Time dilation as an observer in special relativity I've been having a discussion regarding time dilation relating to special relativity and how it should be observed from the FoR (Frame of Reference) of "the person moving" :
I assert- If we have a time is slower in the moving FoR than when I observe the stationary frame of reference (earth) then I should perceive events as occurring faster, i.e., if I observe 15 years passing and 90 years pass in the stationary frame then I should observe events on earth happening at a rate of 6:1. So if I observe a 1 min event on earth I watch that entire event occur in 10 seconds. 
My thought process is similar to the idea that if I was near a massive object and looked out I would observe events further away as moving faster. My teacher seems to believe this line of reasoning is incorrect but will not (or cannot) give me the correct thought process.
Hence my question is, "is my logic correct, and if not why?"   
 A: You're kind of right, in a certain sense.  The problem is that in relativity you need to be very careful about what you mean when you say "observe".
If you were in a rocket ship traveling away from Earth at half the speed of light, and you looked back at the Earth, then even without time dilation you would see events happening in slow motion because the light takes longer to get to you the further away you are.  This is basically the same as the doppler effect.  If there were no time dilation, then for every two seconds of your time you would see one second of Earth time.  Taking time dilation into account, you see one second of Earth time every 1.73 seconds of your time.
On the return trip (traveling towards Earth at half the speed of light) then if there were no time dilation you would see events speeded up: for every two seconds of your time you would see three seconds happening on Earth.  Taking time dilation into account, you see three seconds of Earth time happening in 1.73 seconds of your time.
If you didn't know about relativity, so that you only took the doppler effect into account, it would seem to you that events on Earth were taking place about 16% faster than they ought to.  So in that sense you would see time traveling more quickly on Earth.
[Critics are asked to draw the relevant diagrams for themselves before saying that I'm wrong about this; it's surprising, but if you think about it carefully it has to be true.  Consider someone on the rocket ship receiving "ticks" from a clock on Earth; over the round trip, you have to receive the right number of ticks for the amount of time that passed on Earth, i.e., you receive more ticks over the round trip from the clock on Earth than from your own clock.]
[Addendum: the key point here, which I overlooked previously, is that the naive observer will do his calculations differently depending on whether he thinks that he is moving or that the Earth is; this both explains the asymmetry in the apparent time dilation and makes it clear why it isn't a good description of reality.]
On the other hand, if you were traveling at half the speed of light in a very big circle around the Earth you would see events on Earth taking place 16% slower than you would expect.  So the apparent time dilation (as measured by a naive observer on the rocket ship) varies depending on your relative direction as well as your relative speed.
When doing special relativity properly, we imagine a network of observers all traveling at the same speed but in different places.  This can be (somewhat improperly) simplified by supposing that the observer on the rocket ship has some way of seeing what was happening on Earth "right now" without being limited by the speed of light.  Technical note: when I say "right now" I mean "right now in the observer's frame of reference".  
In this hypothetical case the observer would see things very differently to the naive observer described previously.  On both the outgoing trip and the return trip (and also on a circular trip) events on Earth would be seen taking place 16% slower than normal.  This is the behaviour your teacher (and John) is describing; an observer on the rocket ship and an observer on Earth both see each other in slow motion.  This time dilation depends only on the relative speed, not on the direction of travel, and is real in a sense that the apparent time dilation I described earlier is not.
However, whenever the rocket ship accelerated, something odd would happen.
Suppose you were traveling away from Earth at half the speed of light for a year, took one day to turn around, and then traveled back, also at half the speed of light for a year.  As far as you're concerned you would be away for two years.  During the first year, on the way out, you would see 10.4 months take place on Earth.  During the second year, on the way back, you would also see 10.4 months take place on Earth.  During the turnaround, though, in a single day of your time you would see 6.9 months take place on Earth, making a total of 27.7 months.  At the end of the trip you've experienced 16% less time than the Earth has, but (from your point of view) it all happened when you accelerated; the rest of the time it was the Earth, not you, that was in slow motion.
