Shawn Garsed said:
Thanks for the example and right now I'm letting it sink in, but do you know if there's a visual representation of this?
I would recommend drawing some simple diagrams of the scenario I gave for yourself if you're having trouble following it. You really only need three diagrams:
1. One showing how things look at t=0 in your frame, where the moving rod is 20 light-seconds long and aligned parallel to your ruler, and the back end is lined up with the x=0 mark on your ruler with your clock attached to that mark reading t=0, and the front end lined up with the x=20 mark on your ruler, with your clock attached to that mark reading t=0 as well.
2. Another diagram showing how things look at t'=0 in the rod's rest frame, where the rod is 25 light-seconds long, and the distance between the x=0 and x=20 marks on your ruler is shrunk to only 16 light-seconds. At this moment the front end of the ruler is next to the x=20 mark on your ruler, and your clock at that mark reads t=0; and at the same time in this frame, the back end of the ruler is 9 light-seconds away from the x=0 mark on your ruler, and your clock at the x=0 mark reads t=-12 seconds.
3. A final diagram showing how things look 15 seconds later at t'=15 in the rod's rest frame, when the back end of the rod is lined up with the x=0 mark on your ruler (which means the front end is now 9 light-seconds away from the x=20 mark on your ruler). At this moment, your clock at the x=0 mark reads t=0 seconds, and your clock at the x=20 mark reads t=12 seconds.
Once you have these diagrams, you can verify that since the rod is moving at 0.6c relative to your ruler, the rod is shrunk by a factor of 0.8 in your frame and likewise the distance between the x=0 mark and x=20 mark on your ruler is shrunk by the same factor of 0.8 in the rod's rest frame. You can also see that in the rod's frame, both your clocks are slowed by the same factor of 0.8 (they both tick forward by 12 seconds in the 15 seconds between the first and second diagram), and they are also consistently out-of-sync by 12 seconds.
Shawn Garsed said:
Also, in the thread you referred me too in an earlier post I wanted to know how you calculated the time-coordinates on B's ruler as seen from ruler A's frame.
Mathematically the simplest way is probably just to use the Lorentz transformation, which says that if you know the position x and time t of an event in A's frame, the position x' and time t' of the same event in B's frame must be:
x' = gamma*(x - vt)
t' = gamma*(t - vx/c^2)
with gamma = 1/\sqrt{1 - v^2/c^2}, and v being the velocity that B is moving in A's frame.
Another way would be to use the formulas for length contraction, time dilation and the relativity of simultaneity separately. The length contraction formula says that if two marks on B's ruler are a distance L apart in B's frame, in A's frame the distance between these marks will be only L' = L*\sqrt{1 - v^2/c^2}. The time dilation formula says that if it takes a time T for a specific clock which is at rest in B's frame (say, the clock with the red hand on B's ruler) to tick forward by a certain number of seconds, then in A's frame that clock will tick slower and therefore take longer to tick forward that amount of seconds, taking a time T' = T/\sqrt{1 - v^2/c^2} to do so. Finally, the relativity of simultaneity formula says that if two clocks at rest in B's frame are a distance L apart and synchronized in B's frame, then in A's frame they will be out-of-sync by vL/c^2 (with the clock in 'front' relative to their direction of motion in A's frame being behind the clock in 'back' by this amount).
If you like I can pick two specific clocks in B's frame and show how the way I drew them in the diagram of A's frame matches up with what you should expect based on both the Lorentz transformation and also based on the separate formulas for length contraction, time dilation and the relativity of simultaneity.
