I Relativistic speed of a rocket with constant thrust

[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.

 

[Flisp]

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.

Ah, you might be right. I saw this p-like greek letter and thought it was a p like in momentum. You mean I can use the equation for v(t) = (1-p)..., right, where p = [m(0) - w t]/m(0)...

 

[Flisp]

Ah, you might be right. I saw this p-like greek letter and thought it was a p like in momentum. You mean I can use the equation for v(t) = (1-p)..., right, where p = [m(0) - w t]/m(0)...

At closer inspection... With that equation
$$v(t) = \frac{1-r^{2u}} {1+r^{2u}}$$ where $$r = \frac{m(0) - w t} {m(0)}$$
I can calcualte acceleration up to a certain top speed during the journey, using a certain part of the fuel, but not deceleration (back to zero upon arrival at destiantion), using the rest of the fuel.
How do I do that?

 

[PAllen]

At closer inspection... With that equation
$$v(t) = \frac{1-r^{2u}} {1+r^{2u}}$$ where $$r = \frac{m(0) - w t} {m(0)}$$
I can calcualte acceleration up to a certain top speed during the journey, using a certain part of the fuel, but not deceleration (back to zero upon arrival at destiantion), using the rest of the fuel.
How do I do that?

Well the simplest approach is symmetry. That is, burn e.g. 90% of your mass to half way point, burn 90% of remaining mass decelerating. This means your thrust on the return trip will be a constant 10% of the constant thrust getting to half way point. However, your deceleration profile will be the exact inverse because you are starting with 1/10 the mass. If you want the same thrust for the deceleration phase, you would want to start deceleration much closer to your destination, and the deceleration you experience will much larger than the acceleration. I don't have time now to play with this, but it should be possible to figure out based on the formulas in the link. However, I think the symmetric approach is more sensible.

 

[DrStupid]

Do you know the max specific impulse of that?

With 1 % mass defect (that should be the maximum for fusion reactions) and 100 % efficiency it is the 14 % c I posted above. My calculation works similar to jbriggs444’s description in #63:

Let’s assume a small amount dm of fuel is “burned” with the mass defect d and the released energy is used to accelerate the exhaust with the efficiency $\eta$. If dm is sufficiently small compared to the total mass of the rocket, this results in the total energy

dE^2 = \left[ {\left( {1 - d} \right) \cdot dm \cdot c^2 + \eta \cdot d \cdot dm \cdot c^2 } \right]^2 = \left( {1 - d} \right)^2 \cdot dm^2 \cdot c^4 + dp^2 \cdot c^2

of the exhaust and therefore in the specific impulse

\frac{{dp}}{{dm}} = c \cdot \sqrt {\left[ {2 + \left( {\eta - 2} \right) \cdot d} \right] \cdot \eta \cdot d}

With a mass defect of 0.64 % for the fusion of deuterium to helium and an efficiency of 100 % this results in a specific impulse of 11 % c.

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

The differential equation holds for any thrust profile. It even works with variable specific impulse.

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?)

All you need to do is the transformation of time from the rocket frame (or any other frame where the profiles of mass-flow and specific impulse are given) into the rest frame of the external observer.

 

[Orodruin]

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.

I have never seen anything like this defined before. It does not seem like a very useful property. What observer are you talking about? The word "proper" in relativity typically refers to an invariant quantity that corresponds to the coordinate expression of something measured along the worldline of some observer in that observer's rest frame. "Proper" velocity would then always be zero, which is rather tautological, just saying that the velocity of anything in its rest frame is zero.

 

[DrGreg]

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.

I have never seen anything like this defined before. It does not seem like a very useful property. What observer are you talking about? The word "proper" in relativity typically refers to an invariant quantity that corresponds to the coordinate expression of something measured along the worldline of some observer in that observer's rest frame. "Proper" velocity would then always be zero, which is rather tautological, just saying that the velocity of anything in its rest frame is zero.

"Proper velocity" is, unfortunately, an established name. I really don't like it, for the reasons you give (and the additional reason that it suggests a definition of "proper acceleration" that is wrong). I much prefer the alternative name "celerity".

 

[Flisp]

All you need to do is the transformation of time from the rocket frame (or any other frame where the profiles of mass-flow and specific impulse are given) into the rest frame of the external observer.

I have no idea, how to do that. As I understand the what-ever-you-call-the-speed-measured-by-the-astornaut can become higher than c, due to time dilation. That is now prohibited by the term $\frac{w^2} {c^2}$. Or am I wrong?

