The twins example of the special theory of relativity

[rabs]

I have a question concerning the story of the twins where one stays on earth and the other travels in space ship through space at nearly the speed of light. When the twin on the space ship returns to earth he would be younger than his twin who stayed on earth since traveling at near the speed of light will cause time to slow down.

 

My question is wouldn't they both be the same age? Since relativity says motion is relative to the observer and to the people on earth the twin on the space ship would be in a slower time, wouldn't it be that to the twin on space ship, the earth would be traveling at near the speed of light and therefore those on earth would be in a slower time?

 

Or would it be a matter of perspective? The twin on the earth would think the twin on the ship was younger and the twin on the ship would think the other was younger?

[pgrmdave]

If I understand it correctly, it is because while the motion appears relative, the acceleration only occurs to the space-ship twin. However, I'm not really a physics geek, so you'll have to wait for a more qualified person for a better answer.

[freeztar]

It's known as the Twin Paradox. The wiki for this is quite good! :)

[Jimoin]

Gah, stupid postcount.

 

YouTube - Time Travel for Pre Schoolers http://www.youtube.com/watch?v=PmV664VvpW0

[Boerseun]

I understand your point.

 

Two people are in different frames of reference, equally valid, why does the one age and the other not?

 

Simple layman's answer is that they both started from the same frame of reference, and only one accellerated relative to that particular frame.

 

The older twin never acellerated relative to the original frame, the younger one did.

[modest]

My question is wouldn't they both be the same age? Since relativity says motion is relative to the observer and to the people on earth the twin on the space ship would be in a slower time, wouldn't it be that to the twin on space ship, the earth would be traveling at near the speed of light and therefore those on earth would be in a slower time?

 

I've found this is a pretty common question. If you combine the following two assumptions then the twin paradox seems puzzling for the reasons you give.

 

1. Velocity time dilation is reciprocal. Each moving person sees the other person's clock run slow.
   
2. There is no preferred reference frame (ie there is no "at rest")
   

 

But, this situation is not entirely symmetric. As Pgrmdave notes, one person changes speed and feels acceleration while the other does not. There is also another factor that explains well why there is a net dilation effect.

 

....3. Each person sees the other reference frame as length contracted.

 

Let's say Tom and Flash are twins. Flash is an astronaut. Tom stays on Earth while Flash makes a round-trip to a star 10 lightyears away in a rocket. He travels at eight-tenths the speed of light (.8c) from Tom's perspective and Tom travel's at eight-tenths light speed from Flash's perspective.

 

Tom calculates how long the trip will take from his perspective:

[math]Tom's \; time = distance/velocity[/math]

The distance Tom sees is 20 total (roundtrip) lightyears and velocity is .8c.

[math]T_{(Tom)} = 20 / .8 = 25 \ years [/math]

Tom can calculate how long the trip took for Flash using the equation for

time dilation

, [math]T_{(Flash)} = T_{(Tom)} \sqrt{1-(v/c)^2}[/math],

[math]T_{(Flash)} = 25 \sqrt{1-(.8/1)^2} = 15 \ years[/math]

Tom ages 25 years while Flash only ages 15.

 

Flash calculates things from his perspective:

[math]Flash's \; time = distance/velocity[/math]

The earth and distant star appear to move at .8c for Flash. The distance between the earth and star appear

length contracted

by, [math]D_{(Flash)} = D_{(Tom)} \sqrt{1-(v/c)^2}[/math]

[math]D_{(Flash)} = 20 \sqrt{1-(.8/1)^2} = 12 \ lightyears[/math]

Plugging in these values gives,

[math]T_{(Flash)} = 12 / .8 = 15 \ years [/math]

Flash ages 15 years. This is the same as we found above, only calculated from Flash's perspective this time.

 

This shows how both twins can see the other's clock run slow yet have a net dilation effect when they reunite. The situation is not completely symmetric because the planet and star are motionless relative to the twin that stays on earth, but they move relative to the astronaut. Because of length contraction, this distance is dilated for one but not the other.

