I have been reading the article on the circular sagnac effect, and I think the following formula in the article is wrong:
The Sagnac Effect and Uniform Motion
t’ = 2
pr/(c + v) (equation 3 in the article)
I am only just beginning to think about it now.
Multiplying both sides by (c+v) gives:
t’ (c+v) = 2
pr
Distributing
amount of time t`gives
t’c+vt’ = 2
pr
The problem I have is with the RHS.
The photon hasn't really travelled in a full circle of radius R.
And the reason is certainly clear, the disc is spinning.
To help me understand this better, I am thinking of a tiny marble, on a frictionless disc of ice, with no friction whatsoever.
Then the disc starts spinning, as if someone turned a record player on.
Because there is no friction, the marble remains at rest in an outside frame.
Yet to an observer anchored to the spinnin disc, the marble "appears" to have travelled in a circle.
But, a frame fixed to the ice disc, is a non-inertial frame.
In particular, if an observer fixed to the ice disc completely disregards external objects (which too appear to be orbiting)
The marble appears to have a centripetal acceleration.
But the net force on the marble is zero.
So something strange is going on in the physics of this.
You can see it as you switch from the record frame, to the inertial frame, and back again.
I would be the first to argue that you can do physics in any frame, inertial or otherwise, but you need to have transformation formulas.
At any rate, right now I am only concerned with their equation (3).
Here it is again:
t’ = 2
pr/(c + v) (equation 3 in the article)
The writer also says that t' is the time on the spinning disc, as if time is different there, as opposed to measuring time using a clock, at rest in an inertial frame, which clock is necessarily external to the disc.
t’c+vt’ = 2
pr
I think the simplest way to clear up the problem is to use the precise definition of "relative speed" which ironically is an absolute in all frames.
You know, there is the same problem with measurements made on earth, as is going on on the record player.
The stars don't stay fixed in the sky.
I can do this correctly, but I just need to think for a few moments carefully.
Let me refer to the external frame, as frame F.
When the motor is off, the disc
isn't spinning in frame F.
Now, suppose that the following experiment is performed in frame F.
Somehow, a photon is fired, and made to go around the circumference of the "record player" in time
Dt
0
as measured by a clock at rest in frame F.
Maybe frame F is inertial maybe not, but that is the measurement by the clock.
Now, using a "meter string" the circumference of the record player is measured to be:
2
p R
So we have two solid measurements in the frame, and we define the ratio of them to be the tangential speed of the photon in frame F. That is:
v
t= 2
p R/
Dt
0
Now, we can relate the tangential speed, to d
q/dt as follows:
v
t= R
w
w = d
q/dt
Ok firstly...
The circumference of the record player is stipulated to be an absolute in all frames of reference.
Now, if we turn on the record player, and the marble remains at rest while the record player turns, after some time T has elapsed (as measured in F), the marble will again coincide with the starting point (which is just a mark dug into the record).
WE could also give the marble a "push," and if there is a lip around the record players perimeter, the marble would travel in a circle in frame F (instead of remaining at rest in frame F), in some amount of time which could be measured by a clock at rest in F.
So these are the kinds of measurements we can make.
The case of the Sagnac device is interesting for the following reason:
Not only is the photon traveling in a circle in frame F, but the disc is spinning as well.
Ah HAH!
Ok so now, in symbolic imitation of the article, let us do this:
c = 2
p R/
Dt
0
Now, that isn't necessarily 299792458 meters per second, it is whatever quantity is obtained upon taking the ratio of the measurements (as already described). I chose the letter c, because the author of that article used the letter c for that ratio. At least this way, I can check what they are doing.
So when the device isn't spinning you get whatever you get for c, as measured by devices at rest in frame F.
Now let me glance at the article.
Their very next step is to simply write the following formula:
t’ = 2
pr/(c + v) (equation 3 in the article)
I would much prefer to write things in terms of d
q/dt, and attempt to derive that formula, rather than simply pull it out of thin air.
For what it is worth, that is the first time that symbol v is brought into their argument, and the author calls it "speed of a point on the periphery of the disc."
