# Relativistic Rocket Equation

I am familiar with using Tsiolkovysky's Rocket Equation, $$\Delta v=v_{e} \ln \left(\frac{m_{i}}{m_{f}}\right)$$ However, after trying to work out an exhaust velocity needed in a hypothetical situation, I achieved a result higher than the speed of light. Obviously this is nonsense and a relativistic version of this equation is needed to handle these higher speeds. However after some searching I cant seem to find one.

• There is an extremely good reference on this here. – user21103 Mar 11 '20 at 10:54
• I noticed that you've accepted an answer. Are you 100% sure that you are comfortable with it and it's exactly what you needed, or are you really hoping for an expression that gives the final velocity in the "rest frame" (the frame where the rocket started) as a function of $v_e$ and $m_i/m_f$ (which is what I'd like to see) such that it never exceeds the speed of light? – uhoh Apr 4 '20 at 16:53
• @All The current accepted answer is wrong, and I am deprecating it. Please see my second answer below. I cannot delete accepted answers, so the original remains up for the time being. – Quietghost Apr 8 '20 at 17:18

tldr; If $$\Delta v>c$$, its a reference frame issue. Use rapidity to calculate $$\Delta v$$ instead. If $$v_e > c$$, then there is no solution for the specified conditions, because the exhaust velocity requires more energy input than there is mass-energy in the fuel.

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There are 2 parts to this question that can be addressed. When solving the rocket equation, there are 2 velocities that can be solved for (given the other variables) and either could exceed the speed of light if the right numbers are used.

(1) What if $$\Delta v > c$$?

(2) What if $$v_e > c$$?

The first question is answered by considering reference frames. A $$\Delta v$$ larger than the speed of light does not mean the rocket's velocity relative to the initial inertial frame is greater than $$c$$. Instead, it is the measured $$\Delta v$$ in the frame of reference of the rocket that is. It is similar to the scenario where you can accelerate at 1g for 2 years, but your velocity is not $$2c$$, however your observed (integrated) $$\Delta v=a\Delta t$$ is larger than $$c$$. Here, $$a$$ is the local acceleration you would feel onboard the rocket, however the observer from an inertial frame would observe a different acceleration.

The correct way to compute your actual change in velocity is to consider the correct reference frame and integrate the local change in velocity relative to that reference frame. In all reference frames, $$dp_{ex} = dp_{rocket}$$ In the local frame of the rocket, relativistic effects of the rocket's velocity are ignored. So $$dp_{rocket} = (m-dm)dv$$ and the classical rocket equation results, and $$\Delta v$$ may be greater than $$c$$. Instead, if we consider an inertial frame of reference, $$dp_{rocket} = (m-dm)d(\gamma v)$$ The resulting solution to the rocket equation can be easily put in terms of a term called rapidity. The neat property of rapidity is that it adds just like velocities do in Galilean relativity. $$r \equiv \tanh^{-1}\left(\frac{v}{c}\right)$$ $$\Delta r = \frac{v_e}{c}\ln\left(\frac{m_i}{m_f}\right)$$ This allows you to calculated the actual $$\Delta v$$ relative to the initial inertial frame of reference.

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On to the second question: what if $$v_e>c$$? Returning to first principles, we know that $$dp_{ex} = dp_{rocket}$$ and should the rocket velocity not be relativistic, then $$\gamma_e v_e dm_{ex} = (m-dm_{r})dv_{r}$$ However, there is a hidden nuance with this equation (Thanks to @Litho for catching this). With chemical rockets (low exhaust velocities), the mass of the exhaust leaving the rocket is equal to the change in mass of the rocket according to the principle of conservation of mass. $$dm_{ex} = dm_{r}, \hspace{10pt} \gamma = 1$$ But in relativity, one of the first lessons is that mass is not conserved! Instead, mass-energy is. Therefore if the exhaust is expelled at velocity $$v_e$$, then this kinetic energy comes from its initial rest mass. Starting from the energy-momentum relation $$E^2 = (mc^2)^2 + (pc)^2$$ and using the definition for an object at rest, $$E = m_{rest}c^2$$, one can show (algebra) that for a mass $$m$$ moving at speed $$v_e$$: $$m_{rest} = \gamma m, \hspace{10pt} \gamma = \frac{1}{\sqrt{1-{\frac{v_e^2}{c^2}}}}$$ $$dm_{rocket} = \gamma dm_{ex}$$ This factor of $$\gamma$$ cancels with the $$\gamma$$ in the momentum equation above, and so the solution is actually the classical rocket equation. Therefore, the classical rocket equation holds even for relativistic exhaust velocities.

So what happens when $$v_e>c$$? You are S.O.L. The energy required to achieve this exhaust velocity is larger than the initial mass-energy of the fuel. There is no solution for $$v_e$$ for the $$\Delta v$$ and masses specified.

• +1 Kudos for having the tenacity to continue to pursue this and get it right! I'm a little uncomfortable with seeing "$\Delta v = a \Delta t$ is larger than 𝑐. There is nothing wrong with this" without at least a caveat that acceleration $a$ is funky here. The whole point of this answer is to explain that using $ma = F$ instead of $dp/dt = F$ was the mistake that led to the confusion. Is it possible to clarify that your $a$ isn't really a meaningful acceleration? – uhoh Apr 9 '20 at 0:51
• Absolutely, I can clarify this. In the rocket reference frame you would feel $a$, but in the inertial reference frame $a$ would be different – Quietghost Apr 9 '20 at 16:52
• Look great, thanks! – uhoh Apr 9 '20 at 22:31
• If the rocket is powered by beamed power, can't the energy be more than the rest mass of the fuel? – ikrase Sep 21 '20 at 9:51
• @ikrase If the power is beamed to the rocket, then for the relativistic case the concept breaks down. There are 2 effects, namely the momentum transfer from beaming energy (similar to solar sails) and the additional energy that can be added to the fuel that invalidate the momentum balance principle that makes this a rocket. But if you think about it, why try to add this beamed energy to the fuel when you could just reflect it using a mirror for the momentum advantage anyway and not take the conversion losses? (Additional directional control possibly) – Quietghost Sep 28 '20 at 15:22