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| doug... |
 Posted: Tue Oct 27, 2009 5:40 pm |
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Guest
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Henry Wilson DSc wrote:
[quote]On Tue, 27 Oct 2009 14:10:29 -0700, eric gisse <jowr.pi.nospam at (no spam) gmail.com
wrote:
HW at (no spam) ..(Henry Wilson DSc). wrote:
On Tue, 27 Oct 2009 12:32:03 -0700 (PDT), Darwin123 <drosen0000 at (no spam) yahoo.com
wrote:
On Oct 16, 11:39 am, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p
wrote:
tominlag... at (no spam) yahoo.com> wrote in message
news:to2hd5pm6nvm8qdul7kecjf5do7imn1ng5 at (no spam) 4ax.com...
Correct. Very good analysis. One tiny flaw...
Newton's corpuscles of light model, today called photons, predates
Walter Ritz by 250 years.
The behavior of photons does not resemble the behavior of Newtons
corpusules. Newtons corpusules do not have a frequency associated with
them. The energy in a photons is proportional to the frequency of the
wave.
No it isn't. the correct equation is E = hc/lambda....and nobody even
knows what the term 'wavelength' actually implies in the case of light.
There is no direct 'frequency' associated with light as we observed it.
Any 'oscillation' is intrinsic to the individual photon and is not
directly observeable.
Waveguide. Try again.
Generated microwave, RF, etc....that's not individual photons....
[/quote]
Microwaves and light are the same thing. But ralph will try to
lie his way out of that one again.
[quote]
The difference becomes really important when dealing with
interference effects. Newton did not predict that light could show
interference effects. Neither beats nor diffraction is consistent with
Newton's corpusules.
Diffraction is certainly consistent with Bath. So are beats if coherent
light is used.
No, not really.
Henry Wilson...www.scisite.info/index.htm
Einstein...World's greatest SciFi writer..
Henry Wilson...www.scisite.info/index.htm
Einstein...World's greatest SciFi writer..[/quote] |
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| Androcles... |
| Posted: Tue Oct 27, 2009 6:11 pm |
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"Darwin123" <drosen0000 at (no spam) yahoo.com> wrote in message
news:664e0c5d-cbdd-4aaf-a60f-399f10c3006a at (no spam) r5g2000yqb.googlegroups.com...
On Oct 27, 6:32 pm, "Inertial" <relativ... at (no spam) rest.com> wrote:
[quote]"Darwin123" <drosen0... at (no spam) yahoo.com> wrote in message
news:8726a9ec-2c0d-4d34-a1d5-cedf92ed3415 at (no spam) y23g2000yqd.googlegroups.com...
7) Therefore, Einstein was an idiot.
Why didn't you use this simple argument instead of all
that blather about c+v versus c-v?
Cute.
Does a standing wave have a wavelength?
Yes.
Does a standing wave have a frequency?
Yes.[/quote]
I know full well that frequency times wavelength is the "speed" of
a wave, even though the nodes of the wave aren't going anywhere. I
wasn't presenting this as a real argument <duhh>.
I was pointing out that that the "speed of light" isn't defined as
the actual motion of energy. The speed of light characterizes a
property of the vacuum.
Androcles sees "c+v" and "c-v" and automatically interprets this
as a speed of light. After making this identification, he accuses
Einstein of contradicting himself. Einstein said the speed of light is
c, and then he says it is c+v or c-v. What an idiot!
However, light energy isn't traveling at either "c+v" or "c-v".
The energy in an interferometer is really stored in a standing wave,
which doesn't move at all. So Androcles didn't really think carefully
as to what "speed of light" means.
============================================
It means this, you moron.
http://sci.esa.int/science-e/www/area/index.cfm?fareaid=12
"The Huygens probe has successfully landed on Saturn's largest moon Titan.
At around 11:30 UTC 14 January 2005 the probe touched down on the surface
of this distant world. "
We didn't get to know about it until 12:40 UTC because it takes 1 hour and
10 minutes
for signals to travel from Titan to Earth, or if we heard about it
at 11:30 the touchdown was at 10:20.
So Dorkwin123 didn't really think carefully as to what "speed of light"
means. |
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| doug... |
| Posted: Tue Oct 27, 2009 7:26 pm |
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Androcles wrote:
[quote]"Darwin123" <drosen0000 at (no spam) yahoo.com> wrote in message
news:8726a9ec-2c0d-4d34-a1d5-cedf92ed3415 at (no spam) y23g2000yqd.googlegroups.com...
On Oct 27, 3:16 pm, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
You are a clueless babbling lunatic.
I will now prove Einstein an idiot using your assumptions in
an even clearer manner than you have presented.
1) Standing waves made of light exist in the Michaelson-Morley
interferometer,
==========================================
Dead on arrival. Light is a photon stream, not a standing wave.
[/quote]
Androcles is showing his great ignorance again. He has never
heard of a Fabry-Perot resonant optical cavity. We keep
expecting him to find new ways to show his lack of knowledge
and he never disappoints us.
[quote]
[/quote] |
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| Androcles... |
| Posted: Wed Oct 28, 2009 12:23 am |
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Guest
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"Darwin123" <drosen0000 at (no spam) yahoo.com> wrote in message
news:73e81849-45bc-497b-b260-6460f062bc97 at (no spam) s31g2000yqs.googlegroups.com...
On Oct 27, 6:49 pm, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
[quote]"Darwin123" <drosen0... at (no spam) yahoo.com> wrote in message
news:8726a9ec-2c0d-4d34-a1d5-cedf92ed3415 at (no spam) y23g2000yqd.googlegroups.com...
On Oct 27, 3:16 pm, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
You are a clueless babbling lunatic.
I will now prove Einstein an idiot using your assumptions in
an even clearer manner than you have presented.
1) Standing waves made of light exist in the Michaelson-Morley
interferometer,
==========================================
Dead on arrival. Light is a photon stream, not a standing wave.
[/quote]
If that is so, why does the intensity output of the Michaelson
interferometer change as the mirror is moved?
===========================================
Why doesn't my bathroom scale change colour when I stand on it?
Because it was never designed to, that's why.
Michelson interferometers do not output intensity, you idiot. |
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| ... |
| Posted: Wed Oct 28, 2009 7:11 am |
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On Fri, 16 Oct 2009 16:08:07 +0100, tominlaguna at (no spam) yahoo.com wrote:
[quote]I missed the opportunity to comment on this subject when a thread was
started by Jonah Thomas last month. I hope to continue the discussion
from this new starting point.
