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tommac
2008-Apr-18, 07:46 PM
I am not sure if I am using the term right ... but if I put a light on a rod .. and a receptor on a crossing rod so that the two rods formed an x in I then began to rotate the x quickly in the direction of the light ...

would this light get redshifted in the receptor?

ctcoker
2008-Apr-18, 07:54 PM
I don't think so. As the x rotates, both the light and receptor rotate in the same direction with the same angular speed. Therefore, while the light is redshifting because the light is "moving away" from the detector, the light will also blueshift as the detector is moving toward the light with the same velocity. If I remember my rotational mechanics correctly, this means the two Doppler Shifts will cancel out.

tommac
2008-Apr-18, 08:09 PM
I don't think so. As the x rotates, both the light and receptor rotate in the same direction with the same angular speed. Therefore, while the light is redshifting because the light is "moving away" from the detector, the light will also blueshift as the detector is moving toward the light with the same velocity. If I remember my rotational mechanics correctly, this means the two Doppler Shifts will cancel out.
Yes this is what I thought ... however light is pointing and accellerating constantly at towards the tangent ... The acceleration and velocity of the receptor is in a different angle actually 90 degrees different. I am not sure if this means anything ... that is why I am confused.

blueshift
2008-Apr-18, 08:44 PM
Yes this is what I thought ... however light is pointing and accellerating constantly at towards the tangent ... The acceleration and velocity of the receptor is in a different angle actually 90 degrees different. I am not sure if this means anything ... that is why I am confused. Be careful when wording things.

Both your light source and your detector are located in the same reference frame. There is no redshift or blueshift of any kind. Light source and detector do not have their distances changing at all for any motion to be detected by the detector. You do not count in the experiment. What measurement your eyes make are not where detection takes place.

Light does not accelerate, it free falls effortlessly. Your light SOURCE can be accelerating but not the light. Light does decelerate by refraction in glass, atmospheres and water but it does not get any kick up to speed c.

Angular momentum can cause both a redshift and a blueshift by making your ears the detector and a sound source of a steady pitch ( a buzzer of some sort) located on a string that is twirled around someone else's head, not your head. By listening while the buzzer is being twirled you will hear a shift in frequency of the sound source. The twirler will not.

Jeff Root
2008-Apr-19, 02:10 AM
I'm fairly sure that there is Doppler shift while the apparatus is accelerating
or decelerating, but not while it is rotataing at constant speed. You might
find a lot more about this by looking up "Sagnac experiment", "laser gyro",
"ring laser", and "interferometric fiber-optic gyro", or "IFOG".

-- Jeff, in Minneapolis

grav
2008-Apr-19, 05:02 AM
It can be demonstrated that the speed directly between any two co-rotating points on a rotating disk (or sphere), after the speed of each (which depends upon the distance from the center axis), the instantaneous direction of travel, and the angles between them and their directions of travel have all been accounted for, comes to exactly zero, so there should be no Doppler shift.

By the same token, however, as Jeff mentioned, there is also the Sagnac effect, where two beams of light travel in opposite directions around a rotating apparatus, either by means of optic fibers, or simply by reflecting off of mirrors. This does produce an interference between the two beams which is proportional to the area within the reflecting zone of the mirrors, or to that of the enclosed area of the optic fibers. Ironically, this would seem to disprove a ballistic theory of light, where the light might also travel with the additional speed of the source, and so essentially rotate with the apparatus as well, which would be perceived in the same way as if the apparatus wasn't rotating at all, which would then produce no interference. I am just speculating, but it may have something to do with light travel time, whereas the light would not be perceived as coming from the same point that emitted it, since the apparatus has rotated some during the time of travel of the light.

Hornblower
2008-Apr-19, 10:36 AM
At the risk of sounding pedantic, let me remind everyone that we are not discussing angular momentum here. This merely is a kinematic discussion of the effects of the motions of the source and the receptor.

mugaliens
2008-Apr-19, 06:38 PM
I am not sure if I am using the term right ... but if I put a light on a rod .. and a receptor on a crossing rod so that the two rods formed an x in I then began to rotate the x quickly in the direction of the light ...

would this light get redshifted in the receptor?

I've seen much guessing, here....

A definitive and emphatic answer AND the reason as to why this is so: No. At any given velocity there is no shift at all. It would be blue-shifted during acceleration, and red-shifted during deceleration.

