User:Wwoods/Relativistic rocket formulas

(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) &tau;: 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 &rho;: mass ratio =  M0/M &beta;ex: effective exhaust speed(/c)

Relativistic quantities:


 * $$ \theta(\tau) = \frac{a}{c}\tau

$$ : velocity parameter

a   theta(tau)  =  - tau        : velocity parameter c
 * $$ \beta = \frac vc = \tanh\theta = \tanh\frac ac \tau

$$            v                             a    beta  =  -  =  Tanh{ theta }  =  Tanh{ - tau } c                            c


 * $$ \gamma = \frac{1}{\sqrt{ 1 - \beta^2 }} = \cosh\theta = \cosh\frac ac \tau

$$                    1                                       a    gamma  =  __________________  =  Cosh{ theta }  =  Cosh{ - tau } 2                              c               Sqrt{ 1 - beta  }

Recurring constants:


 * $$ c =  2.99792458 \times 10^8 \mbox{ m/sec}

$$                        8    c  =  2.99792458 × 10  m/sec


 * $$ \mbox{year } = 3.155693 \times 10^7  \mbox{ sec }

( \sim \pi \times 10^7 \mbox{ sec, or }     \sim\sqrt{10^{15}} \mbox{ sec}) $$                         7                 7              15    year  =  3.155693 × 10  sec  ( ~ pi × 10  sec, ~ Sqrt{10 } sec )


 * $$ \mbox{lightyear } = 9.460530 \times 10^{15} \mbox{ metres}

$$                             15    lightyear  =  9.460530 × 10  metres


 * $$ \mbox{for } a = 10 \mbox{ m/sec}^2

= 1.052626 \mbox{ ly/yr}^2 , $$                   2                    2    for a = 10 m/sec  =  1.052626 ly / yr ,


 * $$ \frac ac = 3.33564095 \times 10^{-8} \mbox{ sec}^{-1}

= 1.052626 \mbox{ year}^{-1} $$       a                   -8   -1                -1 - =  3.33564095 × 10  sec  =  1.052626 year c


 * $$ \frac{c^2}{a} = 8.98755179 \times 10^{15} \mbox{ metres }

= 0.9500051 \mbox{ light-years} $$        2        c                   15 - =  8.98755179 × 10  metres  =  0.9500051 light-years a

(C)--

In the ship's frame of reference,


 * $$ v(\tau) =  c \tanh{\frac ac \tau};\quad

v(\tau) \rightarrow c \left( 1 - e^{-2\frac ac \tau}\right) $$                   a                              -2(a/c)tau v(tau) =  c Tanh{ - tau }  ;   v(tau) -> c ( 1 - e         ) c


 * $$ D(\tau) = \frac{c^2}{a} \left(\cosh{\frac ac \tau} - 1\right) ;\quad

D(\tau) \rightarrow \frac{c^2}{2a} e^{\frac ac \tau} $$            2                                       2            c         a                             c   (a/c)tau D(tau) =  - ( Cosh{ - tau } - 1 )  ;   D(tau) ->  __  e            a         c                             2a


 * $$ \tau(D) = \cosh^{-1}\left(\frac{a}{c^2} D + 1\right)   ;\quad

\tau(D) \rightarrow \frac ca \ln\left(\frac{a}{c^2} D \right) $$           c           a                          c      a tau(D)  =  - ArcCosh{ ___ D + 1 }   ;   tau(D) -> - Ln{ ___ D } a           2                         a       2 c                                c


 * $$ d(\tau) = \frac{c^2}{a}

\left( 1 - \mbox{sech} \frac ac\tau \right) ;\quad d(\tau) \rightarrow \frac{c^2}{a} \left( 1 - 2 e^{-\frac ac\tau} \right) $$            2                                      2            c             a                        c         -(a/c)tau d(tau) =  - ( 1 - Sech{ - tau } )  ;   d(tau) -> - ( 1 - 2 e        ) a            c                        a


 * $$ j(\tau) = \frac{c^2}{a} \ln\cosh\left( \frac ac\tau \right) ;\quad

j(\tau) \rightarrow c \times \tau - \frac{c^2}{a} \ln{2} $$            2                                                2            c           a                                    c j(tau)  =  - Ln{ Cosh[ - tau ] }    ;   j(tau) -> c × tau - - Ln{2} a          c                                    a
 * $$ T(\tau) =  \frac ca \sinh{ \frac ac\tau }          ;\quad

T(\tau) \rightarrow  \frac{c}{2a} e^{\frac ac\tau} $$           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 = \frac{a}{\gamma^3}

$$         a A  =  ________ 3       gamma


 * $$ A(v) =  a \left( 1 - \beta^2 \right)^{\frac 32}

$$                      2  3/2 A(v)  =  a ( 1 - (v/c)  )
 * $$ D(T) =  \frac{c^2}{a} \left( \sqrt{ 1 + \left( \frac ac T \right)^2  } - 1 \right)

$$          2          c               a    2 D(T) =  - ( Sqrt{ 1 + ( - T )  } - 1 ) a              c


 * $$ T(D) = \frac ca \sqrt{ \left( \frac{a}{c^2} D + 1 \right)^2  - 1 }

$$         c           a         2 T(D) =  - Sqrt{ (  ___ D + 1 )  - 1 } a           2 c


