# B 2 clocks -- Using orthogonal light path detectors in a space ship

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1. Oct 29, 2017

### Ross B

Hi

I cant see the error in this can someone please explain where I went wrong?

A man is in a space ship travelling at a constant velocity

He makes 2 identical tubes of length L with a mirror at one end, tube a and tube b

He has a single light bulb. Next to the bulb is a detector. He carefully measures that the bulb and the detector are equidistant from the mirrors.

He places the non mirrored ends of the tubes at the globe, with the tubes perpendicular to each other see the figure

The detector detects if 2 flashes of light arrive at the detector at the same time. The bulb flashed for a brief instant and the detector detects if the flash arrives from each tube at the same time.

The analysis for tube “a” is

T = ((L – vt)/c) + ((L + vt)/c)

(L – vt)/c = is the time taken for the light to leave the light source and travel to the mirror.

(L + vt)/c = the time to do the return trip.

T = the time taken for the round trip

The analysis for tube “b” is

T = 2L/square root(c^2 - v^2)

But

2d/square root(c^2 - v^2) does not equal ((L – vt)/c) + ((L + vt)/c)

There are 4 identical space ships, travelling at the same velocity, wrt and external observer, east, north south and west. They all conduct exactly the same experiment, simultaneously and the observer in each space ship all get the same result.

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2. Oct 29, 2017

### Ibix

If you are including the speed of the ship in the analysis then you are not working in the rest frame of the ship. What happens to the tube parallel to the motion of the ship in that case?

3. Oct 29, 2017

### Ross B

I dont understand - is that not the correct analysis for the tube parallel (tube a)?

The analysis for tube “a” is

T = ((L – vt)/c) + ((L + vt)/c)

(L – vt)/c = is the time taken for the light to leave the light source and travel to the mirror.

(L + vt)/c = the time to do the return trip.

T = the time taken for the round trip

Say the globe flashes at t1. If the space ship is moving at a constant velocity v wont the clock be in a different position in space when the flash arrives at the mirror say at t2?

4. Oct 29, 2017

### Ibix

In a frame where theship is moving, yes. But in such a frame, what happens to the length of tube a?

5. Oct 29, 2017

### Ross B

the length of tube a, according to the man at rest wrt tube a is L, as he measured it just before he made the tube?

all the external observer does is measure the velocity and direction of each space ship. They cannot see the experiment

6. Oct 29, 2017

### Staff: Mentor

It does for v=0, which is the case in the ships reference frame.

You can use this fact to derive the length contraction formula in other frames.

7. Oct 29, 2017

### Ross B

but v does not equal zero ...so I sort of fail to see yr point ?

I have no interest in length contraction, as the analysis is the analysis of the observer at rest wrt to the tube ? observer at rest wrt to the tube measured the tube at L just before he made the tube

As there are spaceships going east, north, south and west, according to the external observer, then all parties must agree that at least one space ship v does not equal 0

8. Oct 29, 2017

### Ibix

So v is zero ("at rest with respect to") for the person making the measurements. As @Dale points out, your formulae reduce to the same thing in that case.

9. Oct 29, 2017

### Ross B

but that is confusing. As there are 4 identical space ships, carrying out an identical experiment going north, south, east and west, all at v, simultaneously, they must all agree that at least one space ship v does not equal zero. In which case for any party at rest wrt the tube to assume v = 0 is an erroneous assumption isnt it?

10. Oct 29, 2017

### Ibix

No. They can each consider themselves at rest and the others to be moving. That's what the principle of relativity means, applied to this case.

11. Oct 29, 2017

### Staff: Mentor

In the frame where the tubes are the same length v does equal zero. In other frames the tubes are not the same length.

Where v is zero.

12. Oct 29, 2017

### Ross B

so if spaceship number one says Im at rest, your all moving, the others must concede they are in fact moving. In which case the results of the experiments in space ships 2 - 4 should not accord with the results of the experiment in space ship 1?

13. Oct 29, 2017

### Staff: Mentor

What you are describing is similar to the Michelson Morley experiment and your error was their (failed) hypothesis, which led to the recognition that there was no aether or universal rest frame:
https://simple.wikipedia.org/wiki/Michelson–Morley_experiment
Your basic error here is that you are mixing and matching frames inappropriately when doing the analysis. You can't measure the tube length in the stationary frame and then apply that length to a different measurement done in a moving frame. You must be consistent. And the easiest way to do that is for each ship to declare themselves stationary and then predict their experiment will show no motion.

