(A)------------------------------------------------------------------
Assuming a magical stardrive which allows you to accelerate
continuously away from Earth at 10 m/s^2, the time taken to
reach various distances is:
Earth dist : Earth time speed ship time ship distance
__________ __________ ____________ _________ _____________
.05 ly : 0.31 yr 0.312 c 0.31 yr 0.048 ly
.10 ly : 0.45 yr 0.426 c 0.43 yr 0.091 ly
0.25 ly : 0.73 yr 0.611 c 0.67 yr 0.198 ly
0.50 ly : 1.10 yr 0.755 c 0.94 yr 0.328 ly
1 ly : 1.70 yr 0.873 c 1.28 yr 0.487 ly
2 ly : 2.79 yr 0.9467 c 1.71 yr 0.644 ly
4 ly : 4.86 yr 0.9814 c 2.22 yr 0.768 ly
10 ly : 10.91 yr 0.99623 c 2.98 yr 0.868 ly
25 ly : 25.93 yr 0.99933 c 3.80 yr 0.915 ly
50 ly : 50.94 yr 0.999826 c 4.44 yr 0.932 ly
100 ly : 100.95 yr 0.9999557 c 5.09 yr 0.941 ly
1000 ly : 1000.95 yr 0.99999955 c 7.27 yr 0.949 ly
10000 ly : 10000.95 yr 0.9999999955 c 9.46 yr 0.950 ly
(B)------------------------------------------------------------------
More generally, if a ship can accelerate continuously away from Earth,
at constant acceleration a, as measured onboard the ship,
Values as measured by an observer in Earth's frame of reference:
T: Earth time (since the ship's launch)
D: Earth distance (from Earth to the ship)
A: Earth acceleration (of the ship)
Values as measured by an observer on the ship:
a: ship acceleration (of the ship)
τ: ship time (proper time) (since the ship's launch)
d: ship distance (from Earth to the ship)
j: distance travelled by ship
M: mass of ship
M0: initial mass
ρ: mass ratio = M0/M
βex: effective exhaust speed(/c)
Relativistic quantities:
- : velocity parameter
a
theta(tau) = - tau : velocity parameter
c
v a
beta = - = Tanh{ theta } = Tanh{ - tau }
c c
1 a
gamma = __________________ = Cosh{ theta } = Cosh{ - tau }
2 c
Sqrt{ 1 - beta }
Recurring constants:
8
c = 2.99792458 × 10 m/sec
7 7 15
year = 3.155693 × 10 sec ( ~ pi × 10 sec, ~ Sqrt{10 } sec )
15
lightyear = 9.460530 × 10 metres
2 2
for a = 10 m/sec = 1.052626 ly / yr ,
a -8 -1 -1
- = 3.33564095 × 10 sec = 1.052626 year
c
2
c 15
- = 8.98755179 × 10 metres = 0.9500051 light-years
a
(C)------------------------------------------------------------------
In the ship's frame of reference,
a -2(a/c)tau
v(tau) = c Tanh{ - tau } ; v(tau) -> c ( 1 - e )
c
2 2
c a c (a/c)tau
D(tau) = - ( Cosh{ - tau } - 1 ) ; D(tau) -> __ e
a c 2a
c a c a
tau(D) = - ArcCosh{ ___ D + 1 } ; tau(D) -> - Ln{ ___ D }
a 2 a 2
c c
2 2
c a c -(a/c)tau
d(tau) = - ( 1 - Sech{ - tau } ) ; d(tau) -> - ( 1 - 2 e )
a c a
2 2
c a c
j(tau) = - Ln{ Cosh[ - tau ] } ; j(tau) -> c × tau - - Ln{2}
a c a
c a c (a/c)tau
T(tau) = - Sinh{ - tau } ; T(tau) -> __ e
a c 2a
(D)------------------------------------------------------------------
Alternately, in the Earth's frame of reference,
your acceleration is measured as:
a
A = ________
3
gamma
2 3/2
A(v) = a ( 1 - (v/c) )
2
c a 2
D(T) = - ( Sqrt{ 1 + ( - T ) } - 1 )
a c
c a 2
T(D) = - Sqrt{ ( ___ D + 1 ) - 1 }
a 2
c
aT c
v(T) = ______________________ = ________________________
a 2 a -2
Sqrt{ 1 + ( - T ) } Sqrt{ 1 + ( - T ) }
c c
v(T) 1 1
beta(T) = ____ = _____________________ = ______________________
c c 2 a -2 1/2
Sqrt{ 1 + ( ___ ) } { 1 + ( - T ) }
aT c
a 2
gamma(T) = Sqrt{ 1 + ( - T ) }
c
c a a 2
tau(T) = - Ln{ - T + Sqrt[ 1 + ( - T ) ] }
a c c
a a
A(T) = _______________________ = _____________________
a 2 3 a 2 3/2
Sqrt{ 1 + ( - T ) } { 1 + ( - T ) }
c c
(E)------------------------------------------------------------------
For a trip which goes from standing start to standing finish,
calculate the T, τ, etc. to reach the midpoint, then double.
