Is specific impulse limited by the speed of light?

Question

In most references the specific impulse is the effective exhaust velocity (actual exhaust velocity plus corrections due to pressure difference between exhaust and surrounding atmosphere) divided by a standard gravity, which also incidentally gives $I_\mathrm{sp}$ units of seconds. The rocket equation is then

$$\mathrm{d}v=I_\mathrm{sp}g\ln\left(\frac{m_i}{m_f}\right)=v_e\ln\left(\frac{m_i}{m_f}\right)$$

which makes sense. The thing is I see this exact formulation used a lot when talking about photon rockets or engines with relativistic propellant; it would make sense to me to multiply by the Lorentz factor given by the particular velocity $\gamma=\frac{1}{\sqrt{1-\frac{v_e^2}{c^2}}}$, since the ejected propellant will be observed by those aboard the rocket to have a mass (and thus momentum) scaled upwards by that factor. I never see this in any text though.

This omission would also put an upper limit on specific impulse (and thus on thrust-efficient delta-v, i.e. not spending millions of years to accelerate, and thus on effective space exploration) because without relativistic corrections $v_e$ caps out at $c=299792458$ which is in the grand scheme of the tyranny of the rocket equation not a huge number. Am I wrong for relativistically correcting the rocket equation at least in this sense?

(This is just focusing on relativistic corrections due to the speed of the ejected propellant, not from the ship's speed. I know that there are more complex rocket equations to account for that.)

Answer 1

No specific impulse is not limited by the speed of light. According to Wikipedia

Specific impulse, measured in seconds, effectively means how many seconds a given propellant, when paired with a given engine, can accelerate its own initial mass at 1 g.

One might reason that because you can't throw propellent out the back of a rocket any faster than the speed of light, this would put a hard limit on how long a vehicle could hover in a 1 g gravity field. Because $$thrust=\dot{m}v_{Exhaust}$$

If $v_{Exhaust}$ is limited by the speed of light and is thus finite, and the thrust you need to hover is finite, then the mass flow rate, $\dot{m}$, is a value that you can calculate to be greater than zero. So eventually you're going to run out of fuel. Therefore ISP must be finite and limited by the speed of light.

But what actually happens as you throw the propellent out the back at closer and closer to the speed of light is that it gets heavier. This is why you will often see "$m_0γ$" inside relativistic equations, where $m_0$ is the rest mass and $γ$ is the Lorentz factor that you mentioned above in the question. So the speed of light doesn't really limit the ISP because as your exhaust velocity approaches the speed of light, the propellant gets heavier. At the limit, the effective mass of the expelled propellant is infinite and the rest mass flowrate is zero.

Now we are assuming here that we are using an external power source to accelerate the propellant. If the propellant provided the power that would certainly be very limiting - but it wouldn't be the speed of light limiting the ISP, it would be the finite amount of energy stored in the propellant.