Relativistic speed of a rocket with constant thrust

[DrStupid]

I don't know how to do that.

I'll try to do that as simply as possible: Let't say the rocket starts in its own inertial rest frame, burns a small amount dm of fuel, ejects the exhaust with the specific momentum u and gains the momentum dp and the speed dv according to

$dp = u \cdot dm = \frac{{m \cdot dv}}{{\sqrt {1 - \frac{{dv^2 }}{{c^2 }}} }} \approx m \cdot dv$

For a sufficiently small dm this turns into

$dv = u \cdot \frac{{dm}}{m}$

In another frame of reference, where the rocket initially moves with the speed w the resulting speed is

$w' = \frac{{w + dv}}{{1 + \frac{{w \cdot dv}}{{c^2 }}}}$

That means the additional speed incrases by

$dw = w' - w = \frac{{c^2 - w^2 }}{{c^2 + w \cdot dv}} \cdot dv \approx \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot dv = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m} \cdot dm$

That gives you a differential equation for the speed in this inertial frame of reference:

$\frac{{dw}}{{dm}} = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m}$

To get your specific case you just need to insert u=c.

 

[Orodruin]

burns a small amount dm of fuel

A warning to the OP about signs. This definition means that the mass after ejection is m-dm, not m+dm. This affects the integration of the resulting differential equation. Personally, I always prefer to have m+dm after the ejection, meaning that dm will be negative, but it will make getting the signs right less cumbersome.

 

[jbriggs444]

If you read my original question you see that I talk about mass decreasing all the time.

Right. If mass were to decrease as velocity increases and if velocity increases at just the right rate to keep total energy (mass plus kinetic energy) constant then that is the same as keeping relativistic mass constant. That is because relativistic mass is the sum of rest mass and kinetic energy.

 

[DrStupid]

A warning to the OP about signs. This definition means that the mass after ejection is m-dm, not m+dm.

Or that u is negative because u and dv have opposite directions.

 

[Flisp]

I'll try to do that as simply as possible: Let't say the rocket starts in its own inertial rest frame, burns a small amount dm of fuel, ejects the exhaust with the specific momentum u and gains the momentum dp and the speed dv according to

$dp = u \cdot dm = \frac{{m \cdot dv}}{{\sqrt {1 - \frac{{dv^2 }}{{c^2 }}} }} \approx m \cdot dv$

For a sufficiently small dm this turns into

$dv = u \cdot \frac{{dm}}{m}$

In another frame of reference, where the rocket initially moves with the speed w the resulting speed is

$w' = \frac{{w + dv}}{{1 + \frac{{w \cdot dv}}{{c^2 }}}}$

That means the additional speed incrases by

$dw = w' - w = \frac{{c^2 - w^2 }}{{c^2 + w \cdot dv}} \cdot dv \approx \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot dv = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m} \cdot dm$

That gives you a differential equation for the speed in this inertial frame of reference:

$\frac{{dw}}{{dm}} = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m}$

To get your specific case you just need to insert u=c.

Greate, some equations, I'll try to put them into my sread sheet and come back!

 

[jbriggs444]

But am I not right, that the loss of energy only affects the maximum final speed, not the basic shape of the curve?

If you lose energy faster then you lose mass faster and, for a given fixed [proper] thrust, you gain speed faster. There is no set maximum final speed because you have not set a criterion for termination of the run.

If we terminate when the craft's rest mass reaches zero then clearly the limiting velocity will be the speed of light.

 

[PeterDonis]

constant thrust

What exactly does "constant thrust" mean? In a relativistic scenario, it's important to be precise about this.

Burning constant amounts of fuel, I would expect a linear increase of momentum and an exponential increase of speed

This is impossible in relativity. Momentum can increase without bound but speed can never reach the speed of light. So it's impossible to have both of the things you describe true in a relativistic scenario.

