Electron-positron annihilation as a propellant

[Jarvitä]

So, an electron and a positron colliding at low energies (ie, relatively slow, not in a particle accelerator) tend to produce two photons at gamma-ray energy levels (511 keV). Could this reaction be leveraged for spacecraft propulsion? Consider the following, even a miniscule amount of matter-antimatter (in terms of mass, not current production capacits) - a kilogram - would impart on the spacecraft a momentum of ~150 million kg m/s - a delta-V of 14 km/s for a 100 ton spacecraft, from a single kilogram of propellant, no reaction mass required.

Is there anything obvious I'm missing, other than the fact that we just don't have a way of producing this much antimatter yet?

 

[T.Neo]

The problem is that the energy is released in the form of gamma ray photons and these are pretty much impossible to reflect with any known material (they either just go through stuff and get absorbed, and can only be reflected at very low angles by "grazer" mirrors; highly impractical for a propulsion system).

The only chance of using such gamma rays for propulsion is to invent some sort of gamma ray reflector that somehow works within the laws of physics, or to somehow tune down the wavelength of the gamma rays to one that is reflectable by matter (i.e. at least in the UV range) by doing something with them- such as rely on a process such as compton scattering through a dense medium- and hope you don't need to much of that dense medium so as to make your drive impractical.

I've heard an "electron gas mirror" theoretically being used to reflect the gamma rays emitted from a black hole due to hawking decay (for propulsion), but I'm not sure how such a thing would work or even what it is.

Of course, even such a 'miniscule' amount of mass as a kilogram of antimatter, is an utterly huge amount compared to the truely minute amounts we are currently able to manufacture. Granted, particle accelerators are meant as research platforms and not antimatter factories, and the production rate might be increased if things were optimised, but that's still a very large amount of antimatter. Even if you had a very advanced 'antimatter factory', you'd have to put a huge amount of energy into it.

But maybe, just maybe, if antimatter propulsion were promising enough, there'd be more studies into producing large(r) quantities of it.

 

[Jarvitä]

I thought the photons' momentum itself acts as a reaction force?

ie, in a perfect vacuum and absence of any other forces, lighting a flashlight / laser pointer will eventually propel you in the opposite direction of the beam? Imagine that, with high energy gamma rays instead of visible light.

Even if you had a very advanced 'antimatter factory', you'd have to put a huge amount of energy into it.

The amount of energy needed to produce m antimatter is [math]E\ =\ m\ \frac{c^{2}}{\eta{}}[/math], eta representing the efficiency of the factory as a number between 0 and 1. In the best case, you only need as much energy as you're going to get out of it. And even at lower efficiencies, it's probably still more effective to use this energy on an AM factory on Earth than to bring means to produce it along with the spacecraft.

 

[T.Neo]

I'm not sure if they do (it would make sense, from an intuitive point of view). The problem is of course the fact that if they aren't reflected the photons will spread out in a sphere, not a cone, and would thus be absolutely worthless for thrust.

You need to redirect the photons for things to work, otherwise you just get a powerful gamma-ray source and no engine.

 

[Jarvitä]

But do the photons always spawn isotropically? If positrons and electrons were held in some kind of dipole bubble by an external field, and all annihilated in the same orientation, would the photon burst be more directed?

Otherwise, we could put a thick lead sphere around the "engine", cut a little hole in it and call it [math]\eta{}\ =\ \frac{\Omega{}}{4\pi{}}[/math]. Even at a few percent efficiency, this is still orders of magnitude better than the VASIMR / fusion concepts.

 

[T.Neo]

You lost me with the isotropic photons and the dipole bubbles and stuff. :shifty:

What do the photons react against? They are spawned from a particle and an antiparticle, but how is that energy transferred to the actual body of the ship?

The lead sphere thing won't work because you'd get a huge amount of waste heat entering the sphere, and the thrust would be low (both because you're only getting a few photons to leave the sphere, and because if you release too much energy, the sphere melts).

The thrust would probably be so low that it would easily be beaten by fusion, VASIMR and even ion concepts, which are all (even fusion) far easier to achieve (since they don't need such amounts of antimatter) and are most certainly more efficient (presumably only a few photons make it out the back of the drive).

