# Relativistic rocket equation (intuitive derivation)

• TubbaBlubba
In summary, the conversation discusses the derivation of the rocket equation in Newtonian mechanics and the need for a relativistic equation. Various methods for deriving the equation are mentioned, and the speaker shares their attempt at understanding it in a more intuitive way. However, their reasoning is found to be incorrect and the correct method is explained. The conversation also mentions the textbook used (Rindler's "Relativity: Special, General and Cosmological, 2nd ed") and the book's hint for the equation. The speaker then shares their own method of deriving the equation using conservation of momentum and energy, and the conversation ends with a discussion of different approaches to deriving the rocket equation.
TubbaBlubba

## Homework Statement

In Newtonian mechanics, the rocket equation is derived by solving the simple differential equation -dm U = m dV, where U is the velocity of the expelled material relative to the rocket; a matter of conservation of momentum. In order to get the correct relativistic equation, one wishes to arrive at the differential equation ##-dm U = \gamma^2(V) m dV##. I know of lengthy derivations using velocity addition and so on, but I want to try to understand it in a more intuitve and managable way. My textbook (Rindler) is not exactly helpful by providing the hint of writing M for the rest mass of the rocket and dM for the "relativistic mass" of the expelled material...

## Homework Equations

We wish to arrive at ##-dm U = \gamma^2 m dV## using relativistic mechanics.

## The Attempt at a Solution

My line of reasoning is this: The final velocity V is the velocity relative to Earth (or wherever the rockets starts from). At any given moment, we will observe the rocket of mass m changing its velocity by dV to increase its momentum by ##\gamma(V) m dV##. On the other hand, the expelled material is expelled at a velocity U, constant relative to the instantaneous rest frame of the rocket. So from Earth, the expelled material will be observed to have momentum ##-\frac{U}{\gamma(V)}dm##, giving us the desired differential equation ##-dm U = \gamma^2(V) m dV##..

Is this line of reasoning acceptable, intuitively speaking?

TubbaBlubba said:
My line of reasoning is this: The final velocity V is the velocity relative to Earth (or wherever the rockets starts from). At any given moment, we will observe the rocket of mass m changing its velocity by dV to increase its momentum by ##\gamma(V) m dV##. On the other hand, the expelled material is expelled at a velocity U, constant relative to the instantaneous rest frame of the rocket. So from Earth, the expelled material will be observed to have momentum ##-\frac{U}{\gamma(V)}dm##, giving us the desired differential equation ##-dm U = \gamma^2(V) m dV##..

Is this line of reasoning acceptable, intuitively speaking?

No. Mainly because the expressions you have for the change in the momentum of the rocket and the momentum of the ejected material are not correct. Note that the change in the rocket's momentum will have the contribution you quote from the change in its velocity - but it will also have contributions from the changes in the mass and in the gamma factor. Likewise, the momentum of the ejected material needs to be considered more carefully. Note that the velocity of the ejected material is not necessarily in the opposite direction of the spaceship velocity in the Earth frame. If ##V > U##, then the ejected material will still be going in the forward direction.

A priori, you should be able to do the analysis in the Earth's rest frame, but you need to take more care when deriving the expressions for the momenta.

TubbaBlubba
Orodruin said:
No. Mainly because the expressions you have for the change in the momentum of the rocket and the momentum of the ejected material are not correct. Note that the change in the rocket's momentum will have the contribution you quote from the change in its velocity - but it will also have contributions from the changes in the mass and in the gamma factor. Likewise, the momentum of the ejected material need to be considered more carefully. Note that the velocity of the ejected material is not necessarily in the opposite direction of the spaceship velocity in the Earth frame. If ##V > U##, then the ejected material will still be going in the forward direction.

A priori, you should be able to do the analysis in the Earth's rest frame, but you need to take more care when deriving the expressions for the momenta.
Thank you, I will think a bit more about it.

OK, I worked it out a priori using differential calculus and conservation of momentum/energy. Yeah, it's not that hard as long as you keep various Lorentz-factor identities in mind.

TubbaBlubba said:
My textbook (Rindler)

By the way, Rindler has written several textbooks on relativity. Not that it mattered here, but it will generally be helpful if you quote the title of the book you are referring to.

TubbaBlubba
Orodruin said:
By the way, Rindler has written several textbooks on relativity. Not that it mattered here, but it will generally be helpful if you quote the title of the book you are referring to.
Yes; I'll keep that in mind!

TubbaBlubba said:
Yes; I'll keep that in mind!

OK, so which one ARE you referring to?

Ray Vickson said:
OK, so which one ARE you referring to?
"Relativity: Special, General and Cosmological, 2nd ed".

Rindler suggests as a hint the equation dM U = M du', where du' is the speed of the rocket relative to its instantaneously inertial frame and dM the "relativistic mass" of the jet.

What I eventually did was to set up conservation equations (g for gamma):
MVg(V) = (M + dM)(V + dV)(g(V + dV)) + ((V-U)/(1-UV))g((V-U)/(1-UV)))dM'
and similarly for energy, some manipulations, and you get there eventually.

Yes, it is rather straight forward really once you keep track of everything. I still prefer the derivation where you start by looking at the instantaneous rest frame because there are simply less things to keep track of. The addition of velocities is simply
$$V + dV = \frac{V+dV'}{1+V\, dV'} \simeq (V+dV')(1-V\, dV') \simeq V + dV' (1 - V^2) = V + \frac{dV'}{\gamma^2},$$
directly identifying ##dV' = \gamma^2 dV##.