 

[jartsa]

I try to calculate the speed curve of a relativistic rocket driven by a 100% efficient engine with constant thrust, when traveling to a distant star. All equations I can find consider constant acceleration, which of course is not working, since the ships mass decreases when the fuel is used, resulting in a low acceleration at the beginning and high acceleration/deceleration towards the end.
I tried using the relativistic momentum equation E² = p² + m² (using c = 1), and p = w * m,
(where E is energy, p momentum, m mass, and w proper speed) and assuming that the entire mass of the used fuel is transferred into momentum so that the energy of the rocket is constant. However, the momentum curve I get is steep at start and flat at the end (without deceleration). The resulting speed curve is steep at start, flattens and gets steeper again. I don't understand why. Burning constant amounts of fuel, I would expect a linear increase of momentum and an exponential increase of speed since less and less mass is accelerated by the constant force from the engine.
Where am I thinking wrong, here?

So I guess your reasoning goes like this:

E² = p² + m²

Our accelerating vehicle only does work on itself, so the E² stays constant, in the frame of the road. The m² is the rest mass squared. If the m² decreases by 4 then the p² must increase by 4. In the road frame. And the only reason that m² decreases is that the amount of fuel decreases.

Very good this far, I would say.

... But then for some reason the reasoning goes all wrong. Did you change to the vehicle frame? Well, don't do that

In the vehicle frame a constant rate of fuel consumption will generate a constant thrust. And now back to the road frame, where fuel is consumed at some rate that we know, after shortly visiting the vehicle frame.

 

[m4r35n357]

I try to calculate the speed curve of a relativistic rocket driven by a 100% efficient engine with constant thrust, when traveling to a distant star. All equations I can find consider constant acceleration, which of course is not working, since the ships mass decreases when the fuel is used, resulting in a low acceleration at the beginning and high acceleration/deceleration towards the end.

There is an article here that might help.

 

[PAllen]

There is an article here that might help.

Note to OP - the whole first section of that link is the constant proper acceleration case that you are not interested in. However, the second acceleration profile considered is exactly the simplest reasonable consistent implementation of your desired scenario.

 

[Filip Larsen]

If OP wants to know about relativistic rocket equation (i.e. a constant thrusting rocket) then I am puzzled why a straight forward Wikipedia reference like [1] has not already been mentioned. As far as I can see this equation even holds for optimal efficiency (most delta-v per mass ratio), i.e. a photon rocket. Or did I miss something?

[1] https://en.wikipedia.org/wiki/Relativistic_rocket#Formula_for_Δv

 

[PeterDonis]

Or did I miss something?

Yes, you did: the rocket equation you refer to is for the case of constant proper acceleration, not constant thrust. The two are not the same.

 

[Flisp]

If you lose energy faster then you lose mass faster and, for a given fixed [proper] thrust, you gain speed faster. There is no set maximum final speed because you have not set a criterion for termination of the run.

If we terminate when the craft's rest mass reaches zero then clearly the limiting velocity will be the speed of light.

A rocket contains enigines, passegers, fuselage etc, (called payload) ... and fuel, giving you a fuel-payload ratio. You burn the fuel only, not the craft or payload. Once all fuel is burnt you naturally have to terminate acceleration or thrust. The higher the thrust from fuel ratio, the higher the maximum final speed.

 

[PeroK]

A rocket contains enigines, passegers, fuselage etc, (called payload) ... and fuel, giving you a fuel-payload ratio. You burn the fuel only, not the craft or payload. Once all fuel is burnt you naturally have to terminate acceleration or thrust. The higher the thrust from fuel ratio, the higher the maximum final speed.

One of the problems with interstellar space flight is that to get to relativistic speeds you need a very high fuel-to-payload ratio. To get close to $c$ your rocket must start out as almost all fuel.

It's not so silly, therefore, to extrapolate to the point where the payload is of negligible mass compared to the fuel.

 

[Filip Larsen]

the rocket equation you refer to is for the case of constant proper acceleration, not constant thrust.

For the classical rocket equation the specific acceleration profile is irrelevant, i.e. it covers both constant acceleration and thrust as long as propellant ejection speed remain constant. Or put in other words, the time integral of the (possibly time-varying) acceleration is the total delta-V.

As I understand the Wikipedia page section I referred to, it argues that the same is the case in the relativistic case, that is, the proper time integral of the proper acceleration is still equal to the same delta-V. I must admit the section has a big hand waving step going to the the last equation where it just replaces speed with rapidity, so perhaps my understanding is wrong?

 

[PeroK]

For the classical rocket equation the specific acceleration profile is irrelevant, i.e. it covers both constant acceleration and thrust as long as propellant ejection speed remain constant. Or put in other words, the time integral of the (possibly time-varying) acceleration is the total delta-V.

