Relativistic centripetal force

[starthaus]

You have not obtained \frac {d^2x'}{dt'^2} in your document. All you have done is taken the derivative of x with respect to time twice and then transformed the result by converting dt to dt' using dt = gamma(dt'), What you have failed to do, is convert dx to dx' using dx = f(dx') where f is the long function inplicit in expression (4) of your document.

You need to stop trying to find fault with what I am trying to teach you. The method is showing how to find \frac {d^2x}{dt^2} as a function of \frac {d^2x'}{dt'^2}, \frac {dx'}{dt'}, etc. Looks that after a lot of prodding, you went ahead and you did it yourself.

It is easy to complete expression (6) because it is obvious from the terms you have completed and from the method you use in obtaining expression (5), that all you are doing is dividing each term by \gamma dt'. The completed expressions are (as far as I can tell):

\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - \frac{d}{dt'}R\gamma\omega\sin(\gamma\omega t')\right)

\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') + \frac{d}{dt'}R\gamma\omega\cos(\gamma\omega t')\right)

Good, this is correct.

Note that for constant uniform rotation, the last term of each expression containg R probably vanishes,

False, it doesn't "vanish". This is basic calculus, remember?

 

[Dale]

I would be very interested in your contribution.

Since the vote is split 50/50 I will appoint myself vice president and cast the deciding vote in favor of posting it

So throughout this I will use the convention that the time coordinate is the first coordinate and that timelike intervals squared are positive. So given a wheel centered at the origin and rotating about the z axis in a counter clockwise direction we can describe the worldline for any particle on the rim by:
\mathbf s =(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

The four-velocity is given by:
\mathbf u=c \frac{d\mathbf s/dt}{|d\mathbf s/dt|}=(\gamma c,\; -\gamma r \omega \; sin(\phi + \omega t),\; \gamma r \omega \; cos(\phi + \omega t),\; 0)
where |x| indicates the Minkowski norm of the four-vector x and \gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}

The four-acceleration is given by:
\mathbf a=c \frac{d\mathbf u/dt}{|d\mathbf s/dt|} = (0,\; -\gamma^2 r \omega^2 \; cos(\phi + \omega t),\; -\gamma^2 r \omega^2 \; sin(\phi + \omega t),\; 0 )

So the magnitude of the four-acceleration (which is frame invariant and equal to the magnitude of the proper acceleration) is given by:
|\mathbf a|^2=-\gamma^4 r^2 \omega^4

That is what I had in my Mathematica file, but I certainly can go from there or explain things if either of you have questions or requests.

 

[starthaus]

Since the vote is split 50/50 I will appoint myself vice president and cast the deciding vote in favor of posting it

So throughout this I will use the convention that the time coordinate is the first coordinate and that timelike intervals squared are positive. So given a wheel centered at the origin and rotating about the z axis in a counter clockwise direction we can describe the worldline for any particle on the rim by:
\mathbf s =(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

The four-velocity is given by:
\mathbf u=c \frac{d\mathbf s/dt}{|d\mathbf s/dt|}=(\gamma c,\; -\gamma r \omega \; sin(\phi + \omega t),\; \gamma r \omega \; cos(\phi + \omega t),\; 0)
where |\mathbf x| indicates the Minkowski norm of the four-vector x and \gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}

The four-acceleration is given by:
\mathbf a=c \frac{d\mathbf u/dt}{|d\mathbf s/dt|} = (0,\; -\gamma^2 r \omega^2 \; cos(\phi + \omega t),\; -\gamma^2 r \omega^2 \; sin(\phi + \omega t),\; 0 )

So the magnitude of the four-acceleration (which is frame invariant and equal to the magnitude of the proper acceleration) is given by:
|\mathbf a|^2=-\gamma^4 r^2 \omega^4

Careful, see here. There is the additional condition \gamma=1 that must not be overlooked.

That is what I had in my Mathematica file, but I certainly can go from there or explain things if either of you have questions or requests.

We've been over this, this is exactly the same as pervect's post. The subject of the discussion is the appropiate transforms between frames (see post #3). Lorentz transforms for translation are not appropiate for describing rotation. kev is almost done with learning how to use the Lorentz transforms for rotation.

 

[Dale]

Careful, see here. There is the additional condition \gamma=1 that must not be overlooked.

No. Since,\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2} the condition \gamma=1 would imply \omega=0 which is not true in general.

We've been over this, this is exactly the same as pervect's post. The subject of the discussion is the appropiate transforms between frames (see post #3). Lorentz transforms for translation are not appropiate for describing rotation. kev is almost done with learning how to use the Lorentz transforms for rotation.

