Need some help with space travel and general relativity
Cyde Weys
I got as far as up to special relativity in college physics but I never
got to general relativity. So I was looking for some help on how to
solve this (hopefully) simple problem.
Let's say I have a spacecraft accelerating at A. It is trying to cover
a distance of D. At the halfway point to D the spacecraft is going to
turn around and continue accelerating at A such that it comes to a
standstill upon reaching the end of the distance D such that it can
land on a planet, dock with a space station, whatever. My question is:
given A and D, let's assume A is 9.81m/s^2 and D is 4 lt-yrs for the
purpose of this example, how much time will have elapsed in the
reference frame of the spacecraft? The amount of time elapsed from the
inertial reference frame of either of the endpoints of D should be
calculable using non-relativistic physics, correct? And it should of
course be above 4 years.
What happens if we change the distance D to, say, 80 lt-yrs? Will the
ratio of spacecraft time to outside time change? I suspect it would.
Is there a simple enough equation that will take A and D as variables
and spit out the elapsed time from the accelerating reference frame of
the spaceship? Thanks a lot for the help!
(In case you're wondering what this information is for, I'm thinking of
writing a science-fiction short story involving physically possible
space travel, not magical "warp" technology. I want the timeframes
involved at least to be physically realistic, even if the story ends up
being crap.)
Androcles
"Cyde Weys" <cy...@umd.edu> wrote in message
news:1125090306.1...@g43g2000cwa.googlegroups.com...
|
| I got as far as up to special relativity in college physics but I
never
| got to general relativity.
Unfortunately you never got past the origins, and only ever learnt
tau = (t-vx/c^2)/sqrt(1-v^2/c^2)
xi = (x - vt)/sqrt(1-v^2/c^2)
If that is correct, I can help you write an even better story.
Androcles.
Uncle Al
For constant acceleration then constant deceleration, pick you desired
turning point fraction of lightspeed,
For b = v/c
m/M = (1-b)/(1+b)
Where "m" is total transported mass and "M" is "m" plus matter +
antimatter fuel load for 100% conversion into energy.
For 90% of c, payload is 5.3% of initial mass.
For 99% of c, payload is 0.5% of initial mass.
If you want to identically return, your initial load is
m/M = [(1-b)/(1+b)]^2
For 90% of c, payload is 0.28% of initial mass.
For 99% of c, payload is 0.0025% of initial mass.
For a full trip at 90% of lightspeed, X time at home would be 44% of X
on board. The integrated time dilation for accelerate up and down
will be very modest. You don't need General Relativity. If you don't
have the fuel then the time elapsed hardly makes a difference ven with
SR. The smallest research nuclear submarine, seating 12, dispalces
400 tonnes. The Apollo 12 Command Service Module plus Lunar Module
summed to 44 tonnes. GOT ANTIMATTER? Don't be standing behind it
when the pilot throws it into gear.
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/qz.pdf
Bill Hobba
You do not need general relativity for this. See the FAQ
http://math.ucr.edu/home/baez/physics/
under Can Special Relativity Handle Accelerations
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
Thanks
Bill
Eli Botkin
T = (2*c/A)*arccosh[(A*D/(2*c^2) + 1]
alanm...@yahoo.com
Relativistic Rocket
The Relativistic Rocket
The theory of relativity sets a severe limit to our ability to explore
the galaxy in space ships. As an object approaches the speed of light,
more and more energy is needed to accelerate it further. To reach the
speed of light an infinite amount of energy would be required. It
seems that the speed of light is an absolute barrier which cannot be
reached or surpassed by massive objects (see relativity FAQ article on
faster than light travel). Given that the galaxy is about 100,000
light years across there seems little hope for us to get very far in
galactic terms unless we can overcome our own mortality.
Science fiction writers can make use of worm holes or warp drives to
overcome this restriction, but it is not clear that such things can
ever be made to work in reality. Another way to get around the problem
may be to use the relativistic effects of time dilation and length
contraction to cover large distances within a reasonable time span for
those aboard a space ship. If a rocket accelerates at 1g (9.81 m/s2)
the crew will experience the equivalent of a gravitational field with
the same strength as that on Earth. If this could be maintained for
long enough they would eventually receive the benefits of the
relativistic effects which improve the effective rate of travel.
