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Let's say our villain fires an antimatter rifle from behind a magnetic field over a km from the target. The bullet is traveling over 1000 meters per second. If it reaches its target, the 13 gram bullet will annihilate itself in the target producing a small half megaton nuclear explosion destroying the target (and several city blocks for good measure).

However! Along its flight path, the antimatter bullet must encounter numerous air particles that annihilate with it, producing micro gamma ray explosions and slowing it down. Does this mean that the bullet never reaches its target? Would the shooter be able to fire far enough away to be safe from the gamma rays and the explosive matter–antimatter annihilation?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – Monty Wild Nov 12 at 2:52
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The bullet will not go far

It depends somewhat on the cross-section of the antimatter bullet. Let's assume that the bullet is slightly under 10mm in diameter, giving it a cross-section of 0.0000762 square metres. Passing through 1000m of air it will encounter head-on 0.0762 cubic metres of air, which conveniently is 3 moles.

The molar mass of dry air is 28.9647 grams/mol. So, even without impingement of side-on collisions, the bullet will run into over 13 grams of air in the first sixth of its flight. So the half megaton explosion will be spread out in some fashion over the first 160 metres of the bullet's flight path, causing far more damage to the sniper than the target.

Even if the bullet is a narrow rod with half the assumed diameter (resulting in a quarter of the cross-section), it still is not going to make it to anywhere near the target.

L Dutch is correct that the bullet will be decelerating immediately due to the gamma ray explosions at the front (Relativistic Baseball was the first reference I checked too!), but the deceleration is only relevant for working out how close to the shooter the bullet finishes exploding.

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    $\begingroup$ The bullet doesn't have to collide with it's entire mass worth of air in order to stop. The energy released from colliding (and annihilating with) less than a nanogram of air will bring it to a sudden stop. $\endgroup$ – Surprised Dog Nov 10 at 16:30
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    $\begingroup$ You'll get some interesting effects as the bullet will likely form a (very short lived) plasma sheath around it, and the heat and reaction products will carve out a lower-pressure channel through the air. It'll travel slightly further than you might think, but not enough to make any practical difference to the shooter. $\endgroup$ – Starfish Prime Nov 10 at 22:10
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    $\begingroup$ Anyway, doesnt matter, 1km distance is anyway not safe here... $\endgroup$ – Sascha Nov 11 at 1:51
  • $\begingroup$ Cross-section and shape of the bullet are trivial. These yield a ballistic coefficient, which combined with other factors (gravitational force, angle of inclination / declination, air density (temp and humidity), can be used to calculate the flight path of the bullet. But at 1000 m/s (~3281 fps), well in the velocity range of current rifles (a bit faster than most 200 grain / 13 gram bullets), nothing prevents contact with air, so the bullet undergoes matter/anti-matter annihilation on all surfaces. Assume firing from a rail gun. It would never make it out of a barrel pushed by a propellant. $\endgroup$ – G DeMasters Nov 11 at 8:35
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    $\begingroup$ @Sascha Yeah, the villain really ends up being more of a suicide bomber than a sniper. $\endgroup$ – reirab Nov 11 at 18:05
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Make the antimatter charged and use a magnetic bottle to keep it contained. It will reach its target without touching matter. On impact, the magnetic bottle mechanism breaks and the antimatter escapes.

The bullet may be a little bulky, but some hand waving miniturizations should get it small enough to fire out of a weapon. Depending on the size, you may need something more like a mortar, but with such a weapon, close enough is good enough.

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    $\begingroup$ Another advantage to launching it with a mortar is that the person firing it is farther away from the target and, thus, less likely to be destroyed along with the target. $\endgroup$ – reirab Nov 11 at 18:16
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    $\begingroup$ Magnetic mirrors, says Wikipedia, are “inherently leaky,” only trapping particles at specific speeds and angles. Magnetic bottles are simplified magnetic mirrors, with even less trapping ability. It seems very likely that the accelerations undergone by the bullet/mortar are going to throw the anti-particles out pretty easily, where they will collide with matter forming the (I assume) shell to keep the interior chamber a vacuum, triggering the annihilation early. This answer would be much improved if it could show this wouldn’t happen. $\endgroup$ – KRyan Nov 11 at 18:23
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    $\begingroup$ I came here to suggest this approach! I bet you could contain a tiny piece of antimatter in a .50 caliber round. Once the round is breached upon impact, that tiny piece is enough to destroy its target. $\endgroup$ – Andrew Brēza Nov 12 at 4:39
  • $\begingroup$ @AndrewBrēza Would that make the antimatter pointless then ? How much Antimatter do you need to do sizable damage, more than a 50cal round on its own ? $\endgroup$ – GamerGypps Nov 12 at 9:01
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    $\begingroup$ @gamer the kinetic energy of a .50 cal round 20KJ tops, and projectile energy more or less equates to damage. Antimatter annihilates in contact with matter, meaning mass is converted to energy according to Einstein’s equation E = mc2. The amount of (anti)matter required to liberate 20KJ of energy is about 2 × 10^-10 grams, which is the same mass as 10 human red blood cells. So, an incredibly tiny amount of antimatter packs a lot of punch. For comparison, 1 stick of dynamite delivers about 1MJ of energy, or about 50 times a .50 cal bullet, so about 1 x 10^-8 grams or 500 red blood cells. $\endgroup$ – Bohemian Nov 12 at 12:00
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Let's say the bullet is a standard .308 Winchester round with a diameter of 7.85mm which means the bullet is encountering:

