Nathan123 said:
I get the feeling that very few people really understand how this works.
No,
you don't understand how this works. Relativity physicists understand it just fine. As I've already said, you are confusing yourself by trying to reason about this using ordinary language instead of math.
Nathan123 said:
I am hearing that relativity of simultaneity is the answer (and I'm afraid it might get overused because it seems to answer everything). But I am not sure exactly where it comes into play here.
That's because you aren't doing the math, and, as I've already said, trying to understand how it all fits together without the math is going to be harder than just learning the math.
Here is a quick summary of the math for the Earth and ship scenario.
First, let's look at things in the Earth frame. Let's consider the following events:
A: The ship is 0.6 light-hours from Earth, and is traveling towards Earth at speed ##0.6 \text{c}##. We'll call the time of this event time ##t = -1## in the Earth frame (for reasons which will become apparent below), so the coordinates of this event are ##(x, t) = (0.6, -1)## (we're using light-hours and hours as our units of length and time).
E: This is the Earth (more precisely, a chosen point on the Earth, the one the ship will end up arriving at) at the same time, in the Earth frame, as event A. We are treating this point on the Earth as the spatial origin of the frame, so the coordinates of this event are ##(0, -1)##.
B: The ship arrives at Earth (at our chosen point). This event has coordinates ##(0, 0)## in the Earth frame (as you can see from the coordinates of event A and the speed of the ship).
The distance of the ship from Earth at event A is the spatial distance between event A and event E, which, since they both have the same time coordinate in this frame, is just ##0.6##.
Now, let's look at things in the ship frame. We apply the Lorentz transformation equations, ##x' = \gamma \left( x - v t \right)##, ##t' = \gamma \left( t - v x \right)##. The relative velocity between the frames is ##v = - 0.6## (because the ship is moving in the ##- x## direction in the Earth frame), so we have ##\gamma = 1 / \sqrt{1 - v^2} = 1 / 0.8 = 1.25##. So the coordinates in the ship frame of our three events turn out to be:
A: ##(0, - 0.8)##.
E: ##(- 0.75, -1.25)##.
B: ##(0, 0)##. (Note that, for the Lorentz transformation equations I gave above to work properly, the event that has coordinates ##(0, 0)## in one frame--i.e., the spacetime origin--must also have those coordinates in the other frame.)
Now, observe:
First, in this frame, events A and E do not happen at the same time (whereas they do in the Earth frame). This is relativity of simultaneity.
Second, the time from event A to event B is 0.8 in this frame (whereas it was 1 in the Earth frame). This is time dilation.
But, third, the time from event A to event B is
not "the time it takes the ship to reach Earth" in this frame. In this frame, the ship is motionless; the Earth is moving. So ##0.8## hours is the time it takes Earth to reach the ship. And the question is: how far does the Earth travel to reach the ship in this frame?
To figure that out, we need to find the event on Earth's worldline that is simultaneous with event A in the ship frame--i.e., that has time coordinate ##t' = - 0.8##. Since the Earth is traveling at speed ##0.6## in this frame (in the positive ##x'## direction), then this event, which I'll call event C, has coordinates:
C: ##(x', t') = (- 0.48, - 0.8)##.
So the Earth takes ##0.8## hours to reach the ship in this frame, and travels a distance of ##0.48## light-hours--whereas, in the Earth frame, the ship traveled a distance of ##0.6## light-hours. And
that is length contraction: ##0.48 = 0.8 * 0.6##.
So,
if you do the math, you see how all three things--length contraction, time dilation, and relativity of simultaneity--work together. But leaving out anyone of them leads you astray.
