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rtharbaugh1
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Space under construction: The twin ship problem
The basic problem is that one of a set of twins leaves Earth to travel at near-light speed, encounters time dilation due to relativity, and returns to find the other twin has experienced more time and is now "older." I wish to discover how to calculate this problem under variable conditions of acceleration. I have never worked through this problem before and am not sure of what mathematics is required, but hope to gather the information to solve the problem and present it here. Any helpful comments are solicited.
I will specify flat space and two engines, one, the thruster, providing forward acceleration, and the other, the "kicker," providing a small lateral thrust so that the ship moves in a very large circle.
The ship leaves Earth, travels in a circle, then returns to pass by near-Earth space at a high velocity.
For simplicity, I will assume that space is flat and that the entire trip can be accomplished without correction for extraneous mass or charge.
I will further assume that the thrusters are not limited by fuel considerations, but can continue to provide thrust within their frame of reference at a constant rate for any necessary length of time.
The problem now is to find the correct formulas, and then to determine how to calculate them.
I have the following formulae from DW, who tried to help me with a similar problem in another thread at
https://www.physicsforums.com/showthread.php?t=15826
[tex]v = ctanh(\frac{\alpha \tau}{c})[/tex]
[tex]v = \frac{\alpha t}{\sqrt{1 + \frac{\alpha ^{2}t^{2}}{c^2}}}[/tex]
where:
[tex]v[/tex] = velocity as observed from the rest frame
[tex]c[/tex] = speed of light
[tex]\alpha[/tex] = proper acceleration (as experienced by accelerated frame
[tex]\tau [/tex] = time elapsed in accelerated frame
[tex]t[/tex] = time elapsed in rest frame
and for convenience, the term
[tex](\frac{\alpha \tau}{c}) = \theta[/tex]
DW says [tex]\theta[/tex] is known as rapidity.
I notice that [tex]\theta[/tex] is a dimensionless quantity, with terms for Distance divided by Time above and below the fraction, so they cancel.
I see that in the rest frame formula the term subtracted from [tex]1[/tex] in the square root denominator (bottom part of the fraction) is, in form, very like the term [tex]\theta^2[/tex]. I am curious about this similarity, and wonder if it would help my understanding to think of the phrase
[tex]\sqrt{1 + \frac{\alpha^2 t^2}{c^2}}[/tex]
as Pythagorean, thus following the formula
[tex]A=\sqrt{B^2+C^2}[/tex]
where A is the hypoteneus of a right triangle in which B is equal to one and C is [tex]\frac{\alpha t}{c}[/tex]. Note capital C has no relation to small c.
I notice that the frame of reference which is the rest frame is dependent on the motion of the observer. (Aside, has anyone counted up the ways we on Earth are moving through sidereal space? Sidereal means with reference to the fixed stars, and of course there aren't any. But if we chose to use the Z-10 galaxy (did I get the term right? read about it on pf somewhere, will try to find source) which is the oldest known galaxy as the reference frame, then how many 'proper' motions? )
Setting the acceleration at one gravity reduces the formulae to:
[tex]v = ctanh(\frac{\tau}{c})[/tex]
[tex]v = \frac{t}{\sqrt{1 + \frac{t^{2}}{c^2}}}[/tex]
c might also be taken as unitary, yielding the formulae as
[tex]v = tanh\tau}[/tex]
[tex]v = \frac{t}{\sqrt{1 + t^{2}}}[/tex]
then
[tex]tanh\tau}=v = \frac{t}{\sqrt{1 + t^{2}}}[/tex]
Is this formula useful in calculating the twinship problem? I place it here for contemplation, as it seems to me to reveal something about the nature of time and space.
What is [tex]\tau[/tex]when t is one? [tex]tanh\tau}=\sqrt{2}^{-1}[/tex]
What is t when Tau is one? [tex]tanh1=\frac{t}{\sqrt{1 + t^{2}}}[/tex]
I don't think my hand calculator does hyperbolic tangents. However there is a graph of the hyperbolic tangent on Wikipedia at http://en.wikipedia.org/wiki/Hyperbolic_tangent
and I can get an idea of the behavior of the function from reading the graph. Also, there is a calculator on the web at http://www.karlscalculus.org/cgi-bin/newcalc.pl which does hyperbolic calculations.
Then I should be able to build a table showing the time dilation expected to be observed by the Earth twin in the rest frame and experienced by the twinship as it passes near Earth space as a function of t, time elapsed on earth. The time of passage could be varied by adjusting the circumfrence of the circle due to the kicker engine. I should be able to calculate a view of the ship as if it were passing near Earth at any time t. For example, if the kicker were set so that the ship would pass Earth after a t of one year, how fast would the ship be traveling? How much time would have elapsed on board the ship? And if t were two years, three years, and so on.