$\pi * 3.925^2 * 1000 = 48,400$ cubic mm of air.

The density of air at 20 degrees C is $1.2 KG / m^3$ which means the bullet is encountering around $0.058$ grams of air per meter of its flight path.

Via $E=MC^2$, a total of 0.116 grams of matter is equal to 10440000000000 Joules of Energy which is more than the 3000 or so Joules that was imparted to the bullet on launch. At this rate the bullet won't even make it 1 meter (or 1mm) away from the rifle before the explosion on its tip bring it to a sudden halt and then annihilating itself a terrific mushroom cloud. Maybe next time, we should save the antimatter weapons for Space!

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    $\begingroup$ FWIW, the energy release will be spread out over a volume... much of the radiation will be in the form of 100MeV+ gamma rays, and those have an attenuation length in air of a metre or two (at least). The resulting fireball will resemble a nuclear explosion, and will exert a more or less even force on the bullet in all directions. It can't reasonably be said to have stopped before it simply vapourises into a short-lived ambiplasma cloud. $\endgroup$ – Starfish Prime Nov 10 at 22:05
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    $\begingroup$ @StarfishPrime Half the radiation hits the bullet's front, though, sending its leading atoms careening off into all directions. The front layer of bullet material thus acts like a rocket drive pushing backwards with an extremely high specific impulse. Some highly intelligent minds actually wanted to use this effect to build the mother of all rockets, they called it "Project Orion" ( en.wikipedia.org/wiki/Project_Orion_(nuclear_propulsion) ). $\endgroup$ – cmaster Nov 10 at 22:26
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    $\begingroup$ @cmaster even the attenuation length in lead is somewhat longer than the length of the bullet. The whole bullet will heat up at more or less the same rate and will explode outwards. Orion used fast particle radiation to impart momentum, not EM radiation, and not ablation. It isn't even slightly comparable to this situation. $\endgroup$ – Starfish Prime Nov 11 at 7:51
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    $\begingroup$ It seems most everyone here comes from a physics or other theoretical background. Having been a marksman of varying skill for 11/12th's of my life, I can tell you from experience that the assumption of 0% humidity is folly. The simple ballistic effect of humidity and precipitation is quite evident shooting at a range of 1000 meters. A misty morning would surely significantly increase the rate of matter/anti-matter collision, water being so much denser than air. $\endgroup$ – G DeMasters Nov 11 at 9:00
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    $\begingroup$ @GDeMasters Water (raindrops, say) is much denser than air, but water vapor (the water component of humidity) has a molecular weight of 18, much less than the 29+ of air. Very humid air is actually less dense than dry air at the same temperature. Not much less -- but the vapor pressure of water, times the percentage relative humidity, will give a partial pressure that will tell you what fraction of the air has been replaced by lighter water vapor. $\endgroup$ – Zeiss Ikon Nov 11 at 20:25
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The bullet will only work as intended if flying through high vacuum. In atmosphere, it would be necessary to evacuate its path, or at least a portion of its path directly in front of the bullet. Surprisingly, this might be possible: If an intense laser is fired in the bullet's path, it will heat and ionize the air; this alone will make it less dense, and the remaining ionized (charged) plasma may be swept out of the way with an electric discharge. So, a laser clears an ionized path for what is essentially artificial lightning, and the lightning clears a vacuum path for the bullet. The only problem is, the laser fires in a straight line, and the bullet flies in a ballistic curve. But, for a fast bullet flying a relatively short distance in vacuum, perhaps the deviation is small enough to be compensated for by shaping the laser path as a vertical slit rather than a cylinder.

