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Why do you need to measure the speed of light in two directions?

by goodabouthood
Tags: directions, light, measure, speed
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harrylin
#109
Feb2-12, 05:18 AM
P: 3,187
Quote Quote by Tracer View Post
Wouldn't stellar aberation suffice? Where Vt is the Earth's orbital velocity and theta is the angle of aberation:

c = Vt/tan(theta)
According to SR no "absolute" one-way speed of light determination is possible and it's a usual (convenient) choice of the observer to define it as half the two-way speed.

So, here you effectively propose stellar aberration to break the PoR. However it is expected to obey the PoR as already elaborated in Einstein's 1905 paper, in section 7:
www.fourmilab.ch/etexts/einstein/specrel/www/
Tracer
#110
Feb2-12, 09:59 AM
P: 46
Quote Quote by harrylin View Post
According to SR no "absolute" one-way speed of light determination is possible and it's a usual (convenient) choice of the observer to define it as half the two-way speed.

So, here you effectively propose stellar aberration to break the PoR. However it is expected to obey the PoR as already elaborated in Einstein's 1905 paper, in section 7:
www.fourmilab.ch/etexts/einstein/specrel/www/
Is it that there needs to be experimental proof that light speed is identical for both halves of the round trip measument of c?

If the angle of stellar aberration is the only variable measured for light passing through a measuring apparatus in each direction and the angle of aberration is the same for light travelling in both directions, then the speed of light is confirmed to be the same in both directions without the need to measure the length of an AU or the tangential velocity of the Earth's orbit. Then one clock round trip measurements of the speed of light can be truly said to be the result of equal velocities for each half of the round trip.
_PJ_
#111
Feb2-12, 10:21 AM
P: 80
There's a gravitational effect on the distance the light travels. In a two-way trip, the gravitational effect is negated.
harrylin
#112
Feb3-12, 03:32 AM
P: 3,187
Quote Quote by Tracer View Post
Is it that there needs to be experimental proof that light speed is identical for both halves of the round trip measument of c?
Quite to the contrary: there is experimental proof that we may freely choose an inertial reference system and pretend that it is "in rest", so that the speed of light becomes (or appears) identical for both halves of the round trip measurement. Since we apply the relativity principle to light, this should be true for any inertial reference system that we freely choose.

It is essential to understand that thus for each choice of reference system the speed of light becomes (or appears) according to your definition, anisotropic relative to any object or system that is moving relative to your chosen system.
If the angle of stellar aberration is the only variable measured for light passing through a measuring apparatus in each direction and the angle of aberration is the same for light travelling in both directions, then the speed of light is confirmed to be the same in both directions without the need to measure the length of an AU or the tangential velocity of the Earth's orbit. Then one clock round trip measurements of the speed of light can be truly said to be the result of equal velocities for each half of the round trip.
There is now a thread opened on stellar aberration, so I'll only give a short answer here. If you reflect light inside an apparatus, no effect from motion is predicted by SR even if the apparatus is in inertial motion.
bahamagreen
#113
Feb14-12, 10:27 PM
P: 543
Quote Quote by harrylin View Post
Quite to the contrary: there is experimental proof that we may freely choose an inertial reference system and pretend that it is "in rest", so that the speed of light becomes (or appears) identical for both halves of the round trip measurement. Since we apply the relativity principle to light, this should be true for any inertial reference system that we freely choose.
The synchronization of clocks at A and B where tB-tA=t'A-tB stipulates that the clocks are both stationary with respect to each other...

How can this stipulation apply when one can't verify this prior to synchronization? Actually it would require multiple synchronization successes to verify points A and B are stationary with respect to one another wouldn't it?

... my point is that even if both A and B are inertial reference frames, that does not imply they are stationary relative to each other. They may be moving together or apart at a constant rate, or one or both may be free falling with a net acceleration between them.

If A and B are in either constant or free fall accelerating relative motion, both are inertial frames, but there are possible synchronization measures where tB-1A=t'A-tB will be true, yet a prior or subsequent measure will be false. So a single successful synchronization measure can give a false positive.

The reflecting mirror method stipulates stationary A and B, but what it really stipulates is that A and B happen to be a particular distance apart when the tB event occurs... and B can be moving wrt A when this happens.

The case of free falling inertial A and B may be bringing GR into this, but the case of a net constant rate distance change between A and B does not; yet the possible false positive for a successful synchronization measure still exists (meaning that the synchronization was only for a moment, the clocks were not syched before and continue to be out of synch after in spite of a momentary tB-tA=t'A-tB.
harrylin
#114
Feb15-12, 07:05 AM
P: 3,187
Quote Quote by bahamagreen View Post
The synchronization of clocks at A and B where tB-tA=t'A-tB stipulates that the clocks are both stationary with respect to each other...

