# Relativity question

Simplex0
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Topic 192001

Imagine that you have a continues light beam traveling from A to B on the left side and one traveling from B to A on the right side and that you are traveling from A towards B in a rocket which have a pair of photo detectors mounted on the left side, one in the front and the other in the back, and an other pair of detectors mounted on the right side in the same way. The distance between the 2 detectors on the left side is the same as distance between the detectors on the right side. When a shutter is opened a light beam hits the first detector and starts a clock on the same side and it stops when the light is detected by the second detector. Will the 2 detector clocks always show the same time despite the velocity of the rocket?

Solomon
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### Relativity question

Quote:
Imagine that you have a continues light beam traveling from A to B on the left side and one traveling from B to A on the right side and that you are traveling from A towards B in a rocket which have a pair of photo detectors mounted on the left side, one in the front and the other in the back, and an other pair of detectors mounted on the right side in the same way. The distance between the 2 detectors on the left side is the same as distance between the detectors on the right side. When a shutter is opened a light beam hits the first detector and starts a clock on the same side and it stops when the light is detected by the second detector. Will the 2 detector clocks always show the same time despite the velocity of the rocket?

Yes. Assuming the setups are truly identical, they will always show the same time, no matter what speed you are travelling at.

Odysseus
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### RE: Imagine that you have a

Quote:
Imagine that you have a continues light beam traveling from A to B on the left side and one traveling from B to A on the right side and that you are traveling from A towards B in a rocket which have a pair of photo detectors mounted on the left side, one in the front and the other in the back, and an other pair of detectors mounted on the right side in the same way. The distance between the 2 detectors on the left side is the same as distance between the detectors on the right side. When a shutter is opened a light beam hits the first detector and starts a clock on the same side and it stops when the light is detected by the second detector. Will the 2 detector clocks always show the same time despite the velocity of the rocket?

Yes; in the rocketâ€™s inertial frame of reference, light will always be observed to travel at c. However, if the detectors are equipped with spectrographs the left-hand beam will appear to be red-shifted (with decreased frequency) and the right-hand beam blue-shifted (higher frequency), due to the Doppler effect. This implies that photons colliding with the left-hand detector (from the rear) will impart less energy than those on the right (from the front).

Simplex0
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### Assume that you travel 1

Message 48959 in response to message 48958

Assume that you travel 1 light year and at a speed that will take the light 1 year, relative to A, fore the light emitted from A to go from the detector in the back to the detector in the front.
I can see that you can use the explanation that the time is moving slower on the rocket and that it is compressed in length to explain that the detectors on the rocket will clock the speed of light unchanged in this case.

But the light going from B to A will travel a much closer distance, relative an observer on B, between the detector in the front to the detector in the back and will cover the distance in a fraction of a second.

The clock on each detector pair on the rocket must suffer from the same amount of slowing so what is the explanation that they will measure the same time of the light traveling between the detectors?

Chipper Q
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### RE: Assume that you travel

Message 48960 in response to message 48959

Quote:

Assume that you travel 1 light year and at a speed that will take the light 1 year, relative to A, fore the light emitted from A to go from the detector in the back to the detector in the front.
I can see that you can use the explanation that the time is moving slower on the rocket and that it is compressed in length to explain that the detectors on the rocket will clock the speed of light unchanged in this case.

But the light going from B to A will travel a much closer distance, relative an observer on B, between the detector in the front to the detector in the back and will cover the distance in a fraction of a second.

The clock on each detector pair on the rocket must suffer from the same amount of slowing so what is the explanation that they will measure the same time of the light traveling between the detectors?

Not sure how A could make such an observation of the clock's time on the rocket (or if A observes 1 year, that won't be the elapsed time observed on the rocket); if light was observed to take that long to go from the detector on the back of the rocket to the detector on the front, then A would observe that the length of the rocket (or distance between the detectors) would be different than what would be measured if you were on the rocket. But what A observes is different from what B observes, and both of these observations are different from what is observed in the rocket's inertial frame. So the explanation Odysseus pointed out hasn't changed: in the rocketâ€™s inertial frame of reference, light will always be observed to travel at c.

Simplex0
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### RE: Not sure how A could

Message 48961 in response to message 48960

Quote:

Not sure how A could make such an observation of the clock's time on the rocket; if light was observed to take that long to go from the detector on the back of the rocket to the detector on the front, then A would observe that the length of the rocket (or distance between the detectors) was 1 light year.

