Relativistic centripetal force

[yuiop]

Looking on the internet, it seems to very difficult to find a simple straightforward statement of the equations for relativistic centripetal force. This is my point of view and hopefully with some feedback we can come to a consensus.

Consider a test mass moving in a circle with radius r. The Newtonian equation for the acceleration of the test mass is:

a = \frac{v^2}{r} = r \omega^2 = r \frac{d\theta^2}{dt^2}.

From the point of view of a non rotating inertial observer in flat space at rest with the centre of rotation, the acceleration of the mass in SR terms is identical to the Newtonian equation.

To an observer co-moving with the test mass, the acceleration of the test mass is:

a' = r \frac{d\theta^2}{d\tau^2} \gamma^2 = r \frac{d\theta^2}{dt^2} \gamma^2 = \frac{v^2}{r}\gamma^2 = a \gamma^2

due to time dilation of the co-moving observer's clock. (\gamma = 1/\sqrt{ (1-v^2/c^2) })

To the co-moving observer, no notion of relativistic mass is involved, so the force is simply:

F' = ma' = m \frac{v^2}{r}\gamma^2 = ma \gamma^2

which is the proper force. If the test mass was resting on the inside of a rotating cylinder with radius r and angular velocity v/r, this is the force that would be measured by a set of weighing scales between the test mass and the wall of the cylinder. If the centripetal force is provided by a string of length r and the test mass was attached to the end of a string by a tension gauge, then this is the force that would be directly measured by the tension gauge.

It is known that the Lorentz transformation of transverse force is F = F ' \gamma^{-1}, so it follows that the force measured by the non rotating inertial observer at rest with the centre of rotation is:

F = \frac{F'}{\gamma} = \frac{ma'}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma

This is not just a notional calculation of coordinate force, but is what would be measured by a tension gauge at the axis end of the rope. This is surprising, because the tension at one end of the rope is different from the tension at the other, which is not the case in Newtonian mechanics. It is also a measure of the centripetal force that would be provided by a non rotating electromagnetic field to keep the test mass moving in a circle. In both these cases, it is the force provided from the non-rotating frame to the moving test mass in the rotating frame.

Given that the Newtonian equation for angular momentum is L = mr^2\omega = mrv it is not difficult to work out that the relativistic angular momentum of the test mass in the non rotating frame is m\gamma r v from the above equations.

 

[Jorrie]

To an observer co-moving with the test mass, the acceleration of the test mass is:

a' = r \frac{d\theta^2}{d\tau^2} \gamma^2 = r \frac{d\theta^2}{dt^2} \gamma^2 = \frac{v^2}{r}\gamma^2 = a \gamma^2

I think you are correct and I'm trying to reconcile it with the equation that I arrived at in https://www.physicsforums.com/showpost.php?p=2678507&postcount=17" for 'centripetal acceleration' in a circular orbit:

<br /> a_{cp} = \frac{v_o^2}{r} = \frac{GM}{g_{tt} r^2} = \frac{GM}{r^2}(1+2v_o^2) <br />

where g_{tt} = 1-2GM/r^2, for c=1 and v_o the local circular orbit velocity, measured at r by a momentarily static observer. This implies that the acceleration is also as measured by the static observer, relative to a hypothetical 'tangential track'. If I convert that to the orbiting observer (like you did), it gives

<br /> a&#039;_{cp} = \frac{v_o^2}{r}\gamma^2 = \frac{GM}{r^2}\left(\frac{1+2v_o^2}{1-v_o^2}\right)<br />

This then is what the equivalent accelerometer (in flat spacetime) should read, not what I implied in the other post. It implies that for a massive particle approaching c, the acceleration and force for a circular path tend to infinity, which correlates.

It is known that the Lorentz transformation of transverse force is F = F&#039;\gamma^{-1}, so it follows that the force measured by the non rotating inertial observer at rest with the centre of rotation is: ...

This is surprising, because the tension at one end of the rope is different from the tension at the other, which is not the case in Newtonian mechanics. It is also a measure of the centripetal force that would be provided by a non rotating electromagnetic field to keep the test mass moving in a circle. In both these cases, it is the force provided from the non-rotating frame to the moving test mass in the rotating frame.

Interesting observation, kev!

 

[starthaus]

It is known that the Lorentz transformation of transverse force is F = F &#039; \gamma^{-1}, so it follows that the force measured by the non rotating inertial observer at rest with the centre of rotation is:

F = \frac{F&#039;}{\gamma} = \frac{ma&#039;}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma

I don't think that you can extrapolate the Lorentz transform (derived for inertial frames) to the transform between the observer frame and the proper frame. The proper frame is rotating wrt the observer frame.

 

[yuiop]

I don't think that you can extrapolate the Lorentz transform (derived for inertial frames) to the transform between the observer frame and the proper frame. The proper frame is rotating wrt the observer frame.

Agreed that the proper frame is rotating and therefore an accelerating or non-inertial frame, but we can use the concept of the momentarily comoving inertial frame and the clock hypothesis to treat calculations in the accelerating frame exactly like an instantaneous inertial frame.

This is a quote from the http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html" :

The clock hypothesis states that the tick rate of a clock when measured in an inertial frame depends only upon its velocity relative to that frame, and is independent of its acceleration or higher derivatives. The experiment of Bailey et al. referenced above stored muons in a magnetic storage ring and measured their lifetime. While being stored in the ring they were subject to a proper acceleration of approximately 10^18 g (1 g = 9.8 m/s2). The observed agreement between the lifetime of the stored muons with that of muons with the same energy moving inertially confirms the clock hypothesis for accelerations of that magnitude.

In other words the time dilation experienced by the muons at 10,000,000,000,000,000,000 g is exactly in agreement with the Lorentz transformation for muons moving inertially in a straight line at a constant velocity equal to the instantaneous velocity of the accelerating muons.

The clock hypothesis can be extended to the transformation of all physical measurements in an accelerating frame, so that the accelerating frame obeys Lorentz transformations exactly as if it was a measurement in an instantaneous comoving inertial frame. This is another quote from ahttp://www.edu-observatory.org/physics-faq/Relativity/SR/clock.html" :

On a more advanced note, the clock postulate can be generalised to say something about measurements we make in a noninertial frame. First, it tells us that noninertial objects only age and contract by the same gamma factor as that of their MCIF. So, any measurements we make in a noninertial frame that use clocks and rods will be identical to measurements made in our MCIF--at least, if these measurements are made over a small enough region. This is because, in general, different regions of the noninertial frame have different MCIFs, a complicating factor that makes the construction of noninertial frames very difficult. In fact, it can only readily be done for constant-velocity and constant-acceleration frames; and for the last, that is rather difficult.