Sue posted a link to a Wang & et al paper which describes their fiber
optical gyro (FOG) experiments. That paper has been superseded by:
http://arxiv.org/ftp/physics/papers/0609/0609235.pdf. This latest
paper provides a more detailed account of that work.
Figure 3 of the new Wang paper shows that when a linear section of the
FOG is moved in translation, there is a fringe shift that is
proportional to the length of that section and the speed of its
motion. Most people that I have discussed this with believe that Dr.
Wang has demonstrated that his design can detect translational motion.
I disagree. They measured the acceleration of the fiber section from
zero to some constant velocity.
The Wang paper has lead me to conclude that the "Sagnac effect" is a
phenomenon peculiar to situations when the source and/or receiver are
experiencing acceleration. There are "Sagnac devices" that can detect
that phenomenon, but they should not be confused with the phenomenon
itself. Examples of the devices are: the passive Sagnac
interferometer devices of Sagnac, Pogany, Michelson-Gale, and
Dufour-Prunier; the active Sagnac interferometer devices of
Macek-Davis, Stedmann, modern laser gyros; and finally the "one-way"
Sagnac system of devices known as GPS.
A simple analogy of the phenomenon can be understood by this example:
Assume you have a long freight car, 100 feet long. There is a dueler
located at each end with identical guns, ammo and skill. If the car
is stationary with respect to the rails or moving at a constant
velocity and both fire their guns at the same time, they both die at
the same time. But, if the train happens to accelerate forward while
the bullets are in flight, the guy at the rear of the car dies first.
The same thing would occur if the car was experiencing acceleration
throughout the gun fight. That, in my opinion, is the phenomenon of
Sagnac. Bullets are flying in two directions covering an equal
distance of 100 feet, but one arrives sooner than the other due to the
acceleration of the receiver.
Paul Anderson was describing a type of device while he thought he was
describing the effect. The generalized Sagnac effect does not deal
with enclosed areas and angular velocity; several detection devices
are based on those criteria, but the phenomenon is not exclusive to
them. Saburi in 1976 demonstrated that there was a radio signal
transit time difference east-west between two earth-stationary
receiver/transmitters. The GPS network is corrected each day to
adjust their clocks so that the one-way transmission of signals is
accurate due to the Sagnac effect. Paul also suggested the Wang
experiment was a modified Fizeau experiment. They used both hollow
fibers and solid cross-section fiber and got the same readings. Others
in the past, Pogany and Harress, investigated the use of glass prisms
in the Sagnac set-up to determine if it was a Fizeau effect, and they
concluded it was not. Post has written about this.
Tom Roberts erroneously states that the ballistic model cannot explain
Sagnac. I will acknowledge that the "re-emission" ballistic model is
denied by the Sagnac results. Tolman (1912) and Panofsky and Phillips
(1961) describe three ballistic models. Waldron (1977) describes two
of the three: the ballistic model of Ritz/Waldron and the re-emission
model. The re-emission model fails in explaining Sagnac and a host of
other experiments.
In the Ritz/Waldron model, a mirror is not a new source, and therefore
light may or may not be reflected at c with respect to it. Its speed
after reflection is based on any relative motion between the source
and the mirror. If there is no relative motion, the reflected photon
will be moving at c; if there is relative motion, v, its speed will be
c +/- v… all with respect to the mirror.
Regards,
Tom Miles
[/quote]
One of the more interesting aspects of the Wang experiment was the
discovery that movement of a straight section of the fibre loop, in
translation, produced a Sagnac signal.
At first, it seems counter-intuitive. The length of the fibre loop
does not change, yet moving part of it changes the optical path
length.
My original comments on the mechanism concluded that acceleration was
the mechanism. Maybe that would be too simplistic as Tom Roberts
pointed out. That would be applicable if the source was illuminated
prior to moving the straight section. But what would happen if the
source was not illuminated until after the translational motion was
linear? Would you not get the same result?
I suspect you would get the same signal.
What I see as happening is illustrated in the two diagrams below. In
the top diagram, there is a straight section of fibre of length L.
There is no translational motion (no relative motion between source,
receiver, or conduit). A photon at location A reflects off the inner
walls of the hollow fibre and eventually arrives at location B; from
there it continues to bounce of the walls until arriving at the
detector located to the right.
In the lower diagram, we have induced translational motion to that
section of the fibre by moving it to the left. Now, instead of
arriving at B, it arrives at B'. Since B' is closer to the detector,
that photon will arrive in a shorter duration of time. The opposite
occurs for the photon going in the opposite direction.
|<---------- L --------->|
__________________
¦ /\ /\ ¦
¦ / \ / \ ¦
¦ / \ / \ ¦ --> to detector
¦ / \/ \ ¦
¯¯¯^¯¯¯¯¯¯¯¯^¯¯¯¯¯
A B
|<---------- L ---------->|
__________________
¦ /\ /\ ¦
¦ / \ / \ ¦
¦ / \ / \ ¦ --> to detector
¦ / \/ \ ¦
¯¯¯^¯¯¯¯¯¯¯^¯¯¯^¯¯
A B B'
So how do we generalize the phenomenon? How do we best describe what
is happening? I am looking forward comments.
OFF TOPIC COMMENT: I will be travelling to California on Friday for
10 days and will not have much opportunity to participate while away.
I hope there will still be interest in this topic upon my return so I
can provide follow-up comments. I am very behind at this point having
only opened about half of the messages posted so far. |
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| Androcles... |
| Posted: Wed Oct 28, 2009 7:30 am |
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Guest
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<tominlaguna at (no spam) yahoo.com> wrote in message
news:4uege5hevdieg6gupdcue8hs22hq291e5j at (no spam) 4ax.com...
[quote]On Fri, 16 Oct 2009 16:08:07 +0100, tominlaguna at (no spam) yahoo.com wrote:
I missed the opportunity to comment on this subject when a thread was
started by Jonah Thomas last month. I hope to continue the discussion
from this new starting point.
Sue posted a link to a Wang & et al paper which describes their fiber
optical gyro (FOG) experiments. That paper has been superseded by:
http://arxiv.org/ftp/physics/papers/0609/0609235.pdf. This latest
paper provides a more detailed account of that work.
Figure 3 of the new Wang paper shows that when a linear section of the
FOG is moved in translation, there is a fringe shift that is
proportional to the length of that section and the speed of its
motion. Most people that I have discussed this with believe that Dr.
