Is light affected by inertia?

Question

A very popular way used by teachers to explain Einstein's theory that The speed of light is a Universal Constant, is to use an example as follows:

Two observers moving relative to each other [let's say flying past each other in rockets].

• Constant Relative Speed = $0.8 c$.
• observer A is carrying a photon clock and fires a laser to a mirror 1.5 meters away.
• Light bounces to a detector.
• Each measures how long it takes between "ticks" of the photon clock.

Observer A's Observations  

• Photon Speed = $c$
• Path Length = $3$ m.
• 1 Tick = $10^{-8}$ sec.

Observer B's Observations   

• A's Speed = $0.8c$
• Photon Speed = $c$
• Path Length = $5$ m
• 1 Tick = $1.67\cdot10^{-8}$ sec

Conclusion. There is no absolute time. but the speed of light is Universal.

Now to my question: 

• We know that the reason why an object thrown in a moving vehicle will fall back in my hands, is do to due to inertia.
• light has no mass and therefore that law inertia do not apply.
• Now my conclusion would be, that in the above example, the light will not bounce back to the detector at all, since the detector is moving at $0.8 c$, and the light is coming back from the mirror in a straight line.

I found that the above example is used by a countless number of teachers and online learning programs and books (I wonder if Einstein himself used it) . It's hard to believe that they are all wrong. Can anybody please explain that? 

P. s. all I know about physics is from searching google or asking questions like that, so I'd be happy if you can write in a way I can understand thanks.

Answer 1

I order to address your question I need to get something out of the way first.

As we know, special relativity is well corroborated. There are the well known confirmations. The muons created in the upper atmosphere that make it all the way to the Earth's surface. In particle accelerators we have that unstable particles have a longer half life than they would have when stationary with respect to the observer, in accordance with special relativity.

It is experimental evidence like that that justifies confidence in a theory. This confidence then justifies confidence in the assumptions that underly the theory.

In the case of special relativity there is that crucial underlying assumption that is commonly expressed as the light postulate: for all members of the equivalence class of inertial coordinate systems the speed of light is the same.

My point is: the purpose of the light clock demonstration is to start with assuming that the speed of light is the same for all members of the equivalence class, and proceed to work out the ramifications of that.

It is incorrect to present the light postulate as a conclusion.

To your question.

In my opinion it is always best to make the setup completely symmetrical, as the phenomenon itself is symmetrical. Keeping the symmetry in mind may help you catch an error in your thinking.

We give each spaceship its own light clock. On each spaceship the light remains inside that spaceship. Each observer creates two plots: one plot represents the motion of the light inside his own spaceship, the other plot the motion of the light inside the other spaceship.

The setup is aligned such that the motion of the light inside each spaceship is perpendicular to the direction of the relative velocity between the two spaceships.

From here on I will use 'perpendicular' for the motion direction perpendicular to the direction of the relative velocity between the two spaceships. (And the same thing for 'parallel', of course.)

Each observer will plot the motion of the light in the other spaceship as a zigzag motion, as he is plotting that motion in his own coordinate system. The faster the relative velocity of the two spaceships, the more elongated the plot of the zigzag.

In each plot the velocity of the light can be decomposed in two velocity components: a perpendicular component and a parallel component.

In each plot: as the light is approaching the detector that it is aimed at it does not have a velocity component in the direction parallel to the detector.

So it's unclear what your concern is. This bouncing scenario will work in analogous way if you perform it with actual particles that are set up for perfect elastic bounce.

That is, even with particles with a rest mass an analogous scenario is possible.

By contrast: in a scenario where one ship is emitting light and the other spaceship is receiving that light, that is a whole different ball game.

The scenario that I discussed (light bounces inside each spaceship) is presented in this light clock interactive animation

It is part of a larger collection of interactive animations by Michael Fowler.

Answer 2

Instead of thinking about inertia it is clearer to think about momentum. When you throw a ball upwards on a moving train, it has two components of momentum: horizontal (due to the train's motion) and vertical (because you threw it). These two components are independent (ie motion in one direction has no impact on the motion in a perpendicular direction), the gravitational force causes a change in the vertical momentum, but there is nothing to reduce the balls's horizontal momentum. Hence, the ball continues to move horizontally, staying in line with the detector, but moves up and down because you and the earth have exerted forces on it.

Now moving on to light, although light has no mass it has energy. This energy is related to the photon's momentum as part of general relativity, $E= pc$ where $p$ is momentum and $c$ is the speed of light. The momentum of light obeys momentum conservation. In the train scenario, the light constantly travels horizontally (same speed as the train) and continually reflects of off the mirror (photons have no mass, so gravity doesn't slow them down). If you were on the train the only thing you would see is that photon bouncing up and down in the same spot - because you are moving at the same horizontal speed as the train and photon. But if you were at the station platform you would see the photon taking a zig-zag path, because you are stationary compared to the train (and everything inside it).

I think the cause of your confusion is you are unfamiliar with inertial reference frames and some momentum concepts. I recommend reading some introductory special relativity / google.

Answer 3

It is easy to be confused by this when you first encounter light clocks, but if you think about it in the right way you will get to understand it.

The underlying point to bear in mind is that directions of motion are frame dependent. Imagine that you shine a laser at a mirror above you so that the light bounces back to you. In your frame, the laser is shining vertically. If I travel past you at some speed, then in my frame the light from the laser is taking a zig-zag path. Conversely, if I shine a laser at a mirror above me so that it bounces back to me, then in my frame my laser is shining vertically, but in your frame my laser is taking a zig-zag path.

If you shine a laser at an angle so that it hits the mirror on the ceiling and bounces back to a spot on the ground a little way from you, then it is talking a zig-zag path in your frame. However, if I go past you at just the right speed, your laser will take a vertical path in my frame.

So, you should be able to see that a vertical path in one frame is a zig-zag in another, and vice versa.

This applies to any motion. If I walk north along a street, then relative to a person walking west, my direction of motion will be north east. Conversely, if I walk north west, then relative to a person moving west I am moving north. So, you can see that direction of motion is frame dependent.

Answer 4

I think actually that the light-clock experiment will work if light were confined and restricted to travel in a tube or cable stretched perpendicularly from the light to the detector. Otherwise, the light will miss the detector, for the reasons stated above. The problem might be in the way the experiment is usually described or illustrated.