 

[Flisp]

I'm writing an article about the curious fact, that we do not find any traces of alien civilisations or have proof of them coming to us. One of the possible reasons I look at are travel costs ond travel times. Although aliens in sf-movies and book frequently visit Earth with huge ships, and crossing vast distances in mere seconds, given the low probability of warp-drives and anti-gravity, I don't think that is possible. What I need to calculate is an as realistic scenario as possible, given the possible technological progress of an alien civilisation, (like anti-matter drive) but not breaking any laws we know are unbreakable. Thus, I need the real speed curve of a rocket under constant thrust from an engine that "slows down" when approaching c, that has slow acceleration, when there is plenty of fuel, and rapid deceleration, near the destination. I need to calculate the experienced flight time and acceleration/deceleration curve, test different fuel-payload ratios and filght times as measured on ship and on the home planet. I had math and physics at high school some 30 years ago, but do not work with it now, and have forgotten most. Curious enough, all online flight time calculators and equations I can find work on constant acceleration and/or totaly improbable scenarios and are of no help to me. Since I lack any knowledge of how to differentiate or integrate, I'm extremely grateful for any equation ready for use. The constants I work with are the initial mass of payload and fuel, and thus their ratio, total flight time and the constant amount of fuel burnt per dt as measured on the ship, as well as the efficiency or specific momentum of the engine.

 

[jbriggs444]

So this hypothetical, advanced civilization is able to construct a ship with a fuel to payload ratio of many hundreds to one, but has never come up with the idea of a multi-stage rocket and are instead stuck with a single engine that cannot be turned off or throttled back?

 

[Flisp]

Multistaging reduces the weight of the ship by the weight of the tank. If the tanks are very ligt, compared to fuel and payload the gain in speed is very small. With heavy tanks you can gain a lot. And of course you can throttle back, then you have a longer flight times, especially at low speeds where time dilation is low. You can also turn off the engine at high speeds an cruise near c. All this I will investigate once I have some usefull equations to work with:-)

 

[Dale]

Curious enough, all online flight time calculators and equations I can find work on constant acceleration and/or totaly improbable scenarios and are of no help to me.

Based on your proposed use, I think that the constant proper acceleration is the best approach. If their technology is at least as advanced as ours then the limiting factor will not be the maximum thrust their machines can produce but the maximum acceleration their bodies can survive.

 

[Flisp]

Not really. Constant acceration at a high fuel-payload ratio would require gigantic and heavy engines, leaving no room for any actual astronaut. And the astronaut may just as well be an artificial one, tolerating as high g-forces as any other part of the ship. So if you need to build a large engine you can just as well use it all the way instead of throteling down short time after start or throw it away halfway through the journey. You can of course throw part of it away as part of a stage, together with tanks, but as long as you have a high fuel-payload ratio the problem is the long and slow start of the journey, not its final faster end. Constant thrust (with maybe some cushioning thrust reduced endphase) is the most realistic scenario. Although, I agree, not entirely perfect. But since I am anyways speculating about a number of things, I can live with that imperfection:-)

 

[Dale]

Ok, but I don’t think there is a nice closed form equation for the constant thrust case. I think that will require a separate numerical solution for each possible combination of parameters. You will need to program it yourself so that you can adjust things.

Edit: actually, that is given here:

There is an article here that might help.

 

[jbriggs444]

Not really. Constant acceration at a high fuel-payload ratio would require gigantic and heavy engines

That is something that multi-stage designs deal with. You throw away the huge engines once their huge fuel supply is exhausted.

 

[jartsa]

Here's a little python program, a numerical simulation. A 100 kg rocket annihilates 1 kg of fuel every second of rocket time. Thrust in Newtons happens to be the same number as c in meters per second at that fuel consumption rate.

import math
c=3.0*10**9
mass=100.0
rapidity=0.0
while 1:
    mass=mass-1
    force=c
    acceleration = force/mass
    rapidity=rapidity+acceleration
    speed = math.tanh(rapidity/c)
    print mass, speed

 

[Flisp]

Ok, but I don’t think there is a nice closed form equation for the constant thrust case. I think that will require a separate numerical solution for each possible combination of parameters. You will need to program it yourself so that you can adjust things.

Edit: actually, that is given here:

I don't expect equations that can do everything for me. I was perfectly happy with the one provided by DrStupid, because I could calculate in small steps, adding or substracting dv as needed, until I realized the slow down effect of the engine near c (as measured by an inertial observer). I need an equation like that one, as measured by the astronaut. From that I can calculate the g-force and travel time experienced by the astronaut. And I need a conversion from rocket time to observer time, which, I guess, involves the Lorentz factor based on the proper speed, as well as a conversion from rocket distance traveled to observer distance measured, so we know how far the rocket made it with some fuel in some time.

 

[Flisp]

That is something that multi-stage designs deal with. You throw away the huge engines once their huge fuel supply is exhausted.

