swampwiz said:
so if I take the other term as 0, there is
x'( t = 0 ) = γ x
t'( x = 0 ) = γ t
Try writing out both transform equations for each of the cases above:
Case 1: (x, 0) transforms to (x', t') as follows:
x' = γ x
t' = - γ v x
where v is the relative velocity.
Case 2: (0, t) transforms to (x', t') as follows:
x' = - γ v t
t' = γ t
You will see why this is useful in a moment.
swampwiz said:
the problem is that the time dilation & length contraction use reciprocals of each other
The common formulas for "length contraction" and "time dilation" actually play a trick on you behind the scenes: they change the pairs of events that they refer to behind your back, so to speak. The key is that the transformation equations above transform *intervals*, i.e., *pairs* of events, not single events; but which particular pair of events describes an object or its motion *changes* when we change frames.
To see how this works, consider the two cases above. Case 1 might describe, for example, a ruler at rest in the unprimed frame, with proper length x. The ruler is moving in the primed frame, so we expect it to be "length contracted", but as you note, when we transform the interval describing its proper length in the unprimed frame (from the origin 0, 0 to x, 0) to the primed frame, the difference in x' coordinates comes out *larger* by the gamma factor, instead of smaller. What gives?
What gives is that we are looking at the wrong interval; the ruler's "length" in the primed frame is described by a *different* interval, one which is purely along the x' axis instead of the unprimed x axis. In other words, the correct interval in the primed frame is the one between the origin of the primed frame and whatever event on the worldline of the ruler's other end is simultaneous with the origin in that frame. That is, we need to find the event on the worldline of the ruler's other end that has t' = 0. The easiest way to find that event is to go back to the unprimed frame and write the equation for the x' axis in that frame: it is simply t = vx. The worldline of the ruler's other end in the unprimed frame is just x = x (the other end is at rest at the same x coordinate for all time), so the point we are looking for is the intersection of that worldline with the x' axis, i.e., (x, t) = (x, vx). So we just have to transform the interval (0, 0) - (x, vx) to the primed frame:
x' = γ (x - vt) = γ (x - v^2 x) = γ x (1 - v^2) = x / γ
t' = γ (t - vx) = γ (vx - vx) = 0
Voila! Now we have an interval that is purely along the x' axis, since t' = 0, and as you can see, this interval has a length that is *shorter* than x by the factor gamma, exactly as desired.
A similar procedure can be done for case 2 to show how the correct "time dilation" can be determined by finding the right interval along the t' axis, instead of the interval transformed above, which is along the unprimed t axis.