GPS and Sagnac effect.

O Great One
I've read that the GPS system has to take into account the Sagnac effect when sending signals from the satellites to the Earth. For those of us who are unfamiliar with the Sagnac effect, it is the delay in receiving(or shorter time in receiving) the signal due to the rotation of the Earth while the signal is in transit to the Earth. Now if the speed of light were always constant then this shouldn't be necessary because the amount of time to receive the signal would only depend on the distance at the time of emission. So what is the official explanation why this doesn't indicate that light is traveling at c+v or c-v relative to us?

DocZaius
"distance at the time of emission" ? I would think the amount of time to receive the signal would rather depend on "distance traveled by light from emission to reception." If you fire a laser at a wall that is receding from you, you shouldn't measure the distance between point of emission and the wall BEFORE it moved, since the light will not hit the wall where it was before it moved, but where it has moved to by the time light got there.

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1effect
So what is the official explanation why this doesn't indicate that light is traveling at c+v or c-v relative to us?

No, it doesn't. The speeds "c+v" and "c-c" that you see are "closing speeds", i.e.:

"c+v" is the speed at which a ray of light "closes" the distance to an "incoming" mirror in the Sagnac experiment

"c-v" is the speed at which a ray of light "closes" the distance to a "running away" mirror in the same experiment.

O Great One
"distance at the time of emission" ? I would think the amount of time to receive the signal would rather depend on "distance traveled by light from emission to reception." If you fire a laser at a wall that is receding from you, you shouldn't measure the distance between point of emission and the wall BEFORE it moved, since the light will not hit the wall where it was before it moved, but where it has moved to by the time light got there.

Think carefully and you'll see I'm correct. Let me give an example. Say I have a car that is moving relative to me at 100 mph. It moves so that it's speed relative to me is always 100 mph and it is 1000 miles away. No matter how I move, the car will reach me in 10 hours. Let's say I move away at 200 mph for 5 hours, I will have traveled 1000 miles in 5 hours and the car will have traveled 300 mph for 5 hours or 1500 miles, the car is now only 500 miles away. I turn around and drive towards the car at 50 mph for 5 hours or 250 miles. The car drives 50 mph for 5 hours and also drives 250 miles. We each cover 250 miles or 500 miles total in last 5 hours. So, the car reached me in 10 hours. So if a flash of light is emitted from a distance of 1 light-year, if it approaches me at c it will reach me in 1 year no matter what subsequent movements I make.

O Great One
Thank You

"c+v" is the speed at which a ray of light "closes" the distance to an "incoming" mirror in the Sagnac experiment

"c-v" is the speed at which a ray of light "closes" the distance to a "running away" mirror in the same experiment.

It's good to see that somebody agrees with me.

1effect
It's good to see that somebody agrees with me.

I thought I made it quite clear that I do not agree with you.

Mentor
You do realize that the Sagnac effect is completely compatible with relativity, right? As the Wikipedia http://en.wikipedia.org/wiki/Status_of_special_relativity" [Broken] article says, "The Sagnac effect, a phenomenon that is taken into account in GPS synchronisation procedures, is predicted by both special relativity and Galilean relativity"

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DocZaius
Think carefully and you'll see I'm correct. Let me give an example. Say I have a car that is moving relative to me at 100 mph. It moves so that it's speed relative to me is always 100 mph and it is 1000 miles away. No matter how I move, the car will reach me in 10 hours. Let's say I move away at 200 mph for 5 hours, I will have traveled 1000 miles in 5 hours and the car will have traveled 300 mph for 5 hours or 1500 miles, the car is now only 500 miles away. I turn around and drive towards the car at 50 mph for 5 hours or 250 miles. The car drives 50 mph for 5 hours and also drives 250 miles. We each cover 250 miles or 500 miles total in last 5 hours. So, the car reached me in 10 hours. So if a flash of light is emitted from a distance of 1 light-year, if it approaches me at c it will reach me in 1 year no matter what subsequent movements I make.

Say you are at point A and a ship passes by you going half the speed of light. 299,792,458 of your meters away (2 seconds later for you), the ship hits point B and makes contact with a switch that sets off a laser beam towards you. At that same moment (again, 2 seconds later from when the ship passed you), you accelerate greatly the opposite way of the ship and point B.

I think that you think that no matter how greatly you accelerate, you will receive that light beam exactly one second and no more than one second from when you set off. You believe that even though the trip for the light beam from point B to where you are when it hits you is greater than 299,792,458 meters, you will have received it no more than one second after the light beam set off. Is my interpretation of your understanding correct? If it isn't, then I am missing your point and I am sorry.

