Detecting absolute motion inside a box

Question

This is not a contradiction and I know it is impossible but still consider a thought experiment by me and point out if something is wrong. See the following picture and then the explanation follows.

Rest frames are easy to understand. They are just for clarification. Lets move on to the moving frames. The velocity of the box is $0.1\ c$. Now a photon is emitted (I am not taking a light ray to avoid complication in discussion). After emission the box also moves a certain distance ahead. So the photon takes more than one second to reach the wall. But even the source moves that much distance ahead. When the light is reflected back, the source and the wall again move forward as the box is moving. But after reflection the wall has no role to play. We are concerned about the source. The source moves a certain distance ahead and therefore light takes less than one second to reach back. But the difference of the first case and the second case is not the same. To explain, I will give some equations.

Moving frame 1:

Velocity of the box: $0.1\ c = 30,000km/s$

Time taken by light to reach the wall $=\frac{330,000\ \text{km}}{300,000\ \text{kms}^{-1}}= 1.1s$

Distance moved ahead by the box from original position: $30,000\ \text{km}$

Moving frame 2:

distance moved ahead by the box from original position $>30,000\ \text{km}$ as the box also moves ahead after reflection.

let $d$ be the distance moved ahead by the box after reflection.

Time taken by light to reach the source: $\frac{270,000\ \text{km}\ -\ d}{300,000\ \text{kms}^{-1}} <0.9s$

Therefore we see that this could possibly determine the absolute motion.

So in total we see that the light will take lesser time to reach the source. Please correct me if I am wrong and give your opinion why is it so.

EDIT 2 for clarification

The box is the compartment in space in which you are moving and I am calculating the time the light takes to reach back the source. I am not even adding velocities. Please correct me and tell me where did I add velocities. For clarification just let me give a simple example. Suppose person A is standing still and B is running towards a light source. Obviously he will see the light before person A even though the speed would be the same. So, what I am doing over here is not increasing the speed of light, but instead decreasing the distance to be covered by it.

Answer 1

The key point that you are missing is that the speed of light is constant for all inertial frames of reference. If you are going $0.99\ c$ and you are holding a flashlight and you turn it on, the photons emitted from the flashlight will appear to you to be leaving the flashlight at exactly $c$ (the speed of light). The key point of special relativity is that all inertial frames are the same - the notion of "absolute" motion does not really exist. That's why it's called a "theory of relativity" - the big conclusion is that your motion can only be described relative to another frame of reference.

What you are describing is one of Einstein's classic thought experiments. In his version there is an obverver on a moving train and an observer watching the train go by. Both observers will see the beam of light moving at the speed of light. What the observers will not agree upon, however, is the rate of passage of time and the length of the train carriage. To the observer on the ground (outside "the box"), the moving box will appear shorter in the direction of motion. Time will appear to pass more quickly for the moving observer in the train. It is precisely because the speed of light does not change for observers moving at different speeds that both time and space must grow or shrink to make up the difference.

Answer 2

The problem is that your analysis is all done from the perspective of the frame that measures the box to be moving at 0.1c -- in this frame, it's true that the time for the light to get from the source to the wall is different from the time for the light to get from the wall back to the source. But if these same events are measured by someone inside the box (or just moving alongside it and looking in), using rulers and clocks at rest relative to the box (and with clocks at either end of the box synchronized in the box's rest frame using the Einstein synchronization convention), then the person inside the box measures the same time for the light to get from the source to the wall as he measures for the light to get from the wall back to the source. To see this you must take into account three different ways in which the rulers and clocks of the person inside the box seem "off" from the perspective of the person who sees the box moving at 0.1c: length contraction, time dilation and the relativity of simultaneity.

Let's call the person who sees the box moving at 0.1c "Alice", and the person inside the box (and at rest relative to it) "Bob". Then length contraction means that if the box has a length of 300,000 km in its own rest frame as measured by Bob's rulers, Alice sees the length of the box as shorter by a factor of $ \sqrt{1 - v^2/c^2} $, in this case $ \sqrt{1 - (0.1c/c)^2} $ = 0.9949874 (and she also see's Bob's rulers as shrunken), so she measures the box to have a length of only 0.9949874 * 300,000 km = 298,496.2 km. Time dilation means that Alice would see Bob's clocks ticking more slowly than hers, so the length of each tick of Bob's clock is longer by a factor of $ 1 / \sqrt{1 - v^2/c^2}$, which is equivalent to saying that if t seconds go by on Alice's clock, Alice measures only $ t * \sqrt{1 - v^2/c^2} $ seconds to have gone by on Bob's clock, or 0.9949874 seconds on Bob's clock for every second that goes by on Alice's clock, as measured in Alice's frame. And the relativity of simultaneity means that if Bob had "synchronized" clocks at either end of the box in his frame, and measured them to be at a distance of 300,000 km apart or 1 light-seconds apart (I will assume as you did that the speed of light is exactly 300,000 km/second even though this is not quite right), then in Alice's frame the two clocks are slightly out-of-sync, at any given moment the clock at the back end of the box (where the source is located) shows a reading that's ahead of the clock at the front of the box (where the reflecting wall is located) by (0.1c)(1 light-seconds)/c^2 = 0.1 seconds.

