Relativistic centripetal force

[yuiop]

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?

The difference between the approach taken by Starthaus compared to the rest of us, is that he is considering the case of a co-moving co-accelerating observer such that this co-accelerating observer measures the acceleration of the particle moving in a circle to be zero.

In the MCIF the co-moving inertial observer is momentarily at rest with the test particle, but it has non-zero acceleration. This is the proper acceleration of the test particle obtained by Dalespam and is what an accelerometer attached to the test particle in the rotating frame would measure.

 

[Dale]

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

Yes, my experience working in non-inertial frames is limited, but that looks like a good place to start. If I get a chance to work it out I will post the results here.

 

[starthaus]

When I do that, I get:

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

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

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

OK, yoy are done. Finally.
All you have to do is to reduce the terms in \gamma^2
The nice thing about this method is that you not only get the magnitude but you also get the components of the acceleration. You get the COMPLETE transforms.

When t'=0, the result is:

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

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

Why do you persist in the error of "setting t'=0". You finally got the general transforms, valid for ANY t'.

 

[starthaus]

The difference between the approach taken by Starthaus compared to the rest of us, is that he is considering the case of a co-moving co-accelerating observer such that this co-accelerating observer measures the acceleration of the particle moving in a circle to be zero.

In the MCIF the co-moving inertial observer is momentarily at rest with the test particle, but it has non-zero acceleration. This is the proper acceleration of the test particle obtained by Dalespam and is what an accelerometer attached to the test particle in the rotating frame would measure.

Umm, no. What do you think \frac{d^2x}{dt^2} is?

 

[yuiop]

Umm, no. What do you think \frac{d^2x}{dt^2} is?

That is one component of the COORDINATE centripetal acceleration as measured in the non rotating inertial frame. What do you think it is??

More generally:

The Euclidean norm of the coordinate centripetal acceleration is R\omega^2

This is what is measured in the non rotating inertial frame.

The Euclidean norm of the proper centripetal acceleration is \gamma^2R\omega^2

by the normal definition of proper acceleration, which is equivalent to the four acceleration, or what is measured by an accelerometer.

You have your own unusual definition of proper acceleration as being the acceleration measured by a co-accelerating observer, which always has the magnitude zero.

 

[yuiop]

Why do you persist in the error of "setting t'=0". You finally got the general transforms, valid for ANY t'.

If the transforms are valid for ANY t' then setting t'=0 is NOT an error as you claim. It is just a convenience for working out a particular case that is representitive for magnitude quantities. It so happens that I demonstrated that it is representative for magnitude quantities, by using a diiferent value of t' and getting the same magnitude.

 

[starthaus]

That is one component of the COORDINATE centripetal acceleration as measured in the non rotating inertial frame. What do you think it is??

Good, you finally learned to correct derivation, using the appropiate Lorentz transforms.
So, what does frame S' signify?

 

[starthaus]

If the transforms are valid for ANY t' then setting t'=0 is NOT an error as you claim. It is just a convenience for working out a particular case that is representitive for magnitude quantities. It so happens that I demonstrated that it is representative for magnitude quantities, by using a diiferent value of t' and getting the same magnitude.

Why would you insist on setting t'=0 when I guided you to deriving the general transforms, valid for any t'?
As to the magnitude, you can derive it directly from the general transforms, right? You don't need to insist on making t'=0.

 

[yuiop]

The nice thing about this method is that you not only get the magnitude but you also get the components of the acceleration. You get the COMPLETE transforms.

Yes, it is sometimes useful to have the components and directions and I also like your transform for other reasons. In fact I like it so much that I would to see the derivation steps broken down a bit more so that I confirm the steps you use to obtain it. So far, you have ignored that request, so I can only hope that a better mathematician than myself can verify the steps you take.

However, nothing in your transform disproves anything I said in #1 or anything Dalespam or Jorrie said. We just differ on the definition of proper acceleration.

Unfortunately, you do not seem to understand the physical significance of your own transformations or the four vector quantities that Dalespam is using.

 

[yuiop]

Why do you persist in the error of "setting t'=0". You finally got the general transforms, valid for ANY t'.

Yes, I got the general transforms and then applied them to a specific case. What is wrong with that? That is what general transforms are for.

 

[starthaus]

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.

Thank you, this is a very good reference.
Look at (5.2). What does it tell you?
Specifically, what is the transform for \frac{d^2 \theta}{dt^2}?

 

[starthaus]

Yes, I got the general transforms and then applied them to a specific case. What is wrong with that? That is what general transforms are for.

There is no reason to make t'=0, I have already taught you how to get the general transforms.

 

[yuiop]

Look at (5.2). What does it tell you?
Specifically, what is the transform for \frac{d^2 \theta}{dt^2}?

