Someone invents a portable, incredibly high-powered laser or similar energy weapon. The mechanism of energy production can be hand-waved away. In other words, it will heat up. Lots. Gigawatts lots, or terawatts/petawatts/etc. Handwave any production rate you like into this.

Using only substances that exist in our real world, what is the practical limit on how quickly a portable energy weapon could be cooled-down? In other words, what is the maximum sustainable energy output of such a weapon, assuming cooling rather than energy production is the limiting factor?

The sort of thing I'm thinking of: Suppose you had a larger device nearby (on a truck, say) which could cool down helium to a liquid and pump it into your energy gun via a tube. The liquid helium would have to be pumped around quickly to prevent it from boiling. For sufficiently high weapon wattage, the sheer volume of helium needed, and the size of the cooling apparatus, would exceed what you can put on an accompanying truck. I suspect substances other than helium will have a better heat capacity before boiling, or a better rate of heat transfer, or some other property, but you get the idea.

Other than the internal mechanism of the weapon, there is no significant advance in materials or other technology beyond what we have here today.

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    $\begingroup$ Newton's third law implies the conservation of momentum, not energy. When a bullet leaves a gun, the forward momentum of the bullet is equal to be recoil momentum of the gun, but almost all the energy is in the bullet. $\endgroup$
    – AlexP
    Dec 10, 2016 at 18:02
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    $\begingroup$ The only reason a laser produces any heat (but for the emitted light) are inefficiencies. There is no hard maximum efficiency for lasers to my knowledge. As said, Newton's laws have nothing to do with this. The maximum possible power of a laser is limited in an atmosphere because at some point the atmosphere becomes plasma, which again has nothing to do with Newton. See that xkcd/what if about pointing lasers at the moon. $\endgroup$
    – Nobody
    Dec 10, 2016 at 18:08
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    $\begingroup$ We need a reason to close: "Bad assumptions embedded in question". We can't answer a physics question when the setup is wrong. $\endgroup$
    – SRM
    Dec 10, 2016 at 18:16
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    $\begingroup$ @SRM this is not even physics stack exchange, we can't expect people to get everything right about physics here, bad assumptions are the norm when talking with non-physicists or non-enthusiasts, so I'd think this question is ok. $\endgroup$ Dec 10, 2016 at 18:28
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    $\begingroup$ Mandatory XKCD reference $\endgroup$
    – SJuan76
    Dec 10, 2016 at 19:20

5 Answers 5


Let's imagine that the hottest part of the apparatus is the barrel, so to speak, which is cylindrical with length $l$, radius $r$ and temperature $T_b$. We can surround it with a fluid of temperature $T_f$. Now, the barrel has a heat transfer coefficient of $h$. The change in heat energy of the barrel over time, $\dot{Q}$, is $$\dot{Q}=hA\Delta T=h(2\pi rl)\left(T_b-T_f\right)$$ where $A$ is the surface area of the barrel. Some things immediately spring out:

  • A greater heat transfer coefficient leads to quicker cooling.
  • A larger area over which to transfer the heat leads to quicker cooling.
  • A greater temperature difference leads to quicker cooling.

$h$ depends strongly on the properties of the materials, which can be hard. We can make some estimates for the other quantities, though.

  • The source is portable (which I'm taking to imply that it could fit in a large van, for instance), so I'll estimate that $l=3\text{ m}$ and $r=0.05\text{ m}$ (perhaps the latter is a bit large). This leads to $A=0.942\text{ m}^2$.
  • Let's be extremely generous and say that the laser reaches temperatures of about $T_b=1,000\text{ K}$. This is really stretching it. At any rate, even if $T_f$ is close to $0\text{ K}$, the difference $\Delta T$ cannot be greater than $1,000\text{ K}$. Therefore, let $\Delta T\sim1,000\text{ K}$.

The best transfer coefficients I can find are actually water-to-water. However, air-to-steam can yield an $h=17\text{ W m}^{-2}\text{ K}^{-1}$ through copper (which has a melting point higher than the temperatures involved here). We therefore have $$\dot{Q}\sim\left(17\text{ W m}^{-2}\text{ K}^{-1}\right)\left(0.942\text{ m}^2\right)\left(1,000\text{ K}\right)\sim16,000\text{ Watts}$$ That's pretty nice . . . if the temperatures (and temperature difference, for that matter) don't cause the entire weapon to fall apart.

  • $\begingroup$ have you seen anything on the new electrocaloric cooling. science.sciencemag.org/content/311/5765/1270.full $\endgroup$
    – John
    Dec 10, 2016 at 22:58
  • $\begingroup$ Your linked overall heat transfer co-efficients are for heat exchangers. Inasmuch as the barrel of a weapon acts as a heat exchanger, it would have to be air to air. Water can transfer much more heat rapidly than air due to its convection and high specific heat, steam can also transfer much more rapidly due to its ability to condense on a cold surface, then evaporate taking latent heat of vaporization with it. You should use the air to air co-efficient with steel of 7.9 . $\endgroup$
    – kingledion
    Dec 11, 2016 at 2:32
  • $\begingroup$ This is actually a good example of why projectile shooting guns are better than lasers. A laser that drops 10% of its input energy as waste heat, and has waste heat limited to 16 kW can output 144 kW in damage causing blasts. A .50 cal has muzzle velocity of 18 kJ, and around 8 rounds per second, so it can put out 144 kW in muzzle energy. So this mega laser is exactly as powerful as a .50 cal; and almost assuredly heavier and more expensive. $\endgroup$
    – kingledion
    Dec 11, 2016 at 2:37
  • $\begingroup$ Your table is "for practically still fluids," i.e. not taking into account convection. You can get orders of magnitude better heat transfer via forced-air cooling (i.e. fans). $\endgroup$ Dec 11, 2016 at 3:52