I deliberately didn't do any Lorentz transforms nor did I look at any other reference frame.

Although a boost is not a spatial rotation you can always find a momentarily co-moving inertial frame (MCIF) for any event on any particle's worldline. This frame will be related to the original frame via a Lorentz transform. In such a frame the particle will only be at rest for a moment, hence the term "momentarily co-moving". So you cannot use a Lorentz transform to find a (non-instantaneous) rest frame for a particle in general, but there is a lot of useful parameters you can obtain from the MCIF. I don't know how kev was using it.

 

[Jorrie]

Since the vote is split 50/50 I will appoint myself vice president and cast the deciding vote in favor of posting it

Thanks DaleSpam, I should have voted for it - I have been asking Starthaus for his treatment before.

Anyway, it seems that your magnitude of the proper acceleration answers the original question of the OP: the relativistic centrifugal force, measured in the way that kev described.

To me it does not seem as if Starthaus's treatment answered that question, but rather a different one on coordinate transformations for rotating frames. I am sure that it is useful in its own right and can possibly be extended to answer the original question.

 

[starthaus]

I deliberately didn't do any Lorentz transforms nor did I look at any other reference frame.

True, neither did you answer the question in discussion. You found the expression for four-acceleration (something that we already knew) but you did not answer the question as to how centripetal force transforms. In order to do that , you need to start from the Lorentz transforms for rotating frames.

Although a boost is not a spatial rotation you can always find a momentarily co-moving inertial frame (MCIF) for any event on any particle's worldline. This frame will be related to the original frame via a Lorentz transform. In such a frame the particle will only be at rest for a moment, hence the term "momentarily co-moving". So you cannot use a Lorentz transform to find a (non-instantaneous) rest frame for a particle in general, but there is a lot of useful parameters you can obtain from the MCIF.

Sure but why use this hack when you are being shown how to use the appropiate Lorentz transforms?

I don't know how kev was using it.

With lots of mistakes.

 

[starthaus]

Thanks DaleSpam, I should have voted for it - I have been asking Starthaus for his treatment before.

Physics is not decided by voting.

Anyway, it seems that your magnitude of the proper acceleration answers the original question of the OP: the relativistic centrifugal force, measured in the way that kev described.

No, it doesn't. Look at post #3, the issue being contested is usage of the appropiate transforms in finding out the way the centripetal force transforms between frames.

To me it does not seem as if Starthaus's treatment answered that question, but rather a different one on coordinate transformations for rotating frames. I am sure that it is useful in its own right and can possibly be extended to answer the original question.

kev is not finished with his exercise , he is just one step away from finishing it. If he corrects his last mistake, he's one step away from getting the correct answer using the correct formalism.
Would you like to finish it for him? It is a matter of doing the derivative correctly and setting some conditions.

 

[starthaus]

To me it does not seem as if Starthaus's treatment answered that question, but rather a different one on coordinate transformations for rotating frames. I am sure that it is useful in its own right and can possibly be extended to answer the original question.

The method I am showing solves a much more general question. I will show you how it also solves the question asked by the OP , as a trivial particular case. Just stick with the approach and you will find out the correct answers obtained through the appropiate formalisms.

 

[starthaus]

No. Since,\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2} the condition \gamma=1 would imply \omega=0 which is not true in general.

No, it doesn't mean that, it only means that the comoving frame (for which you are calculating the proper acceleration) is characterized by \omega=0 . Because the frame is co-moving with the rotating object.
So, you should have obtained the proper acceleration magnitude as:

a=r\omega^2

 

[yuiop]

OK, if I understand correctly, because of the t' contained in the cos and sin functions in the last terms, the expressions for (6) should be:

\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\cos(\gamma\omega t')-\gamma\omega\sin(\gamma\omega t') \frac{dx'}{dt'}-\gamma\frac{d^2y'}{dt'^2}\sin(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\cos(\gamma\omega t') - R\gamma^2\omega^2\cos(\gamma\omega t')\right) (Eq6x)

\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}\sin(\gamma\omega t')+\gamma\omega\cos(\gamma\omega t') \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\cos(\gamma\omega t')-\frac{dy'}{dt'}\gamma^{2}\omega\sin(\gamma\omega t') - R\gamma^2\omega^2\sin(\gamma\omega t')\right) (Eq6y)

Is that correct? As I said before, I am no math expert. Should the same be done for the other terms in the expression?