(calculations made with Karl's Calculator, see link in text above)
[tex]\tau=>tanh(\tau)=>t[/tex]
1=>0.761594155955765=>0.605886858731034
2=>0.964027580075817=>
3=>0.995054753686730=>
4=>0.999329299739067=>
5=>0.999909204262595=>
6=>0.999987711650796=>
7=>0.999998336943945=>
8=>0.999999774929676=>
9=>0.999999969540041=>
I am having trouble seeing the meaning. When tau, the elapsed time on board the ship, is one year, the velocity of the ship is [tex]tanh(\tau)[/tex]=.76..., and the elapsed time on Earth is [tex]t[/tex]=.60...
This seems wrong. The ship twin experiences one year, the Earth twin experiences .6 year? Then the ship twin returns older than the Earth twin? I thought it was the other way around, and the ship twin should return younger than the Earth twin.
DW informs me that the correct t should be 1.2. I notice this is a factor of two times the .60... in my calculation. Perhaps this is only a coincidence. More likely I simply do not know how to work Karl's calculator, even though I managed to get the numbers for [tex]\tau[/tex] out of it correctly. I will return to Karl's calculator and try again. If I have no luck I will search for other calculators that might be easier for me to learn to use.
No luck searching for other calculators. Most search results are adverts for stuff I can't afford to buy right now.
Here is a table of V values for variable t:
(<= notation here means t at right put into formula for V shown on left)
[tex]\frac{t}{\sqrt(1+t^2)}[/tex]<=t
0.707106781186547<=1
0.894427190999916<= 2
0.948683298050514<=3
0.970142500145332<=4
0.98058067569092<=5
0.986393923832144<=6
0.989949493661166<=7
0.992277876713668<=8
0.993883734673619<=9
Then I can merge the two tables by intercollating the values for V:
[tex]\tau[/tex]=>V<=[tex]t[/tex]
----0.707106781186547<=1
1=>0.761594155955765
----0.894427190999916<= 2
----0.948683298050514<=3
2=>0.964027580075817
----0.970142500145332<=4
----0.98058067569092<=5
----0.986393923832144<=6
----0.989949493661166<=7
----0.992277876713668<=8
----0.993883734673619<=9
3=>0.995054753686730
4=>0.999329299739067
5=>0.999909204262595
6=>0.999987711650796
7=>0.999998336943945
8=>0.999999774929676
9=>0.999999969540041
Then it seems to me now that between the first and second anniversary celebrated on the ship, the Earth frame celebrates anniversaries two and three, and between the second and third anniversary on the ship, the Earth celebrates anniversaries four through nine. Roughly speaking, when the ship is approaching it's proper third anniversary, Earth is celebrating nine years since ship departure.
Is this correct?
DW has informed me that this is now correct.
I am less than satisfied with my method, since it seems to me I should be able to input a value for t and derive tau without a lookup table, or vice versa. However, for the sake of my own understanding, I will expand the table to include values up to 18 for Tau, since after 18 Karl's calculator returns a value of unity, due to rounding in the last decimal places. Of course the real value is not unity but only approaches unity.
Here is the revised table:
[tex]\tau[/tex]=>V<=[tex]t[/tex]
----0.707106781186547<=1
1=>0.761594155955765
----0.894427190999916<= 2
----0.948683298050514<=3
2=>0.964027580075817
----0.970142500145332<=4
----0.98058067569092<=5
----0.986393923832144<=6
----0.989949493661166<=7
----0.992277876713668<=8
----0.993883734673619<=9
3=>0.995054753686730
4=>0.999329299739067
5=>0.999909204262595
6=>0.999987711650796
7=>0.999998336943945
8=>0.999999774929676
9=>0.999999969540041
10=>0.999999995877693
11=>0.999999999442106
12=>0.999999999924497
13=>0.999999999989782
14=>0.999999999998617
15=>0.999999999999813
16=>0.999999999999975
17=>0.999999999999997
18=>1.0
Then I can input t values until I find where V for t becomes unitary to the available number of decimals.
I find that when t reaches about 10,000,000, V is 0.999999999999995, so I conclude that when about seventeen years have passed on board the ship, the Earth has gone around the sun about ten million times. In other words, a ship accelerating at one gravity for seventeen years with a kicker engine placed to rotate the ship two pi in seventeen years will return to Earth space ten million years later.
At this point I can only wonder where a ship might end up if it went on for the seventy years of a human lifetime. However I think it is already evident that such a journey would be on the order of the lifetime of the Earth itself, or perhaps even of the galaxy. Or of the Universe?
