Can we efficiently reflect or capture photons of that wavelength?

More precisely, $ 2.426×10^{-12} $ meters, or $ 1.236 × 10^{20} Hz$

Preferably reflect, but capture followed by re-emission as photons of other frequencies, like infrared is okay too.

  • for a rough orientation if the numbers are meaningless to you - they fall pretty deep into cosmic radiation spectrum, way deep frequency of gamma rays, nowhere near to mild X rays.

Why such an oddly specific request? This is related to the concept of antimatter drive. In most of answers that looked at efficiencies achieved by a spaceship exploiting antimatter, the answer is qualified with "assuming annihilation energy can be converted to kinetic energy at near 100% efficiency".

Well, it can. And quite easily too. A plain parabolic mirror will suffice.

The one drive of absolutely maximum possible specific impulse (~$3 × 10^7 s$) is the photon drive.

A photon drive is dead easy to make. Everyone's seen a photon drive, most of us have a couple of them installed in our cars. We call them headlights. A source of photons (lightbulb) and a reflector to make them mostly unidirectional. Frequency doesn't matter much; visible, infrared, ultraviolet, whatever we can reflect. And yet, instead of our spiffy photon drives we use the terrible inefficient internal combustion engines for accelerating the cars. That's due to the unfortunate properties of the photon drive which are a pathetic thrust and absolutely terrible wet:dry mass ratio in case of all conventional energy sources - even including nuclear!

The only energy source that assures 100% fuel->reaction mass conversion for a photon drive is antimatter. Reaction mass being photons, fuel being positrons and electrons.

$ 2.426×10^{-12} $ meters happens to be the low-energy annihilation photon wavelength. (we don't care about high-energy; whatever means we find to infuse the positrons with extra energy would be less energy-efficient than what we get out of the annihilation.)

If we can build a mirror that reflects great most of these photons, and shape it as a pretty long parabolic mirror (quite similar to De Laval nozzle in shape actually!) we have it. And even if we can't reflect them, if we can absorb them and let them be re-emitted, say, as infrared=heat radiation, just into the nozzle and not anywhere else (...right placement of radiators) that still works - a photon is a photon, as long as it's emitted "backwards" it still gives us the propulsion at the awesome $1 {c \over g_0}$ seconds of specific impulse. Of course the heat management problems arise, that are mostly absent in case of the mirror, but that's just an engineering issue ;)

Well, other than that we still need ways to produce, store and manipulate antimatter of course. But assuming we can, if this question is positively answerable, the near-100%-efficient antimatter drive is within our reach.


3 Answers 3


Electromagnetic radiation with frequencies above $10^{19}Hz$ are conventionally called Gamma Rays. Gamma rays will penetrate instead of reflect of most materials. Reflection is only possible at very shallow angles of incidence, and even then it works best for lower energies like X-rays, below $10^{19}Hz$. This method is used in the European X-Ray satellite ROSAT, using a Wolter Telescope.

At higher frequencies or higher incidence angles, the radiation will either pass through the mirror or, especially in the case of gamma rays, it will interact with the mirror material. Most of the energy will then be re-radiated, most of it going in the same direction as the incoming gammas, rather than backwards as you require.

These reflectors are also only used at very low energies - they're used to detect radiation from stars and galaxies. In your case you're talking about megawatts or gigawatts. At those energies, the mirror material will likely vaporise.

In other words, there is no material that will do what you require.

  • 1
    $\begingroup$ Most of the energy will then be re-radiated, most of it going in the same direction as the incoming gammas, rather than backwards as you require — are you trying to say it gets scattered preferentially in the forward direction? The word re-radiated reminds of absorption followed by emission, which should be in all directions. $\endgroup$
    – gerrit
    Commented Nov 15, 2016 at 11:58
  • 1
    $\begingroup$ @gerrit: Yes, the gammas get scattered in the forward direction. At the energy scale in question, the Compton effect is the dominant process on the first interactions with matter (electrons). This is like one marble (photon) hitting a second, resting marble (electron). Intuitively, it's clear none of them moves backwards. Though Compton scattering can be seen as single process, particle physicists see it as two processes: Absorption and re-emission of the photon. $\endgroup$
    – sweber
    Commented Nov 15, 2016 at 14:33
  • $\begingroup$ @sweber Right, I'm used to thinking in infrared where matter will absorb infrared radiation and emit according to its temperature, heating up until reaching radiative equilibrium with its environment. Theoretically this could be used for propulsion by placing a heat source in a body behind the spacecraft and putting some mirrors on the rear, but I very strongly suspect the acceleration will be extremely minimal :) $\endgroup$
    – gerrit
    Commented Nov 15, 2016 at 14:49
  • $\begingroup$ Ah, there is already a question on that :-) $\endgroup$
    – gerrit
    Commented Nov 15, 2016 at 15:53

The wavelength you mention corresponds to about 510 KeV. As hdhondt says, conventional optics, such as a parabolic mirror, don't work at this range. Wolter telescopes use grazing incidence reflectors and can work up to about 100 KeV. The NuSTAR X-ray telescope is a Wolter with a maximum energy of 79 KeV. The BAT (Burst Alert Telescope) on Swift uses a coded mask that can image up to 150 KeV. For higher energies, rather than an optical imaging system, the detectors are in stacks of layers, essentially directional detectors. The Fermi Large Area Telescope used this approach to image up to 300 GeV!

The question you bring up is an interesting one. How would you, even inefficiently, convert the gamma rays from matter/anti-matter collisions into useful energy? Since the wavelengths are much smaller that the space between atoms in an ordinary material, just absorbing the gamma rays (to convert to heat) seems a challenging issue.

NuSTAR links:



Swift/BAT links:



Fermi Large Area Telescope links:



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    $\begingroup$ The 510 keV is a poofteenth close to the mass of an electron at 511keV $\endgroup$ Commented Nov 15, 2016 at 16:32
  • $\begingroup$ @MyOtherHead The question does specify 511 keV, the energy of each photon in electron-positron annihilation. This answer changed it to "about 510 KeV". $\endgroup$
    – uhoh
    Commented Nov 16, 2016 at 5:15

As the other answers state, there's not much we can do to reflect or otherwise redirect these gamma rays.

However, there is one thing we can do: Blackbody radiation.

If the gamma rays hit an object made of ordinary matter, they can heat it to incandescence, and it will reemit energy in the easily reflectable infrared, visible, and ultraviolet spectrum.

Your drive might end up consisting of an antimatter annihilation zone, an immense tungsten or ceramic absorbtion-radiation sphere, and then a vastly larger parabolic sail made of flimsy gossamer material.

  • $\begingroup$ I like it. A solar sail spacecraft that carries its own little sun with it. $\endgroup$
    – SF.
    Commented Aug 17, 2020 at 12:13
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    $\begingroup$ Suppose your reradiator is a tungsten sphere of radius $R$ meters just below its melting point $T$ (3700K). It radiates power $4\pi R^2\sigma T^4$ which is roughly 130MW times $R^2$. This gives a thrust of 0.4N (times $R^2$). It seems like the Tungsten absorver would need to be a few cm thick to absorb most of the gammas at that energy, so mass might be about 100 $R^2$ kg, so your acceleration is a about $0.4 mm/s^2$ which is actually quite respectable. Of course it's hard to have a supply of positrons without also having antiprotons.... $\endgroup$ Commented Aug 17, 2020 at 17:58

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