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Twinship, an attempt to understand SR

by rtharbaugh1
Tags: attempt, twinship
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rtharbaugh1
#1
Apr11-04, 01:56 PM
P: 310
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

http://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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DW
#2
Apr12-04, 10:42 AM
P: 328
The equations I gave were for the case that the ship was speeding up under the conditions of constant proper acceleration. Then [tex]t = \tau [/tex] is only the case at t = 0 which is when the ship is instentaneously at rest. The case for arbitrarily proper time dependent accelerations in arbitrary directions is covered by transformation equations 5.4.4 at http://www.geocities.com/zcphysicsms/chap5.htm#BM65. This included the case of circular motion for the prime frame, but not spin. Otherwise the directions and accelerations for the primed frame are completely arbitrary. If you want to add spin to the accelerated frame then You have to a second transformation. Are you wanting the ship to speed up as it travels the circle or to stay at constant speed?
rtharbaugh1
#3
Apr12-04, 12:28 PM
P: 310
Thanks for being patient with me, DW. I am trying to start with the simplest possible case and then if I can understand that to proceed to more interesting ideas. So I am thinking that spin is very minimal in the description so far. I suppose the ship is traveling in a very large circle in flat space, so one might say it completes one turn of spin in the plane of the circle in one turn around the circle. I assume this is negligible? My reason for the circle is that I would like the twin on Earth to observe the twin on the ship in near space rather than at a great distance.

I suppose the acceleration due to the thrusters to be very large compared to the acceleration given by the kicker. In flat space, I think, the ratio of the two accelerations would determine the size of the circle and the length of time for the ship to complete it.

I am assuming a constant acceleration of the ship due mostly to the thruster, which I will set for now at one G. So the ship is still accelerating as it makes its pass by Earth.

I looked at the link you gave but the maths are beyond my current limit. I think with some study I might be able to read them.

By the way, can I set v=v in the equations, so that


[tex]ctanh(\frac{\alpha \tau}{c}) = \frac{\alpha t}{\sqrt{1 + \frac{\alpha ^{2}t^{2}}{c^2}}}[/tex] ?

Or would it be better to write [tex]V_\tau=ctanh(\frac{\alpha \tau}{c}) [/tex] and [tex]V_t=\frac{\alpha t}{\sqrt{1 + \frac{\alpha ^{2}t^{2}}{c^2}}}[/tex]

Or even [tex]V_\tau=ctanh(\theta_\tau) [/tex] and [tex]V_t=\frac{\alpha t}{\sqrt {1 + \theta_t^2}}[/tex]

I have made some corrections to the first page in this thread based on this discussion. Thanks, Richard

DW
#4
Apr12-04, 10:13 PM
P: 328
Twinship, an attempt to understand SR

Quote Quote by rtharbaugh1
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

http://www.physicsforums.com/showthread.php?t=15826


[tex]v = ctanh(\frac{\alpha \tau}{c})[/tex]in the ship


[tex]v = \frac{\alpha t}{\sqrt{1 + \frac{\alpha ^{2}t^{2}}{c^2}}}[/tex] from the Earth

where:

[tex]v[/tex] = velocity

[tex]c[/tex] = speed of light

[tex]\alpha[/tex] = proper acceleration (as experienced by accelerated frame). For the purpose of this simplest problem, I set the proper acceleration at one gravity.

[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 velocity 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 the square of theta, if only t, the rest frame time, is in the place where [tex]\tau[/tex] occurs in the formula for the accelerated frame. I assume now that I can take the square of the rapidity theta ([tex]\theta[/tex] in the rest frame, since at rest [tex]t=\tau[/tex].

The two formulae can then be written as follows:


[tex]v = ctanh(\theta)[/tex] in the ship (accelerated frame of reference)


[tex]v = \frac{\alpha t}{\sqrt{1 + \theta^2}}[/tex] from the Earth (rest frame of reference).
You can't make that replacement in the last equation. The rapidity is defined from a claculation from the inertial frame, not the proper frame. For any time other than 0 it is not a valid replacement and the times are not equal.


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?

I notice that the term [tex]\sqrt{1+\theta^2}[/tex] has the Pythagorean form [tex]z={\sqrt{a^2 + b^2}}[/tex] where I have written z instead of c to avoid confusion with c= speed of light. This leads me to wonder if there is a connection here with space-time geometry. In a dimensionless space with a base length (adjacent side) taken as one and a height taken to be [tex]\theta[/tex], the hypoteneus z varies from a value of one when [tex]\theta=0[/tex] to a value of [tex]\sqrt {2}[/tex] when [tex]\theta=1[/tex].
Again you can't make that replacement, and also the rapidity is not restricted to values between 0 and 1. It can be allowed to be any real number.