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  • $\begingroup$ ...or make the bullet slightly conic and fire it small end forward. The rare annihilations happening mostly on its bottom side will slow it down a bit, but also keep it flying in a straight line. However, you will need ultra high vacuum for the flight path and also somehow persuade the surrounding air not to close back on the flight path. $\endgroup$ – Jirka Hanika Nov 11 at 23:47
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Your villain is just going to commit a very elaborate and probably expensive suicide, and his target will be dead as a collateral effect.

A magnetic field is effective at keeping away only moving charges. Most of the matter we have at sea level is in neutral state, including air, hands of whoever is loading that gun, the gun itself and the building where the sniper is hiding.

Even assuming that somehow the bullet can leave the rifle, we can rephrase this What If to get an idea of what happens (the bullet in your case is non relativistic, but it's generating gamma rays anyway)

I am quoting the content of that page, just editing the relevant part to suit your case (in italics)

These gamma rays and debris expand outward in a bubble centered on the rifle exit. They start to tear apart the molecules in the air, ripping the electrons from the nuclei and turning the air [...] into an expanding bubble of incandescent plasma. The wall of this bubble approaches the target at about the speed of light[...]

The constant annihilation at the front of the bullet pushes back on it, slowing it down, as if the bullet were a rocket flying tail-first while firing its engines.

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  • $\begingroup$ There is no difference between "moving" and "stationary". It all depends on the reference frame, and the bullet's reference frame is going very, very fast relative to everything else (hence everything else is moving relative to the bullet). $\endgroup$ – Brilliand Nov 11 at 3:11
  • $\begingroup$ @Brilliand, the bullet before being fired is stationary with respect to the gun and surrounded by neutral matter $\endgroup$ – L.Dutch - Reinstate Monica Nov 11 at 4:20
  • $\begingroup$ +1 especially for "Most of the matter we have at sea level is in a neutral state" $\endgroup$ – Ghedipunk Nov 11 at 7:35
  • $\begingroup$ @L.Dutch-ReinstateMonica Oh, pre-firing... apologies for the misreading. But isn't most matter made up of charges, which would have to move in order to make contact with the stationary antimatter? (Neutrons ignore magnetism, but they aren't seen without protons very often.) $\endgroup$ – Brilliand Nov 12 at 1:02
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I here assert that the antimatter bullet will make its way through the air to the target thanks to something like the Leidenfrost effect. It will in fact accelerate.

https://en.wikipedia.org/wiki/Leidenfrost_effect

The Leidenfrost effect is a physical phenomenon in which a liquid, close to a surface that is significantly hotter than the liquid's boiling point, produces an insulating vapor layer that keeps the liquid from boiling rapidly. Because of this 'repulsive force', a droplet hovers over the surface rather than making physical contact with the hot surface.

Of course with the ordinary Leidenfrost effect, the vapor molecules touch and interact with both surfaces.

With the bullet, the first molecules encountered will turn into radiation. This front of generated radiation will play the role of the vapor, dispersing the gas away from the bullet as it flies. Occasional contact with lucky molecules able to traverse the front of radiation will regenerate this front.

The bullet will be surrounded by this front of radiation on all sides, pushing it. But the back of the bullet is bigger than the pointed tip - with the result that radiation generated by gas molecules contacting the back of the bullet confer more push than those in the front. The bullet will therefore accelerate faster and faster until its shape changes from mass loss or because it has melted due to the radiation.

Once it has melted there will of course be a big explosion, as will be the case if it decelerates into some non-gas mass.

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    $\begingroup$ From the point of view of the bullet, the air molecules are more or less stationary. As it moves through the air it leaves a vacuum behind it (which causes a sonic boom). Thus, no air molecules will interact with the rear of the bullet to push it. It will, however, experience radiation pressure in the front slowing it down. $\endgroup$ – Surprised Dog Nov 11 at 4:41
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    $\begingroup$ This reminds of a similar concept answer I left on physics SE but never got a definitive analysis of whether or not it was correct: physics.stackexchange.com/questions/491782/… $\endgroup$ – Surprised Dog Nov 11 at 4:43
  • $\begingroup$ @SurprisedDog you get sonic booms with both compression and decompression; if you could extrude a wire at supersonic speeds the tip would produce a sonic boom even though there is no tail and no trailing vacuum. $\endgroup$ – Starfish Prime Nov 11 at 8:05
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    $\begingroup$ By your explanation, any bullet-shaped object would accelerate towards its front. $\endgroup$ – pacholik Nov 11 at 9:08
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    $\begingroup$ The cross-sectional area of a bullet is the same from the front and from the back. It's not bigger from the back. $\endgroup$ – reirab Nov 11 at 18:23
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The answer is not over any distance without an impractically miniature electromagnetic bottle. Think about it in the most basic terms. Matter cancels anti-matter. The distance to be traveled (even in air) contains enough molecules that the +/- cancellation shed factor would start as soon as it leaves the end of your magnetically bottled barrel. The density of your projectile would go poof as the reaction snowballed and new surface area was exposed. It's matter of pure cancellation, so no Leidenfrost effect, No air cavitation and no ice bullet. Just good old fashion poof, the millisecond it cleared magnetic barrel. no scope needed it's not going anywhere.. Shooter would have a heck of a sun burn though.. LOL

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For present-day values of "gun" and "bullet": not a chance!