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DocZaius
Let me come up with a simpler example not involving any acceleration (only inertial frames):

At t=0 seconds a mirror passes by you at 1/2c. At t=2 seconds (when the mirror is 299,792,458 meters away from you) you send a pulse of light to it.

When do you receive the pulse of light back?

The way I would work this out is that the light pulse will meet the mirror at t=4 seconds since at t=4 seconds the light pulse will have traveled 599,584,916 meters (the amount of distance traveled by light in 2 seconds) and so will the mirror going at half the speed of light but having passed you at t=0 seconds.

So at t=4 seconds the light pulse will be 599,584,916 meters away from you and coming back, and will thus travel an additional 2 seconds back to you. At t=6 seconds you will have received it back.

If you agree with me so far, then I don't see where we disagree.

It seems to me that you think that since the distance between you and the mirror at the time of your sending the light beam is 299,792,458 meters, then it will only take one second for the light beam to reach the mirror. You seem to think that the implication of the constancy of the speed of light is that any movement occurring after a photon's emission is irrelevant since the photon's time to travel is already predetermined. In your understanding, it seems that light would hit the mirror one second after emission or at t=3 seconds and, somehow being 299,792,458 meters away, come back to you in one second at t=4 seconds. Is that right?

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O Great One
Let me come up with a simpler example not involving any acceleration (only inertial frames):

At t=0 seconds a mirror passes by you at 1/2c. At t=2 seconds (when the mirror is 299,792,458 meters away from you) you send a pulse of light to it.

OK so far.
When do you receive the pulse of light back?

That is difficult to answer since we are dealing with the 'magic' of SR where nothing makes sense.

The way I would work this out is that the light pulse will meet the mirror at t=4 seconds since at t=4 seconds the light pulse will have traveled 599,584,916 meters (the amount of distance traveled by light in 2 seconds) and so will the mirror going at half the speed of light but having passed you at t=0 seconds.

So relative to the mirror, after 2 seconds light has approached the mirror at a rate of (299,792,458 meters/2 seconds) or 1/2c. So if you are holding the mirror then light is moving relative to you at .5c. You will find that the light is passing you at a speed of .5c and not c. I wil repeat again. IF THE SPEED OF LIGHT IS ALWAYS MEASURED AS BEING C THEN IT DEPENDS ON THE DISTANCE AT THE TIME OF EMISSION.

So at t=4 seconds the light pulse will be 599,584,916 meters away from you and coming back, and will thus travel an additional 2 seconds back to you. At t=6 seconds you will have received it back.

If you agree with me so far, then I don't see where we disagree.

It seems to me that you think that since the distance between you and the mirror at the time of your sending the light beam is 299,792,458 meters, then it will only take one second for the light beam to reach the mirror.

If there is somebody holding the mirror and light must move at speed c relative to him or her, then yes.

You seem to think that the implication of the constancy of the speed of light is that any movement occurring after a photon's emission is irrelevant since the photon's time to travel is already predetermined.

Yes. Can you please show otherwise?

In your understanding, it seems that light would hit the mirror one second after emission or at t=3 seconds and, somehow being 299,792,458 meters away, come back to you in one second at t=4 seconds. Is that right?

Well, I guess if you believe in SR then the light would have to magically 'speed up' so that it reaches the mirror at time t=3 seconds and then since at that point in time the mirror is 1.5 light-seconds away from you it reaches you in 1.5 seconds at time t=4.5 seconds.

JustinLevy
You seem to think that the implication of the constancy of the speed of light is that any movement occurring after a photon's emission is irrelevant since the photon's time to travel is already predetermined.
Yes. Can you please show otherwise?
You are misunderstanding something fundemental.
Special relativity says the speed of light is constant according to measurements using the coordinates of an inertial coordinate system.
Using the coordinate system in which the spatial origin is accelerating according relative to an inertial frame, the speed of light will not be constant. In fact, it is possible for an object to "outrun" light in the sense that a light beam will never catch up with the object, despite the object moving with v<c according to any inertial frame.

As a quick counter-proof that your idea cannot work, take a light bulb and two observers who are initially at the same position. Have the light bulb blink. While the light is traveling towards the observers, have one move away from the light. The light will reach one observer BEFORE the other, despite the light bulb being the same distance away from both observers when it blinked.