As far as I know, the record for watercooled heatsinks goes to... actually several different designs (search "heatsink water MW/m2"), there are several papers on heatsinks that can handle 20+ Megawatts/m2 (that's surface of the sun levels). I remember seeing a design capable of 40MW/m2, basically a Tungsten or Molybdenum plate with a tight bunch of supersonic water jets being blasted at it (the trick is to move the water fast enough and with enough force that it doesn't boil).

But your real problem is the thermal conductivity of the device itself, diamond being the best confirmed thermal conductor at ~2000W/m*K or 2kW/m2 of heat down a 1m long block would result in a 1'K (1'C) temperature difference. The solution? Make said laser very flat and long, thin and high in surface area is the way to get really good cooling.

But the true record holder for most powerful heatsink ever goes to the precooler stage of the SABRE hybrid rocket engine where they claim "The experimental device achieved heat exchange of almost 1 GW/m3", but seeing as it used liquid hydrogen, you could fire off your multi-gigawatt laser device for as long as your supply of liquid hydrogen held up, then you'd have to wait while you condensed some more.

Of course the real limit will be the devices efficiency. If your device is 90% efficient (and some lab-grade lights are getting pretty close to that now) then you can pump out ~10GW continuously per m3 of device until you run out of coolant.

As a side note, I do remember seeing a paper on a carbon nanotube capacitor with a power density of a couple of Megawatts/liter (or a couple of GW/m3) but I can't seem to find the page. (aluminium foil caps can produce even more power, but their energy density is terrible in comparison, by like several orders of magnitude)

So the upper limit for current technology both in heatsinking and in power sources seems to be around the Gigawatt/m3 mark. Solution? Switch out your van for a big rig, a 40' container for the laser, another 40' for the cooling and a final 40' for the power supply and you've probably got a Terawatt class device that could run for some appreciable fraction of a second (and a large road train too)


Avoid Excessive Heating of the Weapon

This is sort of not what your are asking, but it is a way to avoid having to cool your weapon--namely do not overheat the weapon in the first place. One way to achieve this is to have your energy projecting sources seperated from each other enough that any one emitter is easily cooled. Then have each of the beams meet at the target, uniting into one beam with constructive interference.

If you want to, you can incorporate bounce stations/vehicles so that your beams can be generated from many different places and be reflected accordingly to the convergence point with mobile aircraft, satellites, ground vehicles, etc.

If you carry this idea out far enough, you can have enormous power plants stationed around the globe or solar system, either mobile or not, and have a network of bounce stations capable of dividing the beams so that certain percentages can be sent simultaineously to many recipients, i.e. powering many handheld guns, mobile vehicles, airships, destroyers, nanodrones, etc. all at the same time.

  • $\begingroup$ The premise is that the weapon is arbitrarily powerful, so any mitigation strategy will eventually be rendered moot as the power increases and you have to intervene to remove the heat. $\endgroup$
    – spraff
    Dec 10, 2016 at 19:34
  • $\begingroup$ @spraff I'm confused...are you saying that there is no possible way to cool the weapon? You obviously couldn't mean that, so I'm confused. $\endgroup$ Dec 10, 2016 at 19:43
  • $\begingroup$ I'm saying that my premise is that, for arbitrarily increasing power outputs, weapon heat production is unavoidable. In your bounce stations suggestion, there would be heat generated at the point of collection. If you offshore 99% of heat production but then make the weapon 100x as powerful you're back to where you started and heat must be actively disposed-of from within the weapon itself. $\endgroup$
    – spraff
    Dec 13, 2016 at 18:00

We have to take newtons law of cooling into consideration here, Newton's law of cooling states that the rate of heat loss of a body is proportional to the difference in temperatures between the body and its surroundings. As such, it is equivalent to a statement that the heat transfer coefficient, which mediates between heat losses and temperature differences, is a constant. we can go for something cold like liquid nitrogen and give your weapon Liquid cooling system but it will work fine only if your weapon does not fall apart by it's own heat.


Hand-waving away the energy generating part renders this question not very effective.

Efficiency is of paramount importance. If the process is ideal, 100% of energy is converted into the laser beam. Then you don't have a heat transfer problem. The most efficient lasers today can convert about 70% of electricity into light.

For any amount of heat generated one can build big enough of a heat exchanger, to have enough surface area, with millions of heat sink fins, given a bit of temperature differential, to conduct all the heat away.

Therefore I can say there is no limit based on the constrains provided.

In practice there is always constrains, such as size, weight, cost, etc. Therefore in the real world the real answers to real problems are found in balancing all the constrains.


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