But we now choose to extend the clock postulate to include all measurements (though perhaps it can be argued that all measurements only ever use clocks and rods anyway). This idea leads onto "covariance", which is a way of using tensors to write the language of physics in a way that applies to all frames, noninertial as well as inertial.

If a local observer observes an accelerating particle, all he needs to know is its instantaneous velocity and he can work out all the proper measurements of the particle, by using the Lorentz transformations and treating the particle as if it was moving in a straight line in flat spcae. This is a powerful concept and very convenient!

 

[starthaus]

Agreed that the proper frame is rotating and therefore an accelerating or non-inertial frame, but we can use the concept of the momentarily comoving inertial frame and the clock hypothesis to treat calculations in the accelerating frame exactly like an instantaneous inertial frame.

The point was that you can't use results that were derived for frames in relative linear
uniform motion.

 

[yuiop]

The point was that you can't use results that were derived for frames in relative linear uniform motion.

The point is that the clock postulate says you can.

The result I obtained by using that postulate:

F = \frac{F &#039;}{\gamma} = \frac{ma &#039;}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma

is supported by experimental measurements of the force required to deflect an electron from a straight path in a cathode ray tube. They confirm that the deflection force required is greater than the Newtonian prediction by a factor of \gamma = 1/ \sqrt{1-v^2/c^2}. It is also confirmed in hundreds of cyclotrons and particle accelerators all over the world.

 

[starthaus]

The point is that the clock postulate says you can.

The result I obtained by using that postulate:

The clock postulate says nothing about attempting to use a formula derived for one case (linear relative motion between two frames with aligned axes) to a totally different situation (motion with a variable direction velocity between two frames that don't have aligned axes). In order to do things right, you would need to do a derivation from scratch, using the generalization of the Lorentz transforms for rotating frames.

F = \frac{F &#039;}{\gamma} = \frac{ma &#039;}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma

is supported by experimental measurements of the force required to deflect an electron from a straight path in a cathode ray tube.

No, it isn't. The electron deflection is obtained from solving a totally different equation:

\frac{d}{dt}(\gamma m_0 v)=qvxB

The LHS represents the force described as \frac {dp}{dt} and the RHS is the Lorentz force.

You need to contrast the relativistic equation with the Newtonian one:


\frac{d}{dt}(m_0 v)=q v x B

Note that neither equation has nothing to do with any centrifugal force nor does it have anything to do with the way forces (centripetal or not) transform, so, I don't even see how your quote is relevant to the subject.

 

[neyzenilhan]

general relativity is needed for this problem. accelerated frame.

 

[Jorrie]

The clock postulate says nothing about attempting to use a formula derived for one case (linear relative motion between two frames with aligned axes) to a totally different situation (motion with a variable direction velocity between two frames that don't have aligned axes).

But surely one can set up an inertial frame where the center of the circle is static have an inertial observer momentarily comoving with the circling mass, as kev used. The clock postulate and all Lorentz transformations between the static an the comoving observer should hold. Why would what happens after that moment (variable direction) influence the momentary observations?

 

[starthaus]

But surely one can set up an inertial frame where the center of the circle is static have an inertial observer momentarily comoving with the circling mass, as kev used.

This is not good enough. The transformations were derived for the case when the axes of the two inertial frames are parallel and when the relative motion between frames coincides with the common x-axis. Neither of these two conditions is satisfied.
You need to do the derivation correctly, by using the extension of the Lorentz transforms to rotating frames.

The clock postulate and all Lorentz transformations between the static an the comoving observer should hold.

The clock postulate has nothing to do with this exercise.
The Lorentz transforms for linear motion do not hold in the case of rotation, a different set is used.

 

[yuiop]

F = \frac{F &#039;}{\gamma} = \frac{ma &#039;}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma

is supported by experimental measurements of the force required to deflect an electron from a straight path in a cathode ray tube.

No, it isn't. The electron deflection is obtained from solving a totally different equation:

\frac{d}{dt}(\gamma m_0 v)=q v x B

The LHS represents the force described as \frac {dp}{dt} and the RHS is the Lorentz force.

The LHS of your relativistic equation is not "a totally different equation". I can obtain it from my equation showing it is the same. Starting with my equation:

F = \frac{F &#039;}{\gamma} = \frac{ma &#039;}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma

F = ma \gamma = \gamma m \frac{v^2}{r}

Since v^2/r = dv/dt

(See http://theory.uwinnipeg.ca/physics/circ/node6.html, http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration, http://dev.physicslab.org/Document....me=CircularMotion_CentripetalAcceleration.xml and http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cf2 )

it follows that:

F = \gamma m \frac{v^2}{r} = \gamma m \frac{dv}{dt} = \frac{d}{dt} (\gamma mv) = \frac{dp}{dt}

Q.E.D.

\frac{d}{dt}(\gamma m_0 v)=q v x B

...You need to contrast the relativistic equation with the Newtonian one:

\frac{d}{dt}( m_0 v)=q v x B

There is something wrong with RHS of your two equations:

If

\frac{d}{dt}( m_0 v)=q v x B

is true, then by the properties of simultaneous equations, it must follow that:

\frac{d}{dt}(\gamma m_0 v)= (q v x B) \gamma

 

[starthaus]

The LHS of your relativistic equation is not "a totally different equation". I can obtain it from my equation showing it is the same. Starting with my equation:

F = \frac{F &#039;}{\gamma} = \frac{ma &#039;}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma

F = ma \gamma = \gamma m \frac{v^2}{r}

Since v^2/r = dv/dt

(See http://theory.uwinnipeg.ca/physics/circ/node6.html, http://en.wikipedia.org/wiki/Acceleration#Tangential_and_centripetal_acceleration, http://dev.physicslab.org/Document....me=CircularMotion_CentripetalAcceleration.xml and http://hyperphysics.phy-astr.gsu.edu/hbase/cf.html#cf2 )

it follows that:

F = ma \gamma = \gamma m \frac{v^2}{r} = \gamma m \frac{dv}{dt} = \frac{d}{dt} (\gamma mv) = \frac{dp}{dt}

Q.E.D.