 

[Jarvitä]

Isotropic (adj.) i-so-trop-ic 1. omnidirectional, invariant with respect to direction

What I'm asking is, can the electron-positron pairs be oriented such that they produce coherent gamma-ray bursts? Or are the photon directions unrelated to the orientation of the electron-positron pairs?

 

[T.Neo]

I don't know. I wouldn't think so, but I'm not making that statement based on anything I've read or heard.

 

[RisingFury]

I thought the photons' momentum itself acts as a reaction force?

It does, but you won't get yourself moving.

Photons carry momentum, just like a piece of mass. The problem is that linear momentum needs to be conserved. If you have a particle and it's antiparticle at (almost) 0 velocity and they annihilate, then the sum of linear momentum before the event must equal the linear momentum after the event. That means one photon was thrown one way, the other photon the other way. That will *always* happen. The only way to get any thrust would be to reflect a significant portion of gamma radiation.

A solar sail does exactly that... photons strike the solar sail imparting momentum. If they they're reflected, they impart even more momentum.

Edit:
Also, you'd only get the thrust from reflected gamma rays, not from the other ones...

 

[Jarvitä]

So, you do get some momentum even if just one of the gamma-ray photons gets absorbed. So, if you place a lead hemisphere over the annihilation core and it intercepts half the photons (while the other half escape the missing hemisphere), the general momentum will still be linear in the polar direction of the lead hemisphere. The linear momentum, however, is reduced to 1/8 times the efficiency of the lead medium You'd also need some heavy-duty radiators on the other side, though.

 

[RisingFury]

So, you do get some momentum even if just one of the gamma-ray photons gets absorbed. So, if you place a lead hemisphere over the annihilation core and it intercepts half the photons (while the other half escape the missing hemisphere), the general momentum will still be linear in the polar direction of the lead hemisphere. The linear momentum, however, is reduced to 1/8 times the efficiency of the lead medium You'd also need some heavy-duty radiators on the other side, though.

Yea, you could absorb the photons, but reflecting them will give you twice the thrust. And a giant hunk of very dense material isn't a good way to get yourself moving...

Ironic, isn't it. The ultimate power source, but we know no efficient means of exploiting it, even if we found a way to produce and store it en mass...

 

[Brycesv1]

lets be optimistic and hope that if we DO discover gamma ray reflecting unobtainium, it will probably lead to our first interplanetary drives.

 

[tori]

I don't think so, even if we do, we're still looking at 300 MW/N of thrust (as with any other photon drive). This assumes 100% efficiency of both the reaction and the reflection (producing gamma radiation of laser-like directionality).

Assuming one pair gives off 1022 keV of energy, a drive with say 10 kN of thrust would need at least 3 TW of input power in the form of electron-positron pairs. That's 1.833e+25 electron-positron pairs per second. That's some 16.7 mg/s of pure electron/positron gas (a very huge amount).

Another way to look at that is from the electrical point of view: 1 Coulomb of charge is approximately equivalent to 6.241e+18 electrons. Thus the electron supply system would have to constantly output 1.46 MA of current.

There are tons of other designs that are more workable than this.

 

[Jarvitä]

Okay, the 300 MW/N figure is just plain wrong. From what I can tell, its origins can be traced to here:

The momentum of a photon is p = E/c, where E is the energy of the photon. So the thrust delivered by a stream of photons is ∂p/∂t = ∂E/∂t/c. This boils down to:

F = P/c

P = F * c

This guy is treating photons as newtonian particles. But a photon isn't even a "real" particle, it has no mass, therefore you have to calculate its energy/momentum in terms of special relativity.

Momentum: [math]p\ =\ \frac{h}{\lambda{}}[/math]Energy: [math]E\ =\ \frac{h\ c}{\lambda}[/math]
h being the Planck constant and lambda the wavelength.

So you can't calculate thrust like that. You have to get it from power:

[math]P\ =\ \frac{n\ E}{t}[/math], n being the number of photons.

Then, you can calculate the delta-V in a given amount of time:

[math]\Delta{} v\ =\ \sqrt{\frac{2\ n\ E\ t}{m}}[/math]
And the force: [math]dF\ =\ \frac{m\ \partial{}v}{\partial{}t}[/math]
Check the numbers, it's far more efficient than this attempt at newtonian approximation.