If they are correct, then when t' = 0, \sin(\gamma\omega t') = 0 and \cos(\gamma\omega t') = 1 and the equations simplify to:

a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x'}{dt'^2}-\frac{dy'}{dt'}\gamma^{2}\omega- R\gamma^2\omega^2\right) (Eq7x)

a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(\gamma\omega \frac{dx'}{dt'}+\gamma\frac{d^2y'}{dt'^2}\right) (Eq7y)

Now if the particle is stationary in the rotating frame and located on the y'=0 axis, which is the radial line from the centre of rotation to the particle, the result is:

a_x = \frac{d^2x}{dt^2}= -R\omega^2\right)

a_y = \frac{d^2y}{dt^2}= 0

which is a good result, because that is what we expect in the non-rotating lab frame. This is the coordinate acceleration.

Equations (7x) and (7y) can be rearranged to give:

a'_x = \frac{d^2x'}{dt'^2}=\gamma^{2}\left(\frac{d^2x}{dt^2}+\frac{dy'}{dt'}\omega+ R\omega^2\right)

a'_y = \frac{d^2y'}{dt'^2}=\gamma\frac{d^2y}{dt^2}-\omega \frac{dx'}{dt'}

If the particle is released so that it flies off in a straight line from the point of view of the lab frame at time t’=0, its initial velocity in the rotating frame was zero so that upon release dx\prime/dt\prime \approx 0 and dy\prime/dt\prime \approx 0 then:

a&#039;_x = \frac{d^2x&#039;}{dt&#039;^2} =\gamma^{2}\left(\frac{d^2x}{dt^2}+\frac{dy&#039;}{dt&#039;}\omega+ R\omega^2\right) <br /> =\gamma^{2}\left(0+\frac{dy&#039;}{dt&#039;}\omega+ R\omega^2\right)<br /> \approx \gamma^{2}\left(R\omega^2\right)

a&#039;_y = \frac{d^2y&#039;}{dt&#039;^2}=\gamma\frac{d^2y}{dt^2}-\omega \frac{dx&#039;}{dt&#039;}<br /> =0-\omega \frac{dx&#039;}{dt&#039;} \approx 0

So in the lab frame the centripetal acceleration of the particle when it is traveling in a circle is R\omega^2 and when it is released the acceleration of the particle according to the observer in the rotating frame is initially \gamma^{2} R\omega^2. This magnitude of this measurement coincides with with the magnitude of the four acceleration given by Pervect and Dalespam (Thanks for the imput by the way ), which is the proper acceleration as measured by an accelerometer.

Note:The x and x' axes always pass through the center of rotation so a_x and a_x&#039; represent the inward pointing centripetal acceleration at time t'=0, which these calculations are based on. a_y is the angular acceleration directed tangentially.

Conclusion

Coordinate centripetal acceleration:

a_x = R\omega^2 (Proven by Starthaus)

Proper centripetal acceleration:

As measured by an accelerometer in the rotating frame:

a\prime_x = \gamma^2 R\omega^2 (Proven by Dalespam and Pervect)

As measured by initial “fall rate acceleration” in the rotating frame:

a\prime_x \approx \gamma^2 R\omega^2 (Proven by Starthaus)

The above is basically in agreement with everything I said in #1 but the methods of measurement have been more carefully defined now.

This has significance because it tells us how a particle behaves in a gravitational field, by applying the equivalence principle.

 

[Dale]

you need to start from the Lorentz transforms for rotating frames.

I have never heard of the Lorentz transform for rotating frame. AFAIK the Lorentz transform transforms between different inertial frames, not between an inertial frame and a rotating frame.

 

[Dale]

No, it doesn't mean that, it only means that the comoving frame (for which you are calculating the proper acceleration) is characterized by \omega=0 . Because the frame is co-moving with the rotating object.
So, you should have obtained the proper acceleration magnitude as:

a=r\omega^2

The Minkowski norm of the four-acceleration is frame invariant. So it does not matter which frame you calculate it in; it will be the same in all frames. If those factors of 1-r^2\omega^2/c^2 appear in one frame they must appear in all frames.

