A thought experiment: Reference frame: A surface in space with no atmosphere. A rocket ship is approaching just above the level of the surface. Its measured velocity is a constant 0.8c (0.8 times the speed of light) relative to the surface. Special relativity predicts that time in the space ship runs at 0.6 times that on the space surface. The space ship has a very small window which will face the space surface when it arrives, as well as two light detectors, one at the window and a long one 1 m away and parallel to the window. A laser beam projects a beam perpendicular to the space surface and in the path of the coming spaceship window. A very short flash of light falls on the window detector and then on the detector 1m away as the space ship passes. The detectors in the space ship show the beam to take a skewed path because of the ships 0.8c motion relative to the light beam. The length of that path is observed to be (0.8^2+1^2)^1/2 = 1.28 m. Since the time dilation is 0.6 relative to the surface, the time difference at the two detectors measured by a clock in the spaceship should be: 0.6x1.28/c where c is in m/sec. This yields 2.56x10^-9 seconds. Dividing by the distance 1.28 m gives a time it took the beam to travel a meter of 2.00x10^-9, yielding a measured c of 5.0x10^8 m/sec if the time dilation were correct. Of course this is nonsense. (Note that 5.0x10^8/0.6 = 3.0x10^8 suggesting maybe there is no time dilation in the transverse direction.) A person on the space surface, knowing that the window and screen are 1 m apart and recognizing that the very short beam will travel in a straight line calculates that the light beam takes 1/c seconds to traverse the distance between the window and screen from his reference position which is 3.33x10^-9 sec. A person in the space ship sees the light beam traveling 1.28 m in his reference frame. Since he must, according to relativity, measure c at 3.0x10^8 m/sec he should experience a time delay between the two light signals of 1.281/3.0x10^8 or 4.27x10^9 sec. Thus the time dilation factor would be 0.781 in the transverse direction and not 0.6 as special relativity would predict for the direction of travel. How can time be dilated differently in one direction (the direction of travel) from the transverse direction? A clock cannot distinguish directions. Perhaps someone can straighten me out, or time dilation in special relativity crumbles.
No, it's (1.33^2+1^2)^1/2 = 1.67 m. The distance is 0.8 m in the surface frame, from which the rocket appears length contracted.
You are right, of course. I didn't consider the fact that in the rockets reference frame distance in the direction of the surface has to expand by a factor of 1.33 if time slows by a factor of 0.6 so that both observers see the other's reference frame receeding at 0.8 c. Thanks for straightening me out.