As I understand the Wikipedia page section I referred to, it argues that the same is the case in the relativistic case, that is, the proper time integral of the proper acceleration is still equal to the same delta-V. I must admit the section has a big hand waving step going to the the last equation where it just replaces speed with rapidity, so perhaps my understanding is wrong?

My suggestion is to post this in the homework section, using the idea of a photon engine. Your first step is to show the following:

$$E = \frac{m_0^2 + m^2}{2m_0}$$
$$\gamma = \frac{m_0^2 + m^2}{2m_0m}$$
$$v = \frac{m_0^2 - m^2}{m_0^2 + m^2}$$
$$\frac{m}{m_0} = \sqrt{\frac{1-v}{1+v}}$$

Where $m_0$ is the initial mass of the rocket; $m$ is the mass after a certain number of photons have been ejected; and $E, \gamma, v$ are the energy and speed of the rocket in the initial rest frame.

The next step is to turn these equations into a differential equation for the speed of the rocket with respect to proper and/or coordinate time, assuming constant mass conversion to photons.

 

[jartsa]

In the vehicle frame a constant rate of fuel consumption will generate a constant thrust. And now back to the road frame, where fuel is consumed at some rate that we know, after shortly visiting the vehicle frame.

Actually as I was considering a road vehicle, fuel consumption rate in vehicle frame should be proportional to speed, in order to get a constant thrust in vehicle frame. Speed means speed of road in vehicle frame - or speed of vehicle in road frame.

 

[Flisp]

One of the problems with interstellar space flight is that to get to relativistic speeds you need a very high fuel-to-payload ratio. To get close to $c$ your rocket must start out as almost all fuel.

It's not so silly, therefore, to extrapolate to the point where the payload is of negligible mass compared to the fuel.

Yeah, that is exactly what I want to calculate. But since all equations I could find are based on constant acceleration, I got stuck. If you start with a high fuel-payload ratio you need a very powerfull engine to get somewhere at all, and for many years you have no time dilation because you are to slow. Then, toward the end of the flight you need to slow down. Now the rocket is ligher, but you can not decelerate using the full power of the engine, because you get to high g-forces. So I guess, that even with a very powerfull engine and a high fuel-payload ratio there are limits how fast you can get to a distant star.

 

[Flisp]

That gives you a differential equation for the speed in this inertial frame of reference:

$\frac{{dw}}{{dm}} = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m}$

To get your specific case you just need to insert u=c.

I'm not sure I get this right:
I use
${dw} = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m}\cdot{dm}$
A) you write $w$ what would be proper velocity (measured by the astronaut) but you say its in the inertial frame (observer) what used to be called $v$. Which one is it?
B) I get 2 different results depending on how I calculate and both are very low and can not be right.
B1) If I cut the travel into a large number of small steps using the resulting w and m from the previous step as input to the new step I get a max speed for a rocket with a fuel-payload ratio of 1:1 and an u of 1 of 0.6 c.
B2) Making one big leap using an initial w = 0 and m = 2, I get a max w of 0.5 c. Which one should I use?
C) And if I set u = 0.008 what would be the max efficiency that a fusion powered photon engine could produce, I get only a max speed of 0.006 c.
D) Even at a fuel-payload ratio of 1000:1 I only get a max velocity of 0.055 c.
That can't be right?
And in this calculation I haven't even started to include deceleration!

 

[Flisp]

There is an article here that might help.

I'm sorry. It's about 3.5 decades ago I had that kind of math a school and do, frankly, not remember much of it:-)

 

[Flisp]

So I guess your reasoning goes like this:

E² = p² + m²

Our accelerating vehicle only does work on itself, so the E² stays constant, in the frame of the road. The m² is the rest mass squared. If the m² decreases by 4 then the p² must increase by 4. In the road frame. And the only reason that m² decreases is that the amount of fuel decreases.

Very good this far, I would say.

... But then for some reason the reasoning goes all wrong. Did you change to the vehicle frame? Well, don't do that

In the vehicle frame a constant rate of fuel consumption will generate a constant thrust. And now back to the road frame, where fuel is consumed at some rate that we know, after shortly visiting the vehicle frame.

