# What would an antimatter engine be like?

1. Mar 21, 2017

### jostpuur

Antimatter thrust engine would be somewhat like a chemical or nuclear thrust engine, but with the difference that the exhaust would be pure light, and the fuel would the lightest possible fuel currently known to theoretically exist. That is known, but what would it really be like, meaning like how would humans perceive it? What would be an artistic expression of an antimatter thrust engine that would get reasonable close to a real thing?

One way to create an effect of an antimatter thrust engine would be to place some led lights at the bottom of some flying saucer, make the flying saucer hang from some wires, shoot it with camera, and finally remove the wires with CGI. The led lights at the bottom of the flying saucer might create an effect that would be similar to a real antimatter engine.

Would the pleasant glow of the led lights be realistic? Perhaps a real antimatter thrust engine wouldn't be so nice after all? Maybe real antimatter thrust engines would be like super weapons that melt kilometer deep narrow holes into the ground and blind all those humans who are too close to see the engine at action?

How could you know?

2. Mar 21, 2017

### John Morrell

Well, obviously the way it looks will depend on how it is built even more than the type of fuel. Planes and cars both use fossil fuels but obviously they look quite different. The antimatter could be combined with regular matter and the momentum of the light and particles could push the ship, or they could be used to generate electricity first and then power other types of motors.

A few things that you can be sure of, the principle of conservation of energy will be conserved. So unless you are talking about a REALLY huge ship, I don't think you'll be melting "kilometer deep narrow holes into the ground". If there was that much energy in the exhaust, you'd have either a really really big ship or a very inefficient motor.

This is much more speculative, but if you are planning on just "burning" the antimatter at the back of the ship to gather thrust, I don't think that the light would be red. Presumably the radiation would be much more energetic and thus higher wavelengths of light, so the only red light would be from other particles that had been subsequently irradiated and glowed.

3. Mar 22, 2017

### jostpuur

If the light beam was very narrow and sharp, then reasonably small energy might come with very high energy density, and it might have some cutting power.

Yes, shooting water vapor that has been heated by antimatter fuel might be good idea, because it might leak less dangerous radiation to the surroundings. I hadn't thought about that. I guess that water vapor could glow the scifi style too. Anything will glow if it is sufficiently hot.

4. Mar 22, 2017

### rootone

Antimatter annihilation would produce intensely energetic gamma rays.
Those are not visible but probably would cause anything exposed to them to get very hot and glow in the visible range.
(including the eyeballs of anyone attempting to take a peek at them)
I think the big problem is converting enough of the energy of the gamma rays into a directional thrust force

Last edited: Mar 22, 2017
5. Mar 23, 2017

### snorkack

Photon engines ARE inherently extremely inefficient. After all, you are propelling a small reaction mass at 300 million m/s, not a large reaction mass at a few thousand m/s. To produce 1 N thrust (lift just 100 g), you'd need 300 MW power. To produce 1 million N thrust (lift a 100 t ship, like Apollo or Shuttle), you need 300 TW, which would be turned to heat wherever your photons get absorbed.

The inefficiency would be the same if you turned mass to some other low or no rest mass particles, like neutrinos or gravitons.

6. Mar 23, 2017

### Khashishi

It's inefficient in terms of energy, but it's as good as you can do in terms of specific impulse. Carrying reaction mass defeats the point of having an energy dense fuel.

It would not be very nice at all. To accelerate a 1000 kg spaceship from 0 to 100 km/s, it would have to emit exhaust with total momentum of 10^8 kg m/s. If that exhaust is pure radiation, how much radiation would that be?

Well, the energy of that radiation would be 10^8*c = 3*10^16 J ≈ 7 megatons of TNT yield. Not too nice.

Now, if they used some reaction mass as propellant, the exhaust could be much nicer. But the propellant would add a lot of mass to the fuel. Perhaps the aliens could use propellant when near populations, and switch to pure radiation when in deep space.

7. Mar 23, 2017

### jostpuur

I felt unsure about the distribution with respect to frequency (or wavelength), and that's why I didn't start calculating anything, but yes of course, you can convert impulses to tons of TNT without a need for distributions with respect to frequency.

I'll take a second look at some magnitudes. Suppose we want to accelerate 1000kg vehicle to 100km/h. Taking into account that that's 27m/s, we are talking about an impulse of 2.7 * 10^4 kg*m/s. For photons that will be 8*10^12 Joules, and 2 kilotons of TNT. That's too much energy to by released near anything important.

