cam875 said:
so in the end if they both measure the same speed where does time dilation come from? or am i now confusing all this new stuff flooding my brain lol.
Well, if each one is carrying a clock, then each will measure the other one's clock to be running slower, so there is time dilation. If you look at that
thread with the illustration I made, and take note of the clock times, you can see that each individual clock on one ruler (like the red clock at the middle or the left or right green clocks) is ticking at half speed in the frame of the other ruler.
It might help to do the math for a similar example. Say we have two observers A and B who each have a ruler which is 20 light-seconds long in their own rest frame, and each one has 2 clocks attached to either end of their own ruler which are synchronized in their own frame. Suppose the two rulers approach each other and pass side-by-side, and each observer notes when any pair of clocks pass right next to each other. In A's frame, if B is approaching from the right, and B's ruler is shorter than A's ruler in this frame, then first B's left clock will pass next to A's right clock (event 1), then later B's right clock will pass next to A's right clock (event 2), then B's left clock will pass next to A's left clock (event 3), and finally B's right clock will pass next to A's left clock (event 4). It may help to draw a little diagram of this.
Let's say that in A's frame, B is approaching at 0.8c. That means that in this frame B's ruler will be shrunk by a factor of \sqrt{1 - 0.8^2} = 0.6, so in A's frame B's ruler is only 20*0.6 = 12 light-seconds long. Suppose also that when event 1 occurs, both A's right clock and B's left clock (which are passing next to each other at this moment) read t=0 seconds. B's right clock is 12 light-seconds away from A's right clock at this moment, so if B is moving at 0.8c B's right clock will reach A's right clock 12 l.s./(0.8 l.s./s) = 15 seconds later in A's frame, so A's clock reads t=15 seconds at the time of event 2. And since A's ruler is 20 l.s. long while B's is 12 l.s. long in this frame, at the moment their right clocks line up, the left clock of B will be 20 - 12 = 8 light seconds away from the left clock of A. So, it'll take another 8/0.8 = 10 seconds for the two left clocks to line up (event 3), meaning A's clock reads t=15+10=25 seconds at the time of event 3. At this moment B's right clock is 12 light-seconds away from A's left clock, so it takes another 12/0.8=15 seconds for those clocks to line up (event 4), so A's left clock reads t=25+15=40 seconds at the moment of event 4.
Now we can figure out what B's clocks read at each of these passing-events. In A's frame, both of B's clocks are ticking slow, by a factor of 0.6. Also, because of the
relativity of simultaneity, if B's clocks are synchronized and a distance L apart in B's own frame, in a frame where they are moving at speed v they'll be out-of-sync by vL/c^2, with the front clock's time being behind the back clock's time by this amount. In A's frame, v=0.8c, and the distance between the clocks L in
B's frame is 20 l.s., so the two clocks are out of sync by 16 seconds in A's frame; this means at the moment B's left clock reads t'=0 seconds (event 1), B's right clock already reads t'=16 seconds in A's frame. 15 seconds later in A's frame, each of B's clocks has only ticked forward by 15*0.6=9 seconds, so at the moment of event 2 (when B's right clock lines up with A's right clock), B's left clock reads t'=0+9=9 seconds and B's right clock reads t'=16+9=25 seconds. 10 seconds later at the time of event 3 (A's left clock lining up with B's left clock) in A's frame, each of B's clocks has only ticked forward by an additional 10*0.6=6 seconds, so now B's left clock reads t'=9+6=15 seconds and B's right clock reads t'=25+6=31 seconds. Finally, 15 seconds later at the time of event 4 (A's left clock lining up with B's right clock) in A's frame, each of B's clock has ticked forward by another 15*0.6=9 seconds, so B's left clock now reads t'=15+9=24 seconds and B's right clock reads t'=31+9=40 seconds.
So, looking only at the clocks that line up, we have:
event 1: A's right clock reads t=0, B's left clock reads t'=0
event 2: A's right clock reads t=15, B's right clock reads t'=25
event 3: A's left clock reads t=25, B's left clock reads t'=15
event 4: A's left clock reads t=40, B's right clock reads t'=40
Now, look at what B will conclude about this in his own frame, where B's clocks
are synchronized and running at normal speed, and the distance between them is 20 light-seconds rather than 12. Event 1 is A's right clock passing B's left clock, and event 2 is A's right clock passing B's right clock; these events are 25 seconds apart according to B's clocks, and this is the time it took A's right clock to pass from the left end to the right end of B's 20 light-second ruler, so A's right clock must be traveling at 20 l.s./25 s=0.8 l.s./s=0.8c in B's frame. Likewise, event 3 and event 4 represent A's left clock passing from one side of B's ruler to the other, and these events happened 40-15=25 seconds apart in B's frame too, so B also concludes the left clock was moving at 0.8c.
