# “Oh-my-god” particle drive performance

One of theories behind the origin of the esoteric oh-my-god particle was that it's exhaust of an alien spaceship. There are other explanations, some quite plausible, but let us for the purpose of this question assume that that one is true and try to examine performance of such drive.

Known, or plausibly suspected:

• the particle energy was 48J.
• it was a proton. Rest mass: $1.67 \cdot 10^{-27}$kg
• it was moving extremely close to speed of light. Let's round the speed to $3\cdot10^8$ m/s
• the source is outside our galaxy, but not very far from its edges. Source.
• mass equivalent of 48J is $5.28 \cdot 10^{-16}$ kg (calculated using this )

Now, for my assumptions about the craft:

• it doesn't carry its own energy source for this propulsion
• mass of any energy source other than antimatter would make propelling particles this close to the speed of light pointless; it's better to pack more propellant and push it to, say, 90% of speed of light instead of pumping so many joules into mass equivalence of energy without gaining extra Isp.
• if the energy source was antimatter, the trivial, optimal propellant would be photons. There's absolutely no point converting them into energy to propel a proton while they can provide about the same (minimally better) performance directly.
• It abuses energy-mass equivalence to create a large exhaust mass from minimal amount of propellant. Specific impulse is near the absolute maximum $c/g_0$, but thanks to using external energy, it gets far more exhaust gas mass than its fuel flow would provide, circumventing tyranny of the rocket equation — for every $1.67 \cdot 10^{-27}$kg of fuel spent, $5.28 \cdot 10^{-16}$ kg of fuel is expelled! (Of course also mass equivalent of the energy is lost, but it's soon recuperated by harvesting external energy!)

Now, I don't know what kind of energy source would it be. Uhoh ballparked the cosmic background non-uniformities at 1 microwatt per $m^2$ - but it might be a beamed energy from the craft's origin, it might be shine of the (still near) galaxy, or other unknown source.

Regardless — the point where I'm asking your help is coming up with the new, adjusted rocket equation for this drive. Assuming the rocket still moves at speed pretty far from speed of light (no need to apply relativistic equations to it), what would be the equation for $\Delta v$ where mass of exhaust considerably differs from mass of propellant — "created"/provided externally?

Bonus question: What fuel mass fraction would such a rocket need to achieve 0.1c?

This equation might be related to equations for jet engines. In a jet engine, mass of exhaust is much higher than mass of fuel spent to produce it; the extra mass coming from intake air, expelled air continuously recuperated through intakes.

• This is a really beautiful question! – uhoh Mar 1 '17 at 0:22

## 1 Answer

Rocket equation starts with conservation of momentum:

$$\frac{dp}{dt} = m\frac{\partial v}{\partial t} + v\frac{\partial m}{\partial t}$$

But at such a high energy, the rest mass of the proton can be ignored - it's about 1E-11 (as $m_0c^2$) as big as the energy. So drop the second term.

$$\frac{dp}{dt} \approx m\frac{\partial v}{\partial t}$$

$$\frac{dv}{dt} \approx \frac{1}{m}\frac{dp}{dt}$$

$$\Delta v \approx \frac{\Delta p}{m}$$

For a very relativistic particle:

$$p = \frac{E}{c}$$

For $n_{omgp}$ oh-my-god protons:

$$\Delta p = n_{omgp}\frac{E_{omgp}}{c}$$

To get a rocket of mass $m_R$ to a velocity $0.1c$ velocity (non-relativistic approximation for rocket):

$$0.1 c = \frac{n_{omgp} E_{omgp}}{m_Rc}$$

Rearranging

$$n_{omgp} = \frac{0.1 m_Rc^2}{E_{omgp}}$$

$$n_{omgp} = \frac{0.1 \times 1.0 kg \ \times \text{9E+16} \ \text{m}^2/\text{s}^2}{48 Joules} = \text{1.9E+14 protons}$$

or about 30 micro Coulombs, or about 3E-13 kgs of oh-my-god protons per kilogram of rocket.

• Thanks! And btw, the easiest way to store protons without building excessive charge is hydrogen. Only electron mass of overhead. – SF. Mar 1 '17 at 6:59
• Ya, I'm thinking the ones to be accelerated soon would be kept in a storage ring where they can be cooled, collimated, and ready for final acceleration. Of course they could be stored as negative H${}^-$ ions which makes extraction easier. But you might not want to store any at all, and just use an interstellar proton whenever you run into one. – uhoh Mar 1 '17 at 7:41
• @uhoh Yeah, if you can boost them to this kind of speed there's little reason to carry reaction mass. Congratulations, a working version of the Bussard ramjet! :) – Loren Pechtel Mar 2 '17 at 1:31
• @LorenPechtel I have a theory (which I probably read somewhere and then forgot) that new technology is never really invented as much as it is recalled from SF stories once read in old paperbacks. – uhoh Mar 2 '17 at 1:35
• @uhoh: Oh, shucks, you're giving me too much credit! ;-) – SF. Mar 2 '17 at 7:21