Overview
-The following programs use Einstein's special relativity to study uniformly
( and non-uniformly ) accelerated linear motion of a spacecraft.
-In special relativity, the speed is also v = dx/dt , but acceleration
is a = ( dv/dt ) ( 1 - v2/c2 )-3/2
where t = time passed on Earth
-The proper Time T ( passed on board ship ) verifies dT = dt
( 1 - v2/c2 )1/2
-In all these programs, we use the following units:
-Times ( t , T ) are expressed in Julian years (
365.25 days )
-Abcissas and distances ( x , D ) are expressed
in light-years ( one light-year = 9.460,730,472,580,8 1015
m. )
-The unit of accelerations ( a ) is the "standard"
gravity on Earth g = 9.80665 m/s2
-The velocity of light = c = 299,792,458 m/s ( from the definition of 1 meter )
Remark: Integrals are denoted §
1°) A Trip to the Stars
-A spaceship travels to a star ( situated at a distance D from the Earth
) at a constant acceleration "a" , in order to simulate an artificial gravity.
-Halfway to the star, the constant acceleration "a" becomes a constant
deceleration "-a"
-The return journey goes off similarly.
Return Journey
<<<<<Deceleration "a" <<<<< |
<<<<<Acceleration "-a" <<<<<<<
Earth=O-------------------------------D/2------------------------------------D=(Star)--------------->
x
>>>>>Acceleration "a" >>>>> | >>>>>Deceleration "-a" >>>>>>>
Outward Journey
-The graphic representation of x(t) looks like this:
x
|
| D
*
|
*
*
|
*
*
| D/2
*
*
|
*
*
|
*
*
|*---------------|----------------|-----------------|-----------------*-----
t
Takeoff
Spacecraft reaches the star
Landing
Formulae:
Proper Time on board ship = TS = (4c/a) Arccosh ( 1 + a.D/(2c2)
)
Time passed on Earth = tE = (4c/a) ( aD/c2 + a2D2/(4c4)
)1/2
Maximum speed ( when x =
D/2 ) = vmax/c = Tanh µ where µ = a.TS/(4c)
-Times concern the whole journey: Earth-Star-Earth.
-Synthetic register M may be replaced by register R00.
-Line 03: this constant = 365.25*86400*9.80665/(299792458*4)
The "standard" gravity 9.80665 doesn't seem quite
"standardized"
For instance, I've seen here and there the value 9.80616
In this case, replace line 02 by 0.2580609239
| 01 LBL "ETP"
02 X<>Y 03 .2580738189 04 * 05 STO M 06 * 07 ST+ X 08 ENTER^ |
09 ENTER^
10 X^2 11 LASTX 12 ST+ X 13 + 14 SQRT 15 STO Z 16 + |
17 LN1+X
18 0 19 X<> M 20 ST/ Z 21 / 22 R^ 23 1 24 + |
25 1/X
26 ASIN 27 COS 28 2 29 ST/ L 30 X<> L 31 SIN 32 X^2 |
33 ST+ X
34 X<>Y 35 R^ 36 R^ 37 END |
( 64 bytes / SIZE 000 )
| STACK | INPUTS | OUTPUTS |
| T | / | 1-vmax/c |
| Z | / | vmax/c |
| Y | a | tE |
| X | D | TS |
TS = time passed on
board ship
tE = time passed
on Earth.
vmax = maximum speed
( when x = D/2 )
Example1: On 2007/12/25 immediate
boarding for the first interstellar flight in the direction of Toliman
( alpha Centauri ) Distance D = 4.343 light-years.
Fox Mulder remains on Earth but Dana Scully's flying saucer takes off at
10h13m TT , acceleration = 0.7g
0.7 ENTER^
4.343 XEQ "ETP" >>>>
TS = 8.837390060 years
RDN tE
= 13.09998313 years
RDN vmax/c = 0.9211383858
vmax = 92.11383858% * 299792458 m/s = 276150341
m/s
RDN 1-vmax/c = 0.07886161418
-When she steps from the ship, Dana's watch shows 2016/10/26 , 6h46m41s
wheras according to Mulder, it's 4h40m08s o'clock and we are
on 2021/01/30
Example2: With a = 1g and D = 2,200,000 light-years ( approximately the distance of the Andromeda galaxy )
1
ENTER^
2200000 R/S
>>>> 56.71150062 years
RDN 4,400,003.874 years
RDN vmax/c = 1
RDN 1-vmax/c = 3.877715920 E-13 whence
vmax/c was actually 0.999,999,999,999,612,228,408
-Caculating 1-vmax/c is only useful if the maximum
v/c is very close to 1
-If you don't want to compute this value, replace lines 28 to 36 by
STO T RDN
-If you prefer (1-v2max/c2)1/2
, replace lines 26 to 34 by ENTER^ ASIN
COS
-If you don't want to know vmax/c and
1-vmax/c , delete lines 22 to 36.