Regardless the scenario, if you want to go fast you need a high fuel-payload ratio, menaning that engine and tanks only weigh a fraction of the fuel. The more often you "stage" the better the perfonmance, up to the point where you can envision a continuous staging where you loose some constant part of the engine/tanks with each constant part of fuel burnt. If you calculate that, than you can just as well look at the engine/tanks as a part of the fuel where the fuel/engine as a system simply gets a little less efficient per unit weight. If you have a fuel-payload ratio of 100:1 and the engine/tanks ar e 50% procent of the payload than your fuel including continuous staging is simply 0,5% less efficient. If you want to calculate an actual flight that is of course important in order to hit your destination, but for my purpse it's not relevant.

 

[PAllen]

"Proper velocity" is, unfortunately, an established name. I really don't like it, for the reasons you give (and the additional reason that it suggests a definition of "proper acceleration" that is wrong). I much prefer the alternative name "celerity".

Note also, this is just the spatial part of 4 velocity. Its proper time derivative is the spatial part of 4 acceleration. Of course, proper acceleration is the norm of the 4 acceleration and has the normal meaning of proper in SR.

 

[Nugatory]

One of the possible reasons I look at are travel costs ond travel times.

You may be overthinking the question. Peak efficiency is reached with instantaneous fuel burn and infinite acceleration. Those conditions are obviously impractical for a real voyage, but they're easy to calculate with and guaranteed to be less discouraging than any more realistic calculation. So if they're already sufficiently discouraging, you don't need to make the harder calculations.

 

[PeroK]

Based on your proposed use, I think that the constant proper acceleration is the best approach. If their technology is at least as advanced as ours then the limiting factor will not be the maximum thrust their machines can produce but the maximum acceleration their bodies can survive.

I'm not convinced that interstellar space flights are likely to include the beings themselves. I would expect computerisation and robotics to render any human or alien as just along for the ride.

I'm writing an article about the curious fact, that we do not find any traces of alien civilisations or have proof of them coming to us. One of the possible reasons I look at are travel costs ond travel times. Although aliens in sf-movies and book frequently visit Earth with huge ships, and crossing vast distances in mere seconds, given the low probability of warp-drives and anti-gravity, I don't think that is possible. What I need to calculate is an as realistic scenario as possible, given the possible technological progress of an alien civilisation, (like anti-matter drive) but not breaking any laws we know are unbreakable. Thus, I need the real speed curve of a rocket under constant thrust from an engine that "slows down" when approaching c, that has slow acceleration, when there is plenty of fuel, and rapid deceleration, near the destination. I need to calculate the experienced flight time and acceleration/deceleration curve, test different fuel-payload ratios and filght times as measured on ship and on the home planet. I had math and physics at high school some 30 years ago, but do not work with it now, and have forgotten most. Curious enough, all online flight time calculators and equations I can find work on constant acceleration and/or totaly improbable scenarios and are of no help to me. Since I lack any knowledge of how to differentiate or integrate, I'm extremely grateful for any equation ready for use. The constants I work with are the initial mass of payload and fuel, and thus their ratio, total flight time and the constant amount of fuel burnt per dt as measured on the ship, as well as the efficiency or specific momentum of the engine.

It seems to me that with sub-light speeds, the scope for interstellar travel is limited. It's not impossible - but even an "unmanned" ship that can get close to the speed of light is very limited in what it can do' given the scale of the galaxy. The problem, of course, is not so much the proper time on the ship, but the time to get back or report back to Earth.

That's why science fiction tends to ignore or assume a way to get round the light-speed barrier.

And, if you are trying to establish whether $0.5c$ or $0.9c$ is a realistic limit, it almost doesn't matter. You can assume $0.99c$ and you still can't get very far. Maybe one day a deep-space probe will turn up from somewhere, but its home planet may be hundreds or thousands of light years away. We may be able to exchange a few messages hundreds or thousands of years apart - and that would fascinating - but the science-fiction envisaged repeat visits from the aliens themelves is a bit of a fantasy, as far as I can see.

 

[Flisp]

Here's a little python program, a numerical simulation. A 100 kg rocket annihilates 1 kg of fuel every second of rocket time. Thrust in Newtons happens to be the same number as c in meters per second at that fuel consumption rate.

import math
c=3.0*10**9
mass=100.0
rapidity=0.0
while 1:
    mass=mass-1
    force=c
    acceleration = force/mass
    rapidity=rapidity+acceleration
    speed = math.tanh(rapidity/c)
    print mass, speed

Great, that looks managable.
I tried to put that into a spread sheet and get higher speeds than with DrStupids equation. I believe that is correct.
Can I simply t take rapidity - acceleration to decelerate?
It looks as if all mass is transformed into thrust. If I want to work with lower than 1 (100%) efficiency, is it then enough to say acceleration = force/mass * efficiency? I tried that too and get a final speed of 0.62 c with your equation vs 0.566 c with DrStupids at a 100:1 fuel-payload ratio Is that correct?
And how do I convert that proper time and speed into observed time and speed?

