I Relativistic speed of a rocket with constant thrust

[Flisp]

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?

 

[Dale]

My first guess would be that you are mixing proper time and coordinate time at some point. Can you post your math, being very careful to identify everything?

 

[Ibix]

Google "relativistic rocket". Wikipedia has the maths, and there's more detail at mathpages.

 

[jbriggs444]

100% efficient engine with constant thrust [...] and assuming that the entire mass of the used fuel is transferred into momentum so that the energy of the rocket is constant

So although we have constant thrust, we do not have constant acceleration. The ship's mass varies over time so that its relativistic mass ($\gamma \ m$) remains constant.

This definition of 100% efficiency means that it is not a rocket at all. Instead, it is a race car using the internal energy of its fuel reserves to power the thrust it applies on a stationary highway.

Edit to add: There can be no exhaust stream. Any ejected mass or energy would reduce the remaining relativistic mass.

 

[Flisp]

My first guess would be that you are mixing proper time and coordinate time at some point. Can you post your math, being very careful to identify everything?

At start the speed is so low that there is virtually no difference between the two speeds. I would like to show my math, is there some equation editor?

 

[jbriggs444]

is there some equation editor?

https://www.physicsforums.com/help/latexhelp/

You can find it with Info => Help/How to => LaTeX Primer

 

[Flisp]

Google "relativistic rocket". Wikipedia has the maths, and there's more detail at mathpages.

It's all based on constant acceleration a.

 

[SiennaTheGr8]

Constant proper acceleration!

 

[DrStupid]

assuming that the entire mass of the used fuel is transferred into momentum so that the energy of the rocket is constant

That's impossible.

 

[jbriggs444]

That's impossible.

Impossible for a rocket. Not for a race car. (See post #4)

 

[DrStupid]

Impossible for a rocket. Not for a race car.

I was referring to "assuming that the entire mass of the used fuel is transferred into momentum so that the energy of the rocket is constant"

 

[PeroK]

It's all based on constant acceleration a.

You could also search for "hyperbolic motion relativity", to get an analysis of motion with constant proper acceleratio. E.g.

https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

You could also search for "Rindler Coordinates".

 

[Flisp]

So although we have constant thrust, we do not have constant acceleration. The ship's mass varies over time so that its relativistic mass ($\gamma \ m$) remains constant.

This definition of 100% efficiency means that it is not a rocket at all. Instead, it is a race car using the internal energy of its fuel reserves to power the thrust it applies on a stationary highway.

I'm not sure I understand what you mean. With 100% efficiency I meant that the specific impulse I_sp = 1. For instance some alien anti matter drive. When I get this right I will calculate more realistic I_sp :-)

When I calculate momentum, mass should be constant regardless the frame of referrence accoring to Energy–momentum relation on Wikipedia. The mass of the ship, however, decreases when fuel is burnt.

To calculate the new momentum I use $E^2 = p^2+ m^2$, where E is energy, p is momentum, and m rest mass.
Some fuel is used and all its matter transformed into momentum while the ships total energy remains constant: From one moment to the next burning a certain amount of fuel I have:
$p_{1}^2 + m_{1}^2 =p_{2}^2 + m_{2}^2$
solved for p₂ I get
$p_{2} = \sqrt(p_{1}^2 + m_{1}^2 - m_{1}^2)$

Using $w_{2} = \frac {p_{2}} {m_{2}}$ I calculate the new proper speed.

 

[Flisp]

You could also search for "hyperbolic motion relativity", to get an analysis of motion with constant proper acceleratio. E.g.

https://en.wikipedia.org/wiki/Hyperbolic_motion_(relativity)

You could also search for "Rindler Coordinates".

I don't want constant acceleration, I want constant thrust.

 

[jbriggs444]

With 100% efficiency I meant that the specific impulse Isp = 1

Specific impulse of 1 means that the fuel can support its own mass against the acceleration of Earth's gravity for 1 second. Surely that is not what you mean.

 

[PeroK]

I don't want constant acceleration, I want constant thrust.

Have you tried using a photon rocket, as an efficient form of propellant?

 

[Flisp]

Have you tried using a photon rocket, as an efficient form of propellant?

It really does not matter what engine real or imagined I use. All I want to know at this point is how to calculate the speed curve using constant thrust.

 

[Flisp]

Specific impulse of 1 means that the fuel can support its own mass against the acceleration of Earth's gravity for 1 second. Surely that is not what you mean.

