Aer said:
Do the following analysis: Assume our clock that was previously defined at rest in frame K' is stationary at x=0 and the other clocks move at .9c in the negative x direction. What will the markings be on the notepad?
This is in fact the exercise that I suggested yesterday, so I will give you the solution
First of all, since I am a little bit lazy in typing, I shall use g instead of γ . I will also use c = 1., so that all velocities are given in fact in units of c, or if you prefer, instead of writting β i will use v. So that is g is defined to be g = 1/sqrt(1-v<sup>2</sup>).
In the following I will use the Lorentz transformation:
x' = g(x+vt)
t' = g(t+vx).
The inverse Lorentz transformation is
x = g(x'-vt')
t = g(t'-vx')
This assumes that the two referenxce frames have their origin x= 0 and t = 0 located at the same place when t=0 and t'=0.
This also assumes that the K' frames moves in the negative direction wrt the K frame.
We have now three events:
event A is clock 1 (the clock stationnary in the K frame) ticks 0.
event B is clock 2 (the clock stationnary in the K' frame) ticks 0
and event C is when both clocks reach each other.
Furthermore we assume (as a part of the exercise) that event A and event B are simultaneous in frame K (we shall see that they are not simultaneous in frame B). And finnally, we shall assume that the distance between the clocks when event A and B occur (simultaneously) in frame K, they are at a distance L (again, in frame K). Without loss of generality, we can assume that the events A and B occur at t = t' = 0. As you asked, since x'=0 for clock 2 at t'=0, at t=0, clock2 will be located also at x=0 in frame K (remember that the frames origin are located at the same position at t = 0 and t' = 0).
So that at t = 0, the location of clock1 will be at x = -L.
Let us see what happens in the K frame:
Clock 1 is stationnary so that its equation of motion is
x1(t) = -L
We can also write that for event A, xA = -L and tA = 0
Clock 2 moves to the negative direction with velocity -v, at t= 0 it is located at x=0 So that its equation of motion will be:
x2(t) = -vt (we assume that there is no acceleration and that for negative t the equation of motion is the same)
We can also write that for event B, xB = 0 and tB = 0.
Event C is when both clocks are located at the same position so that event C will occur when x1(t) = x2(t) ==> -L = -vt ==> t = L/v.
So that for event C we have xC = -L and tC = L/v.
tC = L/v will be the time written on the notepad by clock 1.
Due to time dilation, the time that will be shown by clock 2 will be tC/g = L/vg.
Let us go now to the K' frame.
In order to find tC' we have two ways, the first will be the easiest, the second will be more instructive, so let us look at both ways.
The first option is to say that we have an event C with coordinates xC and tC in the K frame and just do the Lorentz transformation in order to find xC' and tC'.
xC' will be given by:
xC' = g(Xc + vtC) = g(-L +v*L/v) = 0. That what we were supposed to find since the event C occur at the location of clock 2 which is allways at x'=0 in the K' frame.
Let us look now for tC'. Lorentz transformation will give:
tC' = g(tC + v xC) = g(L/v - vL) = gL/v(1-v<sup>2</sup>) = gL/vg<sup>2</sup> (since g = 1/sqrt(1-v<sup>2</sup>, we have 1-v<sup>2</sup> -= 1/g<sup>2</sup>) so that tC' = L/gv. This is the same result as we have by just using the time dilation.
The second possibility is to transform the events A and B to the K' frame and to transform the equation of motions
First let us look at clock 1.
Its equation of motion is x1(t) = -L, So using the inverse Lorentz transformation we have g(x1' - vt') = -L, from this we get x1'(t') = vt'-L/g.
The equation of motion of clock 2 in K is given by x2 = -vt so that again by using the inverse Lorentz transformation we have
g(x2'-vt')= -vg(t'-vx2'), g cancels on both side and we are left with x2' - vt' = -vt' +v<sup>2</sup>x2'.
-vt' also cancel so that we are left with x2' = v<sup>2</sup>x2' ==> x2'(t) = 0.Let us calculate the time of event C. Event C will occur when both clocks are at the same location so that x1'(t) = x2'(t)
==> vt' - L/g = 0 ==> t' = L/gv. So that tC' will be tC' = L/gv.
Now (and this is for you MacM, maybe this will help you understand SR), what about time dilation?
Let us calculate at what time event A occurs in reference frame K'.
Since A and B are simultaneous in frame K, we shall see that they are not simultaneous in reference frame K'. Time dilation says that if we take the time from event A to event B in reference frame K' and divide by g, this is what time dilation will give.
since for event A we have tA = 0 and xA = -L' tA' and xA' will be related by Lorentz transformation.
Let us first calculate xA':
xA' = g(xA + tA) = -gL.
Now if we take this location and put it in the equation of motion of clock 1 in the K' frame, (I remind you that it is x1' = vt'- L/g) we get -gL = vt'-L/g ==> vt' = L/g-gL = Lg(1/g<sup>2</sup> - 1), but 1/g<sup>2</sup> = 1-v<sup>2</sup>, so that we get vtA' = Lg(1-v<sup>2</sup>-1) = -Lgv<sup>2</sup> ==> tA' = -Lgv. We see that event A occured before event B in the K' frame.
Of course, we could just have taken the Lorentz transformation of tA in order to get:
tA' = g(tA + vxA) and use tA = 0 and xA = -L to get tA' = -gvL, which is of course the same result.
Now in order to use the time dilation formula, we must calculate the elapsed time between event A and event c in the K' frame.
This elapsed time will be given by Dt' = tC' - tA' = L/gv -(-vgL) = L/gv + vgL = gL/v(1/g<sup>2</sup> + v<sup>2</sup>), again by using the fact that 1/g<sup>2</sup> = 1-v<sup>2</sup>, one gets that Dt' = gL/v(1-v<sup>2</sup> + v<sup>2</sup>) = gL/v.
Time dilation in the K' frame POV will state that the time between the two events A and C is gL/v, so that in the K frame the time between the two events should be gL/vg = L/v.
So that according to reference frame K', on the notepads there should be L/gv as the time given by clock 2 and L/v as the time given by clock 1.
All "accumulated" times are consistent in all frames.