Relativistic rocket equation

[SMJB]

Like the rocket equation, but taking relativity into account.

I know there is one, or at least logically there would be, but the search engines have failed me. So could someone help me out?

Also, it'd be nice if you'd render the variables in nice, simple X-Y-and-Z. It's nice being able to tell them apart from the constants.

[Kuroneko]

Yes, there is. It's easy to get confused when deriving it... as I have in the past, until I learned to think fourth-dimensionally geometrically.

First, you need to understand a certain way of thinking about the Newtonian rocket equation. Let's assume the rocket accelerated from rest, for simplicity, to some velocity v. In the instantaneously comoving inertial frame, the infinitesimal change in momentum of the rocket is d(mv)|v=0 = m dv = −u dm, where u is the speed of the exhaust. As both velocity differences and mass are Galilean-invariant, the latter equality holds in the rest frame, in which the Newtonian rocket achieves a velocity of
[1] vNewton = -u Int{m₀,m}[ dm/m ] = -u ln R
where R = m/m₀ is the mass fraction remaining. If the rocket is not accelerating from rest, this is Δv instead of the final velocity.

Now in STR, we should try the same reasoning but with Lorentz-invariant quantities rather than Galilean-invariant ones. Once again, let's assume the rocket accelerated from rest, and has achieved velocity v. Let's also assume units of c = 1, so all speeds are fractions of lightspeed, etc. Using the transformation v = tanh(η), where η is the hyperbolic angle with the rest frame's time axis (aka rapidity)**, it follows that the Lorentz factor is γ = cosh(η) and γv = sinh(η), which is important because relativistic momentum is p = γmv = m sinh(η). Therefore, in the instantaneously comoving inertial frame, the infinitesimal change in momentum of the rocket is d(m sinh(η))|η=0 = m dη = −u dm. Exactly the same as the Newtonian case, and since both mass and rapidity differences are Lorentz-invariant***, it follows that the relativistic rocket has its rapidity in precisely Newtonian form. In other words,
[2] vSTR = tanh(-u ln R),
where again R = m/m₀, u is the exhaust speed, and the speeds as fractions of lightspeed. If the rocket is not accelerating from rest, you should apply the relativistic velocity addition formula.


** The geometry here is simple: just as tan θ = m = Δy/Δx is the slope in Euclidean space, where θ is angle to the x-axis, we have tanh η = v = Δx/Δt is the "slope" in spacetime, i.e., velocity, where η is hyperbolic angle to the t-axis.
*** If this isn't obvious from the geometry of and the previous line, it's easy to check that the relativistic velocity addition formula (v+v')/(1+vv') is just the same as tanh(α+β), where v = tanh α and v' = tanh β.

[SMJB]

Dude, I'm really just interested in an equation I can plug variables into.

[Simon_Jester]

Dude, that IS an equation you can plug variables into.

I'm not sure what the subscript "STR" means ("v under the special theory of relativity)?

But R is the same mass fraction as in the Newtonian rocket equation, u is exhaust velocity measured as a percent of lightspeed, and the function "tanh" is the hyperbolic tangent, which you can look up for yourself on Wikipedia.

[NoXion]

I've found the calculator on this page to be useful.

[Dr Roberts]

Obviously didn't google it.