Wang has demonstrated that his design can detect translational motion.
I disagree. They measured the acceleration of the fiber section from
zero to some constant velocity.
The Wang paper has lead me to conclude that the "Sagnac effect" is a
phenomenon peculiar to situations when the source and/or receiver are
experiencing acceleration. There are "Sagnac devices" that can detect
that phenomenon, but they should not be confused with the phenomenon
itself. Examples of the devices are: the passive Sagnac
interferometer devices of Sagnac, Pogany, Michelson-Gale, and
Dufour-Prunier; the active Sagnac interferometer devices of
Macek-Davis, Stedmann, modern laser gyros; and finally the "one-way"
Sagnac system of devices known as GPS.
A simple analogy of the phenomenon can be understood by this example:
Assume you have a long freight car, 100 feet long. There is a dueler
located at each end with identical guns, ammo and skill. If the car
is stationary with respect to the rails or moving at a constant
velocity and both fire their guns at the same time, they both die at
the same time. But, if the train happens to accelerate forward while
the bullets are in flight, the guy at the rear of the car dies first.
The same thing would occur if the car was experiencing acceleration
throughout the gun fight. That, in my opinion, is the phenomenon of
Sagnac. Bullets are flying in two directions covering an equal
distance of 100 feet, but one arrives sooner than the other due to the
acceleration of the receiver.
Paul Anderson was describing a type of device while he thought he was
describing the effect. The generalized Sagnac effect does not deal
with enclosed areas and angular velocity; several detection devices
are based on those criteria, but the phenomenon is not exclusive to
them. Saburi in 1976 demonstrated that there was a radio signal
transit time difference east-west between two earth-stationary
receiver/transmitters. The GPS network is corrected each day to
adjust their clocks so that the one-way transmission of signals is
accurate due to the Sagnac effect. Paul also suggested the Wang
experiment was a modified Fizeau experiment. They used both hollow
fibers and solid cross-section fiber and got the same readings. Others
in the past, Pogany and Harress, investigated the use of glass prisms
in the Sagnac set-up to determine if it was a Fizeau effect, and they
concluded it was not. Post has written about this.
Tom Roberts erroneously states that the ballistic model cannot explain
Sagnac. I will acknowledge that the "re-emission" ballistic model is
denied by the Sagnac results. Tolman (1912) and Panofsky and Phillips
(1961) describe three ballistic models. Waldron (1977) describes two
of the three: the ballistic model of Ritz/Waldron and the re-emission
model. The re-emission model fails in explaining Sagnac and a host of
other experiments.
In the Ritz/Waldron model, a mirror is not a new source, and therefore
light may or may not be reflected at c with respect to it. Its speed
after reflection is based on any relative motion between the source
and the mirror. If there is no relative motion, the reflected photon
will be moving at c; if there is relative motion, v, its speed will be
c +/- v. all with respect to the mirror.
Regards,
Tom Miles
One of the more interesting aspects of the Wang experiment was the
discovery that movement of a straight section of the fibre loop, in
translation, produced a Sagnac signal.
At first, it seems counter-intuitive. The length of the fibre loop
does not change, yet moving part of it changes the optical path
length.
My original comments on the mechanism concluded that acceleration was
the mechanism. Maybe that would be too simplistic as Tom Roberts
pointed out. That would be applicable if the source was illuminated
prior to moving the straight section. But what would happen if the
source was not illuminated until after the translational motion was
linear? Would you not get the same result?
I suspect you would get the same signal.
What I see as happening is illustrated in the two diagrams below. In
the top diagram, there is a straight section of fibre of length L.
There is no translational motion (no relative motion between source,
receiver, or conduit). A photon at location A reflects off the inner
walls of the hollow fibre and eventually arrives at location B; from
there it continues to bounce of the walls until arriving at the
detector located to the right.
In the lower diagram, we have induced translational motion to that
section of the fibre by moving it to the left. Now, instead of
arriving at B, it arrives at B'. Since B' is closer to the detector,
that photon will arrive in a shorter duration of time. The opposite
occurs for the photon going in the opposite direction.
|<---------- L --------->|
__________________
¦ /\ /\ ¦
¦ / \ / \ ¦
¦ / \ / \ ¦ --> to detector
¦ / \/ \ ¦
¯¯¯^¯¯¯¯¯¯¯¯^¯¯¯¯¯
A B
|<---------- L ---------->|
__________________
¦ /\ /\ ¦
¦ / \ / \ ¦
¦ / \ / \ ¦ --> to detector
¦ / \/ \ ¦
¯¯¯^¯¯¯¯¯¯¯^¯¯¯^¯¯
A B B'
So how do we generalize the phenomenon? How do we best describe what
is happening? I am looking forward comments.
[/quote]
Since is just a short straight part of this, it's nothing remarkable.
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/MechSagnac.gif
(All three gears are rotating in the same direction.)
http://www.androcles01.pwp.blueyonder.co.uk/Sagnac/Sagnac.htm
and scroll down to "Deception:"
[quote]OFF TOPIC COMMENT: I will be travelling to California on Friday for
10 days and will not have much opportunity to participate while away.
I hope there will still be interest in this topic upon my return so I
can provide follow-up comments. I am very behind at this point having
only opened about half of the messages posted so far.
[/quote] |
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| Posted: Wed Oct 28, 2009 10:54 am |
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On Tue, 27 Oct 2009 10:05:54 -0000, "Androcles"
<Headmaster at (no spam) Hogwarts.physics_p> wrote:
[quote]
tominlaguna at (no spam) yahoo.com> wrote in message
news:ir5de51g0kj01pp4t0t4a4534shsdm1lnm at (no spam) 4ax.com...
On Sun, 25 Oct 2009 16:02:15 -0000, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
tominlaguna at (no spam) yahoo.com> wrote in message
news:7un8e5dhsrhnntge3n5dvcupki4hvqgc6u at (no spam) 4ax.com...
On Sun, 25 Oct 2009 10:21:49 -0000, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
tominlaguna at (no spam) yahoo.com> wrote in message
news:rnu7e5p7cpbinrb136espf5pf7onraap2q at (no spam) 4ax.com...
On Sat, 24 Oct 2009 16:43:07 +0100, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
tominlaguna at (no spam) yahoo.com> wrote in message
news:p526e5paf09ncq9nlj0tk7re78cs4egcv7 at (no spam) 4ax.com...
