Einsteins Twin Paradox

Description

Origin: David M. Weingold in HP-67/HP97 User's Library Solutions - Physics

The program is arranged to calculate subjective and real time differential between an observer on Earth and the pilot of a vehicle accelerating near the speed of light.

If you imagine twins at age 21. One becomes an astronaut and volunteers for the first interstellar flight. He takes off and travels at a ponderous speed of say 2.994444444×108 meters per second. In this situation it is accurate enough to call c the speed of light, 3×108. The astronaut travels for what he measures to be a year well past the sun at which time he fires retro and navigational engines, and turns around and heads toward Earth; the journey naturally takes another year. He is now 23 years old but when he steps from the ship his twin is over 37 years old! That over 16 years had passed on Earth.

The explanation as to why this happened involves very complicated non Euclidian geometry and relativistic considerations of accelerating frame of reference too complicated for this discussion. It sufficies to say that in the event of tremendous accelerations such as the turning around of a space craft traveling near the speed of light that the order of magnitude of energy involved is extremely large and the consideration of it as it interelates to space as time is conceived as a fourth physical dimension of space, the Universe is then conceived as a giant four dimensional sphere with a three dimensional surface.

The space craft in its turning travels relative to the Earth, not as far along that fourth dimension and hence the differential between the twins age. The equations for this case are quite simple and adequate for this case. They consist primarily of the Lorentz transform i.e., √(1 − v²/c²),where v is the velocity of the space craft relative to the Earth, and c is the universal constant, 3×108 meters per second, the speed of llight. The program inputs consist of speed of space craft in meters per second, time passed on Earth, time passed on board the craft, and the ages of the twins before the flights. With input TE, time passed on Earth, the equation
             v2
TS = TE√(1 - ——)
             c2

gives TS, time passed on board ship during journey. Input TS, time passed on board ship, and the equation:
         TS
TE = ————————————
     √(1 − v²/c²)

gives you TE, time passed on Earth during journey.
The label A clears and initiates the program. Label B is the input for average velocity of the craft. Label C is the input for time passed on earth in years and outputs time passed on board ship by hitting GSB 1. Label D input time passed on board ship. GSB 3 give appropriate time passed on earth. GSB 2 and 4 give the ages.

Note:
Be certain that you enter the speed of the space craft in meters per second. All time and age entries must be in years. Outputs will be in years. Do not try to make the space travel at the speed of light, = (3.00x108mete[§/second) as this will only show an error as should be and is implied by the theory of relativity

Sample Problem:
Suppose two twins, age 30, take part in this experiment. The velocity of the ship will average 2.999×108.
Solution:
  1. To initiate, clear registers, set display: A9.000 16
  2. Enter ship average velocity in meters/sec: 2.999 EEX 8 B8.994 16
  3. Enter time passed on ship in years : 1 D3.155 07
  4. Calculate time passed on Earth in years: GSB 33.873 01
  5. Calculate how old Earth man is upon end: 30 GSB 46.874 01
  6. Enter time passed on Earth in years: 25 C7.889 08
  7. Calculate time passed on ship in years: GSB 16.454 -01
  8. Find age of Space twin at end of journey: 30 GSB 23.064 01

Always enter time in years, and speed in meters per second, to go to new case, i.e. new speed or different amounts of time or ages press A, and then continue from step #3 with new values.

Program Resources

Labels

Name Description Name Description
 A Initiate  2 Earth twin final age
 B Set average ship velocity  3 Space time to Earth Time
 C Set time passed on Earth  4 Space twin final age
 D Set time passed in Space  9 # Calculate 1 - v²/c²
 1 Earth time to Space time

Storage Registers

Name Description Name Description
 0 3600 (s/h) .0
 1 24 (h/d) .1
 2 365.25 (d/y) .2 Time passed on Earth [s]
 3 Ts [y] .3 Time passed in Space [s]
 4 Te [y]

Program

Line Display Key Sequence Line Display Key Sequence Line Display Key Sequence
000 030 42,21,13 f LBL C 060 42,21, 3 f LBL 3
001 42,21,11 f LBL A 031 45 0 RCL 0 061 45 .3 RCL . 3
002 42 34 f REG 032 20 × 062 32 9 GSB 9
003 42, 8, 3 f SCI 3 033 45 1 RCL 1 063 10 ÷
004 43 35 g CLx 034 20 × 064 45 0 RCL 0
005 3 3 035 45 2 RCL 2 065 10 ÷
006 6 6 036 20 × 066 45 1 RCL 1
007 0 0 037 44 .2 STO . 2 067 10 ÷
008 0 0 038 43 32 g RTN 068 45 2 RCL 2
009 44 0 STO 0 039 42,21, 1 f LBL 1 069 10 ÷
010 2 2 040 45 .2 RCL . 2 070 44 4 STO 4
011 4 4 041 32 9 GSB 9 071 43 32 g RTN
012 44 1 STO 1 042 20 × 072 42,21, 9 f LBL 9
013 3 3 043 45 0 RCL 0 073 45 .1 RCL . 1
014 6 6 044 10 ÷ 074 45 .0 RCL . 0
015 5 5 045 45 1 RCL 1 075 10 ÷
016 48 . 046 10 ÷ 076 16 CHS
017 2 2 047 45 2 RCL 2 077 1 1
018 5 5 048 10 ÷ 078 40 +
019 44 2 STO 2 049 44 3 STO 3 079 11 √x̅
020 3 3 050 43 32 g RTN 080 43 32 g RTN
021 26 EEX 051 42,21,14 f LBL D 081 42,21, 2 f LBL 2
022 8 8 052 45 0 RCL 0 082 45 3 RCL 3
023 43 11 g 053 20 × 083 40 +
024 44 .0 STO . 0 054 45 1 RCL 1 084 43 32 g RTN
025 43 32 g RTN 055 20 × 085 42,21, 4 f LBL 4
026 42,21,12 f LBL B 056 45 2 RCL 2 086 45 4 RCL 4
027 43 11 g 057 20 × 087 40 +
028 44 .1 STO . 1 058 44 .3 STO . 3 088 43 32 g RTN
029 43 32 g RTN 059 43 32 g RTN