any normal mirror with a wavelength shifter. There are many such materials that will absorb incoming short-wavelength photons and re-emit them at longer wavelengths. Often these wavelength shifting layers are transparent to their own light, so no matter which direction the re-emission of the (redshifted) photons happens, it will either directly bounce back or first hit the mirror and thus bounce back as well.

• Wow. I did not know of the existence of wavelength shifters, thanks. I read Wikipedia article about them. The shift is not continuous and will apply to spectral lines only, right? Mar 25 '21 at 17:07
• Both absorption efficiency and emission spectra will depend on wavelength, and that dependence is different for various chemicals used as wavelength shifters. Both spectra can be complicated, typically spanning a 50-300nm or so with a couple bumps.
– rfl
Mar 25 '21 at 17:14
– JEB
Mar 25 '21 at 17:23
• That is just semantics. A mirror coated with wavelength shifter will in effect work like a mirror that shifts wavelength.
– rfl
Mar 25 '21 at 23:07
• @winston: no, it's not going to be a nice mirror at all. the fluorescence from the wavelength shifter goes in all directions. So you loose the directionality of the photons that are thrown back.
– rfl
Mar 25 '21 at 23:15

The receding mirror has already been mentioned in multiple answers but it obviously going to very quickly become useless because it's too far away.

There's another approach that unfortunately requires building it out of unobtainium:

We need a track with two straight sections and two 180 degree bends. (Longer footraces are typically conducted on something of this shape.) The length of the bends (if laid straight) is the same as the length of the straight pieces. On this track there are 4 mirrors going round and round.

Unfortunately, this results in the next mirror occluding the currently receding one. Thus we need to make the mirror out of infinitely thin material and slice it into tiny strips. While receding the strips are perpendicular to the motion, otherwise they turn 90 degrees and the light passes on by so it's only bounced by the active mirror.

The travel of the mirrors is at constant velocity, so long as the bearings that turn it are frictionless there's no energy use. Unfortunately, the flipping of the mirrors needs power--and lots of it.

Obtaining the unobtainium, powering it and keeping it cool are left as exercises for the reader.

• Thanks. But this is basically another version of sending a mirror away and getting it back. Mar 27 '21 at 6:37

Any fluorescent dye will do, as long as you tolerate some scattering (errr, like 100% scattering).

No.

Such a mirror would violate special relativity.

The Doppler shift is not an afterthought, it is fundamental to special relativity (see the Bondi $$k$$-calculus, https://en.wikipedia.org/wiki/Bondi_k-calculus, for a complete rundown).

In lieu of that, consider the following: all objects have a four velocity (here in 1 spatial dimension):

$$u^{\mu} = (\gamma c, \gamma v)$$

with fixed magnitude:

$$\sqrt{u^{\mu}u_{\mu}} = [(\gamma c)^2-(\gamma v)^2]^{\frac 1 2} = c$$

That means the 4-acceleration is always orthogonal to 4-velocity; thus, the 4-momentum transfer from the photon to mirror has to be orthogonal to the (stationary) mirror's 4-velocity:

$$u^{\mu} = (c, 0)$$

The initial and final photon 4-momenta are:

$$p_i^{\mu} = (p, p)$$ $$p_f^{\mu} = (p', -p')$$

given a 4-momentum transfer:

$$q^{\mu}=p_f^{\mu}-p_i^{\mu}=(p'-p, p'+p)\equiv (\Delta E/c, 2\bar p)$$

The last form is just the photon energy difference and the average momentum.

Now: dot that into the mirror's four-velocity:

$$q^{\mu}u_{\mu} = (c\Delta E/c) - 0\cdot \bar p = \Delta E \equiv 0$$

Hence, for a stationary mirror, the energy change must be zero.

(Note that zero in one frame means zero in all frames, a condition that can be used to derive the relativistic Doppler shift in a manner that does not split it into a "wavelength part" and "time-dilation part").

• What about rfl's wavelength shifters? They seem to do the job. Mar 25 '21 at 17:15
• @Winston I'm not going to let some internet product ruin a brilliant derivation. The is physics stack exchange, not QVC.
– JEB
Mar 25 '21 at 17:19
• Is this a joke? I cannot tell. Mar 25 '21 at 17:30
• This answer is incorrect. The devices are possible -- see my answer as to how (as well as more standard transmission-geometry AOMs and EOMs which would also get ruled out by this answer). I'm sorry to say this so bluntly, but in effect, this answer does little more than produce a sea of symbols deep enough that you can bury the logical disconnect where it is no longer visible. Mar 25 '21 at 19:36

When light travels towards Earth then light is redshifted due to the gravitational field. If you put a mirror on a high mountain top, the image is redshifted at sea level. Though you will never be able to look at yourself, someone down will see the image to be redshifted, though not in the amount and way you want to , I guess.

• As mentioned already, this gets ruled out by a symmetry argument. Yes, the light is red-shifted when it climbs up the mountain $-$ and then it gets blue-shifted by an identical amount when it comes back down. If you ignore that and just place the source and the detector at arbitrary heights, then what you get is just the standard gravitational redshift, and the mirror becomes completely superfluous. Mar 25 '21 at 19:46
• @EmilioPisanty Yes, you are right about that. It's not the mirror that causes the redshift. Mar 25 '21 at 19:50
• Interesting nonetheless. Mar 26 '21 at 7:01