From the point of view of an observer on the Earth, mind you, you were in 16% slow motion the entire time; you were in very slow motion while you were accelerating, but because that was only for a short time it didn't contribute significantly to the overall time dilation.  This is hard to get your head around, but that's reality for you. :-)
[If you don't happen to have a faster-than-light telescope, you can replace the above with a thought experiment involving a network of observers traveling in tandem with the rocket ship, all going the same speed and accelerating at the same time (in their mutual frame of reference); later on, you collect notes from all the observers recording when (in their frame of reference) they passed the Earth and what Earth date it was at the time.]
A: If you're sitting in the rocket then you appear to be stationary and it's the Earth that's moving. Therefore the people in the rocket will see time passing slowly on Earth. Of course we here on Earth see time passing slowly on the rocket. 
The situation has to be symmetrical because it's a fundamental part of special relativity that all frames are equal and there are no special frames. If the people in the rocket saw time passing more quickly on Earth that would prove to them that they were moving and the Earth was stationary, and SR forbids this.
You need to be careful about tossing around concepts like time dilation as it's easy to fall into conceptual traps and end up with a paradox. The only reliable way to do things is to sit down with a piece of paper, choose the events you're interested in then do the Lorentz transforms to find out what happens.
A: The universe actually follows general relativity but very closely approximates special relativity in Earth's gravitational field because it has an escape velocity that's much lower than the speed of light of 11 km/s. Suppose the universe exactly follows special relativity. According to special relativity, all objects experience time dilation by a factor of $\sqrt\frac{1}{1 - \frac{v^2}{c^2}}$ and length contraction by a factor of $\sqrt\frac{1}{1 - \frac{v^2}{c^2}}$. Also,the magnitude of relativistic momentum is $v\sqrt\frac{1}{1 - \frac{v^2}{c^2}}$.
Take the following thought experiment. You're in the middle of a long empty cylindrical space ship travelling at half the speed of light in the direction of the cylinder. Suppose you have a gun that simultaneously shoots two bullets in opposite directions at half the speed of light while you're at rest. Now let's suppose you shoot the bullets towards the two ends of the space ship while you're in the middle of the space ship. The one you shot towards the back end of the space ship will have zero speed and the one you shot towards the front end will travel at $\frac{4}{5}c$. The one you shot towards the back end will reach the back end before the one you shop towards the front end reaches the front end. However, the light from the event of the one you shot towards the back end will also take longer to get back to you. The math shows that you will observe the exact same result as you would if the space ship were stationary if you also take into account time dilation and length contraction.
What about if you shoot a bullet towards the front end at $42c$? Surely, the bullet cannot go backwards in time. Then the fact that so little time passed for you between firing the bullet and observing it hit the front end tells you that the space ship was moving in the forwards direction. It turns out that no matter can travel at the speed of light or faster. If an object keeps accelerating uniformly in its own frame of reference, as it gets closer to the speed of light, it undergoes more time dilation and length contraction and will never reach the speed of light for that reason.
If you can't see outside, you can't do any experiment to see whether you're moving or not. For any event you observe, the place and time you deduce them to have occurred under the assumption that the space ship is stationary is said to be the place and time they occurred in the frame of reference of the space ship. The place and time they occurred in the frame of reference of the space ship is completely determined by the place and time they actually occurred and is gotten by a Lorentz transformation. When two events are separated in space by more than the amount they were separated in time times the distance light can travel in that time, it is possible for a Lorentz transformation to change the order of the timing of the events.
There was an experiment with a device that under the assumption that matter follows Newtonian physics and light travels through a medium at fixed speed, it could be predicted that we could measure motion through that medium. That's probably because Earth revolves around the sun at about 30 km/s. However, no deviation was measured in the device. That's pretty strong evidence for the theory of General relativity which can be very closely approximated by special relativity except for extreme cases like near the event horizon of a black hole.
Source: https://en.wikipedia.org/wiki/Michelson%E2%80%93Morley_experiment
A: 
It is widely believed, that the reciprocal symmetry in Special Relativity is due to the isotopy of speed of light in all relatively moving ref. frames.  Indeed, if each observer synchronizes clocks in his reference frame according to Einstein Method, then each of them will find out that a single clock moving relative to his frame measures a shorter interval of time than (at least two) Einstein – synchronized and spatially separated clocks within his frame.
Of course, if measured by means of Einstein – synchronized clocks, the value of one–way speed of light will be equal precisely to c.
