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Let's say that you are put in charge of man-portable weapons development for a civilisation so advanced that almost anything is possible. You decide that a "dial-a-yield" pulse laser system is the most practical "one size fits all" approach. This weapon can fire brief microsecond (or less) pulses imparting petajoules of energy at the maximum setting.

Would this be practical as a weapon? Would it interact with the atmosphere (e.g. causing spontaneous fusion) in a way that would make it infeasible?

EDIT: I found this website that gave me an approximation of what my "laser" (beam weapon) would do at the maximum gain to a close target. This actually seems like it would be biosphere destroying?

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GuestPerson
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    welcome to worldbuilding, make sure you check out the tour – Topcode Oct 29 '20 at 15:39
  • Note that while we can describe the effects of different powers of laser on the atmosphere pretty easily, whether that makes them "practical" weapons depends a lot on factors like who's shooting at whom, what kinds of defenses are in play, and what the desired effect is. A one-size-fits-all weapon has often been promised, rarely delivered. – Cadence Oct 29 '20 at 15:56
  • It partially answers it. What I gathered is that the higher frequencies are better for not being absorbed by the air. That's not a problem in this case as the polity in question has graser technology at their disposal if necessary. As for your other comment, this is going to be fired by extremely far future tech soldiers at other extremely far future tech soldiers from long ranges of a kilometer or more with armour and shielding capable of surviving indirect nuclear strikes. – GuestPerson Oct 29 '20 at 15:58
  • What is "graser" technology, and why isn't it in the question? (And if you simply just increase frequency -- which is not at all clearly possible in our understanding of lasers -- then you can just go to hard gamma ray range and kill the enemy with relatively low power requirements.) (Plus the firs answer to the linked question gives you an upper bound of how much power you can transmit through air using microwave, after which air itself couples with the beam, becomes very hot and melts your weapon.) – AlexP Oct 29 '20 at 16:08
  • https://en.wikipedia.org/wiki/Gamma-ray_laser – GuestPerson Oct 29 '20 at 16:13
  • Petajoule laser pulses? I don't want to be on the same planet as that! Each pulse has energy of a 250Kiloton nuke. But it is a focused beam!! –  Oct 29 '20 at 18:59

1 Answers1

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High powered Lasers are useless

Thermal Blooming is a effect that renders a lot of laser systems useless. Basically, the laser gets deflected by the air and thus defocuses. Which in effect, makes it very inefficient at long ranges.

There are two ways you can counteract this effect. The first being just putting more power into the system. The second is to change the frequency of the laser.

Keep that in mind as we talk about your Laser.

Firstly, the cross section is important. Since it is portable, we can assume a lens with a radius of what 4cm or so. This means you have the Energy of N petajoules concentrated on an area 50cm². Now of course, N*petajoule is not very helpful measure. So lets go with 2 petajoules of energy for 50cm². Which gives you an energy density of 0.04 petajoules / cm².

How much energy is that? How much heat does this generate?

Well, this is where specific heat capacity comes into play. The heat capacity for air is basically 1, which is to be expected. Now we need the volume of air. For this we assume that the laser is a perfect cylinder with the base of 50cm². Light travels at 300,000,000,000 cm/s. Times that by * 0.1 and multiply by 50 and we get: 1,5*10^12 cm³. Which is 1,500,000 m³. The density of air is 1.225kg/m³. So we have 1,837,500 kg of air in our way.

So, the laser has to travel through 1.8 million kg of air.

How hot does it get ?

First we have to convert kg to g. So 1,873,500 * 1000 = 1,873,500,000 g. Then we just divide this by the joules. Again 2 petajoules are around 2e+15 joules. Thus we get 2e+15 / 1,873,500,000 = 1,067,520,683.

We then take this number and finally divide it by the heat capacity of air. It's capacity is 1 so the final heat output of this laser would be: 1.06 million degrees Celsius.

Now, is this realistic? No. This assumes that the laser would just appear and disappear instantaneously. But, no matter how you spin it, this is the ideal heat output. So in reality, it will be less.

Now 1.06 million Celsius might sound like a lot but remember, this is over a volume of 1.5 million m³. So the actual heat generated for each m³ is only 0.706 C. Not even a degree.

Again, this is not how it would look, in reality the laser beam would still cut through anything but the heat would be so stretched out over such a short period of time that not a lot would actually happen to the air around it. Only the tip would be a problem but it is travelling at lightspeed.

However, you can still assume that the tip of the beam would be around 1.06 million degrees.

1.06 Million degrees however, is nothing. Not a single element will start to fuse at those temperatures.

Going back

I mentioned before that thermal blooming and energy are a concern. And it still is. The good range of the laser might be anywhere around 10km or so. But it is very hard to tell. What is important is that just increasing the energy won't help a lot. At some point, you have to ask what sort of reactor people carry around with them to power it.

So in general, I'd say that the current laser is good enough. 2 petajoules will cut through anything.

End Result

The laser wouldn't do to much to the air around it. Everything that gets hit will have a bad day.

IT Alex
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Erik Hall
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  • Thanks for your detailed answer, it was what I was looking for. – GuestPerson Oct 29 '20 at 17:07
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    This completely ignores the time aspect of heat transfer. 1 million degrees applied for less than a nanosecond does practically nothing. – IT Alex Oct 29 '20 at 17:09
  • I mean i did say that this is not realistic. And i also said that i assume that the entire Beam just appears at once and then goes away everywhere at the same time. This way, the Calculations ASSUMED that the Beam would appear, stay there for .1 sec and then go away. And sure this is wrong for the entire beam. Excluding the tip. For the tip, it should be "fine". But you are right, Time is important. – Erik Hall Oct 29 '20 at 18:16