The above has nothing to do with transformation of force between frames.

There is something wrong with RHS of your two equations:

If

\frac{d}{dt}( m_0 v)=q v x B

is true,

It is true only for non-relativistic speeds, it isn't true for relativistic speeds. This is why, the LHS is different from what you wrote above.

then by the properties of simultaneous equations, it must follow that:

\frac{d}{dt}(\gamma m_0 v)= (q v x B) \gamma

Nope, there is nothing wrong, this is basic relativistic electrodynamics. There is no \gamma in the RHS.

 

[yuiop]

The above has nothing to do with transformation of force between frames.

I obtained my final equation by using the clock postulate and doing a Lorentz transformation from the accelerating frame as it was an instantaneous inertial frame (i.e using the concept of a Momentarily Comoving Inertial Frame) and the result I get agrees with the equation you gave, so I must be doing something right.

 

[yuiop]

Nope, there is nothing wrong, this is basic relativistic electrodynamics. There is no \gamma in the RHS.

The rules of mathematics say that one of your two equations is wrong (unless you are saying \gamma is a constant of numerical value 1, which does not agree with relativity).

 

[starthaus]

I obtained my final equation by using the clock postulate and doing a Lorentz transformation from the accelerating frame as it was an instantaneous inertial frame (i.e using the concept of a Momentarily Comoving Inertial Frame) and the result I get agrees with the equation you gave, so I must be doing something right.

You are getting things "right" by accident. To do things truly right you need to have a rigorous derivation. The fact that your derivation lacks rigor is illustrated by your confusion in solving the cyclotron equations, yours are not even stated correctly (see above).
The subject of SR in rotating frames is quite interesting and deserves a lot of attention.

 

[yuiop]

There is something wrong with RHS of your two equations:

If

\frac{d}{dt}( m_0 v)=q v x B

is true...

It is true only for non-relativistic speeds, it isn't true for relativistic speeds. This is why, the LHS is different from what you wrote above.

It follows that q v x B is also only true for non-relativistic speeds and that for relativistic speeeds you must use \gamma(q v x B)

The subject of SR in rotating frames is quite interesting and deserves a lot of attention.

I am sure other members here will agree that I given the subject of SR in rotating frame a LOT of attention in other threads over the years.

 

[starthaus]

It follows that q v x B is also only true for non-relativistic speeds and that for relativistic speeeds you must use \gamma(q v x B)

Nope, this is definitely not true. For a correct treatment, please see attachment #1 (LorentzForce) in my blog. Or, you can consult Griffith , I can't give you the exact page off the top of my head.

 

[yuiop]

Nope, this is definitely not true. For a correct treatment, please see attachment #3 in my blog. Or, you can consult Griffith , I can't give you the exact page off the top of my head.

I think you will find that the root of our differences is that I am using coordinate time as measured in the lab in which the cathode ray tube is at res,t while you and Grffith are probably using the proper time of the electron.

 

[starthaus]

I think you will find that the root of our differences is that I am using coordinate time as measured in the lab in which the cathode ray tube is at res,t while you and Grffith are probably using the proper time of the electron.

Nope. I am using coordinate time as well.

 

[yuiop]

Nope. I am using coordinate time as well.

Actually, my suggestion that you are using proper time does not work. It seems that the Lorentz force devised by Lorentz in 1898 turns out to be relativistically correct even though it discovered before Einstein's published his paper on relativity in 1905. My statement that one of your two equations is incorrect still stands: Your statement would be more accurately stated as:

(Relativistic:)

\frac{d}{dt}(\gamma m_0 v) = q v x B

(Newtonian)

\frac{d}{dt}( m_0 v) \ne q v x B

although the Newtonian expression is approximately true for v << c. The Newtonian equation is only exact when v=0.

Anyway, none of this means that any of the equations I posted in #1 are wrong and all your equations demonstrate, is that relativistic equations approximate Newtonian equations at low velocities.

 

[starthaus]

Actually, my suggestion that you are using proper time does not work. It seems that the Lorentz force devised by Lorentz in 1898 turns out to be relativistically correct even though it discovered before Einstein's published his paper on relativity in 1905. My statement that one of your two equations is incorrect still stands:

This is very simple stuff, I don't understand why you have so much trouble with it.

Your statement would be more accurately stated as:

I suggest that you study Griffith. You can find this sort of stuff in a lot of books on electrodynamics, both classical and relativistic.

although the Newtonian expression is approximately true for v << c. The Newtonian equation is only exact when v=0.

The Newtonian expression always holds within the Newtonian domain. It is simply due to the fact that, in Newtonian physics F=d(m_0v). By definition.

 

[yuiop]

This is very simple stuff, I don't understand why you have so much trouble with it...

The reason I having trouble is that I do not play around with the equations of electromagnetics much. Here is one problem I am still having:

In your relativistic equation:

F = \frac{d}{dt} \gamma mv = q( \mathbf{v} \times \mathbf{B}) = q v B\\ sin \theta

and considering the simple case where the velocity v and the magnetic field (B) are right angles to each other, so that sin \theta = sin (pi/2) = 1 then:

F = \frac{d}{dt} \gamma mv = q v B

It can be seen that when v = c the LHS is infinite and the RHS remains finite indicating that your equation is wrong.

In this http://en.wikipedia.org/wiki/Lorentz_force#Translation_to_vector_notation" it gives this equation:

\gamma \frac{d\mathbf{p}}{dt} = \gamma q (\mathbf{E} + (\mathbf{v} \times \mathbf{B}))

which when E = 0 reduces to:

\gamma \frac{d\mathbf{p}}{dt} = \gamma q (\mathbf{v} \times \mathbf{B})

The RHS of the Wikipedia equation differs from your equation by a factor of gamma. Maybe I am not the only one who is having trouble with this "simple stuff"?

 

[starthaus]

The reason I having trouble is that I do not play around with the equations of electromagnetics much. Here is one problem I am still having:

In your relativistic equation:

F = \frac{d}{dt} \gamma mv = q( \mathbf{v} \times \mathbf{B}) = q v B\\ sin \theta

and considering the simple case where the velocity v and the magnetic field (B) are right angles to each other, so that sin \theta = sin (pi/2) = 1 then:

F = \frac{d}{dt} \gamma mv = q v B

It can be seen that when v = c the LHS is infinite and the RHS remains finite indicating that your equation is wrong.

