# Perceived direction of light emitted in moving reference frame

I was thinking the other day about the simple example used to demonstrate time dilation effects and to derive the Lorrentz factor - where the time it takes for a light pulse to be emitted, bounce of a mirror, and then detected, is seemingly different to someone in the reference frame of the light pulse and somebody for whom that reference frame is moving.

Thinking about this example, though, I noticed something that I couldn't quite figure out. It's been several years since I took a class on relativity, and I'm sure there's an answer, but I couldn't find it.

My issue is this: for the person in the reference frame, the motion of the light makes sense - it is shot out of the emitter, travels in a straight line, bounces off the mirror, and is received by the sensor. But for the outside observer, for whom the reference frame of the light pulse is moving, the motion of the light doesn't make sense. When the light leaves the emitter, it will be at one point. However, by the time the light reaches the mirror, that mirror will be at a different location than where it was when the light was emitted from the emitter. As far as this observer is concerned, the emitter was emitting the light in some direction, say perpendicular to the motion of the reference frame, and yet that light still hit the mirror, a path which would not have been perpendicular to the motion. This would suggest that either the light was emitted at an angle, which it shouldn't be because the direction of the emitter should be unaltered, or that the light somehow carried with it some momentum from its reference frame that allowed it to travel diagonally, which it obviously shouldn't.

So how do we rectify this dilemma? How do we explain the motionless observer seeing light travel in a diagonal direction when it should be going straight?

• "So how do we rectify this dilemma?" It isn't rectified. It is ignored. "How do we explain the motionless observer seeing light travel in a diagonal direction when it should be going straight?" Well, there is nothing to explain. The motionless observer simply cannot see it. The light can be seen/measured only in the frame, where it is detected. So if it is recorded by a detector located in the moving frame, nobody in the stationary - or in any other frame - can see it, moving straight or at an angle. Simply, you cannot see light that does not come to your eye. – bright magus Dec 4 '14 at 9:28
• @brightmagus Granted, they don't SEE it, but they will be able to notice the results of the beam of light travelling diagonally - for instance, if they're able to receive a signal sent by the light emitter and sensor whenever a pulse is sent/received. The point isn't that they see the light itself, but that from their perspective the light is travelling diagonally rather than straight out of the emitter. Why would this be ignored? – MattS Dec 4 '14 at 9:32
• OK, but there is one important notice. If a person in one frame receives a signal sent from another frame, he is not comparing his data to other frame's data. He merely receives data sent from another frame (the signal will be sent by the sensor after the light traveled a given distance at $c$ within local time $t$). What is ignored is the problem with the apparent angle. Why? Perhaps because if one assumes light travels two different distances (straight as measured locally and diagonal as inferred from the outside) the claim of the constancy of $c$ in any frame might become questionable ... – bright magus Dec 4 '14 at 10:00
• A path that is vertical in one frame is, in general, not vertical in another. It really is as simple as that. The path of the light beam is by assumption vertical in frame B, hence diagonal in frame $A$ – WillO Jan 5 '19 at 0:06
• These comments by @bright magus are at best not helpful, and the last one is plain wrong. – PM 2Ring Jan 5 '19 at 2:51

We can imagine that two observers move relatively to each other and play some kind of ping pong with a photon. Each of them possesses a tube. Through this tube observer spits a photon out and another observer catches it. They have to tilt their tubes so that the photon passed through them.

So as a photon would pass through the both tubes, these observers must always tilt their tubes at corresponding angles.

For example, if both observers keep their tubes at right angle, they in no way will be able to see each other.

If observer A keeps his tube straight up and sends a photon, observer B has to keep his tube at oblique angle into front so as to catch it. He keeps his tube into front exactly as astronomers do looking at distant stars.

Astronomers keep their telescopes at small angle because of aberration of light. It takes some time for the light to pass through a tube, and if a tube is not tilted the photon will hit a wall.

That’s why a moving observer sees a source as if in front of him, though actually the source is exactly “below”.

These angles of directions of the tubes are always tied with each other by relativistic aberration formula

https://en.wikipedia.org/wiki/Relativistic_aberration

An observer can keep his tube at ANY arbitrary chosen angle. But, if they want to see each other, another observer must to adjust his tube accordingly. For example, they can ascribe themselves roughly half of relative velocity each. In this case they must tilt their tubes at equal angles. One observer tilts backward, another – forward.