 

[starthaus]

OK, if I understand correctly, because of the t' contained in the cos and sin functions in the last terms, the expressions for (6) should be:

\frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x&#039;}{dt&#039;^2}\cos(\gamma\omega t&#039;)-\gamma\omega\sin(\gamma\omega t&#039;) \frac{dx&#039;}{dt&#039;}-\gamma\frac{d^2y&#039;}{dt&#039;^2}\sin(\gamma\omega t&#039;)-\frac{dy&#039;}{dt&#039;}\gamma^{2}\omega\cos(\gamma\omega t&#039;) - R\gamma^2\omega^2\cos(\gamma\omega t&#039;)\right) (Eq6x)

\frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x&#039;}{dt&#039;^2}\sin(\gamma\omega t&#039;)+\gamma\omega\cos(\gamma\omega t&#039;) \frac{dx&#039;}{dt&#039;}+\gamma\frac{d^2y&#039;}{dt&#039;^2}\cos(\gamma\omega t&#039;)-\frac{dy&#039;}{dt&#039;}\gamma^{2}\omega\sin(\gamma\omega t&#039;) - R\gamma^2\omega^2\sin(\gamma\omega t&#039;)\right) (Eq6y)

Is that correct?

Correct. Finally.

If they are correct, then when t' = 0, \sin(\gamma\omega t&#039;) = 0 and \cos(\gamma\omega t&#039;) = 1 and the equations simplify to:

Incorrect. In the comoving frame you need to realize that \frac{dx&#039;}{dt&#039;}=0 and \frac{dy&#039;}{dt&#039;}=0. Try continuing the calculations from this hint.

 

[starthaus]

I have never heard of the Lorentz transform for rotating frame. AFAIK the Lorentz transform transforms between different inertial frames, not between an inertial frame and a rotating frame.

I gave a couple of references earlier in the thread. Here they are again:

1. R. A. Nelson, "Generalized Lorentz transformation for an accelerated, rotating frame of reference", J. Math. Phys. 28, 2379 (1987);

2. H.Nikolic, “Relativistic contraction and related effects in non-inertial frames”, Phys.Rev. A, 61, (2000)

 

[starthaus]

The Minkowski norm of the four-acceleration is frame invariant. So it does not matter which frame you calculate it in; it will be the same in all frames. If those factors of 1-r^2\omega^2/c^2 appear in one frame they must appear in all frames.

No argument. The argument is about the fact that proper acceleration is DIFFERENT from four-acceleration. It differs exactly by the \gamma factor. What you calculated iis the four-acceleration, not the proper acceleration. But, we are digressing. The discussion is about the correct usage of Lorentz transforms. Please be patient, kev is one step away from finishing the computations.

 

[Jorrie]

... In the comoving frame you need to realize that \frac{dx&#039;}{dt&#039;}=0 and
\frac{dy&#039;}{dt&#039;}=0.

It appears to me as if you have a different definition of comoving frame than the rest of us. We are usually referring to the momentarily comoving inertial frame (MCIF). Or do I read you incorrectly?

 

[starthaus]

It appears to me as if you have a different definition of comoving frame than the rest of us. We are usually referring to the momentarily comoving inertial frame (MCIF).

The comoving frame (S') is moving with the particle, hence \frac{dx&#039;}{dt&#039;}=\frac{dy&#039;}{dt&#039;}=0.

Or do I read you incorrectly?

Apparently you did read it incorrectly. Hopefully the above explanation corrected your view.

 

[Dale]

The argument is about the fact that proper acceleration is DIFFERENT from four-acceleration. It differs exactly by the \gamma factor.

Well, yes, the proper acceleration is a 3-vector and the four-acceleration is a 4-vector, so obviously they are different. But the Minkowski norm of the four-acceleration is equal to the Euclidean norm of the proper acceleration, they do not differ by a factor of \gamma. I.e. the Minkowski norm of the four-acceleration is the magnitude of the acceleration measured by an accelerometer.

P.S. I looked for both of those references. I didn't have access to the first, and I couldn't find the second at all. I can use the usual non-generalized Lorentz transform to answer any questions that can be expressed in terms of a MCIF, but I won't be able to use these non-inertial generalizations.

 

[starthaus]

I.e. the Minkowski norm of the four-acceleration is the magnitude of the acceleration measured by an accelerometer.

The wiki page contradicts you. But, we are digressing, could you please go all the way back to post #3 to get the full scope of the disagreement? It is about the proper usage of Lorentz transforms.

 

[starthaus]

P.S. I looked for both of those references. I didn't have access to the first, and I couldn't find the second at all.

I also gave the arxiv link for the second paper. The author is a very respected physicist (and a mentor on this forum, see here ).

I can use the usual non-generalized Lorentz transform to answer any questions that can be expressed in terms of a MCIF, but I won't be able to use these non-inertial generalizations.