On Sat, 24 Oct 2009 02:34:44 +0100, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
tominlaguna at (no spam) yahoo.com> wrote in message
news:g6n3e5paio3me8frn6onelb3hpe1rb5p98 at (no spam) 4ax.com...
On Fri, 23 Oct 2009 15:02:59 +0100, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
tominlaguna at (no spam) yahoo.com> wrote in message
news:v393e5tbb3uslb0qqihjq4dt8ubip09kg8 at (no spam) 4ax.com...
On Thu, 22 Oct 2009 16:44:03 +0100, "Androcles"
Headmaster at (no spam) Hogwarts.physics_p> wrote:
tominlaguna at (no spam) yahoo.com> wrote in message
news:bqs0e5lqmuqtjqft1lvurh8ui21i974qp0 at (no spam) 4ax.com...
On Thu, 22 Oct 2009 22:49:58 +1100, "Inertial"
relatively at (no spam) rest.com
wrote:
tominlaguna at (no spam) yahoo.com> wrote in message
news:0qc0e5dhbaf4vu8qhib1s6omos1cv3e8m6 at (no spam) 4ax.com...
As I understand, ballistic theory means light behaves light a
ballistic
particle .. so it's velocity in some frame is the sum of the
velocity
of
source and that of the light wrt the source.
That is correct to this extent: If the observer has relative
motion
with respect to the source, he will experience that the light
at
c
+/-
v, where v is the value of the relative motion.
Similarly light would reflect
from a mirror at the same speed as it hits the mirror .. so if
it
hits
a
mirror at c+v, it will leave the mirror at the same speed.
Almost correct. For example, in the situation where a mirror
is
moving normally toward a source at velocity v, the mirror will
experience the light as arriving at c + v. Upon reflection,
the
light
will be traveling at c + 2v with respect to the source; and,
as
you
state, at c + v with respect to the mirror.
With source and reflector relatively at rest, the ray reflects
at
(MINUS) -c, just as a bouncing ball would with a perfectly
elastic
collision.
With the bat/mirror swinging at -v toward the source, the
approach velocity is c+v, the return velocity relative to the
pitcher is -c-v.
That is, the ball leaves the pitcher at c and leaves the bat
at -c-v.
One can imagine the ball leaving the bat at -c and the bat
moving
at -v, so the ball is caught by the pitcher at -c-v. There is
no
2v,
you cannot add v twice.
Just imagine the bat is moving backwards at v = c, so that c-v =
0.
You would not add v twice for that, would you? The ball
would have to go through the bat if you did.
Wrong.
This is the same situation you have with Doppler radar.(#) A
car
is
speeding toward the Doppler radar gun parked aside the road.
The
formula used by the gun's software to calculate the speed of the
car
is:
f = fo(1 + 2v/c*COS(theta))
where:
v*COS(theta) is the component of the car's velocity toward the
gun.
f is the frequency measured by the gun
fo is the frequency emitted by the gun
Guess where the "2" came from?
__________________________________________
(#) C. L. Andrews, "Optics of the Electromagnetic Spectrum",
(1960)
page 30.
====================================================
According to Professor Christian Andreas Doppler,
f = fo * (1-COS(theta).v/c)
which is
= fo * (c/c - COS(theta).v/c)
= fo * (c - COS(theta).v) /c
and for theta = 0,
= fo * (c - v)/c
which was used by Einstein
http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img107.gif
who simply, clumsily and incorrectly divided the whole thing by
sqrt(1-v^2/c^2), which is rather stupid since it assigns a BLUE
shift
to a transversely moving object, INCREASING its frequency, much to
my mild amusement and, I hope, yours too.
Another reference where fortunately Einstein's crap is omitted:
http://tinyurl.com/yfe2hgn
However, I digress, but I don't guess. Where does the 2 come from
in C. L. Andrews' medieval tome?
Is there a derivation from first principles or are you merely
citing
it
authoritatively?
===========================================
Moving target discrimination and resolution in simple CW radar
systems
are based on the Doppler Effect, which is now discussed.
Consider a stationary radar with a frequency of f = 1/To, where To
is
the period of the transmitted wave, and a target moving at a
constant
speed, v, toward the radar, as shown in Figure 13-1. (The speed v
will
be assumed to be positive if the target is approaching the radar,
negative if receding.) At time t = to, let the target be at range
R
=
Ro, and assume that at that time a peak or crest of the wave
(point
A
in Figure 13-1) is emerging from the radar's antenna. At time t =
to
+ To, the next crest of the wave (point B) is emerging; at this
time,
let the target's range be R = R1. The time, dt, necessary for
point
A
on the wave to travel from the radar to the target is the distance
travelled divided by the speed of the wave:
dt = (Ro - vdt)/c
or [equation
(13-2)]
dt = Ro/(c + v)
where c is the wave's speed, that is, the speed of light. The time
necessary for point A to return to the radar is again dt.
Therefore,
the total round trip takes 2dt, and point A returns to the radar
at
time
t1 = to + 2Ro/(c+v) [equation (13-3)]
Similarly, point B returns to the radar at time
t2 = to + To + 2R1/(c+v) [equation (13-4)]
Since t1 and t2 are the times of arrival of two adjacent peaks of
the
wave (points A and B), the period of the received wave, To, is
t2 -
t1, or
T'o = To - 2(Ro - R1)/(c+v) [equation (13-5)]
As can be seen from Eq. (13-5), the period (frequency) of the
received
wave decreases (increases) from the period (frequency) of the
transmitted wave due to scattering from a moving target. This is
the
Doppler Effect.
Now, since vTo = Ro - R1,
or, in terms of the received frequency, f' = 1/T'o,
f' = f(1+v/c)/(1-v/c) [equation (13-7)]
For most cases of interest, v/c << << 1. Thus,
f' = f(1+v/c)(1+v/c+v^2/c^2+...)
[equation(13- ]
f' = f(1+2v/c+2v^2/c^2+...)
is well approximated by
f' = f(1+2v/c) [equation(13-9)]
====================================
"Principles of Modern Radar", Jerry L. Eaves and Edward K. Reedy,
(1987) pages 399-401
Two authors to get it wrong, eh?
T0 is the emission period, given.
(Why numerical zeros (0) become lower case letter 'o's is a mystery,
but never mind, I'll live with it.)
R0-R1 is the distance the target moves in duration T0.
Hence v = (R0-R1)/T0 -- agreed.
