# How Can We Derive the Relativistic Rocket Equation?

• B
• greg_rack
In summary: So\begin{align*}\tanh^{-1} \left( \frac{v_2}{c} \right) - \tanh^{-1} \left( \frac{v_1}{c} \right) & = \frac{u}{c} \ln \left( \frac{m_1}{m_2} \right) \\\implies \tanh \left( \frac{v_2-v_1}{1-v_2 v_1/c^2} \right) &= \tanh \left( \frac{u}{c} \ln \left( \
greg_rack
Gold Member
I am studying through online resources some principles of spacecraft propulsion, since it really fascinates me, and makes me want to know a bit more about it :)
For rockets, thruster, I found the Tsiolkovsky rocket equation:
$$\Delta v=v_e ln(\frac{m_0}{m_f})$$

Of course, rockets can travel up to relativistic speeds, thus a relativistic implementation of the RE must be taken into account for a higher degree of accuracy in calculations.
All implementations I found, rely on this equation which gives the mass ratio in terms of ##\Delta v## and exhaust speed ##v_e##... but I couldn't manage to derive it, since in all papers it is given without demonstration:

Could you please give me a hint on how to get to this identity?

I think you can do it like this. Let ##v## be the velocity of the rocket measured in the initial rest frame of the rocket. At some time ##t##, the rocket has momentum ##p## and energy ##E##. At some later time, after ejecting some amount of mass ##dk##, the rocket has mass ##m+dm##, momentum ##p + dp## and energy ##E + dE##. Note that, unlike energy and momentum, mass is not conserved in this process i.e. ##dm \neq - dk##.

If the speed of the ejected mass ##dk## with respect to the spaceship is ##u##, then its velocity in the initial rest frame is ##\tilde{v}= \dfrac{v-u}{1-uv}##. The ejected mass possesses energy ##-dE## and momentum ##-dp##. The change in momentum of the rocket is ##dp = d(\gamma m v) = \dfrac{-\tilde{v} dk}{\sqrt{1-\tilde{v}^2}}## and the change in energy of the rocket is ##dE = d(\gamma m) = \dfrac{-dk}{\sqrt{1-\tilde{v}^2}}##. Write\begin{align*}
d(\gamma mv) &= \tilde{v} d(\gamma m) = \dfrac{v-u}{1-uv} d(\gamma m) \\
\implies \frac{d(\gamma mv) }{dv} &= \dfrac{v-u}{1-uv} \dfrac{d(\gamma m)}{dv}
\end{align*}Can you solve this differential equation?

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JD_PM, vanhees71 and Dale
I think all one has to do is replace normalized velocity ##\beta = v/c## with rapidity, w.
https://en.wikipedia.org/wiki/Rapidity

Rapidities add in SR, while velocities don't. So, v = at becomes w = a##\tau##.

So, for the non-relativistic case we can say
$$m \frac{dv}{dt} = -v_e \frac{dm}{dt}$$
which reduces to
$$dv = v_e \frac{dm}{m}$$
which gives the solution
$$v =v_e ln \frac{m_i}{m_f}$$

Replacing v by the rapidity w gives:
$$dw = \frac{v_e}{c} \frac{dm}{m}$$
and
$$w = \frac{v_e}{c} ln \frac{m_i}{m_f}$$

where
$$w = \frac{1}{2} ln \frac{1+v/c } {1-v/c }$$
and in the limit when v is small, ##w \approx v/c##.

This can be solved to get v as a function of the mass ratio rather than w as a faction of the mass ratio.

Hopefully I didn't make any silly errors, but I won't guarantee it. I wouldn['t be too surprised if I made some error with a factor of c, I usually use geometric units where c=1.

hutchphd
It is not so hard from first principles, though. From #2, write\begin{align*}
v \frac{d(\gamma m)}{dv} + \gamma m &= \frac{v-u}{1-uv} \frac{d(\gamma m)}{dv} \\

\gamma m &= \left(\frac{v-u}{1-uv} - v \right) \frac{d(\gamma m)}{dv} \\

\gamma m &= \left(\frac{u(v^2-1)}{1-uv} \right) \frac{d(\gamma m)}{dv} \\ \\

\int_{v_1}^{v_2} dv \frac{(1-uv)}{u(v^2-1)} &= \int_{\gamma_1 m_1}^{\gamma_2 m_2} \frac{d(\gamma m)}{\gamma m} \\

\frac{-1}{u} \int_{v_1}^{v_2} dv \frac{1}{1-v^2} - \int_{v_1}^{v_2} dv \frac{v}{v^2-1} &= \int_{\gamma_1 m_1}^{\gamma_2 m_2} \frac{d(\gamma m)}{\gamma m} \\ \\

\frac{-1}{u} \left[ \mathrm{artanh}(v) \right]_{v_1}^{v_2} - \frac{1}{2} \ln{ \left| \frac{v_2^2-1}{v_1^2-1} \right|} &= \ln \left( \frac{\gamma_2 m_2}{\gamma_1 m_1} \right)

\end{align*}Because ##v_1, v_2 < 1##, ##\frac{1}{2} \ln{ \left| \dfrac{v_2^2-1}{v_1^2-1} \right|} = \dfrac{1}{2} \ln{\left( \dfrac{1-v_2^2}{1-v_1^2} \right)} = \ln{\left( \frac{\sqrt{1-v_2^2}}{\sqrt{1-v_1^2}} \right)} = \ln{ \left( \dfrac{\gamma_1}{\gamma_2}\right)} ##. So\begin{align*}

\frac{-1}{u} \left[ \mathrm{artanh}(v_2) - \mathrm{artanh}(v_1) \right] &= \ln \left(\frac{m_2}{m_1} \right) \\

\implies \mathrm{artanh}(v_2) - \mathrm{artanh}(v_1) &= u \ln \left(\frac{m_1}{m_2} \right) \\

w_2 - w_1 &= u \ln \left(\frac{m_1}{m_2} \right)

\end{align*}##w## is the rapidity

JD_PM and vanhees71

## 1. What is the Relativistic Rocket Equation?

The Relativistic Rocket Equation, also known as the Tsiolkovsky Rocket Equation, is a mathematical equation that describes the motion of a rocket in space. It takes into account both Newton's laws of motion and Einstein's theory of relativity to calculate the velocity and mass of a rocket as it travels through space.

## 2. How is the Relativistic Rocket Equation derived?

The equation was developed by Russian scientist Konstantin Tsiolkovsky in the late 19th century. He used a combination of Newton's second law of motion, which states that force equals mass times acceleration, and Einstein's theory of special relativity, which explains how objects move at high speeds.

## 3. What are the variables in the Relativistic Rocket Equation?

The equation includes five variables: the initial mass of the rocket, the final mass of the rocket, the velocity of the rocket, the speed of light, and the change in time. These variables are used to calculate the velocity of the rocket at any given time during its journey.

## 4. How is the Relativistic Rocket Equation used in space exploration?

The equation is used by scientists and engineers to design and plan space missions. It helps determine the amount of fuel needed for a rocket to reach a certain destination, as well as the maximum velocity that can be achieved. It also helps calculate the trajectory of the rocket and the amount of time it will take to reach its destination.

## 5. Are there any limitations to the Relativistic Rocket Equation?

While the equation is very useful in space exploration, it does have some limitations. It assumes that the rocket is traveling in a vacuum and does not take into account external forces such as gravity or air resistance. It also does not account for the effects of relativity on the rocket's mass and velocity at extremely high speeds.

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