8. Mar 23, 2017

### jostpuur

I did some magnitude calculations with massive propellants. Suppose we want to accelerate a 1000kg vehicle to 100km/h by shooting a small drop of vaporized water out of the rear. If the mass of the drop is $\Delta m$, then the energy that needs to be given to it will be $\frac{p^2}{2\Delta m}$, where $p$ is the impulse 2.7*10^4 kg*m/s. If you want to store an energy $E$ in an antimatter tank, that amount of antimatter fuel will have mass $\frac{E}{c^2}$. This means that the total mass of both fuel and propellant will be

$$f(\Delta m) = \frac{p^2}{2c^2\Delta m} + \Delta m$$

We see it makes no sense to use arbitrarily small $\Delta m$, because eventually the weight of the antimatter fuel needed to accelerate that extremely small drop will get greater than the weight of the drop itself.

$$0 = f'(\Delta m) \quad\implies\quad \Delta m = \frac{p}{\sqrt{2}c}$$

According to this formula the optimal size for a water drop will be 6*10^(-5) kg.

The above calculations assumed that the efficiency would be perfect, but I'm not sure how the inefficiency issues would affect the optimal size of the water drop. I guess we can say that the optimal size of the water drop is still somewhere near 0.1g (grams)? If you can accelerate a 1000kg vehicle to 100km/h by shooting an antimatter heated 0.1g drop of water out of the rear, is it good?

Last edited: Mar 23, 2017
9. Mar 23, 2017

### snorkack

Remember that antimatter itself, converted into photons, still has nonzero mass and momentum.
We have the expression:
E2=(pc)2+(mc2)2
Rearranging for the known (goal) momentum, you get
(pc)2=E2-(mc2)2
then it follows that
(pc)2=(E-mc2)*(E+mc2)
Note that the expression (E-mc2) is the kinetic energy of the reaction mass - the energy produced by the engine.
The bigger is m (the rest mass expended in the jet), the smaller the energy can be, for same momentum.

10. Mar 23, 2017

### John Morrell

In practical terms, the higher velocity you can give to the exhaust the more efficient your rocket will be, simply because any fuel that you carry has a weight and needs to be taken into account. A lot of people have mentioned how much energy in radiation you would be releasing to provide a high thrust, but the fact is that a chemical motor has to provide even more energy than that.
"
That's essentially what rockets are doing when they lift off- they are setting off huge explosions directly beneath them. Also not too nice.

The attraction of antimatter is really just it's energy density. You carry a very small mass of fuel with a very high amount of potential energy.

11. Mar 23, 2017

### jostpuur

I'll take a closer look at how much chemical rocket engines release energy. I'm again interested in making a 1000kg vehicle going 100km/h, i.e. 27m/s.

If the exhaust speed is $\nu$, the mass of the vehicle at start is $m(0)$, and the amount of burnt fuel is $m_{\textrm{burnt}}$, the achieved speed is

$$v_{\textrm{final}} = -\nu\ln\Big(1 - \frac{m_{\textrm{burnt}}}{m(0)}\Big)$$

You can solve the burnt mass and it is

$$m_{\textrm{burnt}} = m(0)\big(1 - e^{-\frac{v_{\textrm{final}}}{\nu}}\big)$$

If the chemical fuel has an effective energy density $\mathcal{E}_{\textrm{eff}}$ with respect to mass, meaning that when you multiply this number with the mass of a burnt fuel, you get the amount of energy that gets transformed into kinetic energy of both vehicle and the exhaust, a relation

$$\nu = \sqrt{2\mathcal{E}_{\textrm{eff}}}$$

holds. It could be that this relation is not very well known, but anyway, let's not get into that too much. If you want to talk about this relation more, or believe that I've made a mistake and it is wrong, let's discuss that in another thread somewhere in Classical Physics part of the forums.

With chemical fuels you might have something like 10MJ/kg or 100MJ/kg to start with. I'll assume that the efficiency is something like 0.1 or 0.01, and say that we have $\mathcal{E}_{\textrm{eff}}$ = 1MJ/kg. That gives us $\nu$ = 1.4*10^3 m/s.

We can compute $m_{\textrm{burnt}}$ by substituting in 1000kg, 27m/s and 1.4*10^3 m/s, and the result is that 20kg of chemical fuel has to burnt for this acceleration. Using the above assumption about energy in our chemical fuel this means that something like 2*10^8 or 2*10^9 Joules is going to be released when the chemical fuel is burnt.

Earlier I found:

So it looks like that photon propulsion is going to release something like 1000 or 10000 times more energy than chemical propulsion.

When chemical propulsion releases energy into environment, it comes in form of air pressure and heat, which can be dangerous in their own way. If antimatter fueled photon propulsion releases 1000 or 10000 times more energy in form of gamma rays, it's going to be quite a different kind of damage its going leave behind!

Last edited: Mar 23, 2017