2°) Uniformly Accelerated Motion
-This program converts t ( time on Earth ) , T ( time on the spacecraft
) and x ( distance from the Earth ) for a given constant acceleration "a"
-It also calculates the speed v/c ( and 1-v/c ).
Formulae:
v/c = Tanh µ
where µ = a.T/c , v/c = ( at/c ).( 1 + a2t2/c2
)-1/2
t = (
c/a ) Sinh µ
x = ( c2/a
) ( Cosh µ - 1 ) = ( c2/a ) ( ( 1 + a2t2/c2
)1/2 - 1 )
-Due to these formulae, uniformly accelerated motion is often called "hyperbolic motion"
-Synthetic register M may be replaced by register R00.
-Line 14: this constant is 4 times the value appearing line 02
of the "ETP" listing above.
| 01 LBL "ETP2"
02 LBL A 03 SF 00 04 GTO 01 05 LBL B 06 CF 01 07 GTO 00 08 LBL C 09 SF 01 10 LBL 00 11 CF 00 12 LBL 01 13 X<>Y 14 1.032295276 15 * 16 STO M 17 * |
18 FS? 00
19 GTO 00 20 FS? 01 21 GTO 01 22 X^2 23 STO Y 24 1 25 + 26 SQRT 27 1 28 + 29 / 30 LBL 00 31 ENTER^ 32 ENTER^ 33 X^2 34 LASTX |
35 ST+ X
36 + 37 SQRT 38 FS? 00 39 STO Z 40 + 41 LN1+X 42 RCL M 43 ST/ Z 44 GTO 02 45 LBL 01 46 E^X-1 47 LASTX 48 CHS 49 E^X-1 50 + 51 2 |
52 /
53 STO Y 54 RCL M 55 / 56 RCL Y 57 2 58 + 59 R^ 60 * 61 SQRT 62 RCL M 63 LBL 02 64 / 65 R^ 66 1 67 + 68 1/X |
69 ASIN
70 COS 71 2 72 ST/ L 73 X<> L 74 SIN 75 X^2 76 ST+ X 77 X<>Y 78 R^ 79 R^ 80 CLA 81 END |
( 128 bytes / SIZE 000 )
1°) a , x >>>> T
, t , v/c
| STACK | INPUTS | XEQ A | OUTPUTS |
| T | / | 1-v/c | |
| Z | / | v/c | |
| Y | a | t | |
| X | x | T |
Example:
0.7 ENTER^
10 XEQ "ETP2" or XEQ A or [A] in user
mode >>>> T = 3.870348933
RDN t = 11.29945015
RDN v/c = 0.992583500
RDN 1-v/c = 0.007416499860
2°) a , t >>>> T
, x , v/c
| STACK | INPUTS | XEQ B | OUTPUTS |
| T | / | 1-v/c | |
| Z | / | v/c | |
| Y | a | x | |
| X | t | T |
Example:
0.7 ENTER^
10 XEQ B or [B] in user
mode >>>> T = 3.702700069
RDN x = 8.711423203
RDN v/c = 0.990559778
RDN 1-v/c = 0.009440221722
3°) a , T >>>> t
, x , v/c
| STACK | INPUTS | XEQ C | OUTPUTS |
| T | / | 1-v/c | |
| Z | / | v/c | |
| Y | a | x | |
| X | T | t |
Example:
0.7 ENTER^
10 XEQ C or [C] in user
mode >>>> t = 951.2809258
RDN x = 949.8980538
RDN v/c = 0.999998942
RDN 1-v/c = 0.000001058151320
-If you don't want to compute 1-v/c, replace lines 71 to 79 by
STO T RDN
-If you prefer (1-v2max/c2)1/2
, replace lines 69 to 77 by ENTER^ ASIN
COS
-If you don't want to know vmax/c and
1-vmax/c , delete lines 65 to 79.
