Relativistic Rocketry

Exploring the physics of rockets in Einsteinian relativity.

Maria Nicolae, 21 September 2025

A rocket is a machine that propels itself forwards by ejecting exhaust behind it at high speed. By Newton's third law of motion, any vehicle that is propelled forwards must be pushing something backwards, and a rocket uses its own exhaust for this, rather than something in the surrounding environment like a road surface or an atmosphere, making it suitable for spaceflight.

In quantitative terms, a rocket is continuously consuming its own mass (its fuel) to create exhaust with some momentum. For a given differential change in mass $d m$ of the rocket, a differential momentum $d p$ is imparted on it. A key figure of merit for a rocket, then, is the ratio of these quantities:

v_e = -\frac{dp}{dm}.

This quantity has dimensions of velocity, and so it is commonly called the exhaust velocity of the rocket, but it is not necessarily the actual physical velocity of the exhaust leaving the rocket. Nonetheless, like physical velocities in relativity, it is limited to the speed of light $c$ , as we will see later.

This post will consist of two main parts. First, I will take the exhaust velocity as a given, and derive equations for how the rocket moves, the total change in velocity the rocket can impart. Second, I will look at how the exhaust velocity comes to be, given the reactions and fuels a rocket might use, as well as looking at how to optimise it.

The Rocket Equation

Nonrelativistic Case

First of all, let us consider the nonrelativistic case, in which addition of velocities is simple and linear. In that case, a change in velocity $d v$ is related to a change in momentum $d p$ by

dp = m\:dv

where $m$ is the mass of the rocket at any given moment. Then, the total change in velocity $\Delta v$ that the rocket can achieve is the integral of the differential change in velocity $d v$ over the entire burn of the rocket, from its initial fuelled mass $m_f$ to its empty mass $m_e$ :

\begin{align*} \Delta v &= \int_\text{burn} dv\\ &= \int_{m_f}^{m_e} \frac{dv}{dp} \frac{dp}{dm} dm\\ &= v_e \int_{m_e}^{m_f} \frac{1}{m} dm\\ \Delta v &= v_e \ln \frac{m_f}{m_e}. \end{align*}

This is the Tsiolkovsky rocket equation. We see from this that, in order to get a payload mass $m_e$ up to a change in velocity $\Delta v$ using a (nonrelativistic) rocket, the amount of fuel the rocket needs grows exponentially with $\Delta v$ . This is the so-called "tyranny of the rocket equation".

Relativistic Case

The key difference going from the nonrelativistic case to the relativistic case is that velocities can no longer be simply added together. Instead, relativistic velocities "sum" via the formula

v_\text{sum} = \frac{v_1 + v_2}{1 + \frac{v_1v_2}{c^2}}.

As before, we will need to take the continuous limit of many small additions of velocities, and integrate that. Doing this directly from this expression, however, is very difficult.

Reparameterised Velocity: The Rapidity

To handle the relativistic addition of velocities, we want to find some parameter of velocity $\zeta(v)$ such that

\zeta(v_\text{sum}) = \zeta(v_1) + \zeta(v_2).

Then, $d\zeta$ is a quantity we can integrate over the burn of a rocket.

Such a parameterisation is

v = c \tanh \zeta.

This can be shown by the fact that the addition formula for hyperbolic tangents

\tanh(a+b) = \frac{\tanh a + \tanh b}{1 + \tanh(a) \tanh(b)}

has the same form as the relativistic addition of velocities. This reparameterisation is a common approach in Einsteinian relativity, and the parameter $\zeta$ is typically called the rapidity of motion.

(The fact that hyperbolic trig crops up here might come as a surprise to you. Indeed, it turns out that Einsteinian relativity is intimately connected to hyperbolic (Lobachevskian) geometry. But that's a story for another time.)