- Voyage length
2c a 2
Voyage length = 2 T( D/2 ) = __ Sqrt{ ( ____ D + 1 ) - 1 }
(Earth time) a 2
2c
= Sqrt{ D^2 / c^2 + 4 D / a }
2
D 4 D
= Sqrt{ ____ + _____ }
2 a
c
Voyage length
2c a
Voyage length = 2 tau( D/2 ) = __ ArcCosh{ ____ D + 1 }
(ship time) a 2
2c
=? ( 2 c / a ) ArcSinh{ 2 T(D/2) a / 2 / c }
a -2
Vmax = V( T(D/2) ) = c Sqrt{ 1 - ( ____ D + 1 ) }
2
or 2c
a
Vmax = V( tau(D/2) ) = c Tanh{ ArcCosh( ____ D + 1 ) }
2
2c
- Voyage length
2
2c a
Voyage length = 2 j( tau(D/2) ) = __ Ln{ ____ D + 1 }
(ship's odometer) a 2
2c
2
for instance, for a = 1 kgal ( = 1000 cm/sec ~ 1 "gee" )
15
Distance to Alpha Cen = 4.3 ly = 41 Pm = 41 × 10 m
6
Tau to Alpha Cen = 111 × 10 sec = 3.5 yr
6
Time to Alpha Cen = 187 × 10 sec = 5.9 yr
V_max = 0.95 c
distance travelled = 2.3 ly
(F)------------------------------------------------------------------
For a perfectly efficient photon rocket,
- so
Mo a
theta = ArcTanh{ beta } = Ln{ __ } = - tau, so
M c
-(a/c)tau
M(tau) = Mo e
For a perfectly efficient photon rocket,
accelerating from v = 0 to β(×c),
Mo 1 + beta
rho = __ = gamma ( 1 + beta ) = Sqrt{ __________ }
M 1 - beta
or alternately,
2
rho - 1
beta(rho) = __________
2
rho + 1
1 1 rho
gamma(rho) = - ( rho + ___ ) ; gamma(rho) -> ___
2 rho 2
For an imperfect rocket, with effective exhaust speed(/c) of βex,
- , so
-(a tau)/(c B_ex)
theta = B_ex Ln { rho } , so M(tau) = Mo e
- , or
2B_ex
rho - 1 1 + beta 1/2/B_ex
beta(rho) = ____________ , or rho(beta) = { _________ }
2B_ex 1 - beta
rho + 1
B_ex
1 B_ex -B_ex rho
gamma(rho) = - ( rho + rho ) ; gamma(rho) -> _____
2 2
(G)------------------------------------------------------------------
FAQ page for The Relativistic Rocket:
http://math.ucr.edu/home/baez/physics/Relativity/SR/rocket.html
also
http://math.ucr.edu/home/baez/physics/Relativity/SR/acceleration.html
http://math.ucr.edu/home/baez/physics/Relativity/SR/velocity.html
John Walker's relativity-exploring ship:
http://www.fourmilab.ch/cship/cship.html
The Oh-My-God Particle:
http://www.fourmilab.ch/documents/ohmygodpart.html
Erik Max Francis's frontpage:
http://www.alcyone.com/max/noframes.html
Wayne Throop's frontpage:
http://www.sheol.com/throopw
Chris Hillman's relativity page:
http://www.math.washington.edu/~hillman/relativity.html
Jason Hinson's FAQ site:
http://www.physics.purdue.edu/~hinson/ftl/index.html