seb7 said:
I understand Einstein's equations on time dilation, but this video talks about why, which doesn't seem correct to me.
Wouldn't his reasoning for the time dilation require a fixed ether frame?
No. What he's talking about when he talks about things slowing down because of speed is relative to ANY inertial rest frame. The simplest example to understand it is a light clock. A light clock consists of two parallel mirrors, with a pulse of light bouncing back and forth between the mirrors. The time for a complete circuit for the pulse of light is just T = 2D/c, where D is the distance between the mirrors. Now, if you set the two mirrors in motion (or alternatively, switch to a different rest frame in which the mirrors are already moving), then the time required for a complete circuit is longer. If we assume that the motion is perpendicular to the line connecting the mirrors, and that the distance between the mirrors is still D (a more thorough analysis is needed to conclude this, so for now, it's just an assumption), then for a pulse of light to travel from one mirror to the other and back at speed c would take a time equal to:
T' = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}} 2D/c
where v is the speed of the mirrors.
To see this: let \delta t be the time required to travel from one mirror to the other. During this time, the second mirror moves forward by an amount v \delta t. So the light, in traveling from one mirror to the other travels a distance D in one direction, and a distance v \delta t in the other. The total distance traveled, by Pythagoras, is: D_{total} = \sqrt{D^2 + v^2 \delta t^2}. If light always travels at speed c, then D_{total} = c \delta t. So we have: c \delta t = \sqrt{D^2 + v^2 \delta t^2}. Solving for \delta t gives \delta t = D/c \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. Double that for a round-trip.
So for a mirror in motion, the time for a complete circuit is longer, by a factor of \gamma = \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}. This is
relative to the inertial reference frame in which v is measured. So the amount of slowing is different for different frames. There is no absolute frame relative to which the slowing is to be measured--time dilation works in any frame.