Orodruin said:
Yes, it is rather straight forward really once you keep track of everything. I still prefer the derivation where you start by looking at the instantaneous rest frame because there are simply less things to keep track of. The addition of velocities is simply
$$V + dV = \frac{V+dV'}{1+V\, dV'} \simeq (V+dV')(1-V\, dV') \simeq V + dV' (1 - V^2) = V + \frac{dV'}{\gamma^2},$$
directly identifying ##dV' = \gamma^2 dV##.

Ah, that's a nice trick - using velocity addition with V + dV. Will have to remember that one.

But a differential equation with M as referring to a conserved quantity and dM to an unconserved, yeah, that irks me.

TubbaBlubba said:
But a differential equation with M as referring to a conserved quantity and dM to an unconserved, yeah, that irks me.
Where is this differential equation?

Orodruin said:
Where is this differential equation?
It's just me being confused. -dM U = M dV' relates the rate of change in the rocket's rest mass to the velocities. But dM is not the rest mass of the exhaust jet at that moment. Or is it? Ugh...

TubbaBlubba said:
But dM is not the rest mass of the exhaust jet at that moment. Or is it? Ugh...
No it is not, it is the change in the rest* mass of the rocket. Note that the rest* mass of the exhaust will be zero if ##U = 1##. The equation is the conservation of momentum equation after taking conservation of energy into account. By conservation of energy, you know that
$$M = (M+dM)(1 + d\gamma) + \mathcal E = M + dM + \mathcal E \quad \Longrightarrow \quad \mathcal E = - dM$$
where ##\mathcal E## is the energy of the ejected material in the instantaneous rest frame of the rocket (note that ##d\gamma = \mathcal O(dV'^2)##). The momentum ##p## of the ejected material is related to the energy by ##p = - U \mathcal E##. From there you can go to the conservation of momentum to deduce that ##-U\, dM = M\, dV'##.

Edit: * Physicists normally only talk about mass. Relativistic mass is an antiquated concept - it is just another name for total energy.

TubbaBlubba
Orodruin said:
No it is not, it is the change in the rest* mass of the rocket. Note that the rest* mass of the exhaust will be zero if ##U = 1##. The equation is the conservation of momentum equation after taking conservation of energy into account. By conservation of energy, you know that
$$M = (M+dM)(1 + d\gamma) + \mathcal E = M + dM + \mathcal E \quad \Longrightarrow \quad \mathcal E = - dM$$
where ##\mathcal E## is the energy of the ejected material in the instantaneous rest frame of the rocket (note that ##d\gamma = \mathcal O(dV'^2)##). The momentum ##p## of the ejected material is related to the energy by ##p = - U \mathcal E##. From there you can go to the conservation of momentum to deduce that ##-U\, dM = M\, dV'##.
Ahhh, thank you. That makes it a lot more clear to me.

Edit: * Physicists normally only talk about mass. Relativistic mass is an antiquated concept - it is just another name for total energy.
Yeah, I know. Thankfully I have lecture and course notes that do away with it, but it can still be very confusing to work with textbook material that uses it.

TubbaBlubba said:
Relativity: Special, General and Cosmological, 2nd ed

Just out of curiosity, do you know why you are using that book and not the other book by Rindler (Introduction to Special Relativity) if your course only covers SR and no GR?

Orodruin said:
Just out of curiosity, do you know why you are using that book and not the other book by Rindler (Introduction to Special Relativity) if your course only covers SR and no GR?
Hm. It's a pretty brief course (3/5ths of one, the rest being Lagrangian mechanics) with extensive lecture notes, some homework and a relatively short exam (IIRC, it used to be oral), based on selections from a given set of problems from the book. Maybe the instructor uses it in several courses and is simply familiar with it, or they want students to be able to use it further on.

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## What is the relativistic rocket equation?

The relativistic rocket equation is an equation that describes the motion of a rocket as it travels through space, taking into account the effects of relativity. It relates the velocity of the rocket to the mass of the rocket and the amount of fuel it carries.

## How is the relativistic rocket equation derived?

The intuitive derivation of the relativistic rocket equation is based on the principles of conservation of momentum and conservation of energy. It considers the change in momentum and energy of the rocket as it expels fuel at a constant rate and accelerates in the opposite direction.

## What are the assumptions made in the derivation of the relativistic rocket equation?

The derivation assumes that the rocket is moving in a constant direction, that the fuel is being expelled at a constant rate, and that there are no external forces acting on the rocket. It also assumes that the rocket and the expelled fuel have constant mass ratios.

## How does the relativistic rocket equation differ from the classical rocket equation?

The classical rocket equation, also known as the Tsiolkovsky rocket equation, only takes into account the effects of Newtonian mechanics and does not consider the effects of relativity. The relativistic rocket equation accounts for the increase in mass of the rocket as it approaches the speed of light.

## What are some real-world applications of the relativistic rocket equation?

The relativistic rocket equation is used in the design and planning of space missions, as well as in the development of propulsion systems for spacecraft. It is also used in the study of astrophysics and cosmology, as it helps explain the behavior of objects traveling at high speeds in the universe.

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