I don't have figure 13-1, that data is not available to me, so I'm
going
to have to guess to get this debate moving along.
I take it that A and B are the endpoints of the transmitted
wavelength,
w = |A-B|.
That is, T'0 is the time between point A hitting the target and
point
B
hitting the target.
Hence T0 = w/c and T'0 = w/(c+v).
But w = c/f and f = 1/T0
so w = cT0,
giving
T'0 = T0. c/(c+v)
and by inverting,
f' = f . (c+v)/c in agreement with Herr Doppler.
T'o = To(c-v)/(c+v) = To(1-v/c)/(1+v/c) [equation (13-6)]
http://www.cartoonstock.com/lowres/wpr0087l.jpg
That (c-v) term sure looks like a rabbit to me.
Incidentally, an engineer riding in the car can determine his speed
if he knows the emitted frequency of the radar gun, all he needs
is a receiver (and the correct equations, of course).
A theoretical physicist with a magical (c-v) rabbit in his hat
cannot do this.
This is important to astronomy, it is no good aiming a radar gun
at a star.
If I were you I'd find an empty space in your top hat and feed those
text books into it, alongside your (c-v) rabbit.
Excuse me if I now chortle, but...
There is no 2v, you cannot add v twice. :-)
Didn't anyone ever tell you that when you're stuck in a hole that
your
first task was to stop digging...
Think of the speeding car as having a transponder rather than a
radar
reflecting surface.
The speeding car's receiver detects this frequency:
f1 = f·(1 + v/c)
The speeding car's transponder now emits a signal with frequency f1;
it has become a moving source.
The stationary highway patrolman's radar gun picks up this frequency
from the source speeding toward him and he records this frequency:
f2 = f1 / (1 - v/c)
combining:
f2 = f·(1 + v/c) / (1 - v/c)
Using binomial expansion and dropping off second order terms, you
get
this:
f2 = f·(1+2v/c)
Go here for more confirmation:
http://en.wikipedia.org/wiki/Doppler_radar
2v
2v
2v
2v
2v
It's bloody obvious that if a rifle (radar gun) is fired by the cop
at the car and the bullet (photon) travels at c relative to the cop,
and
then if an identical rifle (transponder) is fired from the car at the
cop with a muzzle velocity of -c relative to the car, the bullet
(photon)
will reach the cop at velocity -c-v = -(c+v) relative to the cop,
penetrating his Kevlar vest which was design to absorb energy
1/2 mc^2 and not
energy 1/2 m(-c-v)^2
= 1/2 m( c^2 +2cv +v^2)
= 1/2 mc^2 + mcv +1/2mv^2
which for v = c gives
= 2mc^2 which is 4 times greater.
Your f2 = f1 / (1 - v/c) that you pulled out of a hat is for aether
theory, not ballistic theory.
Proof:
Let v = c = 1, then
f2 = f1 / (1 - 1/1) = f1/0 = undefined. You've created a lumic boom
identical to a sonic boom, the car hits the cop at the same instant
as the bullet (impossible in emission theory) and his Kevlar vest
just doesn't matter, the mass of a car is 2 million times the mass
of a rifle bullet.
Bullet circa 1 gram, car circa 2 tonnes.
Excuse me if I now lol, but...
Bwhahahahahahahahahahahaha!
There is no 2v, you cannot add v twice.
Didn't anyone ever tell you that when you're stuck in a hole that you
should keep digging all the way to Wilson in Australia who is digging
his way to America to meet Michael Faraday who is flying Benjamin
Franklin's kite? Or so Henry Wilson thinks...
Let's go back to the ballistic analysis represented by Waldron and
reverse the conditions. Have the mirror be stationary and the source
moving toward it head-on. Can we agree that the speed of the photon
arriving at the mirror is c+v?
Relative to the mirror, yes.
At what speed relative to the mirror does it ricochet off the mirror?
1. c
2. c+v
3. some other value
3, -c-v.
c and v are vectors, hence velocities.
If you choose 1., you adopt the re-emission theory.
If you choose 2., you adopt the Ballistic theory.
The third choice is yet to be named.
False. The third choice is called Emission Fact, it's not a theory.
Good I like your choice.
We have the photon with velocity -(c+v) and the source at +v.
The relative velocity of the photon to the source is calculated to be:
Vrel = -(v+c) -v = -(c+2v)
We hold the mirror "stationary" and move the source at v.
Relative to the mirror, the approach velocity is c+v.
Do we agree? Yes, we do.
The approach velocity of the missile is still c relative to
the moving source.
The missile hits a stationary mirror which returns a bounce at -c
and the source runs toward the missile with a closing velocity of
c+v.
Vrel = -(c) -v = -(c+v).
Do we agree? No we don't.
The reason we don't agree is you've switched in your mind to
the c+v the mirror "sees" and away from the c the source "sees".
That's called frame jumping, and is very easy to do.
Think this. The missile is going away from me at c.
If it hits a wall it'll bounce back at c. If the wall moves
it will add its velocity to the missile. There is no way it
will add twice its velocity to the missile.
Let's do the frequency bit.
Two identical arrows with identical RPM and hence frequency,
one travels at c and the other at c+v, like this:
http://www.androcles01.pwp.blueyonder.co.uk/Wave/Arrows.gif
At the finish line, the faster c+v arrow has not completed the same
number of revolutions as the slower c arrow.
To do so, it has to go beyond the finish line and start its return
from further away. Since speed = distance/time, this reduces
its return speed from -(c+2v) to -(c+v). Of course it actually
reflects without going past the finishing post, but begins its return
journey with some partial revolution to make up causing the source
to detect a lesser count of revolutions as it passes.
So the return frequency is
f' = f * -(c+v)/-c and NOT f' = f (c+2v)/c
You seemed to object when I calculated from first principles
with numbers, complaining that its a hot Volvo and not a starship.
So be it, you do the numbers. Just remember that I want it in
terms of distance and time, not some antique paper full of
Lorentz transformations that'll take my time to read and find
the blunder in. I've done that twice for you already.
Isaac Asimov wrote in "Quasar, Quasar, Burning Bright" ,
ISBN 0-380-44610-3
(concerning life after death)
[ If you want to argue the point, present the evidence.
I must warn you, though, that there are some arguments I will not
accept.
I won't accept any argument from authority. ("The Bible says so")
I won't accept any argument from internal conviction ("I have faith it
is so")
I won't accept any argument from personal abuse ("What are you, an
atheist?")