He could se a reflection of the light from B occuring at the same time he se a reflection from the front detector.

Quote:

But what A observes is different from what B observes, and both of these observations are different from what is observed in the rocket's inertial frame. So the explanation Odysseus pointed out hasn't changed: in the rocketâ€™s inertial frame of reference, light will always be observed to travel at c.

The question is why it is. The clock on A is under the same condition as the clock at B and the clock on the left side of the rocket is under the same condition as clock on the right side of the rocket.

Simplex0
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### Is the length of the moving

Message 48962 in response to message 48961

Is the length of the moving rocket, from the rockets pov, depending on in which direction it is messured?

Chipper Q
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### RE: RE: But what A

Message 48963 in response to message 48961

Quote:
Quote:
But what A observes is different from what B observes, and both of these observations are different from what is observed in the rocket's inertial frame. So the explanation Odysseus pointed out hasn't changed: in the rocketâ€™s inertial frame of reference, light will always be observed to travel at c.

The question is why it is. The clock on A is under the same condition as the clock at B and the clock on the left side of the rocket is under the same condition as clock on the right side of the rocket.

The 'same condition' is that light is always observed to travel at c. If you are in a frame of reference that's moving, the measurements you make of objects/events will be different from measurements of the same objects/events made from a stationary frame of reference. Still trying to understand it myself, but if you perform the proper mathematical transformation from both frames of reference to an inertial frame of reference of light's velocity, then the measurements agree. I hope that's correct.

Simplex0
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### Thank you Solomon, Odysseus &

Message 48964 in response to message 48963

Thank you Solomon, Odysseus & Cipper Q. I gues I have take a look at the experimets that have ben done so fare.

debugas
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Message 48965 in response to message 48964

between what is actually observed and what is calculated to constitute your simultanious reality

When you read that moving object's time is slower it means that standing still observer calculates (in simultanious with him reality) the moving object's time to pass slower. What he sees however depends if the moving object is going away from him or is coming to him. In first case the observer sees the object to be living slower and in the second case the object is seen to be living faster than the observer himself.

Relativity is often taught by taking flat 3D projections of real space-time that are compatible with what we are accustomed to see in our ordinary life - namely the flat 3D space extending infinitely and being simultanious with us (one absolute time everywhere ticking simultaniously everywhere). However in real space-time those "simultanious reality" projections (what observer calculates to be simultanious with him - the 3D space consisting of all the "there-now" dots ) this simultanious reality is dependent on observer.
And often in relativity books its strange effects are explained in terms of "simultanious reality" and not in terms of what observers really see with their eyes. For example one says Sun is now 8 light-minutes away but in reality it may have exploded already and we will only know it after 8 minutes passes - then we can compare our calculated reality with what really happened

I dont know why authors (especially of popular science literature) feel the need to explain relativity in terms of ordinary space and almost always do not even explain to the reader that they talk in terms of the imaginary simulatinous reality projections. I find it misleading and creating more misunderstanding than clarifications of what real space-time is

debugas
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Quote:
The clock on each detector pair on the rocket must suffer from the same amount of slowing so what is the explanation that they will measure the same time of the light traveling between the detectors?

Here i will talk about what you actually see when travelling on the rocket from A to B.
First of all you see observer A living slower than you and observer B living faster than you. That means if you all 3 had the same clock mechanism counting seconds you would see that during your own one second time the observer's A clock only passed 1/3 of a second and observer's B have passed already 3 seconds (it depends on your speed in the example i take it to be 0.8c ). What is interesting here is that observer B and observer A see their clocks passing equally because they are at rest relative to each other.

Side note about imaginary simultanious with you reality:
To make it non-contradictable with your intuitive flat 3D space you think to be simultanious with you all over the universe you have to calculate that space has contracted 1.(66) times and that time has to be moving slower for A and B at 0.6 of your own rate

Now your trouble is in defining how much the two light-beams are packed.
Here i mean that you have to define how many seconds of observer's B life are packed in one meter(or e.g. in the lenght of one light-second) of light-beam coming from B and the same about observer A and his light-beam. Your eyes show you that B life is packed more dense than observer's A life.

the light that passes during your on-board one second time
will show 1/3 of observer's A life passed and 3 seconds of observer's B life
but these figures do not define the speed of light