Episode 3 – Rectilinear Motion. The both source and observer are “in motion”

But, we can consider this case either from reference frame of observer at rest or from the reference frame from moving observer.

If observer A is at rest in origin and keeps his tube straight up, a photon will travel along the y axis, and will come to a moving observer at oblique angle. Thus observer B has to keep his tube at oblique angle and to look into front. If observer B has a mirror he must have another tube which is tilted at the same angle back so as a photon would be able to jump out. Then photon travels back to A and comes back at right angle. This is Transverse Doppler Effect, and observer B is a moving clock. Episode 2 of this video.

In observer B frame observer A keeps his tube straight up, but photon always moves together with observer A and passes through a tube. Good to note, that in this case photon hits observer B at the points of closest approach, exactly as in the previous episode. Please look at the Episode 1 of this video

This video graphically demonstrates that angles of direction of tubes are equivalent in both episodes.

Good to note that a photon "brings back the same amount" of energy when it was released. If a photon was released straight up and had "green" frequency, the photon will be blue shifted at mirror. Mirror gains energy when swallows photon and immediately loses the same amount when reflects it. Thus, photon was "blue" at mirror but will red shift and will turn "green" again when will come back to the source.

Indeed, your question has nothing necessarily to do with relativity. Even in Newtonian kinematics/dynamics detecting such an obliquity in the path of emitted objects are inevitable from the viewpoint of many inertial observers.

Instead of photon, assume that a tiny ball bounces between the floor and the ceiling of a moving train such that the observer in the train's rest frame measures the path of the ball perpendicular to the floor/ceiling (motion direction, that is).

From the viewpoint of the observer at rest in the station, the path of the ball is inclined without applying any relativistic amendments. Applying a little imagination, you can omit this dilemma using some geometry plus "motion".

When it comes to light and frame of references and observers, a quick reminder (you all already know) is that: the second postulate states the idea of speed of light is a constant C. This may seem very straightforward and easy to understand based on practical experience of electromagnetic waves including light beams that we have on our daily life. However, you need to bear in mind that distance and time are defined or measured now relative to this constant ratio C regardless of the frame or observers.

Observers in two or more frame of references may plot different trajectory for the path that light has traveled, however the ratio (the rate of change) of distance over time remains constant C in all frames.

This means if observer 1 in frame 1 starts plotting (a point in his co-ordinates) for each tick of his clock for each location that light photon (or wave) has appeared (reached) in his frame, then measures the distance D between two consecutive points [P1, P2] for two consecutive ticks (T), then he should observe same distance traveled for each period
|D| = C . |T| in this frame 1 co-ordinates. Of course if the source of light moves, or frame 1 moves according to the source of light, then the trajectory observed at this frame may not be a straight line anymore. Nonetheless, the distance measured with this frame clock always remain the same value. Nothing affects this ratio (this rate, or speed).

Now if observer 2 in frame 2 had started plotting also (a point in his co-ordinates) for each tick of his clock for each location that light photon (or wave) was appeared (reached) in his frame, then measures the distance D' between two consecutive points [P'1, P'2] for two consecutive ticks (T'), then he should also observe the distance traveled over each period is the same (which C). But the distance measured may not necessarily be the same distance in comparison to frame 1 (D might not be D'). He may also observe different trajectory (or light path) in his co-ordinate when points are plotted. Therefore, when distance measured in this frame co-ordinates, he may see his ruler (or two points distance in space) reads as value D'. Also the time measured may not be the same in this frame. And tick periods may be measured as T' now instead. (weird? yeah compared to classic mechanics that time is assumed universal, but remind that start a habit to not think of time a universal metric, define time now based on speed of light, and distance travels in the frame of reference). So, in frame 2 co-ordinates, we have: |D'| = C . |T'|

So back to your question, these are key things: Distance and Time are must be measured in each frame correspondingly. To synchronize a clock in a frame, speed of light is the key. Time is measured by the distance traveled in that frame. T = D / C Two observers (two frames)'s measurement of the distance traveled or time spent may not evaluate to the same values. However, when it comes to light, light has traveled a straight path space with same ratio (speed). [assume not bent by gravity around it] Two observes also may not experience same trajectory or same paths plotted in their experience either. [This is true for any objects in motion] Time is not universal.