3°) An Example of non-uniformly Accelerated Motion
-The spaceship acceleration a(T) is now supposed to be known
as a function of the proper time T ( on board )
-We have:
a(T) = ( dv/dt ) ( 1 - v2/c2 )-3/2 = ( dv/dT ) ( 1 - v2/c2 )-1 since dT = dt ( 1 - v2/c2 )1/2 therefore, dv/( 1 - v2/c2 ) = a(T).dT
whence v/c = Tanh ( (1/c) F(T) ) where F(T) is the primitive of a(T) such that F(0) = 0 ( we assume x = 0 & v = 0 when t = T = 0 )
-Likewise, dt = dT ( 1 - v2/c2 )-1/2
= dT ( 1 - Tanh2(F(T)) )-1/2 = dT Cosh(F(T))
whence t = §0T
Cosh((1/c)F(u)).du
( § = Integral )
and dx = v.dt = c.Tanh(F(T)).Cosh(F(T)).dT = c. Sinh(F(T)).dT
whence x = c §0T
Sinh((1/c)F(u)).du
-"ETP3" evaluates t ( time on Earth ), x ( abscissa ) and v/c ( speed of the spacecraft ) at a given proper time T,
assuming the primitive F(T) may be expressed by elementary
functions
Data Registers: • R00: F name ( Register R00 is to be initialized before executing "ETP3" )
R01 = 0 R04 thru R08 are used
by "GL3"
R02 = T R12 , R13: temp
R03 = n = number of subintervals used in Gaussian integration.
when the program stops: R09 = t , R10 = x , R11 = v/c
Flags: /
Subroutines: "GL3" ( cf "Numerical
Integration for the HP-41" )
and a program which calculates F(T) assuming T is in X-register upon entry.
| 01 LBL "ETP3"
02 STO 02 03 X<>Y 04 STO 03 05 X<>Y 06 XEQ IND 00 07 1.032295276 08 STO 12 09 * 10 ST+ X 11 E^X-1 12 STO Y 13 2 |
14 +
15 / 16 STO 11 17 CLX 18 STO 01 19 RCL 03 20 LBL 00 21 STO 03 22 RCL 00 23 STO 13 24 "CT" 25 ASTO 00 26 XEQ "GL3" |
27 STO 09
28 "ST" 29 ASTO 00 30 XEQ "GL3" 31 STO 10 32 RCL 13 33 STO 00 34 RCL 11 35 RCL 10 36 RCL 09 37 CLA 38 RTN 39 GTO 00 |
40 LBL "ST"
41 XEQ IND 13 42 RCL 12 43 * 44 E^X-1 45 LASTX 46 CHS 47 E^X-1 48 - 49 2 50 / 51 RTN 52 LBL "CT" |
53 XEQ IND 13
54 RCL 12 55 * 56 E^X 57 LASTX 58 CHS 59 E^X 60 + 61 2 62 / 63 RTN 64 END |
( 113 bytes / SIZE 014 )
| STACK | INPUTS | OUTPUTS |
| Z | / | v/c |
| Y | n | x |
| X | T | t |
where n is the number of subintervals over which the 3-point Gauss-Legendre formula is applied
Example: acceleration = a(T) = 0.7 cos ( 360°(T/40) ) the whole journey will take 40 years on board.
-Here, we can easily find the required primitive: F(T) = (14/pi) sin ( 360°(T/40) ) = (14/pi) sin ( 9°T ) So we load the following subroutine
01 LBL "FF" ( global label , at most 6 characters
)
02 9
03 *
04 SIN
05 14
06 *
07 PI
08 /
09 RTN
- "FF" ASTO 00
-We set the HP-41 in DEG mode and if we choose T = 1 ( year )
and n = 2
2 ENTER^
1 XEQ "ETP3" >>>> t
= 1.088871751 year = R09 ( the difference
between Earth time and ship time is already about 32 days )
RDN x = 0.376425754 light-year = R10
RDN v/c = 0.616685490 = R11
-With T = 10 and n = 4
4 ENTER^
10 XEQ "ETP3" >>>> t = 190.9695910
( in 1 minute )
RDN x = 189.5025503
RDN v/c = 0.999798046
-To have another approximation for the same T but a different n, simply key in the new n-value and press R/S , for instance:
8 R/S >>>> t = 190.9695910
RDN x = 189.5025369
RDN v/c = 0.999798046
-Note that v/c is obtained exactly since no approximate integration
formula is used.
-In the above example, the whole journey Earth-Star-Earth required
40 years on board ship but almost 764 years have passed on Earth !
-The distance Earth-Star was approximately 379 light-years.
-Results may be wrong if F(T) or its derivatives have singularities.
4°) Non-uniformly Accelerated Motion, a more general
case
-We study the same problem as in paragraph 3 , but we suppose the primitive difficult ( or impossible ) to express analytically. We have to evaluate:
t = §0T
Cosh( (1/c) §0u
a(v).dv ).du
x = c §0T
Sinh( (1/c) §0u
a(v).dv ).du
v/c = Tanh ( (1/c) §0T
a(u).du )
Data Registers: • R00: "a" name ( Register R00 is to be initialized before executing "ETP4" )
R01 = 0 R04 thru R17 are used
by "DIN2"
R02 = T R18 , R22: temp
R03 = n = number of subintervals used in Gaussian integration.