However, the assumption that the one-way speed of light must necessarily be isotropic in all relatively moving laboratories seems most uncommon and counter-intuitive, as well as that clock A may be slower than clock B and vice versa. Indeed, one cannot be slower than the other and vice versa, just as the speed of light cannot be isotropic in all reference frames.
The following simple experiment (to prove it we can safely refer directly to 1905 Einstein’s paper) clearly shows that this is not the case.
Consider a “stationary” observer A, who has a laser pointer. He directs this laser pointer strictly perpendicular to the direction of motion of observer B, who has a mirror (see animation). This laser pointer emits green monochromatic light pulses of proper frequency $\nu$.
At that moment, when observer B crosses this beam of light, he moves tangentially to the wavefront. At this instant A and B are at their (frame–independent) points of closest approach and there is no change in distance versus time. Hence, observer B will see a purely relativistic contribution of time dilation to the Doppler effect.
What frequency will observer B measure at this instant? For an answer let's look into the famous 1905 Albert Einstein paper, § 7.
“From the equation for $\omega‘ $ it follows that if an observer is moving with velocity $v$ relatively to an infinitely distant source of light of frequency $\nu$, in such a way that the connecting line “source - observer” makes the angle $\phi$ with the velocity of the observer referred to a system of coordinates which is at rest relatively to the source of light, the frequency $\nu‘$ of the light perceived by the observer is given by the equation":
$$\nu‘=  \nu \frac {(1-\cos\phi \cdot v/c)}{\sqrt {1-v^2/c^2}}$$ 
This is Doppler’s principle for any velocities whatever.”
Hence, according to A. Einstein, at points of closest approach $(\cos\phi = 0)$ the moving observer will measure $\gamma$ times higher frequency of light, or that the clock "at rest" is ticking $\gamma$ times faster than his own.
The moving observer will see the source not in its actual position, but in front of him at the angle $\sin \theta = v/c$. The moving observer can explain blueshift of frequncy by dilation of his own clock and displacement of apparent position of the source by relativistic aberration of light. Notice that the apparent distance to the source will $\gamma$ times increase to him due to contraction of his own measuring-rod. The relativistic aberration of light was clearly explained in the Feynman Lectures, Relativistic Effects in Radiation, 34-8.
The light reflected from a mirror will come back to the receiver (“stationary” observer) at the same frequency as it was once emitted. A moving mirror is nothing but a moving clock; hence, frequency $\nu‘$ reduces $\gamma$ times back and turns into frequency $\nu$.
It becomes clear that relatively moving observers can in no way measure each other's clock as dilated; if one emits light at the right angle, this light pulse will approach the other at an oblique angle and vice versa.
The source and the observer may also consider themselves moving with the same but opposite velocities within a certain frame of reference; it this case the source must turn the laser pointer backward and the observer–receiver forward at equal angles; in this case they will not measure any dilation of each other's clocks; (Since both the emitter and the receiver have the same speed relative to this system of reference, the dilation of both clocks is of the same magnitude and  there is no differential time dilation.)[see fig.4  in this artcle 7
One may argue that in the “moving observer‘s” frame this effect (blueshift of frequency) can be explained by the presence of a longitudinal component due to the source motion. Yes, it can.  So what? First, Einstein clearly stated that the observer was moving in the frame of the source; second, both the “observer” and the “source” would in no way be able to see redshift of frequency; if one measures a higher frequency, the other will definitely see a lower one.
All that taken together indicates the absurdity of Einstein - synchronization in each frame of reference. If observer A “makes” the one-way speed of light isotropic in his frame (in accordance with Einstein’s synchronization), then (so as to bring the measurement result by means of synchronized clock in accordance with the measurement of frequency) “moving” observer B must take into account his velocity in the frame of the first one and re-synchronize clocks in his frame accordingly.
In this way these observers will introduce universal simultaneity; the pair of clocks of observer A will be Einstein–synchronized and the pair of clocks of observer B will be Reichenbach–synchronized. In other words, if the laboratory of observer B moves in the frame of A with the velocity close to that of light, then, having taken into account the Lorentz contraction and slowed down rate of these clocks, measured by these clocks “forward” one–way speed of light will tend to c/2 and the “backward” one to an infinitely large value, thus maintaining the isotropy of the two - way speed of light