Here's why:

If you had described a linear system, with two spacecraft following one another at .01 c, and again at .9 c, and their relative velocities were the same, there would be no redshift, not even due to the increase in mass of the sending ship (because the receiving spaceship would also have increased in mass).

But in your model, let's assume two situations. In the first, the hub is stationary, with the receiver at top dead center (TDC), and the receiver 90 degrees to the right, at the 3 o'clock position.

In all situations, the contraption rotates clockwise, and the sender emits a phtonic pulse such that the receiver gets it when it's at TDC. Our viewpoint is along the hub's axis.

Let's speed it up a bit, to, say, .5 c rotational velocity at the outer rim, where both the sender and receiver are located. Now the sender has to send the pulse earlier, before it reaches the 3 o'clock position, in order for the receiver to see the pulse when it reaches TDC.

Let's assume for the moment that this requires the receiver to be at the 2 o'clock position and look at the geometry for a moment. The relative velocity between the sender at that position and TDC is different than the relative velocity between the receiver and the sent position when the receiver sees the pulse!

As the rotational velocity approaches the speed of light (let's assume it's so close to c that we can't really discern the difference), then the angular solution is such that the straight-line distance between the sent position and TDC is equal to the angular distance (around the rim) between the receiver's position and TDC.

It's an easy trig problem, if you understand trigonomety.

Fortunately, we don't need to go into the math, as I can put this in laymen's terms.

Now for the relative velocity problem.

The exact position need not be known, but we can approximate it. The receiver would be at an angle between 45 deg and 90 to the right of TDC when it sent the pulse, because the straight line distance from it to TDC has to be equal to the curved path distance from the receiver to TDC.

The key is that because it's less than 90 degrees, the distance is shorter. I'll explain why this still wouldn't result in a shift (except during acceleration or deceleration) in a moment.

Let's assume it's 50 degrees, which is close to the precise answer, but easier to explain without working with all the decimal points.

Let's draw a circle. Put a dot at 50 deg to the right of TDC. That's the sender when he sends the pulse. Put a dot at 0 deg (TDC). That's the receiver when he receives the pulse. Draw a line between the center and both dots. Bisect the angle between the center and both dots (25 deg) and draw that line. You have 25 deg either side of the bisection. Draw a line perpendicular to the movement of both dots at their sending and receiving positions. Draw a line between both dots (the path of the light pulse). The angles to that line and the movement lines are the same: 25 deg.

The key is that this geometry is persistant at any given rotational velocity. The distance between the sender and receiver does not change while the contraption is rotating.

The relative velocity at rest is precisely 0.

The relative velocity in motion at .5 c, .999999c, or whatever value you want to choose, is precisely 0.

Back to the distance is shorter, thing.....

As the contraption accelerates it's rotational movement, the distance gets shorter. Train coming towards you? Higher pitch.

Thus, while the contraption is in the process of accelerating it's angular velocity, there's a blue shift. As it's slowing, there would be a red shift.

At any static (unchanging) angular velocity, there would be no change in distance, thus zero relative velocity, and thus no shift at all.

dhd40
2008-Apr-19, 07:47 PM
I am not sure if I am using the term right ... but if I put a light on a rod .. and a receptor on a crossing rod so that the two rods formed an x in I then began to rotate the x quickly in the direction of the light ...

would this light get redshifted in the receptor?

It may depend on which picture you have in mind: a) or b) (b =Sagnac, see Jeff Root´s post)

Jeff Root
2008-Apr-19, 10:24 PM
mugaliens,

It appears that you decided to change the arrangement of the apparatus
midway through describing it, but accidentally left many of the original
descriptions in place. You have two receivers instead of a sender and a
receiver, describe the receiver as sending as well as receiving, and have
the receiver in two different locations in the same setup:



But in your model, let's assume two situations. In the first, the hub is
stationary, with the receiver at top dead center (TDC), and the receiver
90 degrees to the right, at the 3 o'clock position.
Two receivers.


In all situations, the contraption rotates clockwise, and the sender emits a
phtonic pulse such that the receiver gets it when it's at TDC.
If the receiver is at the top center, and the light is aimed in the direction
of rotation, as specified in tommac's original post*, then the sender must
be near the 9 o'clock position, not the 3 o'clock position. I can't tell
where anything is really supposed to be.

Was it your intention to have the sender or the receiver at a known
position? It even looks like maybe you switched between clockwise
rotation and counterclockwise rotation at some point.