Say in Alice's frame the back end of the box is at position x=0 at t=0 in terms of her space and time coordinates, and at that moment it emits light. Since the box has a length of 298,496.2 km in her frame, the front of the box must be at x=298,496.2 km at t=0. The position of the front of the box as a function of time in Alice's frame is 30,000*t + 298,496.2, and the position of the light emitted from the source as a function of time is 300,000*t, so the light will catch up with the source when 300,000*t = 30,000*t + 298,496.2, or t=1.105541 seconds, at position x=331662.3 km in Alice's frame. Then since the position of the back of the box as a function of time is 30,000*t, and the position of the reflected light as a function of time is -300,000*(t - 1.105541) + 331662.3, the reflected light will get back to the source when 30,000*t = -300,000*(t - 1.105541) + 331662.3, or when t=2.010075 seconds. Thus in Alice's frame the light took 1.105541 seconds to get from the source at the back of the box to the wall at the front, and it took (2.011075 - 1.105541) = 0.905534 seconds to get back to the source after being reflected by the front wall.

So in Alice's frame, at the moment the light is emitted by the source, let's say Bob's clock next to the source reads T=0 seconds, in which case Bob's clock at the other end of the box will be behind by 0.1 seconds (due to the relativity of simultaneity discussed above), reading T=-0.1 seconds. Then if Alice measures a time of 1.105541 seconds for the light to reach the front end of the box, each of Bob's two clocks will only have ticked forward by 0.9949874 * 1.105541 seconds = 1.1 seconds (due to time dilation). So in Alice's frame, at the moment the light reflects off the front wall, the clock at the back of the box next to the source reads T = 0 + 1.1 = 1.1 seconds, while the clock at the front wall where the light is striking reads T = -0.1 + 1.1 = 1.0 seconds. So, Bob measures the ship to be 300,000 km long, and he measures the light to have departed the source at T=0 seconds and to have arrived at the front wall at T=1.0 seconds, meaning he measures the light to have traveled at exactly 300,000 km/s in moving from theback wall (where the source is) to the front wall. Likewise, if the light is reflected and returns to the back wall where the source is located at t=2.011075 seconds in Alice's frame, during that time Alice sees Bob's clock to have only ticked forward by 0.9949874 * 2.011075 = 2.0 seconds, so the light returns to the source when Bob's clock there shows a time of 2.0 seconds. So, since Bob measures the ship to be 300,000 km long, and he measures the light to have been reflected from the front wall at T=1.0 seconds and to have returned to the source on the back wall at T=2.0 seconds, that must mean he also measured the light to have taken 1 second to travel 300,000 km when moving from the front wall to the back wall.

Finally, note that we could make the situation completely symmetrical by imagining a similar box at rest in Alice's frame, with Alice measuring it to be 300,000 km long by her rulers, and with her having two clocks at the front and back which were at rest in her frame and synchronized according to the Einstein synchronization convention. Then Bob would similarly measure Alice's rulers to be shrunken, and her clocks running slow and out-of-sync, and if she sent a light beam from one end of the box to the other and back, she would measure the light ray to take 1 second in each direction using her clocks while Bob would measure it to take 1.105541 seconds in one direction and 0.905534 seconds in the other, using his own clocks.

Answer 3

Notice that the movement of light must always be treated as a local phenomenon, which in your thought experiment means the light is travelling only within the moving box. It is not travelling from the moving box to a stationary observer. Therefore, throughout the whole experiment the box is stationary for the light and no wall is moving toward or from it.

This is exactly equivalent to you moving back and forth in a train. You don't need to move faster when you are walking toward the front of the train because the train is always stationary for the passenger.

You might say: "OK, but from the outside (stationary) point of view the walking person is traveling with the velocity of the train +/- the velocity of the walker. If this adding/subtracting does not apply to the light, then how does the outside observer explain it?" Well, he doesn't really need to. The outside observer cannot actually see the light (nobody can see light that is not coming directly to their eyes), so there is no adding/subtracting needed. The outside observer simply cannot measure the light moving inside the box. He can never obtain independent data concerning its speed.

As to your edit (two person experiment): The same explanation applies. Only the person B will see the light. Person A is not able to see it, because the light is travelling toward person B, and therefore only person B can measure its speed. Consequently, there will be no problem here as well.

Answer 4

As noted within my other post at Time dilation in special relativity, here I have basically the same situation yet the box becomes a 300,000 km long spaceship. The basics are explained. However, the only absolute measure that exists within it, is that all objects are constantly traveling at the speed of light within the 4 dimensional environment known as Space-Time continuum, just as explained by physicist Brian Greene.

All of the equations used in Einsteins theory of Special Relativity are valid between a stationary body, meaning a body at rest in space, and a body that is in motion across space. However, the equations are also valid between two spatially moving bodies. Thus there is no way to determine any absolute measure other than the fact that all objects are constantly in motion across the 4 dimensional Space-Time in one direction or an other, at the speed of light.

Answer 5

Your numbers and reasoning are OK, BUT ...
It is false your conclusion: "Therefore we see that this could possibly determine the absolute motion".

Only the mean speed of light in a closed path can be measured.

You have a precise clock and you will wait the return 2 seconds later and conclude the wall is 1 light sec away. What else can you conclude?

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In a different way: Use microwave antennas and watch the CMB periodic and directional frequency value.
We can conclude, as science do, the daily and annual cycles of Earth around the axis and around the Sun (SSB in fact).
A bonus: we find the absolute motion of the Earth and the Solar System to be 369.2 km/s in a specific direction (Leo constellation).

Answer 6

Your calculations are correct. The light does indeed take longer to reach the other end of the box when they are moving in the same direction. From the point of view of the observer at rest.