MMX vs Earths' rotational sagnac

chinglu and cats said:
I asked my cats and no, acceleration is not necessary for the calculation of the rotational sagnac.
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Yo dog, you and your cats need to get back to the basics. Please read the wikipedia entry on Centripetal Force. A rotating system undergoes acceleration even at constant angular velocity. In other words, if it is spinning at any speed it is under acceleration (which by the way does not mean that the speed must be increasing or even changing). This means that ANY rotation involves "acceleration", and since ANY Sagnac involves rotation, we can conclude that Sagnac only occurs in accelerated sources.

As I said in my first post, MMX is predicted to fail under suitable acceleration. GPS apparently provides enough acceleration for MMX to fail, which means you, your cats and your hamster also fail. Kthx bye.
 
RJBeery said:
Yo dog, you and your cats need to get back to the basics. Please read the wikipedia entry on Centripetal Force. A rotating system undergoes acceleration even at constant angular velocity. In other words, if it is spinning at any speed it is under acceleration (which by the way does not mean that the speed must be increasing or even changing). This means that ANY rotation involves "acceleration", and since ANY Sagnac involves rotation, we can conclude that Sagnac only occurs in accelerated sources.

As I said in my first post, MMX is predicted to fail under suitable acceleration. GPS apparently provides enough acceleration for MMX to fail, which means you, your cats and your hamster also fail. Kthx bye.
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Really??? So what are the units of this mysterious "acceleration" that redefines the very definition of the term "acceleration", which means THE RATE OF CHANGE OF VELOCITY.

Perfect example: A beach ball is on the beach, not moving relative to the beach. Q: What is the ball's acceleration towards the center of the earth? A: 0 m/s^2!

BTW, Don't even try to tell me the ball is accelerating towards the center of the earth at 9.8 m/s^2. The distance between the ball and the center of the earth remains unchanged over a duration of time. That means the ball has a zero velocity relative to the center of the earth. Ten minutes later the ball still has a zero velocity towards the center of the earth, so the velocity of the ball didn't change in those ten minutes, hence the acceleration is 0 m/s^2!

If by chance you do try to say the ball is accelerating towards the center of the earth, please tell me the ball's initial velocity towards the center of the earth, and also how much time it will take for the ball to reach the center of the earth at that velocity (including the acceleration, of course.)
 
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Motor Daddy: Your confusion may be alleviated if you recall that velocity is a VECTOR, where speed is a scalar. A vector includes direction. The beach ball's speed does not change but if it's velocity did not it would fly straight out into space.
 
RJBeery said:
Motor Daddy: Your confusion may be alleviated if you recall that velocity is a VECTOR, where speed is a scalar. A vector includes direction. The beach ball's speed does not change but if it's velocity did not it would fly straight out into space.
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So what is the ball's velocity relative to the center of the earth while on the beach?
 
Depends. If it were on one of the poles it would effectively be zero. This means that I believe a vertical MMX experiment conducted there would provide a null result regardless of the length of the apparatus' arms. Good luck getting the GPS satellite to sit still above the pole, though! Remember that the velocity direction is tangential to the Earth's surface; I reckon the speed at the equator would be the diameter of the Earth per 24 hour period, or very roughly 1000mph (this is a pre-lunch, off the cuff consideration).
 
Motor Daddy said:
Really??? So what are the units of this mysterious "acceleration" that redefines the very definition of the term "acceleration", which means THE RATE OF CHANGE OF VELOCITY.

Perfect example: A beach ball is on the beach, not moving relative to the beach. Q: What is the ball's acceleration towards the center of the earth? A: 0 m/s^2!

BTW, Don't even try to tell me the ball is accelerating towards the center of the earth at 9.8 m/s^2.
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First of all the ball on the beach IS feeling the acceleration of gravity and it's velocity is zero relative to the center of the earth.