* Editing to add: I think tommac meant that the X rotates in the direction
of the light's travel. He could instead have meant that the receptor
rotates toward the light source.

-- Jeff, in Minneapolis

Jeff Root
2008-Apr-19, 10:34 PM
dhd40,

setups a and b will behave the same regarding Doppler shift. The mirror
reverses the image of the light, but that isn't relevant. A Sagnac device
can use any number of mirrors and have any shape light path. Of course
the Sagnac apparatus has features that are irrelevant to the original
question, but it is still a pretty simple thing, and descriptions of how it
works should answer the original question.

-- Jeff, in Minneapolis

tommac
2008-Apr-19, 11:31 PM
I meant that the the receptor is on the x on one leg and the source is on the leg next to it. Their distance away from each other doesnt change ...
But the aparatus ... the x ... can rotate near the speed of light ... point is there a red or blue shift?


mugaliens,

It appears that you decided to change the arrangement of the apparatus
midway through describing it, but accidentally left many of the original
descriptions in place. You have two receivers instead of a sender and a
receiver, describe the receiver as sending as well as receiving, and have
the receiver in two different locations in the same setup:


Two receivers.

If the receiver is at the top center, and the light is aimed in the direction
of rotation, as specified in tommac's original post*, then the sender must
be near the 9 o'clock position, not the 3 o'clock position. I can't tell
where anything is really supposed to be.

Was it your intention to have the sender or the receiver at a known
position? It even looks like maybe you switched between clockwise
rotation and counterclockwise rotation at some point.

* Editing to add: I think tommac meant that the X rotates in the direction
of the light's travel. He could instead have meant that the receptor
rotates toward the light source.

-- Jeff, in Minneapolis

Jeff Root
2008-Apr-20, 12:15 AM
Tom,

That didn't clarify whether you meant that the X rotates in the direction
of the light's travel, or that the receptor rotates toward the light source.
I'm fairly sure you meant the former. However, it doesn't really matter
since mugaliens should have made clear the arrangement of the apparatus
in his description, whether he followed your specification or not.

I hope that mugaliens will re-write the description from scratch once he
has decided which direction he wants the apparatus to rotate, whether
the light beam travels with or against the direction of rotation, and
whether it is the position of the source or the receptor that is specified.

-- Jeff, in Minneapolis

tommac
2008-Apr-20, 01:48 AM
It may depend on which picture you have in mind: a) or b) (b =Sagnac, see Jeff Root´s post)

A! the graphics help!

tommac
2008-Apr-20, 01:52 AM
Also, if you are interested ... I am interested in this ... Is:

If galaxies rotate around a central point could the rotation cause a red shift? would it matter where the receptor was placed on the leg of the x? ( would it matter were a galaxie was relative to the center? )

Like would it matter if it was at the tip of the x or near the base of the x?
If it was near the bass then the emmitor would be moving faster than the receptor but their distance never changes ...

grav
2008-Apr-20, 02:16 AM
Also, if you are interested ... I am interested in this ... Is:

If galaxies rotate around a central point could the rotation cause a red shift? would it matter where the receptor was placed on the leg of the x? ( would it matter were a galaxie was relative to the center? )

Like would it matter if it was at the tip of the x or near the base of the x?
If it was near the bass then the emmitor would be moving faster than the receptor but their distance never changes ...I asked a similar question in this thread (http://www.bautforum.com/questions-answers/70962-stationary-frame-reference.html), although I only understood a portion of the replies, and it was never really absolutely resolved.

tommac
2008-Apr-20, 02:29 AM
I asked a similar question in this thread (http://www.bautforum.com/questions-answers/70962-stationary-frame-reference.html), although I only understood a portion of the replies, and it was never really absolutely resolved.


I tried to read through it but I am not sure if anyone really answered your question

Vanamonde
2008-Apr-21, 08:42 AM
You went and made me think of SS433 - I have thought about SS433 in years! The latest description is "an eclipsing X-ray binary system, with the primary most likely a black hole, or possibly a neutron star" or SS 433 is a microquasar [!], the first discovered." I would have thought a microquasar is about the BIGGEST oxymoron possible!

This is from this (http://en.wikipedia.org/wiki/SS_433) Wikipedia article. It has links to several articles which I am sure is full of scary math and great astrojargon. I hope to find time to explore them.