Why is that to be the road frame. I thought it to be the vehicle frame. And as I did the calculations I got momentum and speed above c, which is okay for the vehicle/rocket frame, but not for the road/observer.
I know I'm wrong somewhere, but I still do not know where, and what is right instead.

 

[m4r35n357]

I'm sorry. It's about 3.5 decades ago I had that kind of math a school and do, frankly, not remember much of it:-)

I understand, that guy is at the limits of my capabilities most of the time ;)

But seriously, if that is the case you might have to manage your expectations (why not just try reading the text, you should get something out of it).

 

[PeroK]

I'm not sure I get this right:
I use
${dw} = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m}\cdot{dm}$
A) you write $w$ what would be proper velocity (measured by the astronaut) but you say its in the inertial frame (observer) what used to be called $v$. Which one is it?
B) I get 2 different results depending on how I calculate and both are very low and can not be right.
B1) If I cut the travel into a large number of small steps using the resulting w and m from the previous step as input to the new step I get a max speed for a rocket with a fuel-payload ratio of 1:1 and an u of 1 of 0.6 c.
B2) Making one big leap using an initial w = 0 and m = 2, I get a max w of 0.5 c. Which one should I use?
C) And if I set u = 0.008 what would be the max efficiency that a fusion powered photon engine could produce, I get only a max speed of 0.006 c.
D) Even at a fuel-payload ratio of 1000:1 I only get a max velocity of 0.055 c.
That can't be right?
And in this calculation I haven't even started to include deceleration!

This is getting too complicated. Using a photon engine is simpler because the emitted energy has no mass and the equations are simpler.

If you can't calculate the energy-momentum for a single photon emission, then you are going to struggle with anything more sophisticated.

But, after this calculation you nearly have your answer for speed vs mass of used fuel.

As before, the homework section is a better place to restart this thread.

 

[DrStupid]

A) you write $w$ what would be proper velocity (measured by the astronaut) but you say its in the inertial frame (observer) what used to be called $v$. Which one is it?

In my equations w is the speed in the inertial frame of an external observer and dv is an infinitesimal small increase of speed in the inertial rest-frame of the rocket.

B1) If I cut the travel into a large number of small steps using the resulting w and m from the previous step as input to the new step I get a max speed for a rocket with a fuel-payload ratio of 1:1 and an u of 1 of 0.6 c.

That’s the correct result.

C) And if I set u = 0.008 what would be the max efficiency that a fusion powered photon engine could produce, I get only a max speed of 0.006 c.

D) Even at a fuel-payload ratio of 1000:1 I only get a max velocity of 0.055 c.

I don’t know why somebody would use a fusion powered photon engine but the results are correct. The maximum possible specific momentum of a fusion drive is 0.14 c. That would result in a final velocity of 0.097 c for a fuel-payload ratio of 1:1 and 0.75 c for a fuel-payload ratio of 1000:1.

 

[Flisp]

In my equations w is the speed in the inertial frame of an external observer and dv is an infinitesimal small increase of speed in the inertial rest-frame of the rocket.

OK.

That’s the correct result.

Greate!

I don’t know why somebody would use a fusion powered photon engine but the results are correct.

What engine has a better perfonmance?

The maximum possible specific momentum of a fusion drive is 0.14 c. That would result in a final velocity of 0.097 c for a fuel-payload ratio of 1:1 and 0.75 c for a fuel-payload ratio of 1000:1.

If I set u = 1.4 I get the same results. Thank you very, very much!

 

[jbriggs444]

I don’t know why somebody would use a fusion powered photon engine

What engine has a better perfonmance?

If you throw the expended fuel (e.g. helium) from such a rocket out the back, you can improve the performance dramatically. By comparison, using fusion to power a photon exhaust and then dribbling the waste helium out the side would be silly.

 

[Flisp]

If you throw the expended fuel (e.g. helium) from such a rocket out the back, you can improve the performance dramatically. By comparison, using fusion to power a photon exhaust and then dribbling the waste helium out the side would be silly.

Do you know the max specific impulse of that?

 

[jbriggs444]

Do you know the max specific impulse of that?

It can be calculated. The first step is to figure out what your fusion engine is using for fuel. Let's say that it is fusing Deuterium to Helium. So you go to the periodic table and look up the atomic mass of Deuterium and the atomic mass of Helium.

