# Is there a maximum ISp?

With ISp defined as ${v_e}\over{g_0}$, the naive approach would tell me, the maximum possible specific impulse, with propellant moving at speed of light relative to the craft, is $c\over{g_0}$ or 30,570,322s.

Of course, with relativistic physics naive approach is often wrong and I feel like I'm making some oversimplification here... so, if I am, could someone straighten it out? What happens with Specific Impulse when extreme energies are put into the propulsion, accelerating the propellant to relativistic speeds?

• Great question! Would the limiting case be a photon rocket? – Organic Marble Jul 10 '16 at 21:14

A photon rocket should have a Ve of c, hence ~30 megasecond Isp. The rocket equation would be tricky to apply, of course -- if you have a magic matter-energy reaction that perfectly converts "fuel" into linear-directed photon exhaust, it works fine, but if energy is going elsewhere, the effective exhaust velocity will be different. You'd account for relativistic effects in applying the ∆v using everyone's favorite sigmoid function, the hyperbolic tangent.

what, you don't have a favorite sigmoid function?

• So - I "naively" plug the speed into ISp, but need to go relativistic when using that ISp for anything else? - on a different note: Considering how weird a unit 'megasecond' is, I checked... that's almost a year. ...And I just got a good line for a sci-fi novel. "...and her ISp is measured in weeks." – SF. Jul 11 '16 at 1:27
• But the limiting case isn't a photon rocket, though. Radiation pressure is an extremely inefficient way to conduct kinetic energy, and you could do better using a particle accelerator firing a proton beam. Your new bottleneck would be e = mc^2. – Wesley Adams Apr 4 at 17:05

No. The number you gave is for a photon drive. However, if you throw particles out the back at relativistic velocity they weigh more than the fuel you drew from the tank. Since there's no limit on how much relativistic mass you can pile on the particle there's no limit on the ISP. Note, however, that there's no way you can boost particles like that using onboard energy.

• Wouldn't that just increase the mass flow? Their mass increase comes from equivalent decrease of mass of on-board power supply.due to its energy loss (RTG battery acting as a fuel tank). – SF. Jul 12 '16 at 4:48
• @SF. Note that I said it would not be possible from onboard energy. If you have to supply the boosting energy you're going to lose more mass than this gains you. The only way to actually pull it off is to obtain the energy from some outside source. – Loren Pechtel Jul 12 '16 at 15:33

The highest possible specific impulse could be achieved by using a 100% efficient e = mc^2 mass to energy converter. Specific impulse is defined as the number of seconds that an engine can provide 9.8 newtons of force given one kilogram of propellant. Here's a bit of math to find this value:

c ~ 300,000,000 m/s
c^2 ~ 9 * 10^16 m^2/s^2
Plugging in one kilogram, we get 1kg = 9 * 10^16 kg * m^2 / s^2 or 9 * 10^16 joules. But we have to divide this by two, because you're accelerating both your vehicle and your propellant. This gives ~4.5 * 10^16 joules of usable energy.
Divide that by 9.8 watts, and there you have it:

~4.6 * 10^15 or ~4.6 quadrillion seconds is the highest possible specific impulse. Kinda puts our engines to shame, doesn't it?

• Can you write the equation that you are using separately from the numbers? You are "dividing Joules by 9.8". If 9.8 has units of m/s^2 then your results is kilogram meters, not seconds. If 9.8 is just a number, then the result is still Joules. Either way I don't see your calculation producing seconds as units. Writing your equation first will help a lot. Thanks! – uhoh Apr 2 at 0:23
• 9.8 is the amount of energy needed to produce 9.8 newtons of force per second. The unit is kg m^2 / seconds^3 or watts. The equation is c^2 / watts where c = 300,000,000 and watts = 9.8. – Wesley Adams Apr 2 at 3:01
• Wait! There was a mistake in the original post! Because you're accelerating both your propellant and your vehicle, you can only use half of your e = mc2 energy for increased vehicle velocity. That means that the maximum theoretical specific impulse is only half of what I originally said. The correct equation is c^2 / 2 * watts. My bad! – Wesley Adams Apr 2 at 5:34
• I've added an answer to your question there which may help here. – uhoh Apr 2 at 6:24
• "9.8 is the amount of energy needed to produce 9.8 newtons of force per second. The unit is kg m^2 / seconds^3 or watts." That's not how it works. If a 1 kg weight lies on a table on Earth, the table constantly pushes it with 9.8 newtons of force without spending any energy at all. This answer is wrong because of this sort of misunderstanding. – Litho Apr 2 at 9:23