How can this stipulation apply when one can't verify this prior to synchronization? Actually it would require multiple synchronization successes to verify points A and B are stationary with respect to one another wouldn't it?
Not at all, it's not difficult to establish if clocks have a constant distance between them and are stationary in a frame in which the laws of mechanics hold. And it's not difficult to extend this method to clocks in motion, as is nowadays done.
... my point is that even if both A and B are inertial reference frames,
A point is not a reference frame!
that does not imply they are stationary relative to each other. They may be moving together or apart at a constant rate, or one or both may be free falling with a net acceleration between them. [..]
According to the original definition, free-fall is not an inertial frame; what was meant is a frame that is in uniform, straight line motion according to astronomical observations. Moreover, SR doesn't account for effects from gravitation.

The case of free falling inertial A and B may be bringing GR into this, but the case of a net constant rate distance change between A and B does not; yet the possible false positive for a successful synchronization measure still exists (meaning that the synchronization was only for a moment, the clocks were not syched before and continue to be out of synch after in spite of a momentary tB-tA=t'A-tB.
Sure, a more complex method is required for such cases, as is done with GPS satellites. That is a completely different topic. You could start that topic if you want to discuss it.

Harald
bahamagreen
#115
Feb15-12, 02:35 PM
P: 543
Harrylin, thanks, I'm figuring it out as I go... I made a model to investigate this and found the following:

In the tB-tA=t'A-tB equation the "t's" are not all representing the same kind of thing.

tA is the initial reading on clock A at tA
but tB is not B's initial setting at tA, it is an elapsed time from B's initial clock setting at tA to the time tB
t'A is also an elapsed time, the sum of tB + the time of the trip back to A

tb-tA=t'A-tB only works if the initial times of both A and B clocks at time tA are set to the same time. You can't just have two clocks A and B with different time settings and run the synchronization. That is, if we call the initial clock reading of A and B as Ain and Bin at time tA, then tA=Ain=Bin. This makes it more clear that tB is really tB-Bin...

This is point where I have that "Duh.." moment about what synchronized clocks really means... that at tA both clocks indicate the same value... definition of synchronous. :0

But what is interesting is that the synchronization works fine (tb-tA=t'A-tB is true) when you add a constant rate of distance change (using non-relativistic speed addition) between A and B. The path from A to B is still the same magnitude as the path from B to A at the time tB, even if the path length was different before or after tB.

So the stipulation that A and B be at rest wrt each other must be to omit the possibility of relativistic effects.
harrylin
#116
Feb17-12, 01:26 AM
P: 3,187
Quote Quote by bahamagreen View Post
[..] So the stipulation that A and B be at rest wrt each other must be to omit the possibility of relativistic effects.
It wasn't a stipulation. It was just a description of a common way in those days to set up a reference system. And the following claim that "it is essential to have time defined by means of stationary clocks in the stationary system" was basically wrong: nowadays one uses mostly clocks that are moving in the stationary system. But what does that have to do with the topic?
ghwellsjr
#117
Feb17-12, 09:56 AM
PF Gold
P: 4,745
Quote Quote by bahamagreen View Post
Harrylin, thanks, I'm figuring it out as I go... I made a model to investigate this and found the following:

In the tB-tA=t'A-tB equation the "t's" are not all representing the same kind of thing.

tA is the initial reading on clock A at tA
but tB is not B's initial setting at tA, it is an elapsed time from B's initial clock setting at tA to the time tB
t'A is also an elapsed time, the sum of tB + the time of the trip back to A

tb-tA=t'A-tB only works if the initial times of both A and B clocks at time tA are set to the same time. You can't just have two clocks A and B with different time settings and run the synchronization. That is, if we call the initial clock reading of A and B as Ain and Bin at time tA, then tA=Ain=Bin. This makes it more clear that tB is really tB-Bin...

This is point where I have that "Duh.." moment about what synchronized clocks really means... that at tA both clocks indicate the same value... definition of synchronous. :0

But what is interesting is that the synchronization works fine (tb-tA=t'A-tB is true) when you add a constant rate of distance change (using non-relativistic speed addition) between A and B. The path from A to B is still the same magnitude as the path from B to A at the time tB, even if the path length was different before or after tB.

So the stipulation that A and B be at rest wrt each other must be to omit the possibility of relativistic effects.
Einstein said at the beginning of section 1 of his 1905 paper:
If a material point is at rest relatively to this system of co-ordinates, its position can be defined relatively thereto by the employment of rigid standards of measurement and the methods of Euclidean geometry, and can be expressed in Cartesian co-ordinates.
Points A and B are fixed with a rigid rod between them and the length has been measured. Let's say it is 1000 feet. Let's also stipulate that the speed of light is 1 foot per nanosecond to make the arithmetic easier. At A is a light source and a clock with an arbitrary time on it when the light is flashed (say 1 PM). This is tA. At B is a mirror and a second clock which also reads an arbitrary time on it when the light hits the mirror, say 2 PM. This is tB. Then when the reflected light gets back to A, the time on the clock is t'A. This time will be 2 microseconds after 1 PM or 1:00:00.000002 PM). Now we plug these number into the equation to see if they equal:

tB - tA = t'A - tB
2:00:00.000000 - 1:00:00.000000 = 1:00:00.000002 - 2:00:00.000000
1:00:00.000000 ≠ -0:59:59.999998

Whoops--they're not equal. The clocks are not synchronized according to Einstein's definition.