If the other ships *could* observe each other's experiments, then you could have one ship monitor the experiment and see what results you would get -- but the catch is that they don't just have to measure the time remotely, they also have to measure the length of the tubes remotely. You cannot mix and match measurements in different frames.

14. Oct 29, 2017

### Staff: Mentor

No, the others are free to use whatever frame they prefer. That is the whole point of the principle of relativity. There is no universal "in fact moving". There is only moving relative to a given frame.

15. Oct 29, 2017

### Ross B

yep thats fine the guy in space ship one prefers to assert he is at rest. so the others must accept they are not at rest. It is impossible for them all to claim to be at rest. And in this context what does "at rest" mean. If someone claims to be at rest, and at least one space ship is not at rest, then does that mean the space ship that is at rest is at rest in an absolute sense?

16. Oct 29, 2017

### Staff: Mentor

You can analyze the problem using the frame in which spaceship #1 is at rest and the other three are moving. In that frame the tubes on spaceship #1 are the same length but the tubes on the other three ships are not.

Or you can analyze the problem using the frame in which spaceship number #2 is at rest and the other three are moving. In that frame the tubes on spaceship #2 are the same length but the tubes on the other three ships (including spaceship #1) are not.

Either way, if you do the analysis correctly you will calculate that on all four ships the two light flashes reach the globe together. No matter which frame we choose, we have three ships that are moving and have tubes of different length and one ship at rest and has tubes the same length.

17. Oct 29, 2017

### jbriggs444

The others are not obliged to do any such thing. All the measurements any one guy in a space ship can make is consistent with the idea that he is at rest. All of the measurements reported by the guys in other space ships are consistent with the idea that all the other spaceships are moving and that their measurements are affected by length contraction, time dilation and relativity of simultaneity.
On the contrary. Each one can claim to be at rest and all possible experiments are consistent with each such claim. What would be inconsistent with experiment would be for them to all claim to be at rest with respect to the same inertial frame.

18. Oct 29, 2017

### Janus

Staff Emeritus
Welcome to Einstein's universe. The fact is that all the rocket's have equal claim to being "at rest". Or put another way, there is no test or experiment that any one of them could perform that would tell them that they are "moving". You seem to be operating under the assumption that the speed of light(in a vacuum) is c relative to one unique frame. This is not the case, the speed of light is c relative to whoever is measuring it.
So for example, If you and a friend were moving at 0.5c relative to each other, and at the moment you passed each other you both set off a flash bulb, Each of you would see yourself as remaining fixed with respect the centers of the expanding spheres if light, while seeing the other moving with respect to the center of the expanding light spheres.

This also means that each of your spaceships would measure the round trip time for the light, no matter which direction in is pointed as being 2L/c.

As already noted this was the gist of the M&M experiment. It was trying to measure the Earth's motion by this method. Instead of using two spaceship, they just did the experiment at two different times. This was to guard against the possibility of the Earth's motion by chance being zero when you did your experiment. Since the Earth travels in an orbit around the Sun, at a different time of the year, it would be moving at a different velocity than it was when you first made the experiment. What they found was that they got a null result both times. The experiment failed to detect any motion of the Earth either time.

This was not expected, and there were a number of attempts made to explain it. Einstein's Theory of Relativity came out as the accepted answer. Not only did it explain the result, but it made predictions of a number of effects that could be (and have been) tested by experiment or observation. And while many of the concepts in Relativity appear counter-intuitive( Many people have trouble with abandoning the comfortable idea of absolute space and time), it has proven to be very accurate in describing how the universe works.

19. Oct 29, 2017

### m4r35n357

To misquote Janus badly, welcome to Galileo's universe. You do not need to invoke special relativity to have this discussion, here is how Galileo describes relativity and inertia nearly 400 years ago: "nor could you tell from any of them whether the ship was moving or standing still".

None of this is Einstein's fault ;)

20. Oct 29, 2017

### Ibix

He doesn't have to claim to be at rest. But if he claims not to be at rest then he has to remember that his clocks are time dilated so don't measure time correctly, and his rulers (those that aren't perpendicular to his chosen direction of movement) are length contracted and don't measure distance correctly. It's a bit of a pain to do it but you can do. If you work it out correctly (remember the relativity of simultaneity) then your pulses will return simultaneously however you do it.