:lol: How can you set v=c when you are dealing with massive particles?
Have you made the effort to read the attachment in the blog?

 

[yuiop]

:lol: How can you set v=c when you are dealing with massive particles?

Mathematically you can set v = c and the infinite result for the LHS indicates the impossibilty or uphysical nature of doing that. The RHS of your equation does not indicate that setting v = c is unphysical, indicating that the RHS of your equation is not a relativistically valid term.

Have you made the effort to read the attachment in the blog?

I have and it starts out with assuming the Lorentz force is relativistic with no proof or explanation why you think it is.

Have you read the Wikipedia article? Can you explain why their result differs from yours?

 

[starthaus]

Mathematically you can set v = c and the infinite result for the LHS indicates the impossibilty or uphysical nature of doing that. The RHS of your equation does not indicate that setting v = c is unphysical, indicating that the RHS of your equation is not a relativistically valid term.

At this point I will recommend that you read pages 26-11 thru 26-13 in Feyman's "Lectures on Physics" Volume II.

It contains the same exact thing as my post and blog.
If you read my attachment 1 in the blog you would have found out that it is IMPOSSIBLE to set v->c. The simple reason is that v=v_0 so v=CONSTANT for the Lorentz force.

I have and it starts out with assuming the Lorentz force is relativistic with no proof or explanation why you think it is.

Very simple, because this is the standard equation in the three-fource formalism in SR. Precisely equation 26.24 in Feynman.
You are mixing indiscriminately four-force formalism (from wiki) with the three-force formalism (from your and my posts) and you are confusing yourself in the process.

Have you read the Wikipedia article? Can you explain why their result differs from yours?

Sure:

1. I use the three-fource formalism (corresponding to eq 26.24 in Feynman)
2. Wiki uses the four-force formalism (corresponding to eq. 26.39 in Feynman)

The two equations represent the SAME thing but expressed in DIFFERENT variables.
Feynman notes at the end of the chapter 26-3 : "Although it is nice that the equation can be written in that way (i.e. the eq 26.39) , this form is not particularly useful. It is usually more convenient to solve for particle motions by using the original equations (26.24) , and that's what we will usually do."

That is precisely what I ALWAYS do. I hope that this clears all your confusions.

 

[yuiop]

If you read my attachment 1 in the blog you would have found out that it is IMPOSSIBLE to set v->c. The simple reason is that v=v_0 so v=CONSTANT for the Lorentz force.

I have read your blog again and it actually quite a good presentation and I see no obvious errors. However, your statement here, that v in the Lorentz force term is CONSTANT and therefore can not be set v = c is misleading and technically wrong. Let us consider the case of an electron moving in a circle in a magnetic field. The v variable in the qvB term is the speed of the instantaneous tangential velocity. In other words it is a magnitude without direction and does not change as the electron moves around the circle. This magnitude is VARIABLE and can take any value between 0 and c. Nor is this magnitude a scalar like the speed of light or charge as it changes under coordinate transformation, so it is constant in a very limited sense.

My objection that the LHS side of your equation:

F = \frac{d}{dt} \gamma mv = q v B (Eq1)

becomes infinite while the RHS does not, is wrong, but not for the reason you give.

Here is the correct counter argument to my flawed assertion:

Your equation can re-written as:

F = \gamma \frac{mv^2}{r} = q v B (Eq2)

(See post #1)

The above can be rearranged to isolate r:

r = \gamma \frac{mv}{qB} (Eq3)

When v = c, r becomes infinite at the same time gamma becomes infinite. This is reasonable because it implies that in the limit as the speed of the particle though the magnetic field ,approaches the speed of light the path becomes straight which correlates with the concept that a particle moving at the speed of light would require infinite force to deflect it. The LHS of Eq1 or Eq2 becomes \infty/\infty which is indeterminate and not infinite like I originally claimed. This is even clearer when the Eq2 is rewritten as:

\gamma mv^2 = r q v B (Eq4)

Both sides of the equation become infinite at the same time when v=c which destroys my original objection, (but not because it is impossible to set v=c as you claim).

You are mixing indiscriminately four-force formalism (from wiki) with the three-force formalism (from your and my posts) and you are confusing yourself in the process.

You are right here. Wiki is using dp/dtau which has proper time and we are using dp/dt which has coordinate time.

This is an earlier objection you made:

Note that neither equation has nothing to do with any centrifugal force nor does it have anything to do with the way forces (centripetal or not) transform, so, I don't even see how your quote is relevant to the subject.

As mentioned above, electrons moving at constant speed in a constant magnetic field travel in a circle. (It is obvious from your blog that you know this). Since they are moving in a circle in gravitationally flat space, a centripetal force is required to make them travel in a circle and this centripetal force is provided by the the Lorentz force and in fact they are numerically equal as long as the electric field is zero. This can be seen in Eq2 where v^2/r represents the centripetal acceleration and the LHS is the relativistic equation for centripetal force. In the case of the cathode ray tube, while the electron is being deflected by the magnetic coils it follows the arc of a circle and the deflection force is equal to the centripetal force required to make the electron deviate from a straight path. There is a nice illustration of this on page 3 of this nicely presented document here: http://www.docstoc.com/docs/24317699/MOTION-OF-A-CHARGED-PARTICLE-IN-A-MAGNETIC-FIELD/

So far you have not demonstrated that any of the equations I posted in #1 are incorrect. Your initial objection in #7 that my equations are not in agreement with your electromagnetic equations was shown to be invalid, because I demonstrated in #11 that they can be obtained from each other and are the same thing expressed in different terms. However, you have made a valuable contribution by providing the electromagnetic equivalents of my equations. One aspect we have not thoroughly investigated is the transformation to the rotating frame. That might be interesting. I suspect the electric field E might make a reappearance in the equations of the transformed frame. I might have a go at working that out later.

This is where I think we are at:

In the non-rotating inertial lab frame S:

F = \frac{F&#039;}{\gamma} = \frac{m_0a&#039;}{\gamma} = m_0 \frac{v^2}{r} \gamma = m_0a \gamma = \frac{dp}{dt} = d\frac{\gamma m_0 v}{dt} = d\frac{m_0 v}{d\tau} = qvB

a = \frac{F}{m_o \gamma} = \frac{v^2}{r} = r \omega^2 = r \frac{d\theta^2}{dt^2} = \frac{dv}{dt} = \frac{dv}{\gamma d\tau} = \frac{qvB}{\gamma m_0}.