Let's subtract the time on clock B by one hour and repeat the experiment the next day (now tB = 1:00:00.000000):

1:00:00.000000 - 1:00:00.000000 = 1:00:00.000002 - 1:00:00.000000
0:00:00.000000 ≠ 0:00:00.000002

Still not synchronized. Now let's advance the time on Clock B by 1 microsecond and repeat the next day (now tB = 1:00:00.000001):

1:00:00.000001 - 1:00:00.000000 = 1:00:00.000002 - 1:00:00.000001
0:00:00.000001 = 0:00:00.000001

Hooray, now they're synchronized.

Your statement about the clocks having initial times on them that are the same is meaningless. That's the whole point of defining a synchronization process--we can't tell when or if the times on remotely separated clocks have the same time on them. Saying "tB is really tB-Bin" means you have missed the whole point of what Einstein is saying. You need another "Duh..." moment. You should not think that there is any reality to the times on remote clocks apart from us putting meaning into those times. It's not that we are figuring out what nature is trying to tell us--we can't--instead, we are arbitrarily putting meaning into nature, at our own whim.
bahamagreen
#118
Feb17-12, 03:27 PM
P: 543
Alright, I see that tB is not elapsed time, but I'm still seeing synchronized clocks as indicating identical times in what Einstein called common "time" of A and B, or "the time of the stationary system.”

Let tA=10 and t'A=20
tB-tA=t'A-tB
tB-10=20-tB
tB=20+10-tB
tB=30-tB
2(tB)=30
tB=15

tB-tA=t'A-tB
15-10=20-15
5=5

So A and B are synchronized if tB=15

If A assumes travel time for AB to equal BA then A may calculate what his A clock indicated when tB occurred.

The elapsed round trip t'A-ta=20-10=10, so AB=BA=5

A may conclude that he sees tB 5 seconds after it happens, so may calculate that tB happened when his A clock was reading t'A-5=20-5=15

So, with regard to what you wrote, "You should not think that there is any reality to the times on remote clocks apart from us putting meaning into those times.", are you saying that this calculation of tB's occurrence with respect to A's clock time is meaningless?

When two clocks side by side show the same time we say that those clocks are synchronous. When two clocks are distant but synchronous, they no longer show the same time, but knowing t'A and assuming AB=BA one can calculate what was the A observer's local time that corresponded to the B time of a distant event.

Is this not correct?

"But it is not possible without further assumption to compare, in respect of time, an event at A with an event at B. We have so far defined only an “A time” and a “B time.” We have not defined a common “time” for A and B, for the latter cannot be defined at all unless we establish by definition that the “time” required by light to travel from A to B equals the “time” it requires to travel from B to A."

My interpretation of Einstein's quote is that the common "time" for A and B IS defined when assuming AB=BA... that what is synchronous is the time readings of synchronous clocks in this common time, which takes into account the propagation delay of light. When A calculates that tB=15 and then calculates that A's clock was at 15 when tB occured, is this not the synchronization of the common time of A and B?

Is this common time not the same as the "time of the stationary system"?

"It is essential to have time defined by means of stationary clocks in the stationary system, and the time now defined being appropriate to the stationary system we call it “the time of the stationary system.”
ghwellsjr
#119
Feb18-12, 04:23 AM
PF Gold
P: 4,745
You seem to now understand Einstein's synchronization process but it appears that you are thinking that Einstein was describing a way to "discover" or "determine" the common "time" by that process, rather than the process "creating" the common "time". You don't seem to appreciate that fact that Einstein's synchronization is merely one of many different ways to "create" the common "time".
lugita15
#120
Feb18-12, 02:04 PM
P: 1,583
I just wanted to point out that while ghwellsjr is completely correct that Einstein synchronization is just "defining" a common time, in the sense that questions about this convention can be answered without even looking at what universe you're in or what laws of physics are true, the same cannot be said of other synchronization methods. For instance, if you synchronize clocks with slow transport, then the question of whether the speed of light will be measured to be isotropic cannot be predicted in a universe-independent fashion. So in the sense that there are nontrivial questions about this synchronization method that can only be answered by experiment, we might want to say that we are "discovering" simultaneity, not "defining" it.
yoron
#121
Mar16-12, 03:43 PM
P: 244
"my point is that even if both A and B are inertial reference frames, that does not imply they are stationary relative to each other. They may be moving together or apart at a constant rate, or one or both may be free falling with a net acceleration between them."

It's even worse than that Bahama :)

The definition of something being 'at rest' in relativity is that it has a uniform motion, nothing more.

You don't have any 'acceleration' at all in uniform motion, and your relative 'velocity' (I won't use speed here as that says nothing about a direction) doesn't mean a thing as I understands it for defining yourself as being 'at rest' relative something else.

There is no 'universal resting place', only relative ones. And what differs being 'at rest in a uniform motion relative being 'at rest' in a acceleration is that in a acceleration you know that you have inertia/gravity acting at you locally, constantly or intermediately, if now that is the right word to use?

If you introduce a third reference frame from where you define two comoving uniformly moving objects to be 'moving', you might do it relative a third frame, as the 'universe' at large for example. That doesn't change the fact that both can define themselves as being 'at rest' relative each other.


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