What then, are the appropriate equations for the relativistic rocket?
First of all we need to be clear what we mean by continuous
acceleration at 1g. The acceleration of the rocket must be measured at
any given instant in a non-accelerating frame of reference travelling
at the same instantaneous speed as the rocket (see relativity FAQ on
accelerating clocks). This acceleration will be denoted by a. The
proper time as measured by the crew of the rocket (i.e. how much they
age) will be denoted by T, and the time as measured in the
non-accelerating frame of reference in which they started (e.g. Earth)
will be denoted by t. We assume that the stars are essentially at rest
in this frame. The distance covered as measured in this frame of
reference will be denoted by d and the final speed v. The time
dilation or length contraction factor at any instant is the gamma
factor ã.
The relativistic equations for a rocket with constant positive
acceleration a > 0 are
c a 2
t = - sinh - T = sqrt[ (d/c) + 2d/a ]
a c
{ sinh x = (ex - e-x)/2 }
2 2
c c 2
d = - [ cosh(aT/c) - 1 ] = - ( sqrt[ 1 + (at/c) ] - 1 )
a a
{ cosh x = (ex + e-x)/2 }
a 2
v = c tanh - T = at / sqrt[ 1 + (at/c) ]
c
sinh x
{ tanh x = ------ }
cosh x
c a c 2
T = - arcsinh - t = - arccosh [ ad/c + 1 ]
a c a
a 2 2
ã = cosh - T = sqrt[ 1 + (at/c) ] = ad/c + 1
c
These equations are valid in any consistent system of units such as
seconds for time, metres for distance, metres per second for speeds and
metres per second squared for accelerations. In these units c = 3 x
108 m/s (approx). To do some example calculations it is easier to use
units of years for time and light years for distance. Then c = 1
lyr/yr and g = 1.03 lyr/yr2. Here are some typical answers for a = 1g.
T t d v ã
1 year 1.19 yrs 0.56 lyrs 0.77c 1.58
2 3.75 2.90 0.97 3.99
5 83.7 82.7 0.99993 86.2
8 1,840 1,839 0.9999998 1,895
12 113,243 113,242 0.99999999996 116,641
So in theory you can travel across the galaxy in just 12 years of your
own time. If you want to arrive at your destination and stop then you
will have to turn your rocket around half way and decelerate at 1g. In
that case it will take nearly twice as long in terms of proper time T
for the longer journeys; the Earth time t will be only a little longer,
since in both cases the rocket is spending most of its time at a speed
near that of light. (We can still use the above equations to work this
out, since although the acceleration is now negative, we can "run the
film backwards" to reason that they still must apply.)
Here are some of the times you will age when journeying to a few well
known space marks, arriving at low speed:
4.3 ly nearest star 3.6 years
27 ly Vega 6.6 years
30,000 ly Center of our galaxy 20 years
2,000,000 ly Andromeda galaxy 28 years
n ly anywhere, but see next paragraph 1.94 arccosh (n/1.94
+ 1) years
For distances bigger than about a thousand million light years, the
formulas given here are inadequate because the universe is expanding.
General Relativity would have to be used to work out those cases.
If you wish to pass by a distant star and return to Earth, but you
don't need to stop there, then a looping route is better than a
straight-out-and-back route. A good course is to head out at constant
acceleration in a direction at about 45 degrees to your destination.
At the appropriate point you start a long arc such that the centrifugal
acceleration you experience is also equivalent to earth gravity. After
3/4 of a circle you decelerate in a straight line until you arrive
home.
How much fuel is needed?
Sadly there are a few technical difficulties you will have to overcome
before you can head off into space. One is to create your propulsion
system and generate the fuel. The most efficient theoretical way to
propel the rocket is to use a "photon drive". It would convert mass to
photons or other massless particles which shoot out the back. Perhaps
this may even be technically feasible if we ever produce an
antimatter-driven "graser" (gamma ray laser).
Remember that energy is equivalent to mass, so provided mass can be
converted to 100% radiation by means of matter-antimatter annihilation,
we just want to find the mass M of the fuel required to accelerate the
payload m. The answer is most easily worked out by conservation of
energy and momentum.