THIS SPACE UNDER CONSTRUCTION. Comments and corrections requested. Thanks, Richard
The basic problem is that one of a set of twins leaves Earth to travel at near-light speed, encounters time dilation due to relativity, and returns to find the other twin has experienced more time and is now "older." I wish to discover how to calculate this problem under variable conditions of acceleration. I have never worked through this problem before and am not sure of what mathematics is required, but hope to gather the information to solve the problem and present it here. Any helpful comments are solicited.
I will specify flat space and two engines, one, the thruster, providing forward acceleration, and the other, the "kicker," providing a small lateral thrust so that the ship moves in a very large circle.
The ship leaves Earth, travels in a circle, then returns to pass by near-Earth space at a high velocity.
For simplicity, I will assume that space is flat and that the entire trip can be accomplished without correction for extraneous mass or charge.
I will further assume that the thrusters are not limited by fuel considerations, but can continue to provide thrust within their frame of reference at a constant rate for any necessary length of time.
The problem now is to find the correct formulas, and then to determine how to calculate them.
I have the following formulae from DW, who tried to help me with a similar problem in another thread at
https://www.physicsforums.com/showthread.php?t=15826
[tex]v = ctanh(\frac{\alpha \tau}{c})[/tex]
[tex]v = \frac{\alpha t}{\sqrt{1 + \frac{\alpha ^{2}t^{2}}{c^2}}}[/tex]
where:
[tex]v[/tex] = velocity as observed from the rest frame
[tex]c[/tex] = speed of light
[tex]\alpha[/tex] = proper acceleration (as experienced by accelerated frame
[tex]\tau [/tex] = time elapsed in accelerated frame
[tex]t[/tex] = time elapsed in rest frame
and for convenience, the term
[tex](\frac{\alpha \tau}{c}) = \theta[/tex]
DW says [tex]\theta[/tex] is known as rapidity.
I notice that [tex]\theta[/tex] is a dimensionless quantity, with terms for Distance divided by Time above and below the fraction, so they cancel.
I see that in the rest frame formula the term subtracted from [tex]1[/tex] in the square root denominator (bottom part of the fraction) is, in form, very like the term [tex]\theta^2[/tex]. I am curious about this similarity, and wonder if it would help my understanding to think of the phrase
[tex]\sqrt{1 + \frac{\alpha^2 t^2}{c^2}}[/tex]
as Pythagorean, thus following the formula
[tex]A=\sqrt{B^2+C^2}[/tex]
where A is the hypoteneus of a right triangle in which B is equal to one and C is [tex]\frac{\alpha t}{c}[/tex]. Note capital C has no relation to small c.
I notice that the frame of reference which is the rest frame is dependent on the motion of the observer. (Aside, has anyone counted up the ways we on Earth are moving through sidereal space? Sidereal means with reference to the fixed stars, and of course there aren't any. But if we chose to use the Z-10 galaxy (did I get the term right? read about it on pf somewhere, will try to find source) which is the oldest known galaxy as the reference frame, then how many 'proper' motions? )
Setting the acceleration at one gravity reduces the formulae to:
[tex]v = ctanh(\frac{\tau}{c})[/tex]
[tex]v = \frac{t}{\sqrt{1 + \frac{t^{2}}{c^2}}}[/tex]
c might also be taken as unitary, yielding the formulae as
[tex]v = tanh\tau}[/tex]
[tex]v = \frac{t}{\sqrt{1 + t^{2}}}[/tex]
then
[tex]tanh\tau}=v = \frac{t}{\sqrt{1 + t^{2}}}[/tex]
Is this formula useful in calculating the twinship problem? I place it here for contemplation, as it seems to me to reveal something about the nature of time and space.
What is [tex]\tau[/tex]when t is one? [tex]tanh\tau}=\sqrt{2}^{-1}[/tex]
What is t when Tau is one? [tex]tanh1=\frac{t}{\sqrt{1 + t^{2}}}[/tex]
I don't think my hand calculator does hyperbolic tangents. However there is a graph of the hyperbolic tangent on Wikipedia at http://en.wikipedia.org/wiki/Hyperbolic_tangent
and I can get an idea of the behavior of the function from reading the graph. Also, there is a calculator on the web at http://www.karlscalculus.org/cgi-bin/newcalc.pl which does hyperbolic calculations.
Then I should be able to build a table showing the time dilation expected to be observed by the Earth twin in the rest frame and experienced by the twinship as it passes near Earth space as a function of t, time elapsed on earth. The time of passage could be varied by adjusting the circumfrence of the circle due to the kicker engine. I should be able to calculate a view of the ship as if it were passing near Earth at any time t. For example, if the kicker were set so that the ship would pass Earth after a t of one year, how fast would the ship be traveling? How much time would have elapsed on board the ship? And if t were two years, three years, and so on.