F=ma Force equals mass time acceleration. So if there is a force being exerted by the mass of the ball then it is due an acceleration. You can test this by placing a 400 kg mass on your big toe. The force that you feel on your toe is due to acceleration from gravity.

Another point is that if you were on a windowless space ship accelerating at 9.8 m/s^2 or you were in a closed room on earth's equator, there is no method to detect any difference between those 2 accelerations.
 
origin said:
First of all the ball on the beach IS feeling the acceleration of gravity and it's velocity is zero relative to the center of the earth.
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Origin I'm trying to point out the difference between acceleration due to centripetal force and acceleration due to gravity with no rotation. The Earth's spin does in fact impart an acceleration greater than zero on a beach ball sitting on its surface, given that the beach ball is not at one of the rotational poles.

Hold a bucket above your head and spin in a circle; zero relative velocity. Hold the bucket out and spin in a circle; greater than zero relative velocity. Your arm feels the bucket trying to "get away" as you spin, and must accelerate it ("change its direction") in order to keep it close. The effect gets greater the faster you spin, however, notice that the effect is gone if you again move the bucket above your head.

The fact that Motor Daddy and Chinglu's Animal Farm all think acceleration involves a speedometer that's moving is the misconception I'm clarifying.
 
RJBeery said:
Origin I'm trying to point out the difference between acceleration due to centripetal force and acceleration due to gravity with no rotation. The Earth's spin does in fact impart an acceleration greater than zero on a beach ball sitting on its surface, given that the beach ball is not at one of the rotational poles.

Hold a bucket above your head and spin in a circle; zero relative velocity. Hold the bucket out and spin in a circle; greater than zero relative velocity. Your arm feels the bucket trying to "get away" as you spin, and must accelerate it ("change its direction") in order to keep it close. The effect gets greater the faster you spin, however, notice that the effect is gone if you again move the bucket above your head.

The fact that Motor Daddy and Chinglu's Animal Farm all think acceleration involves a speedometer that's moving is the misconception I'm clarifying.
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I have given you a Cal Berkley mainstream peer reviewed article that does not need acceleration to calculate the earth's rotational sagnac.

So, you are wrong. Perhaps you should spend more time asking questions than giving silly false opinions.
 
This is just not even wrong:

The Sagnac effect "needs" acceleration. Without acceleration there is no phase difference. The reason GPS has to account for it is because the orbiting satellites have an angular acceleration, they "see" light following a curved path because they're following a curved path.

An article that claims acceleration isn't needed to calculate a Sagnac effect is either making an egregious error, or the person reading it is.
 
arfa brane said:
This is just not even wrong:

The Sagnac effect "needs" acceleration. Without acceleration there is no phase difference. The reason GPS has to account for it is because the orbiting satellites have an angular acceleration, they "see" light following a curved path because they're following a curved path.

An article that claims acceleration isn't needed to calculate a Sagnac effect is either making an egregious error, or the person reading it is.
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The ignorance is amazing.

Here is the standard for TIME SYNCHRONIZATION AND DISSEMINATION IN THE SOLAR SYSTEM

http://www.pttimeeting.org/archivemeetings/2006papers/paper24.pdf

Here again is the Cal Berkeley article.
http://www.physics.berkeley.edu/res.../LaserRingGyro/Steadman/StedmanReview1997.pdf

Now, read these and adjust your false views into something that is mainstream. It is not a positive attribute to be ignorant when you have available information.
 
chinglu: are you claiming the solar system "doesn't need" acceleration?

You're saying the authors of your quoted article didn't need it either, to calculate a Sagnac effect?
 
Stupid question from a lurker...What is MMX? other than a multimedia code process for 386's?
 
Arfa Brane said:
This is just not even wrong:

The Sagnac effect "needs" acceleration. Without acceleration there is no phase difference. The reason GPS has to account for it is because the orbiting satellites have an angular acceleration, they "see" light following a curved path because they're following a curved path.