Setting the acceleration at one gravity reduces the formulae to:


[tex]v = ctanh(\frac{\tau}{c})[/tex]in the ship


[tex]v = \frac{t}{\sqrt{1 + \frac{t^{2}}{c^2}}}[/tex] from the Earth


THIS SPACE UNDER CONSTRUCTION. Comments and corrections requested. Thanks, Richard
You might as well use c = 1 too if you are going to do that. Also I don't know what you mean by "in the ship" etc. They are both the same v. They are both the speed of the ship as reckoned from the inertial earth frame.
rtharbaugh1
#5
Apr13-04, 02:16 AM
P: 310
Thanks.
You said:
"the rapidity is not restricted to values between 0 and 1. It can be allowed to be any real number."

and you said:
"I don't know what you mean by "in the ship" etc. They are both the same v. They are both the speed of the ship as reckoned from the inertial earth frame"

I notated the formula as "in the ship" and etc. to help me remember which was which. I now think I know that tau is the primed frame and t is the rest frame, but I wasn't clear on that before.

Now about the rapidity. Alpha t c^-1 appears to me to be a dimensionless ratio, as alpha t is acceleration multiplied by time, giving a velocity, and c is a velocity, so the dimensions cancel. I assume the rule about no travel faster than light is allowed, so alpha t must always be less than c? Then the ratio, theta, varies between zero and one?

I am using alpha = 1 g for a couple of divergent reasons. First, if you were on a ship it would be a comfortable acceleration to maintain. Second, as the acceleration due to gravity, I imagine it will give me a peek at how all this relates through the equivalence principle to GR. Also, it is rather convenient to be able to ignore alpha when trying to get a grip on what all this formulation "really" means. I suppose if you are a pure mathematician then you will not be sympathetic to my percieved need to make pictures in my head, but for my part it seems to help me. So if I do go on to make c = 1 I imagine I may gain some insight about how the other dimensions of the problem relate to each other.

So V is V and no notation is needed to keep the V related to tau apart from the V related to t, and so I can set one equation equal to the other, so

[tex]ctanh\theta=v=\frac{\alpha t}{\sqrt
(1-\frac{\alpha^2 t^2}{c^2})}[/tex]

Is this correct?

Thanks for your help, DW, I feel like I may be making some progress.

Richard.
DW
#6
Apr13-04, 04:14 AM
P: 328
Quote Quote by rtharbaugh1
I now think I know that tau is the primed frame and t is the rest frame, but I wasn't clear on that before.
tau is your accelerated ship's wristwatch time and t is time for the inertial frame.

I assume the rule about no travel faster than light is allowed,
Right.

so alpha t must always be less than c?
No. v is not alpha t. v is
[tex]v = \frac{\alpha t}{\sqrt{1 + \frac{\alpha^2 t^2}{c^2}}}[/tex]

Then the ratio, theta, varies between zero and one?
No, I said it has no such restriction.

So V is V and no notation is needed to keep the V related to tau apart from the V related to t, and so I can set one equation equal to the other, so

[tex]ctanh\theta=v=\frac{\alpha t}{\sqrt
(1-\frac{\alpha^2 t^2}{c^2})}[/tex]

Is this correct?
The sign in the denominator is wrong, but otherwise, yes.
rtharbaugh1
#7
Apr14-04, 12:26 AM
P: 310
Thank you. I will continue to make changes. BTW, I followed your link for the formula reference, but couldn't see how the mathematics there relates to the mathematics here. When I get some time I will go back and try to figure out what all those betas and gammas mean.

Richard
DW
#8
Apr14-04, 10:11 AM
P: 328
Quote Quote by rtharbaugh1
Thank you. I will continue to make changes. BTW, I followed your link for the formula reference, but couldn't see how the mathematics there relates to the mathematics here. When I get some time I will go back and try to figure out what all those betas and gammas mean.

Richard
beta in that equation is the coordinate velocity of the accelerated observer as a function of his proper time divided by c. gamma in that equation is also a function of proper time as it was composed of beta by
[tex]\gamma = \frac{1}{\sqrt{1 - \beta ^2}}[/tex]
In your case
[tex]\gamma = cosh(\frac{\alpha \tau}{c})[/tex]
rtharbaugh1
#9
Apr15-04, 01:18 PM
P: 310
Thanks, DW

I have spent the past couple days of computer time looking at imaginary numbers, quaternions, hyperbolic funtions, Euler identity, etc. and it seems I have tried to assimilate so many new concepts that I have become totally lost again. The math portion of my brain seems to be shutting down in protest.

It doesn't help any that my browser doesnt support some of the symbols used in Wikipedia. I get lots of little square boxes in places that I am sure must be intended as sqrt signs. And in other places that I don't have any idea what they mean. In one area, I can't find the reference now, was it something to do with quaternions, I saw a square box and thought for a while anyway that the author really meant to render a square box as the math symbol. It was a square box with a exponent of 2, and I thought they were using it something like the Laplacian "L" or the Hamiltonian "H," but I may have been confused.

Anyway my sense right now is that I need to go for a long walk in this lovely spring weather. I think I am going to have to get a better grasp of imaginary numbers, the exponential function, and other math ideas before I can make any further progress in relitivity. Little increments.

Meanwhile, thanks for your time and thoughts. I hope to be able to send Greg B. some cash soon, so I can be a contributing member of the forum.

