I Relativistic speed of a rocket with constant thrust

[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...

 

[PAllen]

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.

 

[jartsa]

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.

 

[Flisp]

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.

 

[Dr Whom]

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.

 

[PeterDonis]

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.

 

[l0st]

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.

You really got me worried about perspectives of deep space interstellar travel with this one. I guess that bothered my conscience, so this morning I got up with idea, that potentially we could recapture those photons at the starting point, if the whole trip is planned well.

 

[PeterDonis]

All, the economics of space colonization and interstellar travel are off topic for this thread. Please limit discussion to the physics question asked in the OP.

 

[PeterDonis]

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.

While this in itself is useful background for the physics discussion in this thread, the article topic itself is out of scope for this thread, and this area of PF.

 

[PeterDonis]

Moderator's note: a number of off topic posts (having to do with the Fermi Paradox, economics of space exploration, etc.) have been deleted.

 

[pervect]

I don't know that there is still interest in the mathematics of a case of a rocket with a variable proper acceleration, but there's an approach I find convenient.

One needs shared definitions of three things: proper time, denoted by $\tau$, proper acceleration, denoted by a, and rapidity, denoted by w.

Wiki has articles on proper time, proper acceleration, and rapidity. One needs to understand all three terms to follow the argument.

The key features of rapidities is that, unlike velocities, in special relativity rapiditiy add. Thus one can write

$$w(\tau) = \int dw = \int \frac{dw}{d\tau} d\tau$$

This works with rapidites (and does not work with velocities), because in 1-space, 1-time, special relativity, rapidities add linearly, while velocities add according to the "relativistic velocity addition law".

The remaining step is to note that $\frac{dw}{d\tau} = a/c$, the rate of change of rapidity with respect to proper time is proportional to the proper acceleration. This can be conveniently done by considering the instantaneous frame of the rocket, and noting that in that frame dv=dw/c and $dt=d\tau$. But dv/dt in the instantaneous rest frame of the rocket is just the proper acceleration of the rocket.

So then we can solve the problem of the non-uniformly accelerating rocket by writing

$$w(\tau) = \int \frac{a(\tau)}{c} d\tau$$

This follows from the chain rule, and the additive nature of rapidity.

As a check, we can consider the case where a is constant, where we get $w = a\tau$, We use the basic conversions of rapidity to velocity and vica-versa (see the wiki article for a discussion or look for a clearer one). As the equations imply, there is a 1:1 correspondence between velocity and rapidity, if we know the velocity, we can compute the rapidity, and vica-versa. These basic conversion equations are:

$$w = \tanh \frac{v}{c} \quad v = c \, \tanh^{-1} w$$

substituting $a(\tau) = a$ into the integral, we find

$$w = \tanh v/c = a \tau / c$$

which matches such references as http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html for the constant proper acceleration case.

 

[Flisp]

I don't know that there is still interest in the mathematics of a case of a rocket with a variable proper acceleration, but there's an approach I find convenient.

...

which matches such references as http://math.ucr.edu/home/baez/physics/Relativity/SR/Rocket/rocket.html for the constant proper acceleration case.

Thank you! It is always helpful if one can double-check the results from differnt approaches.

 

[Alan McIntire]

Convert units to where c =1 (light year/year)
Acceleration in units of ( 1 light year per year per year which works out to about 0.97 Earth g)
distance in light year units.

Then with acceleration g light years per year per year
distance in light years
Speed of rocket as a fraction of c=1

hyperbolic cosine = cosh =( e^x + e^-x)/2
hyperbolic sine = sinh =(e^x -e^-x)/2
hyperbolic tangent= tanh = sinh x/cosh x

If you have a scientific calculator on your computer, there'll be a function for HYP SIN, HYP COS , HYP TAN
Suppose g = 1(.97 Earth g) for rocket time 2 years
then EARTH time will be sinh( gt) = sinh(1*2) =3.626... Earth years will have gone by in 1 'rocket' year.
Velocity of the rocket will be sinh(gt)/cosh(gt) = sinh(1*2)/cosh(1*2)= tanh2 =.964 times the speed of light after accelerating at 1 light year per year per year for 2 years.
DISTANCE traveled in light years will be [ cosh(gt) -1] = cosh2 - 1 = 2.762 light years.

For a second example , suppose you accelerate at a rate of 1.5 light years per year per year for 1.5 rocket years.
Earth time passed will be sinh(1.5*1.5)= sinh 2.25 = 4.69 Earth years
speed of rocket relative to Earth will be tanh 2.25 =.978 c
distance traveled from Earth will be [cosh(2.25) -1] =3.796 light years,

 

[thorpie]

So if we
1 - made our spaceship a large parabolic dish, say 1 km across,
2 - with the dish pointed away from our current star towards our destination star,
3 - with a small atmosphere in the dish,
4 - with a solar receiver on the rear that receives concentrated solar energy sufficient to accelerate at 1 g at the start
5 - somehow made a solar concentrator to send said solar energy to space ship
Then after 1 year the spaceship approaches light speed and relativistic effects start reducing the acceleration - i.e. the apparent energy of the solar light being collected is reduced because the apparent wavelength of the solar light is greater.
If the ship rotates and then the same process is applied as the ship approaches its destination star the ship is slowed and becomes stationary at much the same solar orbit as it started. During acceleration and deceleration could get an atmosphere of say 0.1 atmospheres and could go out and enjoy the open air. Wouldn't life be grand!

 

[Ibix]

@Alan McIntire - you've made the same mistake others, including myself, made on this thread. The OP asks about constant thrust, not constant acceleration.

@thorpie - you are basically describing a solar sail. The problem is that they yield very low accelerations and their performance drops off very rapidly as you move away from the sun. Even if you try to use lasers in solar orbits to drive your sails, there are significant problems with delivering much power to a target as small as a solar sail - both beam-spread and simple accuracy are problematic. Half way to the nearest star is two light years, approximately 10¹³km. Hitting a 1km target at that distance (with a 4 year lag on your targetting data by the time the beam gets there) would be very very difficult, even one that wants to be hit.

 

[Flisp]

I finally finished my article, that started started my questions into this matter. http://florianb.se/the-possibilities-and-impossibilities-of-interstellar-flight/, thank's again Physics Forum for your support.