I won't accept any argument from irrelevance ("Do you think you have
been put on this Earth just to exist for a moment of time?)
I won't accept any argument from anecdote ("My cousin has a friend who
went to a medium and talked to her dead husband")
And when all that, and other varieties of non-evidence are eliminated,
there turns out to be nothing.]
You prove you're wrong for your own benefit and you'll have
my admiration, just as I'd admire Roberts if he admitted he was
wrong.
Do NOT plague me with argument from authority or personal
conviction as you've been doing.
PROVE your case or prove to yourself you were wrong.
Fair enough?
I'm going to have to pass on the admiration carrot at this time;
though I hope to achieve that at some future point since I respect and
admire your knowledge, intelligence, creativity and passion.
Hahahahaha! But see, I created this
http://www.androcles01.pwp.blueyonder.co.uk/Wave/bounce.gif
to prove *I* was wrong!
All three situations are identical, they have to be, I copied and pasted
each wall-bead-groundplane object as a group in every reference frame.
In the fixed frame of reference of the ground plane, the return velocity
is clearly -c-2v.
So now I should hang my head in shame?
Not really. I was wrong to begin with as anyone can be, but quickly
realised it. But then I wanted to show how a preposterous argument
can easily be continued, which only a bigot will do. Now I can say
"I was wrong!", apologise to you, and move on.
[/quote]
Thank you. You are a good man, Charlie Brown.
[quote]That is something few are capable of, and certainly not Roberts. I'm
sorry, Tom, I used you to make a point. I hope Wilson picks up on
it, because although he's likeable he's another stubborn so-and-so
who won't admit error, making very difficult to get the real point
across.
I will provide some background so you can see my passion on this
stance. About 10 years ago due to a health issue, I was confronted
with mortality and decided to focus on something I had started in the
70's, which was physics experimentation. I learned that Wallace
Kantor lived in So. Cal., I contacted him and he provided help in
designing an optical bench. About that same time, I started visiting
the library at UCI, and one of the books I checked out was by Waldron.
I scanned through it and read a few chapters; I don't recall being
very impressed with his thesis. After several years of
experimentation which contained a few high points, the net result was
bitter disappointment that I could not detect the Earth's motion.
Wallace had warned me that I was on a fool's mission.
Following months of emotional depression, I decided to go back to the
drawing boards, literally. Using AutoCAD 2000, I constructed models
of reflection using simple optical principles; no relativity, no
ether, and no emission models. Simply straight optical relationships.
I modelled all kinds of possibilities: mirror moving with stationary
source, moving source with stationary mirror, both moving together,
etc. I did it for 180 and 90 degree reflections and for other angles
in between and beyond. AutoCAD allows great accuracy in construction,
and I recall I was either at 6 or 9 decimal places. The results were
strange at first: the angles of reflection appeared odd in terms of
their numerical values, and there was a pesky 2v in every diagram.
Having recalled 2v from reading Waldron, I returned to UCI and checked
out Waldron again... for the second time in its shelf life it saw the
light of day.
Sure enough, Waldron had determined there should be a 2v component
using momentum and energy conservation calculations. I had confirmed
it empirically with my AutoCAD models. The strange angles turned out
not to be so strange after all when I took the sine of them; they fit
exactly into Snell's Law predictions. The other discovery was that
the reflected ray always moved at c with respect to the mirror image
of the source; that confirmed the prediction of Stewart (1911).
If you have access to some drafting software, you might want to do
some modelling on your own. You'll find the 2v, I assure you.
Yes, of course. I use Google Sketchup, it's fast, fun, and free.
http://sketchup.google.com/
But now persuade Roberts that SR was written by anyone other than
an idiot, since that is basis for this newsgroup's existence and he's
the leading moron and chief babbling fool.
My background in optics is a little different to yours, I was involved
in graphics displays for flight simulators (the real ones, not Microsoft).
http://www.link-miles.co.uk/
In the early days we viewed a 26" monitor in a convex mirror with
a beamsplitter in between, so the screen of the monitor faced down.
http://www.link-miles.co.uk/BigPic3.asp?PicName=imageflyxzzjpg
This isn't raster graphics, it's vector graphics.
I'd take out the pilot's seat, put a theodolite in its place at the pilot's
eye-point (which was also the centre of curvature of the convex mirror)
and tweak the monitor to align it while displaying a test pattern.
Computing speed was a real problem back then, you only have
30 ms to display the whole scene or the display will flicker.
Later versions had all round vision:
http://www.link-miles.co.uk/images/shgr7.jpg
Those things were fun to fly in full motion, although I did crack
my head on the canopy of the RN Sea Harrier when I went slowly
up the ramp of the simulated aircraft carrier and tried to turn -the
software wasn't expecting a roll angle with weight on wheels
and the motion platform went wild. I was aligning the visual model
with the dynamics model, pitching and rolling the ship. Tricky to do,
rotating the ship and the plane at the same time.
http://www.military-today.com/navy/invincible_class_l2.jpg
I've also worked in industrial vision, non-contact gauging. Detect two
edges in the image and compute the distance between them at production
line speeds, reject anything that is out of spec. The darn thing worked
until we walked away, then it rejected everything. Come back and it
worked again. I found it was the technician (or me) leaving his shadow
that moved the computer detected edges in the image. Quick fix was
to add a bright light and turn down the contrast.
[/quote] |
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| Darwin123... |
| Posted: Thu Oct 29, 2009 10:31 am |
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On Oct 28, 2:23 am, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
[quote]"Darwin123" <drosen0... at (no spam) yahoo.com> wrote in message
news:73e81849-45bc-497b-b260-6460f062bc97 at (no spam) s31g2000yqs.googlegroups.com...
On Oct 27, 6:49 pm, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
"Darwin123" <drosen0... at (no spam) yahoo.com> wrote in message
news:8726a9ec-2c0d-4d34-a1d5-cedf92ed3415 at (no spam) y23g2000yqd.googlegroups.com....
On Oct 27, 3:16 pm, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
Michelson interferometers do not output intensity, you idiot.
The output of the Michaelson inteferometer is spectral lines, which[/quote]
can be described as power collected at different angles. The measured
output is proportional to the power detected at a certain angle. One
can analyze an intensity normalized to the peak values. However, a
more quantitative description can be made of the output.
which are measured in radiant intensity. The units of radiant
intensity are watts per steradian.