 

Please ask if you're unsure about anything.

 

~modest

[Erasmus00]

Please ask if you're unsure about anything.

 

Calculate Tom's time from Flash's perspective :)

-Will

[modest]

Calculate Tom's time from Flash's perspective :)

-Will

 

:eek:

 

Ok, let's see... Feel like I'm walking into a minefield...

 

[math]T_{(flash)} = 15 \ yrs[/math]

[math]D_{(flash)} = 12 \ lyrs[/math]

 

Tom's distance:

[math]D_{(Tom)} = \frac {D_{(Flash)}}{ \sqrt{1-(.8/1)^2}} = 20 \ lyrs[/math]

Tom's time:

[math]T_{(Tom)} = \frac {T_{(Flash)}}{ \sqrt{1-(.8/1)^2}} = 25 \ yrs[/math]

Tom's speed:

[math]\frac{D}{T} = \frac{20}{25} = .8c[/math]

 

Seems ok, what am I missing?

 

~modest

[Janus]

:eek:

 

Ok, let's see... Feel like I'm walking into a minefield...

 

[math]T_{(flash)} = 15 \ yrs[/math]

[math]D_{(flash)} = 12 \ lyrs[/math]

 

Tom's distance:

[math]D_{(Tom)} = \frac {D_{(Flash)}}{ \sqrt{1-(.8/1)^2}} = 20 \ lyrs[/math]

Tom's time:

[math]T_{(Tom)} = \frac {T_{(Flash)}}{ \sqrt{1-(.8/1)^2}} = 25 \ yrs[/math]

Tom's speed:

[math]\frac{D}{T} = \frac{20}{25} = .8c[/math]

 

Seems ok, what am I missing?

 

~modest

 

What your missing is that have to use the same form on the time dilation equation to find Tom's elapsed time during the two legs of the trip as perceived by Flash, as you used to find Flash's time as perceived by Tom or

 

Tom's time:

[math]T_{(Tom)} = T_{(Flash)} \sqrt{1-(.8/1)^2} = 9 \ yrs[/math]

 

How does this jive with the fact that When Flash returns he finds that Tom has aged 25 years?

 

For this we have to take into account that flash had to stop and reverse course in order to return to Tom. And in doing so, he switches frames of references. (from one going away from Tom at .8c to one heading towards Tom.) So we have to take Relativity of Simultaneity into consideration.

 

For instance, let's assume that there is a clock sitting at the point where Flash turns around, and it is synchronized to Tom's clock according to Tom.

 

According Flash on his outbound leg of the trip, that clock will always be 8 yrs ahead of Tom's clock (when Tom's clock reads 0, the other clock read 8 yrs, when Tom's Clock reads 1 yr, the other Clock reads 9 yrs, etc. This is according the formula:

 

[math]\Delta T = \frac{xv}{c^2 \sqrt{1-\frac{v^2}{c^2}}}[/math]

 

{x is the distance between the two clock as measured by Flash)

 

According to time dilation, both Tom's clock and the Turn around point clock will accumulate 4.5 yrs on the outbound trip, meaning the turn around clock will read 12.5 yrs upon Flash's arrival.

 

Flash turns around. He is now heading back towards Tom. This means that Tom's clock must now, according to Relativity of Simultaneity, be 8 yrs ahead of the turn around point clock(The clock Flash is next to at this point). So from Flash's perspective, Tom's clock jumps forward by 16 yrs when Flash turns around, and as he leaves the turn around point now reads 20.5 yrs elapsed time.

 

During the return trip, Tom's Clock accumulates 4.5 yrs just like it did on the outbound trip, having it(and Tom} accumulate a total of 25 yrs during Flash's trip.

