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Question

If we use two lasers and put it on the equator and shine one of them in the east and the other one in the west and we have a screen on the same distance (10 km to the west and 10 km to the east), will they arrive at the same time? and why?
My main point is one laser shine with the rotation of the earth and the other one on the opposite direction
from what I know that the light doesn't affect by the rotation of the earth so the laser on the west side will arrive faster than the east side since the screen on the west side will come to the laser reference point.
Correct me if I'm wrong. (with mathematics if you can)

M. Enns 3,8141925 · Answer 1 · 2018-10-18 20:06:14Z

It depends on your frame of reference.

If you are positioned by the laser on the Earth's surface you would observe the two laser beams to have the same speed and your would observe the two targets to be stationary and you would observe the light to take the same amount of time to reach each target.

On the other hand if you were not at rest compared to the laser and screens, say for example you are observing from the moon, you would still observe the laser beams having the same speed but you would see the screen as moving at about 460 m/s. From this frame of reference you would see the light hit the screen to the west first.

What is simultaneous in one frame of reference is not necessarily simultaneous in a different frame of reference.

Would a Coriolis force come into effect in the first picture? Light moves at constant speed, but if it takes a longer path... — 1 hour ago
Yes. This is the key concept in relativity - there is no absolute reference frame for time and space. — 58 mins ago
@John No. This is exclusively a relativistic phenomenon - the beams' arrivals at the screens are simultaneous in the rest frame of the screen, but they are not simultaneous as observed from the Moon. Simultaneity, like space and time, is relative. — 57 mins ago
@EmilioPisanty I mean, we're talking about a rotating frame of reference here — 56 mins ago
There's also another component to this: even if you're at rest, if you're standing closer to one of the screens than the other, you will see the light scattered from that screen first. You will only observe simultaneity if you're at rest with respect to the screens and equidistant from both of them. — 55 mins ago

score 0 · Answer 2 · 2018-10-18 21:05:10Z

One - way speed of light relatively to Earth surface is anisotropic, i.e. it is differenet in different directions. We must remember, that equality of one - way speed of light to constant c is a convention.

So as to measure one - way speed of light we need pair of synchronized clocks, and outcome of the measurement depends on convention how to synchronize them.

Special relativity employs Einstein synchrony convention for all inertial frames of reference.

In rotating frames, even in special relativity, the non-transitivity of Einstein synchronisation diminishes its usefulness. If clock 1 and clock 2 are not synchronised directly, but by using a chain of intermediate clocks, the synchronisation depends on the path chosen. Synchronisation around the circumference of a rotating disk gives a non vanishing time difference that depends on the direction used. This is important in the Sagnac effect and the Ehrenfest paradox. The Global Positioning System accounts for this effect.

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

Let us imagine that a short-wave radar is stationed close to the city of Quito, sending a narrow-angle signal in the east direction. Let us also imagine that all over the equator line a great number of reflectors are stationed in such a way that any of the adjacent reflectors is within the field of vision from another. Let the reflectors deflect the radar signal emitted in Quito in such a way that it, propagating zigzag-wise near the EarthÃ¢Â€Â™s surface, circles the Earth along the equator, coming back to the radar in Quito from the west.

Knowing the length of the zigzag line along which the radar signal propagates and the time needed for signal to circle the Earth, an operator of a radio relay station can calculate the propagation speed of the signal circling the Earth from east to west or reverse. That these speeds will not be identical to and differ from the constant C can be upheld by the following:

Let us ideally place a non-rotating detached onlooker at a distant from the Earth point of the imaginary Earth rotation axis. Let this onlooker be immobile in relation to the centre of the EarthÃ¢Â€Â™s mass and be watching the northern hemisphere rotating counter clockwise below us, mentally following the signal propagation.

Within the reference frame of the detached onlooker the speed of light propagating in space zigzag-wise is equal to the fundamental constant C. If the Earth were not rotating, then the signal in order to circle the hypothetically non-rotating Earth would need the time equal to the length of the zigzag line encompassing the Earth along the equator, divided by the constant C.