 

[Flisp]

I'm not convinced that interstellar space flights are likely to include the beings themselves. I would expect computerisation and robotics to render any human or alien as just along for the ride.
It seems to me that with sub-light speeds, the scope for interstellar travel is limited. It's not impossible - but even an "unmanned" ship that can get close to the speed of light is very limited in what it can do' given the scale of the galaxy. The problem, of course, is not so much the proper time on the ship, but the time to get back or report back to Earth.

That's why science fiction tends to ignore or assume a way to get round the light-speed barrier.

And, if you are trying to establish whether $0.5c$ or $0.9c$ is a realistic limit, it almost doesn't matter. You can assume $0.99c$ and you still can't get very far. Maybe one day a deep-space probe will turn up from somewhere, but its home planet may be hundreds or thousands of light years away. We may be able to exchange a few messages hundreds or thousands of years apart - and that would fascinating - but the science-fiction envisaged repeat visits from the aliens themelves is a bit of a fantasy, as far as I can see.

That is exactly what I concluded, when I first started to dig into it. Even more so when I realized that all calculations where based on (impossible) constant acceleration.

 

[DrStupid]

I have no idea, how to do that.

You already know how to calculate the speed in the frame of the distant observer. You don’t even need to integrate my differential equation. There is a ready to use solution in the wikipedia article linked by Filip Larsen in #46. With my symbols it is

w\left( t \right) = c \cdot \tanh \left( {\frac{u}{c}\ln \frac{{m_0 }}{{m\left( t \right)}}} \right)

In order to use it with the thrust profile in the rest frame of the rocket you need to replace $m\left( t \right)$ by $m\left( {\tau \left( t \right)} \right)$ as given in the rocket. The relationship between the time $t$ in the rest frame of the external observer and the proper time $\tau$ of the rocket is given by

d\tau = dt \cdot \sqrt {1 - \frac{{w^2 }}{{c^2 }}}

That needs to be integrated numerically.

[Moderator's note: off topic content deleted.]

 

[Flisp]

$$w\left( t \right) = c \cdot \tanh \left( {\frac{u}{c}\ln \frac{{m_0 }}{{m\left( t \right)}}} \right)$$

I could not use that, because I could not calculate deceleration with it. Your equation was perfect for me.

 

[PAllen]

Taking Mfb's suggestion, to show the inherent issue with effective interstellar travel, you only need show the implausibility of even the implausibly best approach.

Thus, given the desire to travel L light years in Y years per traveler, with instant acceleration and instant deceleration (and direct conversion of mass to unidirectional photons), the mass ratio needed (to very good approximation as long as Y² << L²) is:

starting mass / ending mass = 4L² / Y²

For 1000 light years in 1 year per traveler, this is already 4 million.

 

[Flisp]

Taking Mfb's suggestion, to show the inherent issue with effective interstellar travel, you only need show the implausibility of even the implausibly best approach.

Thus, given the desire to travel L light years in Y years per traveler, with instant acceleration and instant deceleration, the mass ratio needed (to very good approximation as long as Y² << L² is:

starting mass / ending mass = 4L² / Y²

For 1000 light years in 1 year per traveler, this is already 4 million.

I do not understand that argument. Why per traveler? Why in one year, why not in 10 or 100, or for that sake in 10 000?

 

[Flisp]

So is it correct if I simply multiply your equation with $\ \sqrt {1 - \frac{{w^2 }}{{c^2 }}}$ so I get a lesser amount of fuel being burnt, thus staying in the frame of the observer, but taking into account that less fuel is burnt due to time dilation?

 

[PAllen]

I do not understand that argument. Why per traveler? Why in one year, why not in 10 or 100, or for that sake in 10 000?

Every other acceleration profile will perform worse, since traveler time is 'wasted' going less than peak speed.

For direct conversion of mass to unidirectional photons, conservation of energy and momentum give you that to do from speed 0 to speed v you need m1/m0 = √((1-v)/(1+v)). To decelerate, you need the same ratio again, so for combined acceleration and deceleration you have m1/m0 = (1-v)/(1+v).

Then, to cover L light years in Y traveler years, you should be able to derive that you need v² = 1/(1+Y²/L²) with v = light years per year.

Putting these together, with a little algebra and approximations, gives the formula I gave.

 

[Flisp]

Putting these together, with a little algebra and approximations, gives the formula I gave.

I understood the equation, but not the argument. I guess the word "year" was missing after traveler. Thereof my confusion.

 

[Flisp]

In post #98 you quoted my question but never answered...