[modest]

Very cool Janus. We are having some fun now (or at least I am, I love this kind of thing) :)

 

So, I’m guessing it obviously can’t be coincidence that your method (from flash’s perspective) and my method (inverse of Tom’s perspective) both agree. So, I’ve set out to reduce what you’ve done into what I’ve done…

 

You are saying that this:

[math]T_{(Tom)} = T_{(Flash)} \sqrt{1-(.8/1)^2} = 9 \ yrs[/math]

plus this:

[math]\Delta T = \frac{xv}{c^2 \sqrt{1-\frac{v^2}{c^2}}} [/math]

Is the final answer. I agree with this and I see why it is conceptually true. We might write this in one equation and note that x is the total distance (round trip) and the answer is the total time for tom (round trip). We could also substitute [math]T_{(Tom)} = T_A[/math] and [math]T_{(Flash)} = T_B[/math]…

[math] T_A = \left( \frac{x v}{c^2 \sqrt{1-\frac{v^2}{c^2}}} \right) + \left( T_B \sqrt{1- \frac{v^2}{c^2}} \right) [/math]

Substituting x=vt,

[math] T_A = \left(T_B \frac{v^2}{c^2 \sqrt{1-\frac{v^2}{c^2}}} \right) + \left( T_B \sqrt{1- \frac{v^2}{c^2}} \right) [/math]

pulling out [imath]T_B[/imath]

[math] T_A = T_B \left[ \left(\frac{v^2}{c^2 \sqrt{1-\frac{v^2}{c^2}}} \right) + \left(\sqrt{1- \frac{v^2}{c^2}} \right) \right] [/math]

common denominator,

[math] T_A = T_B \left[ \left(\frac{v^2}{c^2 \sqrt{1-\frac{v^2}{c^2}}} \right) + \left(\frac{c^2(1-v^2/c^2)}{c^2 \sqrt{1- \frac{v^2}{c^2}}} \right) \right] [/math]

adding,

[math] T_A = T_B \left(\frac{v^2 + c^2(1- \frac{v^2}{c^2})}{c^2 \sqrt{1- \frac{v^2}{c^2}}} \right) [/math]

distribute the [imath]c^2[/imath] and cancel,

[math] T_A = T_B \left( \frac{1}{\sqrt{1- \frac{v^2}{c^2}}} \right) [/math]

Which is the equation I used, but now fully and honestly derived from Flash’s perspective. Yes, no, pointless?

 

~modest

[Janus]

While is is possible to derive your equation from what I gave, I don't particularly like that approach to the answer of what Tom's time would be as perceived by Flash.

 

The main problem is that while it gives the right answer for the end of the trip, it doesn't for any other point of the trip. So unless you make it clear that this only applies to the end of the trip, it can cause some people to mistakenly believe that Flash perceives Tom's clock as running fast throughout the trip. This in turn leads to the idea of a prefered reference frame in that the observer who see's the other's clock as running fast is the one who is "really" in motion.

[modest]

While is is possible to derive your equation from what I gave, I don't particularly like that approach to the answer of what Tom's time would be as perceived by Flash.

 

I should have been more clear. I don’t advocate my result as an “approach” toward Tom’s answer from Flash’s perspective. Your approach is correct - it is conceptually correct. What I have done is prove that your approach for Tom as calculated by Flash will always agree (for any given x or v) with Tom’s proper time as calculated by Tom at the conclusion of this paradox.

 

In other words: Tom will always find that his reunited twin’s clock is slow by the Lorentz factor (in as far as the constraints of the thought experiment we’ve given - it would be interesting to try and make a proof for a polygonal path). Flash also always finds that his reunited twin’s clock is fast by the Lorentz factor. Thus the paradox is never a paradox.

 

The main problem is that while it gives the right answer for the end of the trip, it doesn't for any other point of the trip. So unless you make it clear that this only applies to the end of the trip, it can cause some people to mistakenly believe that Flash perceives Tom's clock as running fast throughout the trip. This in turn leads to the idea of a prefered reference frame in that the observer who see's the other's clock as running fast is the one who is "really" in motion.

 

It would be interesting to consider flash at rest while the earth turns around - the earth has 2 frames while flash has 1. I don't think such an approach is possible with special relativity, though, if I recall, it can be done with gravitational fields in GR.

 

~modest