Outside of science fiction, photon rockets are positioned as low-thrust, low-acceleration, space-use-only engines. But... do they have to be?

For purposes of this question, a "photon rocket" is anything which emits photons preferentially in one direction, with thrust being a function of total photon power output and combination. If a given photon source can produce a unidirectional beam, awesome; if, like a standard incandescent filament, it can't, then any lenses or reflector structures used to redirect the output and increase would count against the weight budget for calculating thrust-to-weight ratio.

So, ignoring fuel systems or power sources, and considering only the actual device that makes photons and shoots them out the back, how much thrust per weight could an optimized photon rocket really get, without invoking any magic?


  • Assume "weight" as measured on Earth's surface--i.e., mass, multiplied by 9.8m/s^2

Addressing Misconceptions:

  • This question is not asking about a "ship". It's asking about an engine.
  • "Thrust to weight ratio only makes sense for a ship!" No, thrust-to-weight ratio is commonly calculated for isolated engines, independent of any larger vehicle that they might be integrated into. SpaceX engines: "The engine's 150:1 thrust-to-weight ratio is the highest ever achieved for a rocket engine."; Rocketdyne F1 engines: "Thrust-to-weight ratio 94.1" TWR is a critical statistic to know for a engine, because it constrains the types of ships and missions that you might choose to build around that engine or select that engine for.
  • The maximum TWR for a ship designed around a given enginer ship is the TWR of the engine, in the limit as you design for tinier payloads attached to honkin' huge engines. That's why engine TWR is a relevant design factor--your ship's TWR can't be better than that.
  • "Dissipating heat is not negligible because heat is dissipated as photons." Yes, that is exactly correct! But if you want to use those photons as part of your thrust, you need some structure to reflect them non-isotropically, which becomes part of your mass budget. If at affects the engine thrust, it is part of the engine!
  • "There are so many variables, you can just pick a number for your story!" Sure, but some numbers are physically possible and some aren't. It's blatantly obvious that you can't push arbitrarily large quantities of power through an arbitrarily small device without destroying it, so what actually is the physical limit?
  • If the amount of power you shove through device causes it to destroy itself, you don't have a photon rocket--maybe you did, at a lower power level, but you have turned it into a very poor electric thermal rocket. The TWR and max power rating of a Merlin rocket engine isn't measured by blowing it up, and those of an edge-of-the-envelope photon rocket shouldn't be measured that way either.
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    $\begingroup$ Are the power lines to the device superconducting and magic? At what point does heat generation start becoming something that also needs to be dealt with? $\endgroup$
    – jdunlop
    Apr 16, 2023 at 20:17
  • $\begingroup$ @jdunlop The power supply to the device is irrelevant. Heat generation needs to dealt with when it inhibits the function of the device--e.g., an incandescent filament is limited is limited by the point at which the filament melts. If the target thrust requires cooling mechanisms, those mechanisms would count towards the weight budget. $\endgroup$ Apr 16, 2023 at 20:49
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    $\begingroup$ @LoganR.Kearsley How can the power supply and such be irrelevant when we're asked to find T/W? Power supply participates in being weight. I think what you're getting at is the best specific power of the engine, right? AFAIK, T/W usually applies to the whole vehicle, and whether or not it can get off the ground. Doesn't make sense to ask whether an unpowered & unfueled engine can get off the ground. (A bit pedantic, but maybe a source of confusion and the reason for the 1 close vote so far.) $\endgroup$
    – BMF
    Apr 16, 2023 at 22:13
  • $\begingroup$ @BMF Because the power supply and fuel are not part of the engine, and could vary between installations. If the engine itself has a TWR > 1 (like, say, a Rocketdyne F1), then it is theoretically possible to build a rocket can lift off with a sufficiently large bank of such engines, and then the support infrastructure becomes relevant. If it doesn't (like, say, an ion engine or Hall effect thruster), then that is not possible, and there is no point considering support infrastructure. $\endgroup$ Apr 16, 2023 at 23:14
  • $\begingroup$ @BMF is correct, you cannot calculate the thrust-to-weight ratio without knowing the entire vehicle weight. This is because TWR is a measure of efficiency, not capability. See [1]. If we use wikipedia's definition then we could assume a TWR of just the engine... but what would be the point? The TWR changes the second the engine is installed in something (always to the detriment of the number). (*Continued*) $\endgroup$
    – JBH
    Apr 17, 2023 at 6:55

2 Answers 2


About zero

As you probably already know, the thrust of a photon engine is abysmal per unit of energy spent, but the main problem here is light emitting inefficiency of everything that converts electricity into photons. The best devices I am aware that are capable of emitting light in one hemisphere (not even direction, mind you!) have efficiency of less than 50%, meaning that you have to dissipate at least as much heat as you intend to convert into thrust. And since heat dissipators are included as parts of an engine, combined with that the only heat dissipation mechanism you can use in space is thermal radiation, the size of those dissipators depends linearly on the power to dissipate. Thus eventually the weight of the light emitter becomes irrelevant. So we can imagine the photon engine as a rather simple heater, heated to the maximum temperature possible, with a reflector with assumed 100% reflection rate, but with nonzero mass.

Now, the heater can only dissipate as much energy, according to Stefan-Boltzmann law, with constant under T^4 having a 10^(-8) order, thus with even 4000K temperature of that emitter (a tad greater than highest melting point of known solids) it only emits about 14 MW per square meter of surface. This 14MW of energy, if emitted in photons in one direction, provide 1.4e7/3e8 ~= 0.048 N of thrust, thus if that square meter of emitter has a mass of at least 1 kg (it'll be more, since there should be heat transfer systems in its content), its effective acceleration would be less than 0.048 m/s^2. Divide that by G and find the awful truth.

PS: if there would be no way to convert electricity into photon stream of more than 50%, the best photon engine would be the ship's cooling system supplied with reflectors to emit "forward" less heat than "backward". Aka exposed fusion reactor.

  • $\begingroup$ Blue LEDs have a theoretical maximum radiant efficiency of 93%, and emit directionally. But, I don't know what the minimum mass per watt for such an LED would be, to decisively say that it could provide better thrust than a thermal emitter. ¯_(ツ)_/¯ $\endgroup$ Apr 17, 2023 at 16:14
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    $\begingroup$ I was speaking about practical maximum. However I have found this study iopscience.iop.org/article/10.1149/2.0312001JSS/pdf that says that one part of radiant efficiency in their experiments is indeed capped at 93% or so, due to "nonradiant current losses", so I assume that there could be devices that have such efficiency. There is a problem tho, at power beyond 75 W/cm^2 (750 kW/m^2, 20x less than written here) heat losses start exceeding radiant output. So there are more problems with highly efficient media, they are only that efficient at low power unsuitable for the photon engine. $\endgroup$
    – Vesper
    Apr 20, 2023 at 6:46

The limit is only how much power you are willing to put into it.

The thrust power of an engine is equal to effective exhaust velocity times thrust. The effective exhaust velocity of an ideal photon rocket is c. Therefore, the thrust is directly proportional to the power. At any reasonable levels of power, that thrust will be more or less 0. But if you, say, convert an entire solar mass into energy at the same time and release that as a single burst of gamma rays, you'll get quite a bit of thrust.

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    $\begingroup$ The limit is much lower than that. The most straightforward designs will melt, deform, or otherwise cease functioning long before you can push the power of a star through them. $\endgroup$ Apr 17, 2023 at 14:31

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