wxrocks said:
If I am standing on Earth and you are in a spacecraft moving away from Earth, your spaceship will look shorter
Yes and for me in the spaceship the Earth will be observed shorter!
wxrocks said:
, but my observed distance to Pluto is not going to change. You on the other hand will think your spacecraft still has a 3 meter cockpit, but will think Pluto moved a lot closer.
Correct!
With regards to calculating the times in the twin problem.
So we have the Minkowski metric:
ds^2=c^2dt^2 - dx^2 - dy^2 - dz^2
Let's simplify that a bit for the example:
ds^2=t^2 - dx^2
To find the time difference between the twins one has to compare the proper time, which is the time that each person experienced.
Proper time in a Minkowski space is:
<br />
\int d \tau = \int ds<br />
For the person that stayed at home this calculation is trivial since it's velocity was 0, only time increased. So basically we can simply look at dt only.
For the traveler the problem is more complicated because dt and ds become interrelated.
Perhaps this is going a bit to deep into the matter but what happens is that the 3D space of the traveler is tilted proportionally to his velocity but in an opposite way. The extreme being at the speed of light (but of course he cannot reach it), were space is tilted to a maximum. The tilting allows him to take a "shortcut" in space, the more tilt the shorter distances in the direction of travel are observed.
Still the way to calculate his proper time is the same, we take
\int d\tau = \int ds
Now in order to compose the total proper time for him we have to add the following parts:
a) The initial acceleration
b) The travel with a constant speed up to the deceleration near the halfway point
c) The the deceleration to zero
d) The the acceleration for the return
e) The travel with a constant speed up to the deceleration near the return point .
f) The the deceleration to 0
No how do we calculate the proper time?
For the constant speed parts it is simple:
\int d\tau = \int {\sqrt{1 - (dx/dt)^2 } dt
The acceleration and deceleration parts are more complicated, it depends if a is constant or variable. I would have to look it up.
So in summary, the traveler noticed that the distance became shorter the faster he traveled because of the tilt in 3D space. However his clocks were running as fast as before the trip. For the person who stayed behind there were no noticable changes. He did observe that the traveler's clock slowed down, but of course in reality it did not. It would be kind of silly to suggest that physical properties are changed because someone observes it from another frame of reference.
Hope this helps
This latex feature is nice but it took me quite some edits to get it right since I was not familiar with it.