Since the radiant intensity is normalized, one can describe the
output in other ways depending on the detector used. The intensity of
the output can be decribed as a radiance (watts/srm^2) or spectral
irradiance (Watts per m^2 per nm^2).
The ouput of a Sagnac interferometer can also be described in
terms of intensity. The common implementation would be proportional to
the radiant flux, which is output power. One would measure the power
output as a function of time. However, The radiant flux can also be
normalized to the peak radiant flux. So again, there are several
different quantities that may describe the output of the Sagnac
cavity.
Regardless, you don't seem to have a clear idea of what is
measured from these interferometers. |
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| Darwin123... |
| Posted: Thu Oct 29, 2009 10:56 am |
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Guest
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On Oct 27, 10:02 am, Tom Roberts <tjroberts... at (no spam) sbcglobal.net> wrote:
[quote]Jonah Thomas wrote:
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
A light source emits periodic waves, but as I have discussed before, one
must consider a short light pulse.
One can validly describe the system in terms of wave packets. This[/quote]
is what you are describing. However, the wave packet will itself have
a spectrum one can find by Fourier analysis. These components can and
are Doppler shifted. The clockwise pulse will have components shifted
differently from the counterclockwise pulse.
I think there is some pedagogical confusion with your
representation because the short pulse has a spectral bandwidth. This
would complicate the analysis. The spectral bandwidth of each pulse is
not effected by Doppler shift. I think this spectral bandwidth is what
you mean by saying the "frequency is not changed." However, the
carrier wave frequency will be changed by Doppler shift. So will the
phase when they meet again. You may mean "wave front" instead of
"short pulse".
I think Jonah's problem is the following: His right brain is
Fourier analyzing that short pulse without telling his left brain that
a Fourier analysis is being performed. His left brain reads your post
that says that the frequency of the pulse doesn't change. His left
brain then asks you how come the Fourier components don't change in
frequency.
Now your problem is that your left brain has Fourier analyzed
that short pulse without telling the right brain that it has done so.
So your right brain still thinks there has been no frequency change.
Your left reads his post that the frequency must change, but delegates
the answer to his question to your right brain. Your right brain tell
him that there is no frequency change, just look at that short pulse.
This is why I have never tried to become a psychologist | |
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| Androcles... |
| Posted: Thu Oct 29, 2009 2:51 pm |
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Guest
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"Darwin123" <drosen0000 at (no spam) yahoo.com> wrote in message
news:d7ddf893-256c-4aa1-b9a8-b3fa79ce0106 at (no spam) y32g2000prd.googlegroups.com...
On Oct 28, 2:23 am, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
[quote]"Darwin123" <drosen0... at (no spam) yahoo.com> wrote in message
news:73e81849-45bc-497b-b260-6460f062bc97 at (no spam) s31g2000yqs.googlegroups.com...
On Oct 27, 6:49 pm, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
"Darwin123" <drosen0... at (no spam) yahoo.com> wrote in message
news:8726a9ec-2c0d-4d34-a1d5-cedf92ed3415 at (no spam) y23g2000yqd.googlegroups.com...
On Oct 27, 3:16 pm, "Androcles" <Headmas... at (no spam) Hogwarts.physics_p> wrote:
Michelson interferometers do not output intensity, you idiot.
The output of the Michaelson inteferometer is spectral lines, which[/quote]
can be described as power collected at different angles.
=======================================
Learn to spell the man's name, it is "Michelson".
Interference fringes look like this, they are not spectra.
http://upload.wikimedia.org/wikipedia/en/9/90/Michelson_Interferometer_Laser_Interference_Fringes-Red.jpg |
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| Tom Roberts... |
| Posted: Fri Oct 30, 2009 8:24 am |
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Darwin123 wrote:
[quote]On Oct 27, 10:02 am, Tom Roberts <tjroberts... at (no spam) sbcglobal.net> wrote:
Jonah Thomas wrote:
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
A light source emits periodic waves, but as I have discussed before, one
must consider a short light pulse.
One can validly describe the system in terms of wave packets.
[/quote]
Using wave packets can be problematical, if the dispersion is large within the
bandwidth of the packet.
I misspoke slightly: "short light pulse" can be replaced by "short region of a
light wave".
[quote]This
is what you are describing.
[/quote]
No. I described successive wavecrests of a monochromatic light wave.
[quote]The spectral bandwidth of each pulse is
not effected by Doppler shift.
[/quote]
For very short pulses with very large bandwidths, Doppler shift does affect the
bandwidth; this is usually minor.
[quote]I think this spectral bandwidth is what
you mean by saying the "frequency is not changed."
[/quote]
Not at all. I meant the frequency of a monochromatic light wave. Not its bandwidth.
[quote]However, the
carrier wave frequency will be changed by Doppler shift.
[/quote]
Yes, if Doppler shift applies. For the case I described it doesn't apply. That
was the point: I EXPLICITLY showed that the frequency at the detector is the
same as the frequency at the source. This is true for any rigid apparatus that
constrains the light paths to not change in length, no matter how it moves as
long as accelerations are small. But for an idealized circular Sagnac
interferometer, as long as the light paths remain rigid the acceleration due to
rotation can be large, as it is orthogonal to the light paths.
Tom Roberts |
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| Darwin123... |
| Posted: Sun Nov 01, 2009 11:49 am |
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On Oct 30, 9:24 am, Tom Roberts <tjrob... at (no spam) sbcglobal.net> wrote:
[quote]Darwin123 wrote:
On Oct 27, 10:02 am, Tom Roberts <tjroberts... at (no spam) sbcglobal.net> wrote:
Jonah Thomas wrote:
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
A light source emits periodic waves, but as I have discussed before, one
must consider a short light pulse.
One can validly describe the system in terms of wave packets.
Using wave packets can be problematical, if the dispersion is large within the
bandwidth of the packet.
I misspoke slightly: "short light pulse" can be replaced by "short region of a
light wave".
This
is what you are describing.
No. I described successive wavecrests of a monochromatic light wave.
The spectral bandwidth of each pulse is
not effected by Doppler shift.
For very short pulses with very large bandwidths, Doppler shift does affect the
bandwidth; this is usually minor.
I think this spectral bandwidth is what
you mean by saying the "frequency is not changed."
Not at all. I meant the frequency of a monochromatic light wave. Not its bandwidth.
However, the
carrier wave frequency will be changed by Doppler shift.
