Relativistic speed of a rocket with constant thrust

In summary: There can be no exhaust stream. Any ejected mass or energy would reduce the remaining relativistic mass.
  • #71
PAllen said:
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)...
 
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  • #72
Flisp said:
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?
 
  • #73
Flisp said:
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.
 
  • #74
Flisp said:
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

[itex]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[/itex]

of the exhaust and therefore in the specific impulse

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

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.

Flisp said:
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.

Flisp said:
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.
 
  • #75
Flisp said:
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.
 
  • #76
Orodruin said:
Flisp said:
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".
 
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  • #77
DrStupid said:
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?
 
  • #78
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.
 
  • #79
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?
 
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  • #80
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:-)
 
  • #81
Flisp said:
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.
 
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  • #82
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:-)
 
  • #83
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:
m4r35n357 said:
There is an article here that might help.
 
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  • #84
Flisp said:
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.
 
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  • #85
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.
Code:
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
 
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  • #86
Dale said:
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.
 
  • #87
jbriggs444 said:
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.
 
  • #88
DrGreg said:
"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.
 
  • #89
Flisp said:
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.
 
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  • #90
Dale said:
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.

Flisp said:
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.
 
  • #91
jartsa said:
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.
Code:
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?
 
  • #92
PeroK said:
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.
 
  • #93
Flisp said:
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

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

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

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

That needs to be integrated numerically.

[Moderator's note: off topic content deleted.]
 
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  • #94
DrStupid said:
$$w\left( t \right) = c \cdot \tanh \left( {\frac{u}{c}\ln \frac{{m_0 }}{{m\left( t \right)}}} \right)$$
I could not use that, because I could not calculate deceleration with it. Your equation was perfect for me.
 
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  • #95
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.
 
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  • #96
PAllen said:
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?
 
  • #97
DrStupid said:
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?
 
  • #98
Flisp said:
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.
 
  • #99
PAllen said:
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.
 
  • #100
In post #98 you quoted my question but never answered...
 
  • #101
Flisp said:
I understood the equation, but not the argument. I guess the word "year" was missing after traveler. Thereof my confusion.
The idea is that you want to get e.g. 1000 light years with traveler aging no more than one year for the trip. The formula I gave is the theoretical best possible - starting mass 4 million times payload mass. Every realizable alternative will have worse mass ratio than this. My formula gives the implausible best case for any distance, and any desired traveler aging for the trip.
 
  • #102
Flisp said:
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?
The simulation is inaccurate because of the large time step of one second. Dividing both the mass decrease and the force by 100 should correct that. (Time step becomes 1/100 s)

And dividing the force by two while keeping the mass decrease the same should mean the same as halving the efficiency of the rocket.

At each step the rocket time increases by the time step. And the launchpad time increases this much: gamma * time step.

Gamma can be calculated like this: gamma = math.cosh(rapidity/c)

And deceleration should work as you said.
 
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  • #103
jartsa said:
The simulation is inaccurate because of the large time step of one second. Dividing both the mass decrease and the force by 100 should correct that. (Time step becomes 1/100 s)

And dividing the force by two while keeping the mass decrease the same should mean the same as halving the efficiency of the rocket.

At each step the rocket time increases by the time step. And the launchpad time increases this much: gamma * time step.

Gamma can be calculated like this: gamma = math.cosh(rapidity/c)

And deceleration should work as you said.
Thank you.
 
  • #104
As far as the rockets occupants are concerned it carries on accelerating at a constant speed. To an inertial observer the accelerating slows asymptotically as the speed approaches that of light.

To the occupants distance contraction reduces the distance they have to travel, enabling them to travel at apparently linearly increasing speed towards their destination. To the inertial observer time is slowing for those on board.

For example, accelerating at 1g for a year and then decelerating at 1g for another year will take them almost anywhere in the universe. Time dilation means that while they have experienced only a couple of years the rest of the universe has aged approximately one year plus the distance covered divided by the speed of light.

There remains the slight problem of the fuel. Solve that and off you go to anywhere you like but don’t expect to get back anytime soon!

It's covered thoroughly in an old book called 'Reverse Time Travel', Cassell press and on the web site:

http://drwhom.eu3.biz/pages/RS - 1 Rocket Science.htm.
 
  • #105
Dr Whom said:
As far as the rockets occupants are concerned it carries on accelerating at a constant speed.

First, I think you mean "constant acceleration", not "constant speed", since the latter makes no sense.

Second, as you will see if you go back and read the OP and responses carefully, the scenario is not constant proper acceleration (i.e., acceleration as felt by the rocket occupants), it is constant thrust. They're not the same thing.
 
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