Twin Paradox: Extended

[Atashi]

Twin Paradox:

Lets assume infinite space. Each twin is on a platform situated side by side to each other.

In the original paradox, one twin(b) is propelled at near-lighspeed away from the other twin(a).

As twin(a) you are stationary, and 50years has passed, twin(b) returns and has only aged 25 years.

 

But when in reference you can also claim that:

twin(b) was stationary, and twin(a) had been the one that left and return.

as twin(b) you will have experienced waited 50years for twin(a)'s return.

 

How can this be?

In one world we have twin(a) aging twice as much at the time of reunion

and in the other we have the opposite, though both events are one and the same.

Perception determines reality exclusive only to the observer.

 

 

but at the same time, as twin(a), you could state that twin(b) had been stationary.

and upon your arrival after 25years, twin(b) would have aged 50.

twin(a) would see an older twin(b) at 25years, and a younger twin(b) at 50years...

Or would only 1 Reunion take place?

 

If both had happened, would there be two differently aged twin(b)'s simultaneously?

 

if this is possible, then twin(a) could also claim that both twin(a) and twin(b) were traveling at 1/2 the speed of light away from each other, bringing both of them back to the original location and both having aged the same amount.

and any variation of that would be possible (90%vs10%, 80%vs20%...),

and result in an infinite amount of twin(b)'s being produced.

 

 

 

 

 

 

Let's step this experiment up.

Now lets account for the 3rd observer twin©, or rather triplet, but lets not stress the details.

 

 

 

twin© is put between twin(a) and twin(b)

both twins will propel themselves from twin© at near-lightspeed, and return.

 

 

 

twin© will will have aged 50years while twin(a) and twin(b) will return at the same time, both having aged 25years.

 

 

What will this look like in the perspective of twin(a)?

 

 

 

it would be impossible for twin(b) to have been going twice the speed of light.

so would twin(b) and twin© both be traveling along side each other?

If thats the case then would both twin(b) and twin© return having aged 25years during 50years of twin(a)'s time?

Or would twin© have aged 25years while twin(b) having only aged 12?

Edited September 9, 2011 by Atashi

[Schrödinger's hat]

I whipped up a quick demo for this here -- bear in mind it's still rather alpha so the interface is rather clunky, if it doesn't work on your browser or acts funky I'd like to know so I can fix it.

Click on twins' paradox, un-pause it and watch the demos, should explain better than words, especially if you look at the space-time diagram below.

 

 

Twin Paradox:

Lets assume infinite space. Each twin is on a platform situated side by side to each other.

In the original paradox, one twin(b) is propelled at near-lighspeed away from the other twin(a).

As twin(a) you are stationary, and 50years has passed, twin(b) returns and has only aged 25 years.

 

But when in reference you can also claim that:

twin(b) was stationary, and twin(a) had been the one that left and return.

as twin(b) you will have experienced waited 50years for twin(a)'s return.

 

How can this be?

In one world we have twin(a) aging twice as much at the time of reunion

and in the other we have the opposite, though both events are one and the same.

Perception determines reality exclusive only to the observer.

They will both agree that the travelling twin is the younger one when he gets back.

The key is that the stationary twin did not accelerate.

When the travelling twin went from moving at 0.866c (the speed to age half as much) to -0.866 c, the event he considered 'now' changed dramatically.

When he was travelling away, twin a was aging younger than him, and had only aged 12.5 years at the point twin b decided to turn around.

Then when twin b turned around, the twin a that he considered to be in his present was suddenly 12.5 years older than him. The twin a that is 12.5 years younger than him is suddenly 25 years in his past.

 

but at the same time, as twin(a), you could state that twin(b) had been stationary.

and upon your arrival after 25years, twin(b) would have aged 50.

twin(a) would see an older twin(b) at 25years, and a younger twin(b) at 50years...

Or would only 1 Reunion take place?

 

If both had happened, would there be two differently aged twin(b)'s simultaneously?

 

if this is possible, then twin(a) could also claim that both twin(a) and twin(b) were traveling at 1/2 the speed of light away from each other, bringing both of them back to the original location and both having aged the same amount.

and any variation of that would be possible (90%vs10%, 80%vs20%...),

and result in an infinite amount of twin(b)'s being produced.

 

 

 

 

 

 

Let's step this experiment up.

Now lets account for the 3rd observer twin©, or rather triplet, but lets not stress the details.

 

 

 

twin© is put between twin(a) and twin(b)

both twins will propel themselves from twin© at near-lightspeed, and return.

 

 

 

twin© will will have aged 50years while twin(a) and twin(b) will return at the same time, both having aged 25years.

 

 

What will this look like in the perspective of twin(a)?

 

 

 

it would be impossible for twin(b) to have been going twice the speed of light.

so would twin(b) and twin© both be traveling along side each other?

If thats the case then would both twin(b) and twin© return having aged 25years during 50years of twin(a)'s time?

Or would twin© have aged 25years while twin(b) having only aged 12?

Velocity doesn't add linearly in a relativistic world. If twin b and c both leave a at 0.866c, then from twin b's point of view a will be moving at 0.866c and aging at half the rate. twin c will be moving closer to 0.99c and aging at 1/7th the rate.

Edited September 9, 2011 by Schrödinger's hat

[mathematic]

There is no symmetry. Twin(a) is always in one inertial frame. Twin (b) starts in the same frame as twin (a), then changes to a frame going away from (a), next changes to a frame going back to (a), and finally stopping in the frame of (a). Go through the Lorentz transforms for distance and time for (b) and get the expected result.

Edited September 9, 2011 by mathematic