In the rotating non-inertial frame S' :

F&#039; = m_0 a&#039; = m_0 \frac{v^2}{r}\gamma^2 = m_0 a \gamma^2 = \gamma \frac{dp}{dt} = \frac{dp}{dt &#039;} = d\frac{\gamma^2 m_0 v}{dt} = d\frac{\gamma m_0 v}{d\tau}

a&#039; = \frac{F &#039;}{m_0} = r \frac{d\theta^2}{dt^2} \gamma^2 = \frac{v^2}{r}\gamma^2 = a \gamma^2 = \gamma^2 \frac{dv}{dt} = \frac{\gamma dv}{ d\tau}.

 

[starthaus]

I have read your blog again and it actually quite a good presentation and I see no obvious errors. However, your statement here, that v in the Lorentz force term is CONSTANT and therefore can not be set v = c is misleading and technically wrong.

Attachment 1 shows why this is correct. This is basic electrodynamics, textbook stuff.

Let us consider the case of an electron moving in a circle in a magnetic field. The v variable in the qvB term is the speed of the instantaneous tangential velocity. In other words it is a magnitude without direction and does not change as the electron moves around the circle. This magnitude is VARIABLE and can take any value between 0 and c.

The magnitude is equal to the initial speed of injection v_0. This is many orders of magnitude smaller than c.

Nor is this magnitude a scalar like the speed of light or charge as it changes under coordinate transformation, so it is constant in a very limited sense.

I explained to you that it is equal to v_0. This is exactly the meaning of constant, i.e. it isn't VARIABLE so you can't make it to go to c.

My objection that the LHS side of your equation:

F = \frac{d}{dt} \gamma mv = q v B (Eq1)

becomes infinite while the RHS does not, is wrong,

OK, so we are done now. I am glad that you understood the issue.

 

[starthaus]

Here is the correct counter argument to my flawed assertion:

Your equation can re-written as:

F = \gamma \frac{mv^2}{r} = q v B (Eq2)

(See post #1)

The above can be rearranged to isolate r:

r = \gamma \frac{mv}{qB} (Eq3)

This is not good enough, I am very precise in attachment 1, the correct formula is:

r = \gamma(v_0) \frac{m_0v_0}{qB}

You can see that \gamma(v_0) can never become infinite because v_0 can never reach c.

 

[starthaus]

This is where I think we are at:

Maybe you "are at". I have no idea what is the numerology that you put up. If you want to get things right, you need to learn how to work out the acceleration in rotating frames.

So far you have not demonstrated that any of the equations I posted in #1 are incorrect.

As an example of an obvious error you wrote :

F&#039;=m_0 a&#039; (correct) where, from your chain of equalities, it results that the proper acceleration a' is equal to: \gamma \frac {d}{dt} (\gamma v) . This is clearly incorrect since the proper acceleration a' is equal, by definition to: \frac {d}{dt} (\gamma v)=\gamma^3 \frac{dv}{dt} .

The second row of equalities is just as bad, since you managed to get yet another erroneous result by botching the proper acceleration a different way:

a&#039;=\gamma^2 \frac{dv}{dt}

 

[yuiop]

The magnitude is equal to the initial speed of injection v_0. This is many orders of magnitude smaller than c.

This is petty and wrong in principle and wrong in practice. In the LHC the speed of protons at injection to the LHC 299 732 500 m / s (99.9998% the speed of light). That is NOT many orders of magnitude less than c. See http://www.scienceknowledge.org/?p=6625 and http://lhc-machine-outreach.web.cern.ch/lhc-machine-outreach/beam.htm OK, these are protons and not electrons, but the principle is the same. For electrons, see http://en.wikipedia.org/wiki/Large_Electron–Positron_Collider "In the Large Electron positron collider At a Lorentz factor ( = particle energy/rest mass = [104.5 Gev/0.511 Mev]) of over 200,000, LEP still holds the particle accelerator speed record, extremely close to the limiting speed of light."

I explained to you that it is equal to v_0. This is exactly the meaning of constant, i.e. it isn't VARIABLE so you can't make it to go to c.

The meaning of constant is context dependent. It can mean unchanged under transformation to a different coordinate system or it can can mean unchanging over time in a given experiment or coordinate system.

The speed of light is a constant. If you ask me what the speed of light is I can tell you that it has the value 299 792 458 m/s. The charge of a electron is a constant and I can tell you it has a has the value of 1.602176487(40)×10−19 Coulombs. If I ask you what is the value of the Starthaus Constant v_0, you are not able to tell me, because it depends on the particular experiment and it is therefore a variable.

The Lorentz force is qvB. In an particle accelerator, when the particles are accelerated the magnetic field B has to be increased to maintain the particle trajectories within the confines of the accelerator ring and v increases. The only thing that remains constant during the speed up process is the charge q because that is a true constant.

 

[Jorrie]

This is not good enough. The transformations were derived for the case when the axes of the two inertial frames are parallel and when the relative motion between frames coincides with the common x-axis. Neither of these two conditions is satisfied.

Your objection is also not quite good enough. To make it clearer, shift the central inertial observer to the edge of the circle, where it sits static relative to the center. Let this be the origin of the "rest frame". Now let the circling mass (on the string) fly past it and at the same moment, let the inertial comoving observer fly past it, along the x-axis of the rest frame (hence, 'standard configuration'). For the static and comoving observers, both conditions that you mentioned are satisfied.

The only objection is that the comoving observer is not equivalent to the orbiting observer in terms of proper acceleration. However, the comoving observer can observe the orbiting mass and determine its coordinate acceleration according to the comoving frame (by its clocks and rulers). In the same way the static observer can determine the acceleration of the orbiting mass in the static frame. So, why won't the LTs work for transforming observations between these two inertial frames in standard configuration?

 

[starthaus]

Your objection is also not quite good enough. To make it clearer, shift the central inertial observer to the edge of the circle, where it sits static relative to the center.

I already answered that, if you do that two things will happen:

1. The axes of the two frames will be misaligned
2. The direction of the relative velocity between the two frames will change continously

Therefore, you are not "entitled" to lift the transform derived for frames with aligned axes and declare it the valid transformation of force. You need to use the appropiate transforms and rederive the force transformation from scratch. It is not very difficult to do things the right way.

