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?

• In the professor's frame, events A and B have separation $\Delta d = 0$ because they both take place at the location of the professor. In the professor's coordinates system, each event is located at the origin. This indeed yields $\Delta s = 90$ ns, as desired. – user35736 Jun 16 '16 at 23:28
• @user35736 That's silly. The answer how much time passes in the earth frame can't depend on what frame you decide to calculate with. The time that passes in the earth frame is 150 ns, not 90 ns. – user7348 Jun 17 '16 at 2:27
• Right, but the professor's frame and the earth's frame are different. The time passing in the professor's frame is $90$ ns, while the time passing in the earth's frame is $150$ ns. This yields no contradiction. – user35736 Jun 17 '16 at 2:30
• @user35736 OK, but since the Professor moves relative to the earth, the earth moves relative to the Professor with the same magnitude. Why the different answers then? Doesn't that pick out a preferred frame? – user7348 Jun 17 '16 at 2:34
• It doesn't pick out a preferred frame because there is no way of saying one frame is more "correct" than the other. The only thing that is invariant between frames is the spacetime interval $\Delta s^2 = \Delta t^2 - \Delta d^2.$ The problem gives you $\Delta t$ and $\Delta d$ in the Earth's frame, from which you can compute $\Delta s.$ Since we know that $\Delta d = 0$ in the professor's frame, and $\Delta s$ must be the same, this formula lets us compute $\Delta t$ in the professor's frame. In worrying about questions of "who is moving relative to whom," you might overcomplicate the problem. – user35736 Jun 17 '16 at 2:38

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.