Thanks for Being,

Richard
rtharbaugh1
#10
May1-04, 03:39 PM
P: 310
Quote Quote by rtharbaugh1
Space under construction: The twin ship problem (extract from page one in this thread)

The problem now is to find the correct formulas, and then to determine how to calculate them.

I have the following formulae from DW

[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.

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]

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]

...there is a calculator on the web at http://www.karlscalculus.org/cgi-bin/newcalc.pl which does hyperbolic calculations.

[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.


THIS SPACE UNDER CONSTRUCTION. Comments and corrections requested. Thanks, Richard

Perhaps DW or someone else who understands this stuff can help me out here?


Thanks,

Richard
DW
#11
May1-04, 11:19 PM
P: 328
Quote Quote by rtharbaugh1
Space under construction: The twin ship problem

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.
Thats because that is wrong. When [tex]\tau[/tex] is 1 the velocity is 0.76, but the elapsed time on earth t is not 0.60. It is 1.2.
rtharbaugh1
#12
May4-04, 03:16 PM
P: 310
Quote Quote by rtharbaugh1
Space under construction: The twin ship problem
(exerpt from page one this thread)
...
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.


THIS SPACE UNDER CONSTRUCTION. Comments and corrections requested. Thanks, Richard
. Is this correct?

I notice that a value of 1.2 entered into the formula for t results in a value very nearly .76, as DW said it would.

Thanks,
DW
#13
May4-04, 06:46 PM
P: 328
Quote Quote by rtharbaugh1
. Is this correct?

I notice that a value of 1.2 entered into the formula for t results in a value very nearly .76, as DW said it would.

Thanks,
Yes it is correct now.
rtharbaugh1
#14
May15-04, 12:02 AM
P: 310
DW, as I recall, you said that there would be problems with forces at near light speeds. I think I see one problem with the centripedal forces. At near light speeds, the kicker engine will have to kick much harder to achieve the course correction needed to keep the ship in a circular path. Is this what you meant? Also, in practical terms, the ship would not be able to change course to avoid any potential obstacles. In fact, it seems to me now that any motion with a lateral component on board the ship would require extravagant forces. I have always assumed in the past that these forces would be leveled by being spread out over the longer proper time, but now I wonder if that is right. Perhaps you could enlighten me.

Also, it has occurred to me that the ship could reverse thrust at the midpoint of the trip and decelerate back to the rest frame as it approaches Earth again. I was taught in physics that there is no difference between acceleration and deceleration, so is that true also in SR? Would the ship's time still experience the same effect?

Thanks for your help.
DW
#15
May15-04, 11:15 AM
P: 328
Quote Quote by rtharbaugh1
DW, as I recall, you said that there would be problems with forces at near light speeds. I think I see one problem with the centripedal forces. At near light speeds, the kicker engine will have to kick much harder to achieve the course correction needed to keep the ship in a circular path. Is this what you meant? Also, in practical terms, the ship would not be able to change course to avoid any potential obstacles. In fact, it seems to me now that any motion with a lateral component on board the ship would require extravagant forces. I have always assumed in the past that these forces would be leveled by being spread out over the longer proper time, but now I wonder if that is right. Perhaps you could enlighten me.

Also, it has occurred to me that the ship could reverse thrust at the midpoint of the trip and decelerate back to the rest frame as it approaches Earth again. I was taught in physics that there is no difference between acceleration and deceleration, so is that true also in SR? Would the ship's time still experience the same effect?

Thanks for your help.
I'm not sure what it is you are referring to that I said. Direct quotes with reference would be helpful. But I will say that the force provided by the kicker does become significant. The force it must provide according to the ship frame is
[tex]F'^{r} = mA'^{r} = \gamma ^{2}ma^{r} = - \gamma ^{2}m\frac{v^{2}}{r}[/tex].
The force that the thruster must provide according to the ship frame is
[tex]F'^{tan} = mA'^{tan} = \gamma^{3}ma^{tan} = \gamma ^{3}m\frac{dv}{dt}[/tex].
According to other frames, the force is different, but fortunately you have the ship providing the thrust so these ship frame values are all that matter.
rtharbaugh1
#16
May17-04, 02:37 PM
P: 310
Thanks, DW

The superscript on the first F is not easy to read on my screen. Is it ir, the imaginary radius? I see the same symbol, is it ir, superscript on A, but I am not sure what A is. Then there is Euler's gamma, of which I have only a vague notion, but am trying to study.

In the second formula, F is raised to i tan? Also A to i tan? And of what angle does one take the tangent?

Thank you,

R
DW
#17
May17-04, 06:35 PM
P: 328
They aren't exponents they are superscripts. Its not an i, but a prime to indicate the ship's frame. r means radial component, tan means tangential component. A is four vector acceleration, so for example [tex]A'^{tan}[/tex] means the tangential component of four-vector acceleration according to the ship frame. [tex]\gamma[/tex] is just the special relativistic quantity
[tex]\gamma = \frac{1}{\sqrt{1 - \frac{v^{2}}{c^{2}}}}[/tex].


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