In the Dominion War arc of Star Trek: Deep Space 9, we see several scenes involving ground battles between Federation troops and Jem'Hadar soldiers. One thing that always strikes me as odd is the tendency for the Federation guys to fire their phasers as if they were guns, aiming very quickly and firing a very short blast.

With a gun, that's advantageous to keep recoil under control, but if your weapon fires an energy beam, and your enemies are in the open, standing up and not behind cover, the way Jem'Hadar commonly were, it would seem that the obvious tactic would be to aim at chest height, hold down the firing button, and sweep left-to-right. (Or right-to-left, of course.) But I don't believe we ever see anyone using a phaser in that way.

Was any explanation ever given for this?

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    We did see them at least implying that power was a factor, in one episode they were passing out spare power packs, short bursts use less power than large sweeps. – geewhiz Jan 23 '15 at 23:45
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    I would imagine, as is the case with many video game energy weapons, they tend to heat up. Firing in short bursts with real guns serves an additional purpose of keeping the barrel cool (I'm no expert, but I know some MGs' barrels get RED HOT). The same could assumed to be true of phasers, but I have no canon evidence – Premier Bromanov Jan 23 '15 at 23:46
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    Both comments are true, and I'll add one more: quick, individual bursts can be hard to trace back to their source and allow you move and fire again, but a bright sweeping line that lasts 4-5 seconds will give away your position like a giant neon sign. – Nerrolken Jan 23 '15 at 23:53
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    A better question: why didn't they set the phasers to a wide beam and alternate firing? – Xantec Jan 24 '15 at 11:54
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    @Einer That is entirely correct, except it was Klingons – Tritium21 Jan 24 '15 at 20:23

In real life, marines and soldiers rarely use full auto. It's inaccurate (recoil and spread) and quickly depletes ammo, with minimal enemy harm. Short controlled bursts are preferred. Laser weapons like phasers are not immune to basic physics and still suffer from recoil. As Tom has mentioned, overheating is also a concern. Your weapon needing a multiple second cool down or heat sink (like mass effect 1 and 2 showed) is not an good tactic.

Why they didn't use that towards bad enemy tactics is probably due to out of world filmography reasons. Had they employed sweeping laser tactic, the fight would quickly be over, and would contradict previous phaser usage in both TOS and TNG.

That said, the most common weapon is a type 1 and type 2 hand phasers, gun equivalent. The rifle phasers do have a wide spread fire mode, but that was rarely used (plot or theatrical reasons)

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    Laser weapons like phasers are not immune to basic physics and still suffer from recoil. Basic physics says that recoil happens due to Newton's 3rd law: the kick of the weapon is due to the force of the bullet being suddenly pushed out. No bullet, no recoil. – Mason Wheeler Jan 24 '15 at 10:00
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    @MasonWheeler physics.stackexchange.com/questions/149864/… – user16696 Jan 24 '15 at 16:00
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    @cde So basically no recoil at all - as the answer demonstrated that is hardly a recoil you could even feel. Even a tiny tin soldier (if he could be equipped with such a gun) wouldn't be tipped over by the recoil. – Einer Jan 24 '15 at 16:21
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    @cde I do see your point but [at this point both our sides fails to have a frame of reference] a phaser does not shoot photons but nadion particles. They travel at c so all we know is that they have no inertial mass. Everything beyond that (what recoil they produce) is pure speculation since nadion particles don't exist outside star trek. – Einer Jan 24 '15 at 16:32
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    @Einer - Even though photons have zero inertial mass, they do have momentum--all quantum particles are though to obey the de Broglie relation which says p=h/lambda, where p is momentum, his planck's constant, lambda is wavelength. So assuming "nadion particles" obey this general quantum law they should have some momentum, although the momentum may be rather small even for a large energy. Another de Broglie relation says that f=E/h where f is frequency and E is energy, and for anything moving at the speed of light, f=c/lambda. – Hypnosifl Jan 24 '15 at 19:50

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