In terms of the rapidity, and using some hyperbolic trig identities, the Lorentz factor $\gamma(\zeta)$ is

\begin{align*} \gamma &= \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ &= \frac{1}{\sqrt{1-\tanh^2\zeta}}\\ \gamma &= \cosh\zeta. \end{align*}

This is the amount by which the rocket's time is dilated, and so the "effective travel velocity" $\gamma v$ , the distance the rocket covers in the "stationary" frame divided by the time elapsed in its own frame, is

\begin{align*} \gamma v &= \cosh(\zeta) \:\cdot\: c\tanh(\zeta)\\ \gamma v &= c \sinh \zeta. \end{align*}

The Relativistic Rocket Equation

We can relate $d\zeta$ to the earlier nonrelativistic expression for $d v$ by recognising that that expression is still valid in the instantaneous rest frame of the rocket. This relation is

\left.\frac{d\zeta}{dv}\right|_{v=0} = \frac{1}{c},

and so the relativistic rocket equation is

\begin{align*} \Delta\zeta &= \int_\text{burn} d\zeta\\ &= \int_{m_f}^{m_e} \left(\frac{d\zeta}{dv} \frac{dv}{dp} \frac{dp}{dm}\right)_{v=0} dm\\ &= \frac{v_e}{c} \int_{m_e}^{m_f} \frac{1}{m} dm\\ \Delta\zeta &= \frac{v_e}{c} \ln \frac{m_f}{m_e}. \end{align*}

If we assume that the rocket starts out stationary, the rapidity at the end of the burn is $\zeta = \Delta\zeta$ . Then, the velocity achieved is

v = c \tanh\left(\frac{v_e}{c} \ln \frac{m_f}{m_e}\right).

Here, we have a logarithmic function ( $\ln$ ) inside an even slower-growing function ( $\tanh$ ). Effectively, the tyranny of the rocket equation is compounded by the "tyranny of relativity", in which it is impossible to reach $c$ , and it becomes harder and harder to accelerate as you get closer to $c$ . No surprises there. What about the time-dilated perspective of the crew of the rocket, though? Their effective travel velocity is

\gamma v = c \sinh\left(\frac{v_e}{c} \ln \frac{m_f}{m_e}\right).

Here, the natural logarithm is inside the exponentially-growing $\sinh$ function. That is to say, the tyranny of the rocket equation is cancelled out by an "anti-tyranny of time dilation". Indeed, in the best-case scenario of $v_e = c$ and the "fuel-dominated" limit $m_f \gg m_e$ , the effective travel velocity is linear to the fuel mass ratio:

\gamma v \approx \frac{c}{2}\frac{m_f}{m_e}.

Recall, however, that this result is assuming we use the entire burn of the rocket to speed up. In practice, we will often instead need to use half of our $\Delta\zeta$ to speed up, and the other half to slow down at our destination, meaning that the maximum "cruising" rapidity we achieve is only $\zeta = \Delta\zeta/2$ . In that case, the exact cruising velocities are

\begin{align*} v &= c \tanh\left(\frac{1}{2}\frac{v_e}{c} \ln \frac{m_f}{m_e}\right)\\ \gamma v &= c \sinh\left(\frac{1}{2}\frac{v_e}{c} \ln \frac{m_f}{m_e}\right), \end{align*}

and the best-case scenario from before becomes just

\gamma v \approx \frac{c}{2} \sqrt{\frac{m_f}{m_e}},

meaning that the amount of fuel we need to achieve a given $\gamma v$ is quadratic rather than linear. This is still quite a lot better than the exponential growth of the nonrelativistic case, though!

Exhaust Velocities

Since we are not concerned in this section with changes over time, I drop the differential notation from this point on. In a rocket, a mass $M$ of fuel undergoes some reaction which creates $\eta Mc^2$ kinetic energy, $\eta$ being the mass-energy conversion efficiency of the reaction, and leaves behind reaction products with mass $m = (1-\eta)M$ . The momentum produced by this reaction can then be found by

\begin{align*} (Mc^2)^2 = E^2 &= (mc^2)^2 + (pc)^2\\ p^2c^2 &= M^2c^4 - m^2c^4\\ p &= c \sqrt{M^2 - m^2}\\ p &= Mc \sqrt{1-(1-\eta)^2}, \end{align*}

and the exhaust velocity is therefore

v_e = \frac{p}{M} = c\sqrt{1-(1-\eta)^2}.

We can see from this result that $v_e \leq c$ , with the maximum exhaust velocity $v_e=c$ being achieved when $\eta=1$ , corresponding to massless exhaust, i.e. a photon rocket.

Multiple Exhaust Species

In many cases, the fuel reaction products consist of multiple species, which could each have different momenta. Let us assume there are $n$ species, each with mass $m_k$ and momentum $p_k$ such that

\begin{align*} m &= \sum_{k=1}^{n} m_k\\ p &= \sum_{k=1}^{n} p_k \end{align*}

and the total energy $E$ is

Mc^2 = E = \sum_{k=1}^{n} \sqrt{(m_kc^2)^2 + (p_kc)^2}.