An article that claims acceleration isn't needed to calculate a Sagnac effect is either making an egregious error, or the person reading it is.
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Yes, this nails it.
Chinglu's link said:
In the Sagnac effect, this difference is proportional to the angular velocity.
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This quote is from the abstract of the Cal Berkeley article. The greater the angular velocity = the greater the acceleration = the greater Sagnac effect. Sagnac is produced by rotating systems (as I said), which necessitate acceleration (as I and Arfa Brane have said). Recall that acceleration occurs even with constant angular velocity. Therefore, Sagnac only occurs under acceleration; MMX would fail under suitable acceleration. There is no contradiction. How else shall I say it so that you understand?
 
RJBeery said:
Origin I'm trying to point out the difference between acceleration due to centripetal force and acceleration due to gravity with no rotation. The Earth's spin does in fact impart an acceleration greater than zero on a beach ball sitting on its surface, given that the beach ball is not at one of the rotational poles.

Hold a bucket above your head and spin in a circle; zero relative velocity. Hold the bucket out and spin in a circle; greater than zero relative velocity. Your arm feels the bucket trying to "get away" as you spin, and must accelerate it ("change its direction") in order to keep it close. The effect gets greater the faster you spin, however, notice that the effect is gone if you again move the bucket above your head.

The fact that Motor Daddy and Chinglu's Animal Farm all think acceleration involves a speedometer that's moving is the misconception I'm clarifying.
Click to expand...

I wasn't disagreeing with your point that a change in direction of a moving object is by definition an acceleration. I was trying to answer a specific question of Motor Daddy.
 
arfa brane said:
chinglu: are you claiming the solar system "doesn't need" acceleration?

You're saying the authors of your quoted article didn't need it either, to calculate a Sagnac effect?
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No the acceleration is not needed, the circular orbit is contained in the angular velocity in the equations.
 
RJBeery said:
Yes, this nails it.

This quote is from the abstract of the Cal Berkeley article. The greater the angular velocity = the greater the acceleration = the greater Sagnac effect. Sagnac is produced by rotating systems (as I said), which necessitate acceleration (as I and Arfa Brane have said). Recall that acceleration occurs even with constant angular velocity. Therefore, Sagnac only occurs under acceleration; MMX would fail under suitable acceleration. There is no contradiction. How else shall I say it so that you understand?
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1) Given a fixed area, the greater the angular velocity the greater the sagnac.

2) I am afraid you are wrong about your "acceleration".

http://www.mathpages.com/rr/s2-07/2-07.htm

You can find here that the circular path and the linear velocity is sufficient to calculate the sagnac. Angular velocity is not necessary.
 
chinglu said:
You can find here that the circular path and the linear velocity is sufficient to calculate the sagnac. Angular velocity is not necessary.
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A circular path is linear? Why does the article in your link have a picture of a loop that rotates? Rotation means there is an angular velocity, so "it is necessary" for the Sagnac effect. Angular velocity means acceleration towards the centre of rotation.

You appear to be choosing to preferentially misunderstand the difference between linear and angular velocity. It's sort of funny, in a strange way.
 
arfa brane said:
A circular path is linear? Why does the article in your link have a picture of a loop that rotates? Rotation means there is an angular velocity, so "it is necessary" for the Sagnac effect. Angular velocity means acceleration towards the centre of rotation.

You appear to be choosing to preferentially misunderstand the difference between linear and angular velocity. It's sort of funny, in a strange way.
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It may be funny it you can't understand what is going on. It is obvious you do not understand the relationship between linear velocity and angular velocity in a loop.

Let me know when you have this figured out.
 
chinglu said:
It is obvious you do not understand the relationship between linear velocity and angular velocity in a loop.
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One thing that is rather obvious is that the word "accelerate" in various forms, appears more than half a dozen times in that link. The word "linear" appears zero times.

How much linear velocity can anything traveling in a loop have?
"Let me know when you have this figured out. "
 
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