 

[starthaus]

This is petty and wrong in principle and wrong in practice. In the LHC the speed of protons at injection to the LHC 299 732 500 m / s (99.9998% the speed of light). That is NOT many orders of magnitude less than c. See http://www.scienceknowledge.org/?p=6625 and http://lhc-machine-outreach.web.cern.ch/lhc-machine-outreach/beam.htm OK, these are protons and not electrons, but the principle is the same. For electrons, see http://en.wikipedia.org/wiki/Large_Electron–Positron_Collider "In the Large Electron positron collider At a Lorentz factor ( = particle energy/rest mass = [104.5 Gev/0.511 Mev]) of over 200,000, LEP still holds the particle accelerator speed record, extremely close to the limiting speed of light."

Not a very good example, LHC is not a cyclotron. Particles do not get injected at .9999c, they get accelerated in the ring to such high speeds. The particles travel at variable speed at LHC. Totally different and much more complex equations govern the functioning of LHC compared with the simple cyclotron in our discussion.

The meaning of constant is context dependent. It can mean unchanged under transformation to a different coordinate system or it can can mean unchanging over time in a given experiment or coordinate system.

The speed of light is a constant. If you ask me what the speed of light is I can tell you that it has the value 299 792 458 m/s. The charge of a electron is a constant and I can tell you it has a has the value of 1.602176487(40)×10−19 Coulombs. If I ask you what is the value of the Starthaus Constant v_0, you are not able to tell me, because it depends on the particular experiment and it is therefore a variable.

The Lorentz force is qvB. In an particle accelerator, when the particles are accelerated the magnetic field B has to be increased to maintain the particle trajectories within the confines of the accelerator ring and v increases. The only thing that remains constant during the speed up process is the charge q because that is a true constant.

I can see that you still have problems with the relativistic equation of motion for cyclotrons. Are you still arguing that the equation is not correct because the LHS can go unbound? Aside from the Feynman lectures, I can recommend a few good books on accelerator design. They all share the same equations I showed you earlier, more or less.

 

[yuiop]

Not a very good example, LHC is not a cyclotron. Particles do not get injected at .9999c, they get accelerated in the ring to such high speeds. The particles travel at variable speed at LHC. Totally different and much more complex equations govern the functioning of LHC compared with the simple cyclotron in our discussion.

This is quote from one the links I posted, that you probably did not read:

The particles in the LHC are ultra-relativistic and move at 0.999997828 times the speed of light at injection and 0.999999991 the speed of light at top energy

Before being injected into the LHC ring the particles are pre-accelerated

in at least six different accelerators: the source duoplasmatron 90 keV to 750 keV the RFQ, Linac 2 of 50 MeV, the Source injector (PS Booster or PSB) of 1.4 GeV proton synchrotron (PS) 25 GeV, and finally the Super Proton Synchrotron (SPS) of 450 GeV

While it is true that after being injected into the LHC ring at 0.999997828c the particles are accelerated for another 20 minutes before they get to the top speed of 0.999999991c, but after that, they are kept at a constant speed of 0.999999991c for many hours and during that extended holding period the LHC is acting very much like a simple synchrotron.

I can see that you still have problems with the relativistic equation of motion for cyclotrons. Are you still arguing that the equation is not correct because the LHS can go unbound?

No, I do not think the equations are incorrect and I thought I made that clear in my last post. I am just disagreeing with you on your definition of "constant", but I will put it down to semantics. If you want to define v_0 as a constant that has a variable value depending on the apparatus and experiment used, then I will agree we dissagree on semantics.

 

[yuiop]

Maybe you "are at". I have no idea what is the numerology that you put up. If you want to get things right, you need to learn how to work out the acceleration in rotating frames.

You need to be less aggressive and chill out a bit. Your posts are bordering on personal attacks rather than logical arguments.

As an example of an obvious error you wrote :

F&#039;=m_0 a&#039; (correct) where, from your chain of equalities, it results that the proper acceleration a' is equal to: \gamma \frac {d}{dt} (\gamma v) . This is clearly incorrect since the proper acceleration a' is equal, by definition to: \frac {d}{dt} (\gamma v)=\gamma^3 \frac{dv}{dt} .

The second row of equalities is just as bad, since you managed to get yet another erroneous result by botching the proper acceleration a different way:

a&#039;=\gamma^2 \frac{dv}{dt}

This is more like it. A factual argument. Unfortunately it is you that made the botch here. The definition for proper acceleration that you gave with a factor of gamma^3 is for linear or parallel acceleration and does not aplly in the case we are talking about for the deflection of a charged particle with constant speed in a magnetic field. For circular motion as in a cyclotron, the proper acceleration due to Lorentz force is transverse or orthogonal to the instantaneous velocity of the particle and has the value I gave with a factor of gamma^2.

See the bottom of page 268 in the this book http://books.google.co.uk/books?id=J4glh_8RQlMC&pg=PA268#v=onepage&q&f=false

 

[starthaus]

This is more like it. A factual argument. Unfortunately it is you that made the botch here. The definition for proper acceleration that you gave with a factor of gamma^3 is for linear or parallel acceleration and does not aplly in the case we are talking about for the deflection of a charged particle with constant speed in a magnetic field.

I find this particularly amusing, especially in the light of explaining (see also attachment 1) that
the proper acceleration is, by definition, \frac {d}{dt}(\gamma \vec{v}). In the case of the cyclotron, \gamma is constant so \vec{a}=\gamma \frac {d \vec{v}}{dt}.
Both your expressions for proper acceleration are wrong.
But that's besides the point since none of these cyclotron equations are relevant wrt OP. Do you want to learn how to calculate the transformation of force in rotating frames or not?

 

[Jorrie]

I already answered that, if you do that two things will happen:

1. The axes of the two frames will be misaligned
2. The direction of the relative velocity between the two frames will change continously

And both times, your answer did not fit the question.

1. The 'static' observer at radius r and the single inertial observer, momentarily comoving with the orbiting observer, are inertial and in 'standard configuration', their respective frames having the same origin and with spatial axes aligned, forever.

2. You seem to think that the comoving observer follows the circle, which it does not. Once I've got one point of measurement, I know the magnitude of the centripetal acceleration for all time, provided that the angular velocity of the experiment does not change.