I meant specific impuls = 1 if using speed of light c = 1 light years / year.

 

[PeroK]

It really does not matter what engine real or imagined I use. All I want to know at this point is how to calculate the speed curve using constant thrust.

Constant in which reference frame?

 

[Flisp]

Constant in which reference frame?

The rockets frame. For an astronaut on the rocket the engine works the same regardless the current speed.

 

[DrStupid]

Some fuel is used and all its matter transformed into momentum while the ships total energy remains constant

No, it doesn't. The rocket loses energy with the exhaust (which consists of radiation in case of a specific momentum of c).

 

[PeroK]

The rockets frame. For an astronaut on the rocket the engine works the same regardless the current speed.

Ejecting photons at a constant rate should give you that. Although, as @Dale mentioned above, perhaps the problem will be relating the ship's proper time to the coordinate time in the initial rest frame.

The notes I have on this just give the speed when the ship reduces to a certain mass. Do you already know how to calculate this?

 

[jbriggs444]

I meant specific impuls = 1 if using speed of light c = 1 light years / year.

The speed of light is 1 light year per year. But what does that have to do with the question at hand?

Are you trying to say that the exhaust velocity is c? If so, that goes counter to the original stipulation that the relativistic mass of the craft never decreases.

 

[Flisp]

No, it doesn't. The rocket loses energy with the exhaust (which consists of radiation in case of a specific momentum of c).

Yes, I know, but that would not change my basicly faulty speed curve that has a typical square root shape when reasoning tells me it should have a tsiolkovsky-rocket-equation shape (flat at start, geting steeper all the time) since the engine pumps the same amount of energy per dt into the rocket, while the rocket gets lighter and lighter. Taking losses into account only changes the final max speed, not the basic shape of the curve.

 

[Orodruin]

I suggest you write down the energy-momentum relations in the instantaneous rest frame before and after ejecting a small amount of matter with a fixed speed u relative to the ship. This should give you the change in velocity in the instantaneous rest frame. You can relate this to the change in velocity via relativistic velocity addition and it should give you a differential equation for the velocity as a function of proper time.

 

[sweet springs]

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.

If the rocket has a chemical engine, its mass is decreasing due to exhaust gas of burned fuel.

 

[Flisp]

The speed of light is 1 light year per year. But what does that have to do with the question at hand?

Are you trying to say that the exhaust velocity is c? If so, that goes counter to the original stipulation that the relativistic mass of the craft never decreases.

Did I say mass was constan? Well, it is not. Fuel is burnt. And while all the fuels matter is transformed into pure energy, momentum goes upp, mass goes down. Energy is constant.
Yes exhaust velocity is c.

 

[Orodruin]

If so, that goes counter to the original stipulation that the relativistic mass of the craft never decreases.

I think this was stipulated by you from the OP’s ”100% efficient”. I suspect the OP is not aware of this implication but is really imagining a rocket with an exhaust.

 

[jbriggs444]

Did I say mass was constan?

Yes, you said that [relativistic] mass is constant:

so that the energy of the rocket is constant

Relativistic mass is another name for total energy.

 

[jbriggs444]

I think this was stipulated by you from the OP’s ”100% efficient”. I suspect the OP is not aware of this implication but is really imagining a rocket with an exhaust.

Not so.

 

[Orodruin]

Not so.

I guess I missed the fine print, but I think this is rather a sign of the OP’s misunderstanding. With an exhaust you cannot have constant rocket energy since energy is lost to the exhaust.

 

[Orodruin]

And while all the fuels matter is transformed into pure energy

There is no such thing as ”pure energy”.

Edit: Let me again suggest that you try to write down what I suggested in my first post.

 

[Flisp]

Yes, you said that [relativistic] mass is constant:

Relativistic mass is another name for total energy.

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

 

[Flisp]

I suggest you write down the energy-momentum relations in the instantaneous rest frame before and after ejecting a small amount of matter with a fixed speed u relative to the ship. This should give you the change in velocity in the instantaneous rest frame. You can relate this to the change in velocity via relativistic velocity addition and it should give you a differential equation for the velocity as a function of proper time.

I don't know how to do that. I did take the energy-momentum relations as you can see from the equations I posted earlier, but I did not get changes in velocity but, proper velocity. What is wrong with the equations I used??

 

[Flisp]

I guess I missed the fine print, but I think this is rather a sign of the OP’s misunderstanding. With an exhaust you cannot have constant rocket energy since energy is lost to the exhaust.

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

 

[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?