 * $$ v(T) = \frac{aT}{\sqrt{ 1 + \left( \frac acT \right)^2   }}

= \frac{c }{\sqrt{ 1 + \left( \frac acT \right)^{-2} }} $$                 aT                          c v(T)  =  ______________________  =  ________________________ a   2                     a    -2 Sqrt{ 1 + ( - T ) }       Sqrt{ 1 + ( - T )    } c                         c


 * $$ \beta(T) = \frac{v(T)}{c}

= \frac{1}{\sqrt{ 1 + \left( \frac{c}{aT} \right)^2 }} = \frac{1}{\sqrt{ 1 + \left( \frac acT \right)^{-2} }} $$            v(T)             1                         1 beta(T) =  ____  =  _____________________  =  ______________________ c                   c   2                a    -2   1/2 Sqrt{ 1 + ( ___ )  }     { 1 + ( - T )    } aT                  c
 * $$ \gamma(T) = \sqrt{ 1 + \left( \frac acT \right)^2 }

$$                         a    2 gamma(T) =  Sqrt{ 1 + ( - T )  } c


 * $$ \tau(T) = \frac ca \ln\left( \frac ac T + \sqrt{ 1 + \left( \frac ac T \right)^2 } \right)

$$           c     a                 a    2 tau(T) =  - Ln{ - T + Sqrt[ 1 + ( - T )  ] } a    c                 c


 * $$ A(T) = \frac{a}{\sqrt{ 1 + \left( \frac ac T \right)^2 }^3}

= \frac{a}{\left( 1 + \left( \frac ac T \right)^2 \right)^{\frac32}} $$                  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, &tau;, etc. to reach the midpoint, then double.


 * Voyage length $$ =  2 T( D/2 )  = \frac{2c}{a} \sqrt{ \left( \frac{a}{2c^2} D + 1 \right)^2  - 1 }

= \sqrt{ \frac{D^2}{c^2} + \frac{4 D}{a} } $$                                 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 $$ =  2 \tau( D/2 )  = \frac{2c}{a} \cosh^{-1}\left( \frac{a}{2c^2} D + 1 \right) =? \frac{2c}{a} \sinh^{-1}\left( 2 ( T(D/2) \frac{a}{2 c} \right) $$                                   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 }
 * $$ V_{\mbox{max}} =  V( T(D/2) )  =  c \sqrt{  1 - \left( \frac{a}{2c^2} D + 1 \right)^{-2}}

$$                                         a          -2 Vmax =  V( T(D/2) )  =  c Sqrt{  1 - ( ____ D + 1 )    } 2 or                                      2c
 * $$ V_{\mbox{max}} =  V( \tau(D/2) )  =  c \tanh{ \cosh^{-1}\left( \frac{a}{2c^2} D + 1 \right) }

$$                                             a Vmax  =  V( tau(D/2) )  =  c Tanh{ ArcCosh( ____ D + 1 ) } 2                                             2c


 * Voyage length $$ = 2 j( \tau(D/2) )

= \frac{2c^2}{a} \ln \left( \frac{a}{2c^2} D + 1 \right) $$                                        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,


 * $$ \theta =  \tanh^{-1}{ \beta }  = \ln \frac{M_0}{M} = \frac ac\tau,

$$ so                                   Mo       a theta  =  ArcTanh{ beta }  =  Ln{ __ }  =  - tau,  so                                        M       c
 * $$    M(\tau)  =  M_0 e^{\frac ac\tau}

$$

-(a/c)tau M(tau) =  Mo e

For a perfectly efficient photon rocket, accelerating from v = 0 to &beta;(×c),


 * $$ \rho = \frac{M_0}{M}  = \gamma ( 1 + \beta )

= \sqrt{ \frac{1+\beta}{1-\beta} } $$        Mo                                   1 + beta rho =  __  =  gamma ( 1 + beta )  =  Sqrt{ __________ } M                                  1 - beta

or alternately,
 * $$ \beta(\rho) = \frac{\rho^2-1}{\rho^2+1}

$$                      2                    rho  - 1 beta(rho) =  __________ 2                   rho  + 1


 * $$ \gamma(\rho) =  \begin{matrix}\frac12\end{matrix}

\left( \rho + \frac1\rho \right)  ;\quad   \gamma(\rho) \rightarrow \frac\rho2 $$               1          1                          rho gamma(rho) =  - ( rho + ___ )   ;   gamma(rho)  ->  ___ 2        rho                          2

For an imperfect rocket, with effective exhaust speed(/c) of &beta;ex,


 * $$ \theta =  \beta_{ex} \ln\rho$$,  so  $$

M(\tau) =  M_0 e^{ -\frac{a\tau}{c\beta_{ex}} } $$                                                 -(a tau)/(c B_ex) theta =  B_ex Ln { rho },  so    M(tau)  =  Mo e


 * $$ \beta(\rho) = \frac{ \rho^{2\beta_{ex}} - 1 }{ \rho^{2\beta_{ex}} + 1 }

$$, or $$ \rho(\beta) = \left(\frac{1+\beta}{1-\beta}\right)^{\frac1{2\beta_{ex}}} $$                 2B_ex rho   - 1                          1 + beta  1/2/B_ex beta(rho) =  ____________,  or   rho(beta)  =  { _________ } 2B_ex                            1 - beta rho   + 1


 * $$ \gamma(\rho) = \begin{matrix}\frac12\end{matrix}

\left( \rho^{\beta_{ex}} + \rho^{-\beta_{ex}} \right) ;\quad \gamma(\rho) \rightarrow \frac{\rho^{\beta_{ex}}}2 $$                                                               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

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