First: conservation of energy
The total energy before blast-off is (in the Earth frame)
Einitial = (M+m)c2
At the end of the trip the fuel has all been converted to light with
energy EL, so the total energy is now
Efinal = ãmc2 + EL
By conservation of energy these must be equal, so here is our first
conservation equation:
(M+m)c2 = ãmc2 + EL ........ (1)
Second: conservation of momentum
The total momentum before blast-off is zero in the Earth frame.
pinitial = 0
At the trip's end the fuel has all been converted to light with
momentum of magnitude EL/c, but in the opposite direction to the
rocket. So the final momentum is
pfinal = ã mv - EL/c
By conservation of momentum these must be equal, so our second
conservation equation is:
0 = ã mv - EL/c ........ (2)
Eliminating EL from equations (1) and (2) gives
(M+m)c2 - ãmc2 = ãmvc
so that the fuel:payload ratio is
M/m = ã(1 + v/c) - 1
This equation is true irrespective of how the ship accelerates to
velocity v, but if it accelerates at constant rate a then
M/m = ã(1 + v/c) - 1
= cosh(aT/c)[ 1 + tanh(aT/c) ] - 1
= exp(aT/c) - 1
How much fuel is this? The next chart shows the amount of fuel needed
(M) for every kilogramme of payload (m=1 kg).
d Not stopping, sailing past: M
4.3 ly Nearest star 10 kg
27 ly Vega 57 kg
30,000 ly Center of our galaxy 62 tonnes
2,000,000 ly Andromeda galaxy 4,100 tonnes
This is a lot of fuel--and remember, we are using a motor that is 100%
efficient!
What if we prefer to stop at the destination? We accelerate to the
half way point at 1g and then immediately switch the direction of our
rocket so that we now decelerate at 1g for the rest of second half of
the trip. The calculations here are just a little more involved since
the trip is now in two distinct halves (and the equations at the top
assume a positive acceleration only). Even so, the answer turns out to
have exactly the same form: M/m = exp(aT/c) - 1, except that the proper
time T is now almost twice as large as for the non-stop case, since the
slowing-down rocket is losing the ageing benefits of relativistic
speed. This dramatically increases the amount of fuel needed:
d Stopping at: M
4.3 ly Nearest star 38 kg
27 ly Vega 886 kg
30,000 ly Center of our galaxy 955,000 tonnes
2,000,000 ly Andromeda galaxy 4.2 thousand million tonnes
Compare these numbers to the previous case: they are hugely different!
Why should that be? Let's take the case of Laurel and Hardy, two
astronauts travelling to Vega. Laurel speeds past without stopping,
and so only needs 57 kg of fuel for every 1 kg of payload. Hardy
wishes to stop at Vega, and so needs 886 kg of fuel for every 1 kg of
payload. Laurel takes almost 28 Earth years for the trip, while Hardy
takes 29 Earth years. (They both take roughly the same amount of Earth
time because they are both travelling close to speed c for most of the
journey.) They travel neck-and-neck until they've both gone half way
to Vega, at which point Hardy begins to decelerate.
It's useful to think of the problem in terms of relativistic mass,
since this is what each rocket motor "feels" as it strives to maintain
a 1g acceleration or deceleration. The relativistic mass of each
traveller's rocket is continually decreasing throughout their trip
(since it's being converted to exhaust energy). It turns out that at
the half way point, Laurel's total relativistic mass (for fuel plus
payload) is about 28m, and from here until the trip's end, this
relativistic mass only decreases by a tiny amount, so that Laurel's
rocket needs to do very little work. So at the halfway point his
fuel:payload ratio turns out to be about 1.
For Hardy, things are different. He needs to decrease his relativistic
mass to m at the end where he is to stop. If his rocket's total
relativistic mass at the halfway point were the same as Laurel's (28m),
with a fuel:payload ratio of 1, Hardy would need to decrease the
relativistic mass all the way down to m at the end, which would require
more fuel than Laurel had needed. But Hardy wouldn't have this much
fuel on board--unless he ensures that he takes it with him initially.
This extra fuel that he must carry from the start becomes more payload
(a lot more), which needs yet more fuel again to carry that. So
suddenly his fuel requirement has increased enormously. It turns out
that at the half way point, all this extra fuel gives Hardy's rocket a
total relativistic mass of about 442m, and his fuel:payload ratio turns
out to be about 29.