 

[tori]

Well I don't know, those numbers appear to match pretty well . If the momentum of a single particle released is p, then in order to get a thrust of F (Δp/Δt), you're bound to release n (F/(p/t)) particles.

h = 6.626e-34 (J*s, planck)
c = 299792458 (m/s)
L = 5e-12 (m, 5 pm gamma photon wavelength)
t = 1 (s)
F = 1 (N, required thrust)
----
p1 = h/L = 1.3252e-22 (kg*m/s, momentum of a single photon)
F1 = p1/t = 1.3252e-22 (N, average thrust of 1 photon/s)
n = F/F1 = 7.5460e+21 (required number of photons/s for thrust F)
E = h*c/L = 3.9729e-14 (J, energy of one photon)
P = n*E = 299792458 (W, input power required by this engine)

... which still boils down to P = F * c. The dimensions check out: N * (m/s) = (kg * m / s²) * (m/s) = (kg * m²)/(s³) = W.

Can you provide a proof that this line of reasoning is incorrect?

---------- Post added at 02:11 PM ---------- Previous post was at 02:05 PM ----------

And furthermore, as it is clear that the total thrust is independent of the operating frequency, why use gamma (difficult to produce, direct, and let's face it, it's gonna be a lethal knife for several AUs downrange, cutting through the star system)? Instead of having to produce and store exacoulombs of charge (hint: electron gas is not exactly compressible, coulomb repulsion is pretty strong) and carry a handwavium™ gamma director apparatus, why not just cut out the middle man and produce the photons more directly, in a band that's easier to work with.. say visible, or infrared?

 

[T.Neo]

Ironic, isn't it. The ultimate power source, but we know no efficient means of exploiting it, even if we found a way to produce and store it en mass...

You can always bombard a dense material (like a block of tungsten) with antimatter to heat it up, and then use that to drive a heat engine for example.

There's a related concept that basically works like an NTR, except it replaces the nuclear reactor with said block of tungsten. It doesn't offer any magical increase in performance and it requires exceedingly-difficult-to-get antimatter, so it probably doesn't have much of a practical use.

Power source? Antimatter's a pretty potent energy sink... unless you stumble upon a planet of antimatter that magically exists somewhere.

 

[Jarvitä]

And furthermore, as it is clear that the total thrust is independent of the operating frequency

But that's wrong. The momentum and energy of photons clearly depend on the wavelength/frequency.

 

[RisingFury]

Let's not get into a discussion of what is a particle and what is not, since Physics still struggles with that. The math works and it gives us a good understanding of how... particle-waves behave under different circumstances, but we'll probably need a new mental picture to describe these...


The link between energy and momentum is:
[math]E\ =\ \sqrt{m^2 c^4 + p^2 c^2}[/math]but for a massless "particle":
[math]E = p * c[/math]
And force is defined as change of momentum over time:
[math]F\ =\ \frac{dp}{dt}[/math]
Combine the two and you get:
[math]F\ =\ \frac{dE}{dt} * \frac{1}{c}[/math]
[math]F\ =\ \frac{P}{c}[/math]

This will work in the example of a laser emitting a stream of photons, or a block of led absorbing them.

And the force: [math]dF\ =\ \frac{m\ \partial{}v}{\partial{}t}[/math]

Is that dF a differential d? If so, then this equation stinks...


Edit: I hate LaTeX

 

[Jarvitä]

Power source? Antimatter's a pretty potent energy sink... unless you stumble upon a planet of antimatter that magically exists somewhere.

I never claimed it's an energy source. But it's the most mass-efficient energy storage there is, and that's got to be worth something in space.

EDIT: RisingFury, thanks, I see the P-c relation now. But I still don't understand how the momentum itself can be independent from wavelength, when it's clearly defined as [math]\frac{h}{\lambda}[/math]?

 

[RisingFury]

But that's wrong. The momentum and energy of photons clearly depend on the wavelength/frequency.

Well, that's true... but in terms of Physics alone... if you pump a certain amount of power, you can get lots of low energy photons with little bit of momentum each, or one photon with a lot of momentum.