What is the net change in mass due to the reaction?
What is that change as a fraction of the mass of the two Deuterium atoms that went into the reaction?

The relativistic mass of the ejected helium atom is equal to its rest mass times the relativistic gamma factor. $\gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$. The relativistic mass of the ejected helium atom is also equal to the total mass of the two Deuterium atoms that went into the reaction. That means that the relativistic gamma is equal to the fraction that you calculated above.

Solve for v.

Note that you asked for specific impulse. What you should have asked for was exhaust velocity. They are not the same thing.

If you insist on converting to specific impulse, that can be done. But first you have to look up the definition of specific impulse.

 

[Flisp]

If you insist on converting to specific impulse, that can be done. But first you have to look up the definition of specific impulse.

What I need is the specific momentum u for the equation
${dw} = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m}\cdot{dm}$

 

[Flisp]

That gives you a differential equation for the speed in this inertial frame of reference:
$\frac{{dw}}{{dm}} = \left( {1 - \frac{{w^2 }}{{c^2 }}} \right) \cdot \frac{u}{m}$

I have one more question: This equation gives me the speed as measured in the inertial frame of the observer / home planet, assuming constant thrust. However, as the rocket approaches c time will slow down and the engine on board will burn less fuel per second or hour as measured by the observer. The observer will measure lower thrust, the astronauts will not. That means that this equation calculates on thrust values that are increasingly to high compared to the "real" thrust produced by the engine. How can I adjust the equation so that it calculates with constant proper thrust resulting in proper speed (that then can be transformed into observed speed using the Lorentz factor gamma from that speed, correct?)

 

[PAllen]

I have one more question: This equation gives me the speed as measured in the inertial frame of the observer / home planet, assuming constant thrust. However, as the rocket approaches c time will slow down and the engine on board will burn less fuel per second or hour as measured by the observer. The observer will measure lower thrust, the astronauts will not. That means that this equation calculates on thrust values that are increasingly to high compared to the "real" thrust produced by the engine. How can I adjust the equation so that it calculates with constant proper thrust resulting in proper speed (that then can be transformed into observed speed using the Lorentz factor gamma from that speed, correct?)

The link given in post #44 gives exactly this case. Even if you don’t follow all steps, you can follow the overall logic, taking certain things as given, and use the resulting equations. Specifically, they cover constant thrust as measured by the astronaut, for any chosen fixed exhaust speed, with fixed mass loss rate for the rocket, again, per the astronaut, resulting in fixed thrust.

 

[Flisp]

The link given in post #44 gives exactly this case. Even if you don’t follow all steps, you can follow the overall logic, taking certain things as given, and use the resulting equations. Specifically, they cover constant thrust as measured by the astronaut, for any chosen fixed exhaust speed, with fixed mass loss rate for the rocket, again, per the astronaut, resulting in fixed thrust.

What I can find is the equation after "Another interesting acceleration profile is the one that results from a constant nozzle velocity u and constant exhaust mass flow rate w = dm0/dt,", but I hav no idea how to transform it into something I can use.

 

[PAllen]

What I can find is the equation after "Another interesting acceleration profile is the one that results from a constant nozzle velocity u and constant exhaust mass flow rate w = dm0/dt,", but I hav no idea how to transform it into something I can use.

The next several paragraphs do it for you. What question do you want answered? Every question I would think of is anwered in one of the following derived equations.

 

[Orodruin]

This equation gives me the speed as measured in the inertial frame of the observer / home planet, assuming constant thrust.

Yes, but as a function of burned fuel. Not as a function of time. Of course, if you burn a constant amount of fuel per proper time, there is a linear correspondence to proper time.

To relate proper time and time you can use the time dilation formula.

proper speed

What do you mean by ”proper speed”? Speed is relative.

 

[Flisp]

Yes, but as a function of burned fuel. Not as a function of time. Of course, if you burn a constant amount of fuel per proper time, there is a linear correspondence to proper time.

To relate proper time and time you can use the time dilation formula.

What do you mean by ”proper speed”? Speed is relative.

Speed as measured by the astronaut. That would be distance as measured by the observer divided by time as measured by the astronaut,... I guess.