Yes, if Doppler shift applies. For the case I described it doesn't apply. That
was the point: I EXPLICITLY showed that the frequency at the detector is the
same as the frequency at the source. This is true for any rigid apparatus that
constrains the light paths to not change in length, no matter how it moves as
long as accelerations are small. But for an idealized circular Sagnac
interferometer, as long as the light paths remain rigid the acceleration due to
rotation can be large, as it is orthogonal to the light paths.
Tom Roberts
[/quote]
You are right and I was wrong. However, I am not sure where my
mistake was.
Some articles on the Sagnac effect refer to "transverse" and
"longitudinal" Doppler shifts. Supposedly, the "transverse Doppler
shift" only occurs in Lorentz invariant systems, not in Galilean
invariant systems. I will look at these articles more closely to see
what exactly they meant by "Doppler shift." |
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| Androcles... |
| Posted: Sun Nov 01, 2009 6:57 pm |
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Guest
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"Darwin123" <drosen0000 at (no spam) yahoo.com> wrote in message
news:b965c0ed-b117-4d5d-9a6a-cc7ad5a5dff0 at (no spam) r24g2000yqd.googlegroups.com...
On Oct 30, 9:24 am, Tom Roberts <tjrob... at (no spam) sbcglobal.net> wrote:
[quote]Darwin123 wrote:
On Oct 27, 10:02 am, Tom Roberts <tjroberts... at (no spam) sbcglobal.net> wrote:
Jonah Thomas wrote:
I haven't seen an argument yet why there shouldn't be frequency
differences in Sagnac.
A light source emits periodic waves, but as I have discussed before,
one
must consider a short light pulse.
One can validly describe the system in terms of wave packets.
Using wave packets can be problematical, if the dispersion is large within
the
bandwidth of the packet.
I misspoke slightly: "short light pulse" can be replaced by "short region
of a
light wave".
This
is what you are describing.
No. I described successive wavecrests of a monochromatic light wave.
The spectral bandwidth of each pulse is
not effected by Doppler shift.
For very short pulses with very large bandwidths, Doppler shift does
affect the
bandwidth; this is usually minor.
I think this spectral bandwidth is what
you mean by saying the "frequency is not changed."
Not at all. I meant the frequency of a monochromatic light wave. Not its
bandwidth.
However, the
carrier wave frequency will be changed by Doppler shift.
Yes, if Doppler shift applies. For the case I described it doesn't apply.
That
was the point: I EXPLICITLY showed that the frequency at the detector is
the
same as the frequency at the source. This is true for any rigid apparatus
that
constrains the light paths to not change in length, no matter how it moves
as
long as accelerations are small. But for an idealized circular Sagnac
interferometer, as long as the light paths remain rigid the acceleration
due to
rotation can be large, as it is orthogonal to the light paths.
Tom Roberts
[/quote]
You are right and I was wrong. However, I am not sure where my
mistake was.
Some articles on the Sagnac effect refer to "transverse" and
"longitudinal" Doppler shifts. Supposedly, the "transverse Doppler
shift" only occurs in Lorentz invariant systems, not in Galilean
invariant systems. I will look at these articles more closely to see
what exactly they meant by "Doppler shift."
===============================================
"if an observer is moving with velocity v relatively to an infinitely
distant
source of light of frequency \nu, in such a way that the connecting line
``source-observer'' makes the angle phi with the velocity of the observer
referred to a system of co-ordinates which is at rest relatively to the
source of light, the frequency \nu' of the light perceived by the observer
is given by the equation
\nu' = \nu ( [1-cos(phi).v/c] / sqrt(1-v^2/c^2)) "
Ref: http://www.fourmilab.ch/etexts/einstein/specrel/www/figures/img107.gif
Doppler's equation is \nu' = \nu [1-cos(phi).v/c]
For the case where phi = 0, cos(phi) = 1 and we have
\nu' = \nu [1-v/c]
That is, the observer is moving AWAY from the source. By the PoR,
this should be the same as the source moving away from the observer.
For the case where phi = pi (180 degrees), cos(phi) = -1 and we have
\nu' = \nu [1+v/c]
That is, the observer is moving TOWARD the source. By the PoR,
this should be the same as the source moving toward from the observer.
For the case where phi = pi/2(90 degrees) or 3pi/2 (270 degrees),
cos(phi) = 0 and we have \nu' = \nu [1+0] = \nu
That is, the observer is moving transverse to the source. By the PoR,
this should be the same as the source moving transverse to the observer.
However, such transverse motion cannot be maintained except the
observer travels in a circle around the source or the source around the
observer, for normally the angle phi is constantly changing.
The emergency vehicle approaching you makes an angle that passes
through 90 degrees going negatively,
\nu is high pitched as the vehicle approaches and changes to low pitch
as it recedes.
So it is going from 180 through 90 toward 0 degrees.
Ok so far.
Now the crank Einstein divides that by his fetish, sqrt(1-v^2/c^2), and
creates a shift toward the blue because if you divide by something less
than one you increase the quotient.
Homework:
For NO shift,
\nu' = \nu ( [1-cos(phi).v/c] / sqrt(1-v^2/c^2))
\nu' / \nu = ( [1-cos(phi).v/c] / sqrt(1-v^2/c^2))
1 = [1-cos(phi).v/c] / sqrt(1-v^2/c^2)
Solve for v.
Einstein shift for v = 0.866c changes from blue to red at angle = 54 degrees
Get the stumblebum Roberts to help you with your homework, he's
hopeless at math and physics. |
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| Tom Roberts... |
| Posted: Mon Nov 02, 2009 9:43 am |
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Darwin123 wrote:
[quote]Some articles on the Sagnac effect refer to "transverse" and
"longitudinal" Doppler shifts. Supposedly, the "transverse Doppler
shift" only occurs in Lorentz invariant systems, not in Galilean
invariant systems. I will look at these articles more closely to see
what exactly they meant by "Doppler shift."
[/quote]
Any sensible article will use "Doppler shift" to mean the change in
frequency or wavelength of a light ray (or EM signal) between its
emission from a moving source and its observation by a detector at rest
in a given inertial frame. This depends not only on the speed of the
source, but also on its direction of motion relative to the
line-of-sight from source to detector.
Transverse Doppler shift refers to the fact that in SR, the Doppler
shift does not go to zero when the source is moving exactly
perpendicular to the line-of-sight to the detector. Indeed, it is
essentially the same as "time dilation". This is absent in pre-SR
theories, which give zero Doppler shift for such transverse motion.
Tom Roberts |
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