In short, your problem is that you are trying to define velocity with the relation:
time of light pulse flight from emission to reception = distance between emitter and receiver AT TIME OF EMISSION / velocity
when the actual relation is:
time of light pulse flight from emission to reception = distance between emitter position AT TIME OF EMISSION and receiver position AT TIME OF RECEPTION / velocity

O Great One
You are misunderstanding something fundemental.
Special relativity says the speed of light is constant according to measurements using the coordinates of an inertial coordinate system.
Using the coordinate system in which the spatial origin is accelerating according relative to an inertial frame, the speed of light will not be constant. In fact, it is possible for an object to "outrun" light in the sense that a light beam will never catch up with the object, despite the object moving with v<c according to any inertial frame.

Are you referring to the fact that the Sagnac effect is present on a rotating disk? The GPS satellite is not resting on the Earth at the time of emission, not to mention that in the short time that the signal is in transit the motion of an object on the Earth will be virtually indistinguishable from a straight line.

As a quick counter-proof that your idea cannot work, take a light bulb and two observers who are initially at the same position. Have the light bulb blink. While the light is traveling towards the observers, have one move away from the light. The light will reach one observer BEFORE the other, despite the light bulb being the same distance away from both observers when it blinked.

So then you agree that the speed of light is not always measured at c by all observers?

In short, your problem is that you are trying to define velocity with the relation:
time of light pulse flight from emission to reception = distance between emitter and receiver AT TIME OF EMISSION / velocity
when the actual relation is:
time of light pulse flight from emission to reception = distance between emitter position AT TIME OF EMISSION and receiver position AT TIME OF RECEPTION / velocity

Mentor
Are you referring to the fact that the Sagnac effect is present on a rotating disk? The GPS satellite is not resting on the Earth at the time of emission, not to mention that in the short time that the signal is in transit the motion of an object on the Earth will be virtually indistinguishable from a straight line.
'Virtually indistinguishable" doesn't cut it here. The temporal accuracy and precision needed for GPS is incredibly high, so these "virtually indistinguishable" features of the geometry become significant. The satellites are in orbit and the surface of the Earth is rotating, both effects are large enough that they must be accounted for in order to have precise GPS measurements.

These effects are accounted for using special and general relativity, so I don't know how you think they constitute evidence that the second postulate is wrong. As I said before, the Sagnac effect is actually predicted by special relativity.

JustinLevy
So then you agree that the speed of light is not always measured at c by all observers?
Speed is not an observer issue, it is a coordinate system issue. When you refer to an observer measuring a speed, you need to say what coordinate system the observer is using.

The speed of light is c when using inertial coordinate systems. So if you are considering coordinate systems other than inertial coordinate systems, then yes, the speed of light is not necessarily measured to be c.

Special relativity says the speed of light is c according to all inertial coordinate systems. It doesn't say the speed of light is c according to ALL coordinate systems.

And the sagnac effect can be explained just fine using special relativity and an inertial coordinate system. You can even use a non-inertial coordinate system if you wish (but it only complicates things).

Golfer
GPS website

For a complete discussion on GPS [including all relativistic effects] go to the following site.

physics.syr.edu/courses/PHY312.03Spring/GPS/GPS.html

Stoonroon
The speed of light in opposite directions with respect to rotating observers is greater or less than c. Because the effect involves rotation and therefore non-inertial reference frames, there is no conflict with Special Relativity. As this discussion demonstrates, however, something still strikes some people as "funny" about it, if I may put it that way. One of the funny things is how some authors like to avoid saying the "speed" is not c, that it's a change of path length or time interval, etc. I prefer those authors who don't do that funny dance, but come right out and admit that for rotating observers the speed of light is generally not c.

A key to the problem is the fact that the light path returns to itself without "reflection," but rather by maintaining its "withness" or "againstness" with respect to the rotation, from emission to return. So, in a sense, it is a "one-way" speed measurement...actually an AVERAGE of a one way speed. A Michelson interferometer mounted on the rim of the rotating body will always get a null result because it measures a TWO-WAY light path by reflection.

Another funny thing is how, even though the general, global circumstance demonstrates the non-c speed of light for rotating observers, the special, LOCAL circumstance (demonstrated by the Michelson interferometer) is that the speed of light is isotropically c. What's funny is that logic usually works the other way around. If something is generally, globally so, then specially, locally it may SOMETIMES not be so, but cannot always be so. In Relativity, the Special Theory is regarded as always locally valid regardless of the global circumstance. I'm not saying that it's not, of course. I'm just saying that the logic leading to this conclusion is a little bit funny. Seeing this, one can more readily understand why this subject comes up over and over again.