Thread: Rotation + Speed of Light

1. Rotation + Speed of Light

If I have a rotating very large merry go round and I placed a light at 12 o'clock and a receptor at 9 o'clock ( 90 degree angle ). The light does not point to the receptor but to the point on the circumference that is directly in between the light and the receptor.

Then I sped up the merry go round to a speed where the receptor would catch up to the point where the light was shown.

Picture this as the outside of the merry go round ( obviously would be on a curve ) :
R-----LS-----S

Receptor
Point at which light is shone to
Source

Now if the merry go round was rotating at the Speed of light the time that the light would taked to get to LS ... would be the exact time that R would get to LS.

My question is does this light red shift?

The idea is that the source is moving away from the LS where the light would ultimately be detected. However the receptor is moving towards where the light was originall shone to.

The key here is the direction of movement of the light, the direction of movement of the source, and the direction of movement of the receptor.

the light travels at 45 degrees from the source
The receptor travels towards the spot where it is detected following the curve of the circle
The source pulls away following the curve of the circle.

However since the source is pulling away from the light faster than the receptor is moving towards the light right because of the angle of the light and the path that the source is travelling. Is this correct?

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Well, the merry-go-round can't rotate at the speed of light.

For speeds less than the speed of light:

Redshifted.
You can forget the merry-go-round and just look at the frame in which the light was emitted, and the frame in which it is received. The two frames have the same speed, but different velocity vectors. Each sees the other with a transverse velocity and a recessional velocity.

Grant Hutchison

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Bingo!

Originally Posted by grant hutchison
Well, the merry-go-round can't rotate at the speed of light.

For speeds less than the speed of light:

Redshifted.
You can forget the merry-go-round and just look at the frame in which the light was emitted, and the frame in which it is received. The two frames have the same speed, but different velocity vectors. Each sees the other with a transverse velocity and a recessional velocity.

Grant Hutchison
Very concise, Grant. Wow, we haven't had a relatvistic wheel question for a month or so. Your answer should be filed away for the next one!

Regards, John M.

4. Originally Posted by grant hutchison
Well, the merry-go-round can't rotate at the speed of light.

For speeds less than the speed of light:

Redshifted.
You can forget the merry-go-round and just look at the frame in which the light was emitted, and the frame in which it is received. The two frames have the same speed, but different velocity vectors. Each sees the other with a transverse velocity and a recessional velocity.

Grant Hutchison
Yes but the vectors are pointing in different directions and the source of the light was much closer to where the light will be detected when the light was released.

you have 3 vectors right at any point of time.
1) the motion of the source
2) the mostion of the receptor
3) the direciton that the light is shown ( x SOL ).

Would this create a constant redshift?

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Originally Posted by tommac
My question is does this light red shift?
Yes.

6. Originally Posted by mugaliens
Yes.
OK ... hypothetically ... if the universe revolved around a central point. Wouldnt all galaxies be redshifted in a similar way to the question above?

Basically we would be seeing light from a distant galaxy that is now somewhere else in its orbit of the center of the universe etc ...

Wouldnt that cause a red shift for all galaxies if all of the galaxies are in orbit?

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Originally Posted by tommac
Yes but the vectors are pointing in different directions ...
Yes, I said that.

Originally Posted by tommac
you have 3 vectors right at any point of time.
1) the motion of the source
2) the mostion of the receptor
3) the direciton that the light is shown ( x SOL ).

Would this create a constant redshift?
Yes. At any moment, the relationship between the source and receiver frames is the same, as measured in either frame.

Grant Hutchison

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Originally Posted by tommac
OK ... hypothetically ... if the universe revolved around a central point. Wouldnt all galaxies be redshifted in a similar way to the question above?

Basically we would be seeing light from a distant galaxy that is now somewhere else in its orbit of the center of the universe etc ...

Wouldnt that cause a red shift for all galaxies if all of the galaxies are in orbit?
Work your example for other positions around the wheel. The frequency shift is different for different angles. That's not what we see in the Universe.

Grant Hutchison

9. Originally Posted by tommac
Wouldnt that cause a red shift for all galaxies if all of the galaxies are in orbit?
I haven't worked it out, but that sounds to me like it would only apply to the same orbit.

Think of lights coming from the other horses in the merry go round. I don't think the amount of redshift would coincide with how far away they are.

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Originally Posted by mugaliens
Yes.
That was a quick and dirty answer... Let me add some content, hopefully, to convey some better understanding.

First, let's restate the problem:

Givens:

1. Merry-go-round (MGR) of radius r that rotates clockwise at an angular velocity such that it perimeter is moving at c

2. Point S on the MGR is at 12 o'clock when t=0

3. Point R on the MGR is at 9 o'clock when t=0

4. At t=0, the light at point S emits a pulse

Questions:

Q1. Does this light red-shift?

Statements not in contention:

NCS1: The key here is the direction of movement of the light, the direction of movement of the source, and the direction of movement of the receptor.

Statements in contention:

CS1. The light travels at 45 degrees from the source.