If you are rigidly attached to the disc, then all points on the disc have no speed relative to you, so that v which they introduced, has to be the speed of a point fixed to the edge of the disc, and that speed is defined in frame F, which is external to the disc.
And that speed is appropriately the tangential speed of the disc, in frame F.
So forget about the photon, and focus on the disc.
The center of inertia of the disc is at rest in frame F.
But a fixed point on the periphery of the disc completes one revolution in T seconds.
Thus, the tangential speed of a periphery point in frame F is given by:
v
t= 2
p R/T = R
w
where
w = d
q/dt
Now, what is really going on is this...
At the moment in time that the photon has completed one circular revolution in frame F, a fixed point on the perimeter has advanced through some total angle
Q
And the amount of time this happened in, is the amount of time it took the photon to make one revolution, in frame F, which is the period of the photon.
The period of the photon in frame F was denoted by:
Dt
0
This is different than the period of the spinning disc, which was denoted by T.
Now, the moment in time at which the event began was denoted by t1.
And, the moment in time at which the photon completed one revolution in frame F was denoted by t3.
And so the amount of time of event [t1,t3] is
Dt
0
Now, the tangential speed of the disc was denoted by
v
t
v
t= 2
p R/T = R
w
And
w = d
q/dt
Therefore:
v
t= 2
p R/T = R d
q/dt
So, let us stipulate that the tangential speed of a fixed point on the record player is constant. (this has nothing whatsoever to do with the photon)
So we have this:
v
tdt= 2
p R/T dt= R d
q
Now, we can integrate from moment in time t1=0,
to moment in time t3=t1+
Dt
0
The total change in angle, the integral of dtheta, has been denoted by:
Q
So, we have:
(2
p R/T)
Dt
0 = R
Q
Now, I need to connect this to what they call t`. Here is their quote:
Quote
Referring to Figure 2, and taking t0 as the time observed when the disc is stationary, i.e. the path length divided by the speed of light:
t0 = 2pr/c (2)
The time t’, as observed aboard the spinning disc, for the counter-clockwise beam to complete a circuit, is
t’ = 2pr/(c + v) (3)
where v is the speed of a point on the periphery of the disc.
The difference between the two equations (2) and (3) is:
So the difference in time (dt’) between the stationary and the anticlockwise cases is:
dt’ = (t0 - t’) = t0v/(c + v) (4)
For the other direction, where t” is the time for the clockwise beam to arrive back at point S”, the difference (dt”) between the stationary and the clockwise case is similarly:
dt” = (t” - t0) = t0v/(c - v) (5)
So t` is the amount of time it takes the counterclockwise moving photon to travel from the detector around and back to the detector.
And this has to be
less than the amount of time
Dt
0, you can just figure that out from their diagram.
So my question now, is how do I figure it out. In other words, I now need the time of the following event [t1,t2] using the variables already introduced.
I see very clearly their solution, but I am not the type to use something I cannot either outright understand, or derive myself. What bothers me is that they have the distance traveled as being 2piR, but that isnt the case.
There is something significant here about spinning frames, but I can'tquite put my finger on it.
If you permit yourself to switch between the two frames, then you can make sense out of things, but you have to be very careful about using the definition of speed.
I'msure if you've already solved the problem it's a piece of cake, and I've solved this one before, probably over 15 years ago. Its been so long I have to redo it.I think it was in a book on classical mechanics by Kleppner/ Kolenkow, a book used at MIT. They solved this or something similiar.
It probably didn'tmake sense the first time, which would explain why it didn't keep.
If you are willing to define the speed of the photon, in a frame fixed to the record player, then the photon will have a greater speed in that frame, then in frame F.
And the reason is obvious. You will use 2piR for the distance traveled, and the same amount of time as measured in frame F for event [t1,t2] but in frame F, the distance traveled by the photon is less than 2 pi R, in the same time, as passed in the spinning frame. And hence the speed of the photon in frame F, will be less than the speed of the same photon in the spinning frame. But of course, this has turned distance traveled into something meaningless.
Special relativity is the least of my concerns right now.
Well let me see if I can at least get their formula for t`.