Another way of looking at this odd situation is that both travellers
know that they must take fuel on board initially to push them at 1g for
the total trip time. They don't care about what's happening outside.
In that case, Laurel travels for 28 Earth years but ages just 3.9
years, while Hardy travels for 29 Earth years but ages 6.6 years. So
Hardy has had to sit at his controls and burn his rocket for almost
twice as long as Laurel, and that has required more fuel, with even
more fuel required because of the fuel-becomes-payload situation that
we mentioned above.
This fuel-becomes-payload problem is well known in the space programme:
part of the reason the Saturn V moon rocket was so big was because it
needed yet more fuel just to carry the fuel it was already carrying.
Other fuel options
Well, this is probably all just too much fuel to contemplate. There
are a limited number of solutions that don't violate energy-momentum
conservation or require hypothetical entities such as tachyons or worm
holes.
It may be possible to scoop up hydrogen as the rocket goes through
space, using fusion to drive the rocket. This would have big benefits
because the fuel would not have to be carried along from the start.
Another possibility would be to push the rocket away using an
Earth-bound grazer directed onto the back of the rocket. There are a
few extra technical difficulties but expect NASA to start looking at
the possibilities soon :-).
You might also consider using a large rotating black hole as a
gravitational catapult but it would have to be very big to avoid the
rocket being torn apart by tidal forces or spun at high angular
velocity. If there is a black hole at the centre of the Milky Way, as
some astronomers think, then perhaps if you can get that far, you can
use this effect to shoot you off to the next galaxy.
One major problem you would have to solve is the need for shielding.
As you approach the speed of light you will be heading into an
increasingly energetic and intense bombardment of cosmic rays and other
particles. After only a few years of 1g acceleration even the cosmic
background radiation is Doppler shifted into a lethal heat bath hot
enough to melt all known materials.
For the derivation of the rocket equations see "Gravitation" by Misner,
Thorne and Wheeler, section 6.2.
the softrat
The best exposition I know of is in _Spacetime Physics_ by Taylor and
Wheeler, Amazon Books
http://www.amazon.com/exec/obidos/tg/detail/-/0716723271/qid%3D1125117792/002-3425454-2259223
HTH
the softrat
Sometimes I get so tired of the taste of my own toes.
mailto:sof...@pobox.com
--
If you want the rainbow, you gotta put up with the rain. --
Steven Wright
dlzc1 D:cox T:net@nospam.com N:dlzc D:aol T:com (dlzc)
"Cyde Weys" <cy...@umd.edu> wrote in message
news:1125090306.1...@g43g2000cwa.googlegroups.com...
...
> (In case you're wondering what this information is for, I'm
> thinking of
> writing a science-fiction short story involving physically
> possible
> space travel, not magical "warp" technology. I want the
> timeframes
> involved at least to be physically realistic, even if the story
> ends up
> being crap.)
"The Forever War", already been done. NOT crap. The author
still had to use magical shields and "infinite" power sources.
David A. Smith
Autymn D. C.
sad lack of thoughtful incentive in all engineering and scientific
parties. How would a Stargate-level civilisation plan and power their
deathships? They would first move everyone toward the galactic core
and set and build up a new solar sustem amongst the accretion disc,
sacrificing some globular clusters to custom-make a Kerr black hole
gravity shield-pump around their new worlds just outside the event
horizon. Then they build another fleet in little outside time, make
all the antimatter they want from the pump, and go around the galaxy
pillaging the newly-made heavy fissiles everywhere from stellar
corpses. They /do not/ need yet more fuel to decelerate because they
would've devised a near-equipotential from their electrocolor fuel so
that their ships and their fuels travel in near-parabolic orbits about
each other: After a finite time, their fuel loops around with its
relativistic mass and brakes the ships to their initial speed: Their
momenta are regenerated, with low husteresis loss. For the last
affording, they may need to highjack a neutron star to polarise as a
momentum reservoir; by then, they have already experimented with
quantum gravity, seen its neutron excitation levels, and know what
dials to turn on their gravity pump to ration the star's heft-shot like
honey from a beehive. The neutronium ball may need to travel faster
than the ship though, for it to catch up as it rounds the event
horizon. All of this doesn't take into account the civilisation's
cheating the laws through another, but alike, route though.
-Aut
better life through new maths