Statements in giber (gibberish):

SG1: The receptor travels towards the spot where it is detected following the curve of the circule.

SG2: However since the source is pulling away from the light faster than the receptor is moving towards the light right because of the angle of the light and the path that the source is travelling.

Unfortunately, as stated, the whole problem is a non-sequitor for the following reasons:

- CS1: The point where R receives the signal sent from S is not at a 45 degree angle. Thus, if R sends its signal at 135 degrees clockwise from it's velocity, it will never be received by R! The reason for this is the R is travelling a circular distance at velocity c, while the signal is travelling a linear distance at velocity c. The solution is to find the angle where the distance from S to LC on the circumference of the circle equals the arc-length from R to LC. That angle is not 45 degrees.

- SG1: I've surmised what you're trying to say, but I would have stated it as I did in the previous paragraph.

- SG2: That's just not English. Well, it's a bunch of English words strung together. What you need to understand is that light red/blue-shifts commensurate with the net velocity component to the line between where the light was sent and where it was received. You cannot calculate that until you first calculate the angle as mentioned two paragraphs before! Do that first (simple geometry), and we'll move on from there into the harder stuff.

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There's an additional problem no one here caught: How much the light would be red or blue-shifted.

If the source is moving at c, it would be blue-shifted to a wavelength of 0 in all directions in front of it, and it would be red-shifted down to a frequency of 0 in all directions behind it.

Thus, the receiver would not detect light. Rather, it would detect a discrete jump in the E-M field from one level to a different level.

This is analogus to the fact that for the source, moving at c, time is stopped. Of course this would also mean that you wouldn't even get the step out of it...

12. Originally Posted by NEOWatcher
I haven't worked it out, but that sounds to me like it would only apply to the same orbit.

Think of lights coming from the other horses in the merry go round. I don't think the amount of redshift would coincide with how far away they are.
The further away they are the more redshift we would see ... Bacause we would have moved that much further away. Also moving something in towards the center or out away from the center would speed up their relative movement ( taking the angle of light into account ).

13. C is somewhat arbitrary here and used it more because it made calculations easier ( which I screwed up anyway ). So lets assume a lesser velocity ...

What I am trying to see is if we get redshift as the source is always moving away from us.

The idea is that the light travels via a straight line path while both the source and the receptor are travelling a curved path.

Originally Posted by mugaliens
There's an additional problem no one here caught: How much the light would be red or blue-shifted.

If the source is moving at c, it would be blue-shifted to a wavelength of 0 in all directions in front of it, and it would be red-shifted down to a frequency of 0 in all directions behind it.

Thus, the receiver would not detect light. Rather, it would detect a discrete jump in the E-M field from one level to a different level.

This is analogus to the fact that for the source, moving at c, time is stopped. Of course this would also mean that you wouldn't even get the step out of it...

14. Yes I see my flawed math.
At t0 a photon is emmited in the direction of 10:30.
At t1 the photon is detected at 10:30 by the receptor.

Again the direction of the photon emision is arbitrary based on the speed that the merry go round is travelling ... which is for this problem is also arbitrary ... as long as the speed is enough to produce a noticable redshift.

Originally Posted by mugaliens
Statements in contention:

CS1. The light travels at 45 degrees from the source.
.

15. Originally Posted by tommac
The idea is that the light travels via a straight line path while both the source and the receptor are travelling a curved path.
I don't think the curve matters, the only thing that matters is the direction travelling (tangentally) at the moment the receptor and photon meet. If its a stream of photons, then each one being met will be met at a different location and with different vectors (both) of travel.

16. Originally Posted by NEOWatcher
I don't think the curve matters, the only thing that matters is the direction travelling (tangentally) at the moment the receptor and photon meet. If its a stream of photons, then each one being met will be met at a different location and with different vectors (both) of travel.
I thought it has to do with both the sender and the receptors movement at the moment of release of the photon.

In this case I think that the movement in the direction of the light is greater for the sender than it is for the receptor. Just cant figure out the math.

17. If all of our visible universe is in orbit around a central point. Then wouldnt we see redshift for all galaxies?

Also the further away the galaxy is would have a greater redshift correct?

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Originally Posted by tommac
If all of our visible universe is in orbit around a central point. Then wouldnt we see redshift for all galaxies?

Also the further away the galaxy is would have a greater redshift correct?
No, we see blueshift in the direction of rotation, redshift in the opposite direction, a bit of transverse Doppler radially, and nothing at all directly above and below our rotation plane.
You just need to work it through by plotting velocity vectors and comparing them.

Grant Hutchison

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Even If It Could . . .

Originally Posted by tommac

If all of our visible universe is in orbit around a central point
For various reasons given on other Q&A threads, it is not, and cannot be.

Originally Posted by tommac

Then wouldnt we see redshift for all galaxies?
No.

Originally Posted by tommac

Also the further away the galaxy is would have a greater redshift correct?
No. R&B shifts?

Tom, you might benefit from reading through the numerous Wiki articles on relativity. They are available at all math levels, from none to astonishing.

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