If you are patient, kev is only one step away from finishing the calculatons using the correct "generalizations". Why not be patient and learn something new? Let's give him a little more time, I am quite sure that using the latest hint he'll come up with the correct transformations at his next iteration. He's converging on the correct derivation.

 

[Dale]

The wiki page contradicts you.

I think you must be misunderstanding something. The wikipedia page clearly says that "the magnitude of the four-acceleration (which is an invariant scalar) is equal to the proper acceleration". Although technically it should say "magnitude of the proper acceleration" since the proper acceleration is a vector.

http://en.wikipedia.org/wiki/Four-acceleration

 

[starthaus]

I think you must be misunderstanding something. The wikipedia page clearly says that "the magnitude of the four-acceleration (which is an invariant scalar) is equal to the proper acceleration". Although technically it should say "magnitude of the proper acceleration" since the proper acceleration is a vector.

http://en.wikipedia.org/wiki/Four-acceleration

...for \gamma=1. A very important detail.
Look at it this way: your formula produces an infinite acceleration for \omega-&gt;\frac{c}{r}. This is not only unphysical but also diverges from the Newtonian prediction r\omega^2=\frac{c^2}{r}. So, your application of the method for calculating 4-acceleration does not recover the finite Newtonian predictions in the low speed limit. This should be a clue that it is invalid.
The discussion is about the usage of appropiate Lorentz transforms. I have shown the appropiate Lorentz transforms for rotating frames, with the associate references from peer - reviewed journals. Would you please let kev finish the derivation?

 

[Dale]

...for \gamma_u=1. A very important detail that I have pointed out to you.

And a detail which I have pointed out to you is wrong. We are going around in circles, which I suppose is appropriate given the subject of the thread. To avoid further circulation and to determine where our disagreement stems could you please identify which of the following you disagree with:

1) In the inertial reference frame where the center is at rest and the mass is moving in the x-y plane along a circle of radius r at an angular speed w the four-acceleration is given by:
\mathbf a=c \frac{d\mathbf u/dt}{|d\mathbf s/dt|} = (0,\; -\gamma^2 r \omega^2 \; cos(\phi + \omega t),\; -\gamma^2 r \omega^2 \; sin(\phi + \omega t),\; 0 )
where
\gamma=(1-\frac{r^2 \omega^2}{c^2})^{-1/2}

2) The Minkowski norm of this four-acceleration is given by:
|\mathbf a|^2=-\gamma^4 r^2 \omega^4

3) The Minkowski norm of any four-vector is frame invariant meaning that if you calculate it in one frame it will be the same in any other frame whether inertial or non-inertial.

4) The magnitude of the Euclidean norm of the proper acceleration is equal to the magnitude of the Minkowski norm of the four-acceleration.

5) The Euclidean norm of the proper acceleration is given by:
a=\gamma^2 r \omega^2

Look at it this way: your formula produces an infinite acceleration for \omega-&gt;\frac{c}{r}. This is unphysical.

No, not only is that not unphysical it is correct. For any finite r, the acceleration measured by an accelerometer goes to infinity in the limit as the tangential velocity goes to c.

Why do you keep diverting the discussion towards the derivation of proper acceleration? The discussion is about the usage of appropiate Lorentz transforms. Would you please let kev finish the derivation?

I'm sure our discussion won't prevent kev from finishing the derivation if he wants to.

 

[yuiop]

If they are correct, then when t' = 0, \sin(\gamma\omega t&#039;) = 0 and \cos(\gamma\omega t&#039;) = 1 and the equations simplify to:

Incorrect. In the comoving frame you need to realize that \frac{dx&#039;}{dt&#039;}=0 and \frac{dy&#039;}{dt&#039;}=0. Try continuing the calculations from this hint.

I do set \frac{dx&#039;}{dt&#039;}=0 and \frac{dy&#039;}{dt&#039;}=0 in the very next step.

Also \sin(0) = 0 and \cos(0) = 1 are indesputable mathematical facts.

 

[starthaus]

If they are correct, then when t' = 0, \sin(\gamma\omega t&#039;) = 0 and \cos(\gamma\omega t&#039;) = 1 and the equations simplify to:
[\quote]

I do set \frac{dx&#039;}{dt&#039;}=0 and \frac{dy&#039;}{dt&#039;}=0 in the very next step.

Not in the correct way. You are also setting t&#039;=0 for no reason whatsoever. I have already pointed out this error in the previous post.
New hint: what can you conclude about \frac{d^2x&#039;}{dt&#039;^2} and \frac{d^2y&#039;}{dt&#039;^2}?

