Symmetry, Space-Time interval, and Coordinate Time I had the following question I was working from a book: 

Question: A physics professor runs across the hallway covering 120 ns
  of distance in 150 ns of time as measured in the frame of the earth.
  Assume the professor travels at a constant velocity, how much time
  does the professor's watch indicate has elapsed during the trip?

Solution: Let A be the event that the professor enters the hallway and B be the event that he leaves the hallway. The metric equation is $\Delta s^2 = \Delta t^2 - \Delta d^2$. If you plug in the given data you will get $\Delta s$ = 90 ns. This is the correct solution as given in the book. 
However, I'm confused why I can't analyze the situation in the frame of the professor. I cannot say that from the Professor's frame, events A and B have separation $\Delta d$ and take place $\Delta t$ apart so that the time passing in the earth frame is $\Delta s$ = 90 ns. These would contradict each other, but I cannot see why. It seems to me that the symmetry implies that I should be able to analyze this in either frame since they are both inertial. Furthermore, doesn't the fact that less time ticks off the professor's clock indicate a violation of relativity in that it distinguishes the results in the earth frame from the professor's frame which are both taken to be inertial in the problem? 
 A: In general, it's more useful to think of special relativity problems in terms of the spacetime interval $\Delta s^2$ than in terms of the question of "who is moving relative to whom." A lot of those explanations (in terms of things like time dilation, length contraction, etc) are very ad-hoc, and are designed to make the idea of the spacetime interval more palatable to our brains, which are used to thinking in non-relativistic terms.
The only invariant between reference frames is the spacetime interval $\Delta s^2 = \Delta t^2 - \Delta d^2.$ In the Earth's frame, you are given $\Delta d = 120$ ns and $\Delta t = 150$ ns. From this, you correctly computed $\Delta s = 90$ ns.
This is an invariant, meaning that the professor will measure the same spacetime interval, even though they might measure different distances and times. In fact, you are explicitly given $\Delta d$ in the professor's frame of reference—since events $A$ and $B$ both occur at the location of the professor, the distance between them in the professor's frame of reference is $\Delta d = 0.$ Using this fact, along with the invariant $\Delta s = 90$ ns, you can compute the time elapsed in the professor's frame of reference, which is $\Delta t = 90$ ns.
