Re: Simple Sagnac
- From: sal <pragmatist@xxxxxxxxxx>
- Date: Fri, 29 Jul 2005 10:31:24 -0400
On Fri, 29 Jul 2005 08:25:11 +0000, Dirk Van de moortel wrote:
>
> "Bilge" <dubious@xxxxxxxxxxxxxxxxxxxxxxxxxxxx> wrote in message
> news:slrndeisl0.7ma.dubious@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
>> sal:
>>
>> >Again, "classical" doesn't seem to be defined here. As far as I
>> >can tell he's saying length contraction doesn't play a role, and
>> >you don't need wicked fast velocities. One thing he certainly is
>> >_not_ saying is that a fiber optic ring gyro can be analyzed
>> >using "classical" (non-relativistic) physics.
>>
>> Why not? If the ring rotates with an angular velocity, w, then
>> the light in the direction of rotation has to travel a distance:
>>
>> s = 2\pi r + wrt_1
>>
>> Where t is the time required for the light to reach the point on
>> the ring that it started, since the ring rotated by a distance wrt
>> in that time. Similarly, in the opposite direction, the distance
>> traveled is s = 2\pi r - wrt_2. The speed of light in the ring is v
>> = c/n, so it travels a distance s = vt_1 in the direction of
>> rotation and s = vt_2 in the opposite direction. then,
>>
>> vt_1 = ct_1/n = 2\pi r + wrt_1 => t_1 = 2\pi r/[(c/n) - wr]
>>
>> vt_2 = ct_2/n = 2\pi r - wrt_2 => t_2 = 2\pi r/[(c/n) + wr]
>>
>>
>> t_2 - t_1 = 2 n\pi r [(1/(c - nwr)) - (1/(c + nwr))]
>>
>>
>> = 2pi r [ 2nwr/(c^2 - (nwr)^2)]
>>
>> = 4\pi r^2 [ (n^2 w)/(c^2 - (nwr)^2 ]
>>
>> >Again, I'd be more impressed with the quotes if you explain how
>> >you can use anything other than k+v and k-v for the velocities in
>> >the "classical" case, if you don't happen to have a perfect
>> >vacuum on tap in which to run the experiment.
>>
>> The index of refraction for air at STP for 590 nm is about,
>> 1.00029. Rearranging the above gives:
>>
>>
>> t_2 - t_1 = 4\pi w r^2/[(c/n)^2 - (wr)^2]
>>
>> for n = 1.00029. 1/n^2 = 0.99942 or 99.942% c.
>>
>> The index of refraction is irrelevant. The only point that it would
>> enter the calculation differently than just replacing c by c/n, is
>> if the ring was rotating fast enough that the frequency dependence
>> of n == n(w) mattered.
>>
>> [...]
>> >Again, if you disagree, please explain how such an analysis could
>> >work. (Henri would love to know!) (Sagnac didn't assume fiber optic
>> >loops, of course, since they hadn't been invented yet.)
>>
>> Replace c with c/n.
>
> The thing that puzzled me, and still does, is that he says that both
> the "classical" u- = k - v and the relativistic u- = (k-v)/(1-k v)
> are 'dragged'. Yet, if we take the case where k = 1, the
> relativistic speed reduces to u- = 1, which is something I wouldn't
> call dragged. And it's no closing speed either. But again, I might
> be mistaken. Need more from the sideline.
The problem may be that the word "dragged" isn't exactly a precise
technical term.
What I meant by "dragged" is the signal speed is "k" as viewed by an
observer moving with the ring, and we need to transform that velocity
to obtain the velocity seen from the stationary (lab) frame.
If the signal's in a moving medium (glass or gas) that seems like a
reasonable starting point, classically as well as relativistically.
In Newtonian mechanics you'd then transform that velocity to the
stationary frame by adding or subtracting the speed of the ring; so,
the guy sitting at the lab bench would see the signal moving at k+v or
k-v, instantaneously, at each point on the ring.
In relativity, you use composition of velocities to figure out how
fast the signal seems to be moving in the laboratory frame. And in
that case, if k = c, then the signal's moving at k (= c) in the lab
frame too and you'd like to say it's not being dragged after all. As
Bilge points out the refractive index of air is so low that,
relativistically, there is (essentially) no dragging going on so the
results match the "classical" assumption that light is carried by the
aether and isn't dragged by the ring.
On the other hand, when you use a high-index glass fiber with, say,
N=2, then k = (1/2)c, and the "relativistic drag effect" is
substantial. It's really not going the same _speed_ both ways around
the ring any more as viewed from the stationary frame. Just to make it
really clear, let's assume we have a ludicrously fast ring, and v = c/10.
Relativistic "dragging" gives:
(k+v)/(1+kv) = 0.6/1.05 = 0.57c
(k-v)/(1-kv) = 0.4/0.95 = 0.42c
while Newtonian "dragging" gives:
k+v = 0.6c
k-v = 0.4c
The signal's "dragged" either way, but the results are still different
because the composition of velocities is different classically and
relativistically.
If, instead, you make the (bizarre?) assumption that in a "classical"
world no signal-dragging takes place no matter what, then in the case
where the signal is carried by a glass fiber with N=2, you get yet a
third case. In that case you do see a Sagnac effect of sorts, but the
magnitude won't match the values predicted by relativity.
In conclusion, in the absence of relativity (or some "aether clone" of
relativity) I don't see any way to predict the correct Sagnac fringe
shift when the signal is carried by an optical fiber with relatively
high index of refraction, rather than by vacuum or air.
Make sense?
I should mention I haven't looked at the experiments that have been done
so I don't really know if tests have been done with enough precision to
rule out the "no-drag-in-glass-with-k=N/2" case but I'd be surprised if
they haven't.
>
> Dirk Vdm
--
Nospam becomes physicsinsights to fix the email
I can be also contacted through http://www.physicsinsights.org
.
- Follow-Ups:
- Re: Simple Sagnac
- From: Dirk Van de moortel
- Re: Simple Sagnac
- From: sal
- Re: Simple Sagnac
- References:
- Re: Simple Sagnac
- From: Bilge
- Re: Simple Sagnac
- From: Dirk Van de moortel
- Re: Simple Sagnac
- Prev by Date: Re: Light & very small distances - paradox??
- Next by Date: Re: Simple Sagnac
- Previous by thread: Re: Simple Sagnac
- Next by thread: Re: Simple Sagnac
- Index(es):
Relevant Pages
|
