Special relativity: twin paradox

[jk4]

I am just a little bit unsure about something I read on the "twin paradox".
It talks about Dick and Jane who are twins, each 20yrs old. Dick departs at a speed of 0.80c to a star 20 light years away. So I'm sure you all know how the story goes, Dick is 50yrs old when he returns and Jane is 70.

Ok, then it asks a question:
Dick and Jane each send out a radio signal once a year while Dick is away. How many signals does Dick receive? Hown many does Jane receive? using the same info as above.

I'll leave out a lot of the discussion the book uses and the details, but I'm interested it the fact that the signal rates are different on the return trip then from the outward trip.

Using the doppler effect we find that on the outward trip Dick receives 5 signals from Jane. But, he receives 45 signals from Jane on the return trip.

Now this confuses me because I was under the impression that time would appear to be moving slower for someone moving relative to you regardless of if they are approaching or receding. However, according to this argument, wouldn't it seem that on the return trip that time is moving extremely fast for Jane relative to Dick?

Also, Jane receives 15 signals in 45 years (corresponding to Dicks outward trip), on the return trip Jane receives 15 signals from Dick in the last 5 years of her 50yr wait (the 15 signals Dick sends out on his return), so that means she must see him moving really slow on the outward trip, but then wouldn't it appear to her that he travels 20 light years in just 5 years!

I know I'm wrong because then the theory wouldn't hold, I just want to know what I'm looking at the wrong way.

 

[Fredrik]

This seems to be one of those months where half the new threads are about the twin paradox in one way or another.

I made http://web.comhem.se/~u87325397/Twins.PNG [Broken] space-time diagram for a discussion on another forum. I didn't draw any radio signals, but it would be easy to add them. They would be lines at a 45 degree angle with the time and space axes. Note that if you send a radio signal from Earth that arrives to the rocket at the turnaround event, it would have to be emitted very early.

I'm calling the twin on Earth "A" and the twin in the rocket "B".
Blue lines: Events that are simultaneous in the rocket's frame when it's moving away from Earth.
Red lines: Events that are simultaneous in the rocket's frame when it's moving back towards Earth.
Cyan (light blue) lines: Events that are simultaneous in Earth's frame.
Dotted lines: World lines of light rays.
Vertical line in the upper half: The world line of the position (in Earth's frame) where the rocket turns around.
Green curves in the lower half: Curves of constant -t²+x². Points on the two world lines that touch the same green curve have experienced the same time since the rocket left Earth.
Green curves in the upper half: Curves of constant -(t-20)²+(x-16)². Points on the two world lines that touch the same green curve have experienced the same time since the rocket turned around.

 

[jk4]

I think I understand most of the image. However, wouldn't A not see B's rocket turn around until about 15 years after it happens? because there is about a 15 light year difference in their positions.

 

[granpa]

so??

 

[jk4]

well then that means that A would see B's entire homeward journey in just a few years, like in my original post. Wouldn't Jane see Dick travel 20light years in just 5 years? Which is impossible of course... That's what I'm confused about. Is that statement correct? Jane would see Dick travel 20 light years in just 5 years. Or is that statement false? That's my question, and if it's false, why? I just assumed it was false and that I'm misunderstanding something because he can't travel faster than light.

 

[Fredrik]

It doesn't matter what people see. If you hook up a video camera to a telescope, it would "see" something, but that obviously isn't what's actually happening. If you want to see what really happened you'd have to have a computer show the pictures to you at a rate that compensates for the arrival times of the light.

And if you just imagine a few lines at a 45 degree angle in my diagram, you should understand why Jane would see Dick travel a pretty short distance during the first half of her time, but she should know that that's not what's actually happening.

 

[jk4]

wow, so I was right after all. Thank you all for helping me realize that, or I would have spent a long time thinking I was missing something vital.

 

[yuiop]

...

Also, Jane receives 15 signals in 45 years (corresponding to Dicks outward trip), on the return trip Jane receives 15 signals from Dick in the last 5 years of her 50yr wait (the 15 signals Dick sends out on his return), so that means she must see him moving really slow on the outward trip, but then wouldn't it appear to her that he travels 20 light years in just 5 years!
...

Jane is 65 when she sees Dick turn around. Since Dick is 20 light years away at the turn around event Jane realizes she is seeing an event (the turnaround) that happened 20 years earlier due to light travel time. So when Dick arrives 5 years later after Jane seeing the turnaround event she knows the interval between trunaround and arrival is 25 years of her time so she has no reason to think Dick has been exceeding the speed of light.