(calculations made with Karl's Calculator, see link in text above)
[tex]\tau=>tanh(\tau)=>t[/tex]
1=>0.761594155955765=>0.605886858731034
2=>0.964027580075817=>
3=>0.995054753686730=>
4=>0.999329299739067=>
5=>0.999909204262595=>
6=>0.999987711650796=>
7=>0.999998336943945=>
8=>0.999999774929676=>
9=>0.999999969540041=>
I am having trouble seeing the meaning. When tau, the elapsed time on board the ship, is one year, the velocity of the ship is [tex]tanh(\tau)[/tex]=.76..., and the elapsed time on Earth is [tex]t[/tex]=.60...
This seems wrong. The ship twin experiences one year, the Earth twin experiences .6 year? Then the ship twin returns older than the Earth twin? I thought it was the other way around, and the ship twin should return younger than the Earth twin.
DW informs me that the correct t should be 1.2. I notice this is a factor of two times the .60... in my calculation. Perhaps this is only a coincidence. More likely I simply do not know how to work Karl's calculator, even though I managed to get the numbers for [tex]\tau[/tex] out of it correctly. I will return to Karl's calculator and try again. If I have no luck I will search for other calculators that might be easier for me to learn to use.
No luck searching for other calculators. Most search results are adverts for stuff I can't afford to buy right now.
Here is a table of V values for variable t:
(<= notation here means t at right put into formula for V shown on left)
[tex]\frac{t}{\sqrt(1+t^2)}[/tex]<=t
0.707106781186547<=1
0.894427190999916<= 2
0.948683298050514<=3
0.970142500145332<=4
0.98058067569092<=5
0.986393923832144<=6
0.989949493661166<=7
0.992277876713668<=8
0.993883734673619<=9
Then I can merge the two tables by intercollating the values for V:
[tex]\tau[/tex]=>V<=[tex]t[/tex]
----0.707106781186547<=1
1=>0.761594155955765
----0.894427190999916<= 2
----0.948683298050514<=3
2=>0.964027580075817
----0.970142500145332<=4
----0.98058067569092<=5
----0.986393923832144<=6
----0.989949493661166<=7
----0.992277876713668<=8
----0.993883734673619<=9
3=>0.995054753686730
4=>0.999329299739067
5=>0.999909204262595
6=>0.999987711650796
7=>0.999998336943945
8=>0.999999774929676
9=>0.999999969540041
Then it seems to me now that between the first and second anniversary celebrated on the ship, the Earth frame celebrates anniversaries two and three, and between the second and third anniversary on the ship, the Earth celebrates anniversaries four through nine. Roughly speaking, when the ship is approaching it's proper third anniversary, Earth is celebrating nine years since ship departure.
Is this correct?
DW has informed me that this is now correct.
I am less than satisfied with my method, since it seems to me I should be able to input a value for t and derive tau without a lookup table, or vice versa. However, for the sake of my own understanding, I will expand the table to include values up to 18 for Tau, since after 18 Karl's calculator returns a value of unity, due to rounding in the last decimal places. Of course the real value is not unity but only approaches unity.
Here is the revised table:
[tex]\tau[/tex]=>V<=[tex]t[/tex]
----0.707106781186547<=1
1=>0.761594155955765
----0.894427190999916<= 2
----0.948683298050514<=3
2=>0.964027580075817
----0.970142500145332<=4
----0.98058067569092<=5
----0.986393923832144<=6
----0.989949493661166<=7
----0.992277876713668<=8
----0.993883734673619<=9
3=>0.995054753686730
4=>0.999329299739067
5=>0.999909204262595
6=>0.999987711650796
7=>0.999998336943945
8=>0.999999774929676
9=>0.999999969540041
10=>0.999999995877693
11=>0.999999999442106
12=>0.999999999924497
13=>0.999999999989782
14=>0.999999999998617
15=>0.999999999999813
16=>0.999999999999975
17=>0.999999999999997
18=>1.0
Then I can input t values until I find where V for t becomes unitary to the available number of decimals.
I find that when t reaches about 10,000,000, V is 0.999999999999995, so I conclude that when about seventeen years have passed on board the ship, the Earth has gone around the sun about ten million times. In other words, a ship accelerating at one gravity for seventeen years with a kicker engine placed to rotate the ship two pi in seventeen years will return to Earth space ten million years later.
At this point I can only wonder where a ship might end up if it went on for the seventy years of a human lifetime. However I think it is already evident that such a journey would be on the order of the lifetime of the Earth itself, or perhaps even of the galaxy. Or of the Universe?
THIS SPACE UNDER CONSTRUCTION. Comments and corrections requested. Thanks, Richard
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