I now optimise this, finding the $p_k$ that maximise $p$ under the constraint of $E^2$ . To do this, I use the Lagrange multiplier method, where we find critical points where $p$ and $E$ have parallel gradients

\begin{align*} \frac{\partial p}{\partial p_k} &= \lambda \frac{\partial E}{\partial p_k}\\ 1 &= \frac{\lambda c^2 p_k}{\sqrt{(m_kc^2)^2 + (p_kc)^2}}\\ \lambda^2 c^2 p_k^2 &= m_k^2c^2 + p_k^2\\ (\lambda^2 c^2 - 1) p_k^2 &= m_k^2c^2\\ p_k &= \frac{m_kc}{\sqrt{\lambda^2 c^2 - 1}} \end{align*}

and where the constraint is satisfied:

\begin{align*} E &= \sum_{k=1}^{n} \sqrt{(m_kc^2)^2 + (p_kc)^2}\\ &= \sum_{k=1}^{n} \sqrt{m_k^2c^4 + \frac{m_k^2c^4}{\lambda c^2 - 1}}\\ M &= m\sqrt{1 + \frac{1}{\lambda c^2 - 1}}\\ \frac{1}{\sqrt{\lambda c^2 - 1}} &= \sqrt{\frac{M^2-m^2}{m^2}}\\ p_k &= m_kc\sqrt{\frac{M^2-m^2}{m^2}}. \end{align*}

What we see here, then, is that the optimal distribution of momentum among the species is one where they all have the same velocity, and their momenta are proportional to their masses. The total momentum, given this optimal distribution

\begin{align*} p &= \sum_{k=1}^{n} p_k\\ &= \sum_{k=1}^{n} m_kc \sqrt{\frac{M^2-m^2}{m^2}}\\ &= m c\sqrt{\frac{M^2-m^2}{m^2}}\\ &= c\sqrt{M^2-m^2}, \end{align*}

gives us exhaust velocity

\begin{align*} v_e &= \frac{p}{M}\\ &= \frac{c}{M} \sqrt{M^2-m^2}\\ &= c \sqrt{1 - \frac{(1-\eta)^2M^2}{M^2}}\\ v_e &= c \sqrt{1 - (1-\eta)^2}, \end{align*}

the same as in the prior single-species analysis.

Separate Propellant Species

Until now, I've been using the word "fuel" quite loosely, to refer to both the substance(s) used to generate kinetic energy, and for the reaction mass that that kinetic energy is imparted on. However, in principle, these can be separate, with the words "fuel" and "propellant" referring to the former and latter respectively. For example, there is the concept of the nuclear thermal rocket, in which a nuclear reaction is used to heat a hydrogen propellant which is then vented through the rocket nozzle. We can model this propellant as an additional "spectator species" of mass $m_p$ , by which we modify our fuel/propellant masses:

\begin{align*} M &\to M + m_p\\ m &\to m + m_p. \end{align*}

Since this increases our initial mass $M$ without increasing the amount of kinetic energy $(M-m)c^2$ generated, this can only ever decrease the mass-energy efficiency $\eta$ , and therefore the exhaust velocity $v_e$ . Proposals like the nuclear thermal rocket exist to address engineering challenges, like that of ejecting spent nuclear fission fuel at high speeds, not to improve performance per se.

Examples

Below is a table of various reactions, in ascending order of performance. This includes nuclear fission, nuclear fusion, and matter-antimatter annihilation. For nuclear reactions, speculative reactions producing ⁵⁶Fe, the tightest-bound nucleus, are shown as examples of the theoretical best nuclear fission and fusion reactions. Note, of course, that these show the maximum theoretically possible $v_e$ for the reaction, ignoring the challenges of optimally distributing kinetic energy among the product species.

Reaction	$M$	$m$	$\eta$	Maximum $v_e$
²³⁵U → n + ⁹⁵Sr + ¹³⁹Xe	235.04 Da	234.85 Da	0.084%	4.09% c
56²³⁹Pu → 239⁵⁶Fe	13387 Da	13368 Da	0.138%	5.25% c
²H + ³H → n + ⁴He	5.030 Da	5.011 Da	0.38%	8.66% c
4¹H → ⁴He	4.031 Da	4.003 Da	0.71%	11.9% c
56¹H → ⁵⁶Fe	56.44 Da	55.93 Da	0.89%	13.3% c
e⁺ + e^- → 2γ	2 $m_e$	0	100%	c