An "antimatter bullet" is going to have to be a future-tech device with an internal extremely high vacuum and containment field. That containment field is going to have to be strong enough to keep the AM confined under the acceleration produced by being fired out of a gun, if that word retains its present meaning.

The deceleration of the bullet hitting the target may be great enough to break the confinement. But more probably, the high-tech innards also arm the bullet as it is being fired such that any significant impact causes detonation.

I'm actually sceptical that anything as low-tech as a gun and a "bullet" would be used by anybody that can create and contain antimatter in militarily useful quantities. At the very least, these projectiles will be tiny self-guided (AM-powered?) missiles, such as the "smart bullets" that feature in Vernor Vinge's "The Peace War". The "gun" will be more like a miniature RPG launcher (with target acquisition and guidance features).

But following this line of argument, why restrict the range? Such devices could self-power themselves to anywhere on the globe, using a tiny fraction of the energy packed in their AM payload to do so. Nuclear missiles the size of a bullet. Yuk.

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  • $\begingroup$ I get the impression from the question that the proposed gun is firing literal bullets made out of anti-lead, rather than an antimatter containment chamber holding the antimatter. $\endgroup$ – nick012000 Nov 12 at 11:29
  • $\begingroup$ @nick012000 Indeed: so "not a chance"! $\endgroup$ – nigel222 Nov 12 at 11:30
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I want to focus my answer on maximum (im)possible bullet travel distance. This can happen to relativistic anti-bullet - heat and explosion would not have enough time to affect bullet speed and radiation will not stop it completely before total annihilation (what would happen next nanosecond is out of scope of this answer). So it is simple "total annihilation path".

Lets our bullet be anti-assault rifle bullet, i.e. 5.56×45mm NATO SS109, made of 62gr of anti-copper & mostly of anti-lead, with the bullet having a cross-sectional area of 25.5 mm^2. I do not know exact proportions, but lets say it is 10gr (0.16 moles) anti-copper and 52gr (0.25 moles) anti-lead.

This would give us 6.022e23*(0.16*29 + 0.25*82) = 6.022e23*25.4 = 15e24 anti-protons (and positrons) and 6.022e23*(0.16*34.5 + 0.25*125) = 6.022e23*36.77 = 22e24 anti-neutrons to annihilate.

Air density is 1.2 mkgr/mm^3 or 30.6 mg/m of bullet flight distance. Since both nitrogen and oxygen have (almost) equal proton/neutron ratio, we can take average air mole mass 28.98 g/mole. This would give us about 4.6e21/m = 4.6e24/km of protons, same numbers of electrons and neutrons.

So first thing we can see - anti-protons (and anti-electrons) would deplete much earlier than anti-neutrons and subsequent explosion would have two distinct zones:

  1. 3.3 km - path of total air annihilation. All matter would turn into gamma radiation
  2. +1.5 km - path of neutron-only annihilation. Air would turn into into super-heated plasma (protons+electrons+lots of gamma radiation) - a perfect conditions for good old thermonuclear reaction

Total path would be about 5 km.

So, please, do not fire bullets made from antimatter with relativistic speed in my neighborhood (i.e. Earth, or even my solar system).

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  • $\begingroup$ The antimatter equivalent of electrons is positrons. Also, the question says the bullet's speed is 1,000m/s, which is not relativistic. (This makes Soong-type (Star Trek) androids and Asimov three-law robots particularly dangerous, given their positronic brains...) $\endgroup$ – Ghedipunk Nov 11 at 18:30
  • $\begingroup$ Technically, it just says "over 1000 meters per second", which a bullet traveling at relativistic speeds certainly is! ;) $\endgroup$ – nick012000 Nov 12 at 11:36
  • $\begingroup$ The main point I want do deliver here, that explosion of antimatter is not a simple kaboom with a lot of gamma radiation. It is more complex thing. For example, antimatter ball in air will not instantaniosly explode. When outer layers annihilate with air - there would be vacuum and runaway air - annihilation would stop. Then , when air implode (up to seconds), the process would repeat. This can take many iterations and time before everything ends. $\endgroup$ – ksbes Nov 12 at 13:40

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