when the program stops: R19 = t , R20 = x , R21 = v/c
Flags: F04
Subroutines: "DIN2" ( see the bottom
of this page - paragraph 5 )
"GL3" ( cf "Numerical Integration for the HP-41" )
and a program which calculates a(T) assuming T is in X-register upon entry.
| 01 LBL "ETP4"
02 1.032295276 03 STO 18 04 CLX 05 STO 01 06 RDN 07 STO 02 08 X<>Y 09 LBL 00 10 STO 03 11 RCL 00 12 STO 22 13 "T" 14 ASTO 00 15 "U" |
16 ASTO 04
17 "V" 18 ASTO 05 19 "CH" 20 ASTO 17 21 SF 04 22 XEQ "DIN2" 23 STO 19 24 "SH" 25 ASTO 17 26 XEQ "DIN2" 27 STO 20 28 RCL 22 29 STO 00 30 XEQ "GL3" |
31 RCL 18
32 * 33 ST+ X 34 E^X-1 35 STO Y 36 2 37 + 38 / 39 STO 21 40 RCL 20 41 RCL 19 42 CLA 43 RTN 44 GTO 00 45 LBL "T" |
46 X<>Y
47 XEQ IND 22 48 RCL 18 49 * 50 RTN 51 LBL "U" 52 CLX 53 RTN 54 LBL "V" 55 RTN 56 LBL "SH" 57 E^X-1 58 LASTX 59 CHS 60 E^X-1 |
61 -
62 2 63 / 64 RTN 65 LBL "CH" 66 E^X 67 LASTX 68 CHS 69 E^X 70 + 71 2 72 / 73 RTN 74 END |
( 156 bytes / SIZE 023 )
| STACK | INPUTS | OUTPUTS |
| Z | / | v/c |
| Y | n | x |
| X | T | t |
where n is the number of subintervals over which the 3-point Gauss-Legendre formula is applied
Example: Let's take again a(T) = 0.7 cos ( 360°(T/40) ) as if we couldn't calculate the primitive ( it will be a test case for our program )
-We load the following subroutine
01 LBL "AA" ( global label , at most 6 characters
)
02 9
03 *
04 COS
Store 9 and 0.7 into registers R23 and R24 and replace lines
02 & 05 by RCL 23 & RCL 24 respectively
05 .7
It will speed up execution.
06 *
07 RTN
- "AA" ASTO 00
-We set the HP-41 in DEG mode and with T = 1 and n = 2
2 ENTER^
1 XEQ "ETP4" >>>> t
= 1.088871751
RDN x = 0.376425755
RDN v/c = 0.616685490
-With T = 10 and n = 4
4 ENTER^
10 XEQ "ETP4" >>>> t = 190.9695915
( in 8mn17s )
RDN x = 189.5025508
RDN v/c = 0.999798046
-To have another approximation for the same T but a different n, simply key in the new n-value and press R/S , for instance:
8 R/S >>>> t = 190.9695911
RDN x = 189.5025369
RDN v/c = 0.999798046
-Unlike "ETP3" , "ETP4" re-calculates v/c for each new n-value.
-When n is multiplied by 2 , execution time is approximately multiplied
by 4
5°) Double Integrals §ab
G [ §u(x)v(x)
f(x,y) dy ] dx
-"ETP4" requires to evaluate integrals of this form, where G is a function
of 1 variable.
-Modify the program listed in "Double Integrals for the HP-41" as follows:
-Replace line 01 by LBL "DIN2"
-Delete line 96
-Add 3.6 / FS? 04
GTO IND 17 after line 89
-In other words, the new listing looks like this:
01 LBL "DIN2"
.......................
the same lines as "DIN"
.......................
87 RCL 14
88 RCL 11
89 *
90 3.6
91 /
92 FS? 04
93 GTO IND 17
94 RTN
95 LBL 04
96 RCL 06
97 RCL 09
98 *
99 3.6
100 /
101 STO 06
102 END
( 148 bytes / SIZE 018 )
-Set flag F04, store G name in register R17, and follow the same instructions as "DIN"
Example: Compute §12 Ln( 1 + §xx^2 dy/(x+y) ) dx
-Load these routines in program memory:
01 LBL "T"
02 +
03 1/X
04 RTN
05 LBL "U"
06 RTN
07 LBL "V"
08 X^2
09 RTN
10 LBL "W"
11 LN1+X
12 RTN
"T" ASTO 00 "U" ASTO 04 "V"
ASTO 05 "W" ASTO 17 SF 04
1 STO 01 2 STO 02
-With n = 2 2 STO 03 XEQ "DIN2"
>>>> 0.191218819
-With n = 4 4 STO 03
R/S >>>>
0.191218775
-If flag F04 is clear, "DIN2" performs the same double integral as "DIN"
( G is not taken into account )