I agree that a general method, using rotating frames are better for some purposes, but that does not mean we must discard a simpler, valid method for the problem stated by the OP.

 

[starthaus]

And both times, your answer did not fit the question.

1. The 'static' observer at radius r and the single inertial observer, momentarily comoving with the orbiting observer, are inertial and in 'standard configuration', their respective frames having the same origin and with spatial axes aligned, forever.

The direction of the x-axis of the comoving observer coincides with \vec {v}, so it is changing continously. That is why it is called "comoving" observer. Therefore , contrary to what you claim, it will only be aligned with the x-axis of the 'static' observer only once per revolution. Same thing about the y axis. Only the z-axis is aligned.

2. You seem to think that the comoving observer follows the circle, which it does not.

Of course he does, this is why the instantaneous speed between the frames equals the speed of the revolving particle.

Once I've got one point of measurement, I know the magnitude of the centripetal acceleration for all time, provided that the angular velocity of the experiment does not change.

This already presuposes that \vec {F&#039;} has constant magnitude, independent of position along the circle. But this is what you have been asked to establish. So, you can't use the conclusion in choosing just one frame.

Besides, even if you establish the magnitude of the force, using your method, you can't establish its direction.

 

[Jorrie]

The direction of the x-axis of the comoving observer coincides with \vec {v}, so it is changing continously.

You are continuously missing the fact that kev and myself are talking about a momentarily comoving inertial observer, a common method employed when dealing with accelerating frames. We need only one measurement...

You are referring to a non-inertial comoving observer.

 

[starthaus]

You are continuously missing the fact that kev and myself are talking about a momentarily comoving inertial observer, a common method employed when dealing with accelerating frames. We need only one measurement...

You are referring to a non-inertial comoving observer.

"Comoving" has only one meaning: moving with the particle. Around the circle. For a short period of time, \vec {v} changes by an infiniresimal amount \vec {\Delta v} , so the frame of the comoving observer can be considered inertial. This is standard fare in treating accelerated motion.

 

[Jorrie]

"Comoving" has only one meaning: moving with the particle. Around the circle. For a short period of time, \vec {v} changes by an infiniresimal amount \vec {\Delta v} , so the frame of the comoving observer can be considered inertial. This is standard fare in treating accelerated motion.

And so has 'monetarily comoving inertial observer' only one (different) meaning, IMO. I will rest this until one of the science advisors clarifies.

 

[starthaus]

And so has 'monetarily comoving inertial observer' only one (different) meaning, IMO. I will rest this until one of the science advisors clarifies.

Read post #38 to the end, your approach has severe flaws. You can't pick only one frame, the one that suits you.

 

[Vanadium 50]

Sorry for jumping in so late, but I am afraid this is a solved problem. It's called Thomas precession, and it's covered in a hand-wavy fashion in Taylor and Wheeler, and in a http://bohr.physics.berkeley.edu/classes/221/0708/notes/thomprec.pdf" by Robert Littlejohn.

Unfortunately, the literature on this is a bit of a mess. Definitions are not always consistent, and there's a tendency for some authors to mix and match. One key problem is that "the rest frame of the particle" which is not so trivial for an accelerated particle - this frame is a function of time.

One key point that Littlejohn brings up is "for boosts in an arbitrary direction, the axes of the moving frame, as seen in the stationary frame, are not even orthogonal". This causes all sorts of woe when people try and calculate this - particularly when they try and "pick a convenient direction", as the direction may not be as convenient as it looks.

 

[starthaus]

Sorry for jumping in so late, but I am afraid this is a solved problem. It's called Thomas precession, and it's covered in a hand-wavy fashion in Taylor and Wheeler, and in a http://bohr.physics.berkeley.edu/classes/221/0708/notes/thomprec.pdf" by Robert Littlejohn.

Unfortunately, the literature on this is a bit of a mess. Definitions are not always consistent, and there's a tendency for some authors to mix and match. One key problem is that "the rest frame of the particle" which is not so trivial for an accelerated particle - this frame is a function of time.

One key point that Littlejohn brings up is "for boosts in an arbitrary direction, the axes of the moving frame, as seen in the stationary frame, are not even orthogonal". This causes all sorts of woe when people try and calculate this - particularly when they try and "pick a convenient direction", as the direction may not be as convenient as it looks.

Yes, the problem is much more complicated than the OP made it to be. I put together a new attachment that shows how it needs to be solved rigorously. See my blog.

 

[yuiop]

Sorry for jumping in so late, but I am afraid this is a solved problem. It's called Thomas precession, and it's covered in a hand-wavy fashion in Taylor and Wheeler, and in a http://bohr.physics.berkeley.edu/classes/221/0708/notes/thomprec.pdf" by Robert Littlejohn.

Unfortunately, the literature on this is a bit of a mess. Definitions are not always consistent, and there's a tendency for some authors to mix and match. One key problem is that "the rest frame of the particle" which is not so trivial for an accelerated particle - this frame is a function of time.

One key point that Littlejohn brings up is "for boosts in an arbitrary direction, the axes of the moving frame, as seen in the stationary frame, are not even orthogonal". This causes all sorts of woe when people try and calculate this - particularly when they try and "pick a convenient direction", as the direction may not be as convenient as it looks.

The OP is not about Thomas precession That is a red herring. Thomas precession is about the spin of an extended object about its own axis as it orbits. I am not concerned about the spin of the object as I am essentially considering point particles such as an electron moving in a cylcotron. What I am asking about is the magnitude of the centripetal force. The direction of the force may be changing constantly in the lab frame, but because the particle is moving in a circle, the magnitude of the force is constant and always at right angles to the motion by definition. Imagine you are inside a large hollow cylinder in gravitationally flat space that spun to relativistic speeds. A form of artificial gravity is created that allows you stand on the inside curved wall of the cylinder. Imagine you are standing on some (strong) bathroom weighing scales. What is the reading on those scales? That is one of questions Jorrie and I are asking. If the spin rate of the cylinder is constant and the space is perfectly gravitationally flat and if you mass is constant, is there any reason why the reading on the bathroom scales should be changing over time? The answer is no.