Also \sin(0) = 0 and \cos(0) = 1 are indesputable mathematical facts.

True but totally useless for completing the derivation.

 

[yuiop]

P.S. I looked for both of those references. I didn't have access to the first, and I couldn't find the second at all. I can use the usual non-generalized Lorentz transform to answer any questions that can be expressed in terms of a MCIF, but I won't be able to use these non-inertial generalizations.

This is the second reference Starthaus uses: http://arxiv.org/PS_cache/gr-qc/pdf/9904/9904078v2.pdf

 

[starthaus]

Since the vote is split 50/50 I will appoint myself vice president and cast the deciding vote in favor of posting it

So throughout this I will use the convention that the time coordinate is the first coordinate and that timelike intervals squared are positive. So given a wheel centered at the origin and rotating about the z axis in a counter clockwise direction we can describe the worldline for any particle on the rim by:
\mathbf s =(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

Both Nelson and Nikolic would disagree with the above, I think that this is the origin of your error. Nikolic gives the correct expression for s.
May I suggest that you open a different thread where we can discuss this?

 

[yuiop]

Gron defines a non-inertial metric for a rotating reference frame on Page 89 of his book http://books.google.co.uk/books?id=IyJhCHAryuUC&pg=PA89&lpg=PA89#v=onepage&q&f=false that might be of interest. He uses cylindrical coordinates which seems more natural in this situation, rather than the rectangular coordinates used by Starthaus.

 

[Dale]

we can describe the worldline for any particle on the rim by:
\mathbf s =(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

Both Nelson and Nikolic would disagree with the above, I think that this is the origin of your error. Nikolic gives the correct expression for s.

So what? I can cite people who agree with me:
http://arxiv.org/PS_cache/hep-th/pdf/0301/0301141v1.pdf
http://physics.bu.edu/~duffy/py105/SHM.html
Plus since I don't have access to those references I cannot check the likely situation that you are misunderstanding what Nelson and Nikolic actually claim.

Instead of an appeal to authority why don't you show what is wrong with it? I mean, the worldline of a particle undergoing uniform circular motion is a circular helix and that is the equation of a helix. So what could possibly be incorrect? Also, according to Nikolic what is the correct expression for the worldline of a particle undergoing uniform circular motion in some inertial frame?

 

[yuiop]

Incorrect. In the comoving frame you need to realize that \frac{dx&#039;}{dt&#039;}=0 and \frac{dy&#039;}{dt&#039;}=0. Try continuing the calculations from this hint.

When I do that, I get:

a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(\frac{d^2x&#039;}{dt&#039;^2}\cos(\gamma\omega t&#039;)-\gamma\frac{d^2y&#039;}{dt&#039;^2}\sin(\gamma\omega t&#039;)- R\gamma^2\omega^2\cos(\gamma\omega t&#039;)\right)

a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(\frac{d^2x&#039;}{dt&#039;^2}\sin(\gamma\omega t&#039;)+\gamma\frac{d^2y&#039;}{dt&#039;^2}\cos(\gamma\omega t&#039;) - R\gamma^2\omega^2\sin(\gamma\omega t&#039;)\right)

What are you claiming that tells us?

By the way, if I set the acceleration terms in the rotating frame to zero, the result is:

a_x = \frac{d^2x}{dt^2}=\gamma^{-2}\left(- R\gamma^2\omega^2\cos(\gamma\omega t&#039;)\right)

a_y = \frac{d^2y}{dt^2}=\gamma^{-2}\left(- R\gamma^2\omega^2\sin(\gamma\omega t&#039;)\right)

When t'=0, the result is:

a_x = \frac{d^2x}{dt^2}= -R\omega^2

a_y = \frac{d^2y}{dt^2}= 0

\sqrt{a_x^2 + a_y^2} = R\omega^2\right)

and when t' is set to a value such that (\gamma\omega t&#039;)=\pi/2 (ie a quarter of a turn later) when the x' axis is aligned with the y axis, the result is:

a_x = \frac{d^2x}{dt^2}= 0

a_y = \frac{d^2y}{dt^2}= -R\omega^2

\sqrt{a_x^2 + a_y^2} = R\omega^2

That tells me I can set t' to any value I like and the magnitude of the centripetal acceleration remains constant. Since it is the magnitude we are interested in, in this thread, I am free to set t'=0 for convenience (as I did) and consider that as representative of any arbitary value of t'. This also follows trivially from the circular symmetry as jorrie has been trying to tell you.

P.S. @Dalespam: Did you try the link I gave in #116?