PatPwnt said:
When you say the that two clocks are sychronized, do you mean that if I am in the middle of a train compartment and there is a clock at both ends of the compartment that they both read the same times to me standing mid way(x/2) between them? And that they will appear out of sync with each other by ux/c^2 if I am at rest watching the train compartment drive by?
It's not about visual appearances, but about how different observers assign time-coordinates to events. In SR the concept of an inertial reference frame is usually introduced by assuming each observer has a network of rigid rulers and clocks at rest with respect to them, and that the time and position coordinates of distant events are based on
local readings on this system. For example, if when my clock reads 2008 I notice a distant explosion in my telescope, and I see that it happened right next to the 4-light year mark on my ruler, and that a clock attached to that mark showed a reading of 2004 as the explosion was happening, then I would assign the event of the explosion a time-coordinate of t=2004, not t=2008. But in order to do things that way, each observer must have a way of
defining what it means for clocks at different positions along their rulers to be "synchronized", so this is where Einstein suggested that each observer could define clock synchronization based on the
assumption that light moves at the same speed in all directions in that observer's frame, so that an observer could synchronize two clocks by setting off a flash at the midpoint between them, and having them both show the same reading at the moment the light from the flash reaches each one. A consequence of this definition is that different observers will disagree about simultaneity. For instance, if I'm on a ship with clocks at each end, I can synchronize them in the ship's rest frame by setting off a flash at the midpoint, and setting them to read the same time when the light reaches them. But if the ship is moving forward in your frame, then since your frame's coordinates are constructed around the assumption that light moves at the same speed in both directions relative to
you, then naturally since the back of the ship is moving towards the point where the flash was set off while the the front of the ship is moving away from that point, in your frame the light will catch up with the back clock before it catches up with the front clock, so the back clock's reading will be ahead.
We can see how much it'll be ahead by using the time dilation and length contraction equations. Say my ship is moving forward along your x-axis at speed u in your frame, and in my frame it has a length of x', so using the Lorentz contraction equation, in your frame the length is only x' * \sqrt{1 - u^2/c^2}. Say a flash is set off at the midpoint of the ship when the midpoint is next to the x=0 mark on your ruler, so at that moment in your frame, the back clock is at position x= \frac{-x' * \sqrt{1 - u^2/c^2}}{2} while the front clock is at position x = \frac{x' * \sqrt{1 - u^2/c^2}}{2}. Since both clocks are moving foward at speed u, the back clock's position as a function of time is given by:
x(t) = \frac{-x' * \sqrt{1 - u^2/c^2}}{2} + ut
and the front clock's by:
x(t) = \frac{x' * \sqrt{1 - u^2/c^2}}{2} + ut
Meanwhile, the light beam going in the -x direction has position as a function of time x(t) = -ct, while the light beam going in the +x direction has x(t) = ct. So, to figure out when the light hits the back clock, we solve for t in this equation:
-ct = \frac{-x' * \sqrt{1 - u^2/c^2}}{2} + ut
...which gives t = \frac{x' * \sqrt{1 - u^2/c^2}}{2*(c + u)}
And to figure out when light hits the front clock, we solve for t in:
ct = \frac{x' * \sqrt{1 - u^2/c^2}}{2} + ut
which gives t = \frac{x' * \sqrt{1 - u^2/c^2}}{2*(c - u)}
If you subtract the first time from the second, you get:
\frac{x' (c + u) * \sqrt{1 - u^2/c^2}}{2*(c - u)*(c + u)} - \frac{x' * (c - u) \sqrt{1 - u^2/c^2}}{2*(c + u)*(c - u)} = \frac{x' * u * \sqrt{1 - u^2/c^2}}{c^2 - u^2}
So, this shows that in your frame, the light doesn't hit the front clock until this amount of time has passed since the light hit the back clock. If both clocks were set to read some time T when the light hit them, then the back clock has been ticking forward past T in this time, but it is slowed down by factor of \sqrt{1 - u^2/c^2} thanks to time dilation, so when the light hits the front clock and it reads T, the back clock will read T + \sqrt{1 - u^2/c^2} * \frac{x' * u * \sqrt{1 - u^2/c^2}}{c^2 - u^2} = T + \frac{x' * u * (1 - u^2/c^2)}{c^2 - u^2}. And since (1 - u^2/c^2) = (1/c^2)*(c^2 - u^2), this works out to T + \frac{x' * u * (c^2 - u^2)}{c^2 * (c^2 - u^2)} = T + \frac{x' * u}{c^2}. So, this shows that in your frame, the back clock is ahead of the front clock by x'*u/c^2, where x' is the distance between the clocks in their own rest frame and u is their speed in your frame.
Let me know if you follow this, and if so I can then explain how to derive the time transformation equation from it.