Notice that the question "what is the reading on the bathroom scales?" does not even require us to know what the rest frame of the particle is because all observers inertial or non-inertial will agree what the reading is. In order to transform the force to the non rotating frame we do need some notion of the rest frame of the particle, and as Jorrie and I have pointed out, the Clock Postulate brings about a significant simplification of the analysis.

As the originator of this thread, may I respectfully request we ignore Thomas precession as an unnecessary distraction (and much smaller effect) and just concentrate on the more significant forces?

 

[starthaus]

As the originator of this thread, may I respectfully request we ignore Thomas precession as an unnecessary distraction (and much smaller effect) and just concentrate on the more significant forces?

I attached a new file to the blog that gives the rigorous treatment, I suggest you read it, if you haven't already done so. It shows the correct form of the transforms in rotating frames and how to derive the transformation of speed and acceleration from them. You can't use the Lorentz transforms derived for linear motion, you should be using the appropiate transforms.

 

[yuiop]

I attached a new file to the blog that gives the rigorous treatment, I suggest you read it, if you haven't already done so. It shows the correct form of the transforms in rotating frames and how to derive the transformation of speed and acceleration from them. You can't use the Lorentz transforms derived for linear motion, you should be using the appropiate transforms.

I have read it. There seems to be an error in the expression (6) where you have dy/dt but I suspect it should read d^2 \frac{y}{dt^2}

Also the equations in (6) are incomplete. Can you complete them and then we will see if we can make sense of them?

 

[atyy]

http://arxiv.org/abs/physics/0405090

No string mentioned in the above article though.

 

[yuiop]

I find this particularly amusing, especially in the light of explaining (see also attachment 1) that
the proper acceleration is, by definition, \frac {d}{dt}(\gamma \vec{v}). In the case of the cyclotron, \gamma is constant so \vec{a}=\gamma \frac {d \vec{v}}{dt}.
Both your expressions for proper acceleration are wrong.

The expression you derived in attachment 1 of your blog starts with the statement "dv/dt and v have the same direction and sense" and that makes it clear that you are deriving the parallel acceleration which does not apply for centripetal acceleration in a cyclotron. Not only that, but you get parallel acceleration wrong too.

You start with F = m_0 \gamma^3 dv/dt (the coordinate parallel force) and jump to the conclusion that the coordinate parallel acceleration is a = F/(m_0 \gamma^2) = d(\gamma v)/dt which is unjustified and probably wrong.


In this paper http://www.hep.princeton.edu/~mcdonald/examples/mechanics/matthews_ajp_73_45_05.pdf they give the parallel (longitudinal) acceleration transformation (Eq29) as:

a_x = \frac{a_x &#039;}{\gamma^3 (1+u_x &#039; V/c^2)}

which reduces to:

a_x &#039; = \gamma^3 a_x

when u_x &#039; is set to zero for the case where the accelerating observer is at rest with the test particle. This does not agree with equation you gave in your blog.

The transverse acceleration transformation (the one we require) is given as:

a_y = \frac{a_y &#039;}{\gamma^2 (1+u_x &#039; V/c^2)}- <br /> \frac{a_x &#039; u_y &#039; V/c^2}{\gamma^2 (1+u_x &#039; V/c^2)}

which when u_x &#039; = 0 and u_y &#039; = 0 reduces to:

a_y &#039; = \gamma^2 a_y

which is agreement with the equations I gave in #1.

In this #18 of this old thread: https://www.physicsforums.com/showthread.php?t=187041&page=2,
pervect (with over 5000 posts and respected former pf staff) comes to the conclusion that relativistic coordinate centripetal (transverse) force is: \gamma^2 v^2/r which agrees with the equations I gave in #1.

This article http://www.dragfreesatellite.com/sr_accel_tt.pdf gives the parallel acceleration when the particle is moving in the x direction as:

a_x &#039; = \frac{dv_x &#039;}{dt} = \gamma^3 \frac{dv_x}{dt}

In other words your definition of proper acceleration (unqualiifed) is actually the definition of proper parallel acceleration, which is invalid when we are talking about constant centripetal acceleration acting at right angles to the instantaneous velocity of a particle moving in a circle with constant speed.

It does not explicitly give the acceleration for the y and z directions but it can be worked out from the transformation matrix they give as:

a_y &#039; = \frac{dv_y/dt + v_y v (dv/dt)}{(1-v^2/c^2)}

which when the particle is moving exactly in the x direction, v_y =0 and the equation for the transverse acceleration transformation reduces to:

a_y &#039; = \frac{dv_y/dt}{(1-v^2/c^2)} = \gamma^2 d\frac{v_y}{dt}

which is in agreement with the equations I gave for transverse (centripetal) acceleration and in the dv/dt form you explicitly disputed.

None of you blog entries or posts indicate that you are aware that there is difference between parallel and transverse force. You are aware of that distinction, right?

 

[yuiop]

The direction of the x-axis of the comoving observer coincides with \vec {v}, so it is changing continously. That is why it is called "comoving" observer. Therefore , contrary to what you claim, it will only be aligned with the x-axis of the 'static' observer only once per revolution. Same thing about the y axis. Only the z-axis is aligned.

Jorrie and I did not call it the comoving observer. We explicitly called it the Momentarily Comoving Inertial Frame (MCIF). You are deliberately ignoring the word "momentarily". Momentarily means instantaneous and so it does not apply to an observer that continually follows the particle around the circle as you implied here:

The direction of the x-axis of the comoving observer coincides with , so it is changing continously. That is why it is called "comoving" observer. Therefore , contrary to what you claim, it will only be aligned with the x-axis of the 'static' observer only once per revolution. Same thing about the y axis. Only the z-axis is aligned.

and here:

"Comoving" has only one meaning: moving with the particle. Around the circle. For a short period of time, changes by an infiniresimal amount , so the frame of the comoving observer can be considered inertial. This is standard fare in treating accelerated motion.

You are also ignoring the word "inertial" which also does not apply to a comoving observer moving around the circle in gravitationally flat space, because such a comoving observer would measure a force acting on them, so by definition they would not be inertial.

When an inertial observer moving in a straight line momentarily aligns with an accelerated observer the clock rates and rulers of the two frames all agree in an infinitesimal region of space and time and by the clock postulate the two are equivalent. For the case of a particle moving in a circle, the MCIF observer that instantaneously coincides with accelerating particle at one infinitessimal point of the circle is representive of an infinite number of MCIF observers all the way around the circle by the circular smymetry of the situation.