However, the Earth does rotate!

When the signal arrives at the initial point in the space of the detached onlooker, the radar of Quito will move approximately 62 meters east, and the signal arriving from the west would need extra time equal to 0.2 microseconds to return to the radar.

If the operator turned the antenna 180 degrees and directed the signal westward, the signal would need two microseconds less to circle the Earth and return to the radar because during the signalÃ¢Â€Â™s travel around the Earth the radar would move 62 meters eastward and the signal coming from the east would have no need to cover these 62 meters. The signal delay is a first order effect in relation to the value of v/ÃÂ¡, where v is the linear speed of the rotating EarthÃ¢Â€Â™s surface, and this delay is large enough compared to relativistic effects of the second order of smallness.

Considering the Lorentz contraction of the equator and the slowdown of the rate of the clock moving together with the surface of the rotating Earth, the average speed of light on the way there and back would have been precisely equal to the constant C.

The clock synchronization considering the inequality of the speeds of light from west to east and from east to west will indeed give the same result as the clock synchronization through a synchronizing signal transmitted by a detached onlooker from a point on the imaginary EarthÃ¢Â€Â™s axis of rotation to all point on the equator. The readings of the clocks synchronized with regard to the inequality of the speeds there and back are conceived by a detached onlooker as the same.

The issue of synchronization becomes even more entertaining, if we ideally replace the Earth with a giant ring of an arbitrarily large diameter, accommodating a transmitter/receiver and a system of reflectors there. In this case, at a given linear speed v of the ring and an arbitrarily small angular speed of the ring rotation, the deviation of the signal propagation speed in one of the directions from the constant C will, as a first approximation, equal v.

If we imagine that an inertial laboratory is tangentially flying to the ring at a speed equal to the linear speed of this ring, and that this laboratory, given an arbitrarily large diameter of the ring, during an arbitrarily long period of time finds itself beside a nearby sector of the ring, then during this period of time this sector of the ring and the laboratory will find themselves practically immobile one to another. If the speed of signal propagation in one direction in relation to a sector of the ring is different from C, then why the speed of that same signal (in the same direction) in relation to the inertial laboratory should unquestionably be thought of as equal to constant C?

A short note with math in this book, p. 42, The Sagnac effect

https://www.uio.no/studier/emner/matnat/fys/FYS4160/v06/undervisningsmateriale/kompendium.pdf

M. Enns 3,8141925 · Answer 3 · 2018-10-18 20:06:14Z

It depends on your frame of reference.

If you are positioned by the laser on the Earth's surface you would observe the two laser beams to have the same speed and your would observe the two targets to be stationary and you would observe the light to take the same amount of time to reach each target.

On the other hand if you were not at rest compared to the laser and screens, say for example you are observing from the moon, you would still observe the laser beams having the same speed but you would see the screen as moving at about 460 m/s. From this frame of reference you would see the light hit the screen to the west first.

What is simultaneous in one frame of reference is not necessarily simultaneous in a different frame of reference.

Would a Coriolis force come into effect in the first picture? Light moves at constant speed, but if it takes a longer path... — 1 hour ago
Yes. This is the key concept in relativity - there is no absolute reference frame for time and space. — 58 mins ago
@John No. This is exclusively a relativistic phenomenon - the beams' arrivals at the screens are simultaneous in the rest frame of the screen, but they are not simultaneous as observed from the Moon. Simultaneity, like space and time, is relative. — 57 mins ago
@EmilioPisanty I mean, we're talking about a rotating frame of reference here — 56 mins ago
There's also another component to this: even if you're at rest, if you're standing closer to one of the screens than the other, you will see the light scattered from that screen first. You will only observe simultaneity if you're at rest with respect to the screens and equidistant from both of them. — 55 mins ago

score 0 · Answer 4 · 2018-10-18 21:05:10Z