And remember that what we call a "red shift" or a "blue shift" is the optical version of the Doppler effect which is the apparent change in frequency and wavelength of a wave that is perceived by an observer moving relative to the source of the waves. As in Johann Christian Andreas Doppler in 1842, verified in sound waves by John Scott Russell in 1848. First observed with light in a 1938 experiment performed by Herbert E. Ives and G.R. Stilwell, called the Ives-Stilwell experiment.

tommac
2008-Apr-21, 02:05 PM
This was a great post!!! thank you!

Now for some follow up questions ... what if the x formed an ellipse when it was rotated, Like if the legs were not all the same legnth?

Also what if the aparatus spun on a perfect x but not around the center ...

What I am fishing for is a way to show a red or blue shift.

The direction I am going with all of this is to see if the rotation of galaxies around a central location could cause a percieved red shift over a temporary span.

Now even with what you wrote above .... if the aparatus is accelerating we would see a shift. As gravity pulls on objects in its orbit the orbit will tend to tighten and accelerate right? ( I want to think more about the movements of galaxies not the movement of the galaxy itself ) but it is easier to show on either a galactic level or a solar system level as that is better understood.

Do I make sense here ... I can try to clarify.




I've seen much guessing, here....

A definitive and emphatic answer AND the reason as to why this is so: No. At any given velocity there is no shift at all. It would be blue-shifted during acceleration, and red-shifted during deceleration.

Here's why:

If you had described a linear system, with two spacecraft following one another at .01 c, and again at .9 c, and their relative velocities were the same, there would be no redshift, not even due to the increase in mass of the sending ship (because the receiving spaceship would also have increased in mass).

But in your model, let's assume two situations. In the first, the hub is stationary, with the receiver at top dead center (TDC), and the receiver 90 degrees to the right, at the 3 o'clock position.

In all situations, the contraption rotates clockwise, and the sender emits a phtonic pulse such that the receiver gets it when it's at TDC. Our viewpoint is along the hub's axis.

Let's speed it up a bit, to, say, .5 c rotational velocity at the outer rim, where both the sender and receiver are located. Now the sender has to send the pulse earlier, before it reaches the 3 o'clock position, in order for the receiver to see the pulse when it reaches TDC.

Let's assume for the moment that this requires the receiver to be at the 2 o'clock position and look at the geometry for a moment. The relative velocity between the sender at that position and TDC is different than the relative velocity between the receiver and the sent position when the receiver sees the pulse!

As the rotational velocity approaches the speed of light (let's assume it's so close to c that we can't really discern the difference), then the angular solution is such that the straight-line distance between the sent position and TDC is equal to the angular distance (around the rim) between the receiver's position and TDC.

It's an easy trig problem, if you understand trigonomety.

Fortunately, we don't need to go into the math, as I can put this in laymen's terms.

Now for the relative velocity problem.

The exact position need not be known, but we can approximate it. The receiver would be at an angle between 45 deg and 90 to the right of TDC when it sent the pulse, because the straight line distance from it to TDC has to be equal to the curved path distance from the receiver to TDC.

The key is that because it's less than 90 degrees, the distance is shorter. I'll explain why this still wouldn't result in a shift (except during acceleration or deceleration) in a moment.

Let's assume it's 50 degrees, which is close to the precise answer, but easier to explain without working with all the decimal points.

Let's draw a circle. Put a dot at 50 deg to the right of TDC. That's the sender when he sends the pulse. Put a dot at 0 deg (TDC). That's the receiver when he receives the pulse. Draw a line between the center and both dots. Bisect the angle between the center and both dots (25 deg) and draw that line. You have 25 deg either side of the bisection. Draw a line perpendicular to the movement of both dots at their sending and receiving positions. Draw a line between both dots (the path of the light pulse). The angles to that line and the movement lines are the same: 25 deg.

The key is that this geometry is persistant at any given rotational velocity. The distance between the sender and receiver does not change while the contraption is rotating.

The relative velocity at rest is precisely 0.

The relative velocity in motion at .5 c, .999999c, or whatever value you want to choose, is precisely 0.

Back to the distance is shorter, thing.....

As the contraption accelerates it's rotational movement, the distance gets shorter. Train coming towards you? Higher pitch.

Thus, while the contraption is in the process of accelerating it's angular velocity, there's a blue shift. As it's slowing, there would be a red shift.

At any static (unchanging) angular velocity, there would be no change in distance, thus zero relative velocity, and thus no shift at all.