# How should mass specific impulse calculated for relativistic exhaust?

Optimising for Isp only is problematic, as it's simply:

$$I_{sp} = \frac{v_e}{g}$$

Which is the same as optimising for exhaust velocity.

With no constraints on thrust, particle accelerations can achieve velocities arbitrarily close to the speed of light (The LHC is 3 m/s close). That's an Isp 30.6 million seconds, which can't be directly used in the usual rocket equations since you will have to account for relativistic effects.

I disagree that that a value for ISP calculated in such a way correct. For mass-specific impulse, One measures thrust as a force and divides by mass flow rate. Just because the units are the same as velocity does not mean this is an actual velocity of something!

From Wikipedia's Specific impulse:

By definition, it is the total impulse (or change in momentum) delivered per unit of propellant consumed and is dimensionally equivalent to the generated thrust divided by the propellant mass flow rate or weight flow rate.

In normal chemical rockets ISP does tend to be close to the mass-averaged exit velocity of the exhaust and that is convenient for those who choose to write snappy or clever-sounding SE answers, but once you bring up particle accelerators it all goes sideways.

Question: How is mass specific impulse calculated for relativistic exhaust? If I had 1.0 microamp of 6.5 TeV protons coming out of my tail, what propulsive force would I experience, what ISP would I demonstrate, and how far wrong would those trusting the quoted passage be?

Related:

• See my answer to Could the helical engine work? where the punch line is that when going relativistic, $$\mathbf{F} = d\mathbf{p}/dt$$ and not $$m\mathbf{a}$$.
• Considering that photons have zero rest mass yet plenty of momentum, I rest-mass my case.
• Any time you have relativistic exhaust, the internal power vs external power question becomes very important. With external power, the momentum of your power beam will have a significant fraction of the momentum of your exhaust – ikrase Oct 27 '20 at 3:52
• @ikrase I never thought about that but ya, it's E=pc for the photons and almost the same for the particles. Hey there's a new question in that as soon as I get to a keyboard. – uhoh Oct 27 '20 at 5:30
• See here. If the energy to accelerate the exhaust comes from the spacecraft, the $I_{sp}$ really is $v_e/g$ (assuming no losses, otherwise it's lower). For example, if the spacecraft produces some photons with total momentum $p$, it has to spend energy $pc$, so its mass decreases by $\Delta m=p/c$, so $p=\Delta m \cdot v_e$, just like in the non-relativistic case. – Litho Oct 27 '20 at 8:59
• @Litho that's for photons so I think you are commenting on the previous comment, and not the question which is about protons. The block quote references the Large Hadron Collider, which is large and collides hadrons, but thankfully does not collide large hadrons – uhoh Oct 27 '20 at 9:16
• – uhoh Oct 28 '20 at 1:41