5.2 says t=T which tells me that Gron is using an unusual clock synchronisation method where the clocks in the rotating frame are articificailly sped up so that stay in sync with the non rotating inertial clock at the centre. Unsual but not invalid if you know what you doing. t is NOT proper time in the normal sense, so you can NOT directly draw any conclusions about proper acceleration by directly using those coordinates. \frac{d^2 \theta}{dt^2} is NOT proper acceleration in the normal sense.

 

[starthaus]

5.2 says t=T which tells me that Gron is using an unusual clock synchronisation method where the clocks in the rotating frame are articificailly sped up so that stay in sync with the non rotating inertial clock at the centre. Unsual but not invalid if you know what you doing. t is NOT proper time in the normal sense, so you can NOT directly draw any conclusions about proper acceleration by directly using those coordinates.

Sure you can. So, you don't like your own reference any more, eh?
This is a pretty good book, pretty recent as well.

\frac{d^2 \theta}{dt^2} is NOT proper acceleration in the normal sense.

Huh?

Anyways, the question was : can you derive its transform between RF and IF? It is not a trick question.

 

[Jorrie]

However, nothing in your transform disproves anything I said in #1 or anything Dalespam or Jorrie said. We just differ on the definition of proper acceleration.

Yes, the differences in definitions (of comoving frame and of proper acceleration) have caused a lot of unnecessary bickering in this thread. I have sympathy with students trying to follow what was going on.

By definition, the centripetal force is always radial, so polar (or cylindrical) coordinates seem to be more appropriate than the Cartesian coordinates that Starthaus used. It reminds me about CarlB's efforts of some years ago to cast the Schwarzschild coordinate accelerations into Cartesian form. He succeeded, but the resultant equations were cumbersome and according to my testing, actually used more fp-calcs than the polar coordinate equations. His motivation was to make orbit simulations faster...

I guess there will be applications where the Cartesian components of centripetal accelerations or forces will be very useful, but the mass swinging at the end of a string is probably not one of them.

 

[starthaus]

Yes, the differences in definitions (of comoving frame and of proper acceleration) have caused a lot of unnecessary bickering in this thread. I have sympathy with students trying to follow what was going on.

By definition, the centripetal force is always radial, so polar (or cylindrical) coordinates seem to be more appropriate than the Cartesian coordinates that Starthaus used.

It is not about that, it is about teaching kev how to use the appropiate transforms and how to derive the correct value for the proper acceleration. His OP was wong on both accounts. It took 9 pages and 100+ posts to get the correct results for what should have taken two iterations of derivatives.

 

[Jorrie]

It is not about that, it is about teaching kev how to use the appropiate transforms and how to derive the correct value for the proper acceleration. His OP was wong on both accounts.

No, I think his OP is perfectly valid, by his (and AFAIK, all around here, but yourself) definition of proper acceleration - measured as he stated, by a tension gauge oriented radially (implied by 'centripetal force'). He has not shown a rigorous derivation, but that has been done before by pervect, amongst others.

It took 9 pages and 100+ posts to get the correct results for what should have taken two iterations of derivatives.

I do not quite see how the equations that kev posted in https://www.physicsforums.com/showpost.php?p=2693883&postcount=120" and which you said are correct, produce what the tension gauge (or an accelerometer) measures on the circling mass. Can you please enlighten us on that?

 

[starthaus]

I do not quite see how the equations that kev posted in https://www.physicsforums.com/showpost.php?p=2693883&postcount=120" and which you said are correct, produce what the tension gauge (or an accelerometer) measures on the circling mass. Can you please enlighten us on that?

Give it some time, it will sink in.

 

[Dale]

Hi starthaus, if you think that the value you have had kev derive in #120 represents the proper acceleration of the particle then you are simply wrong.

 

[starthaus]

Hi starthaus, if you think that the value you have had kev derive in #120 represents the proper acceleration of the particle then you are simply wrong.

I have guided kev in doing the correct derivation of the acceleration measured in the inertial frame S as a function of the coordinates in the rotating frame S' by using the appropiate Lorentz transforms. This closes the argument started at post #3. Nothing more , or less.

 

[Dale]

That is correct, it is the coordinate acceleration in the original inertial frame transformed to the rotating frame. It is not the proper acceleration.

 

[starthaus]

That is correct, it is the coordinate acceleration in the original inertial frame transformed to the rotating frame.

Thank you.

It is not the proper acceleration.

May I suggest that you open a different thread that discusses how to obtain the proper acceleration from the correct line element (see Gron's book)? The line element "s" that you used in your derivation is incorrect.

 

[Dale]

May I suggest that you open a different thread that discusses how to obtain the proper acceleration from the correct line element (see Gron's book)? The line element "s" that you used in your derivation is incorrect.

That equation was not the line element, it was the worldline of the particle. A line element is a scalar, the worldline is a four-vector parameterized by some arbitrary scalar.

 

[starthaus]

That equation was not the line element, it was the worldline of the particle. A line element is a scalar, the worldline is a four-vector parameterized by some arbitrary scalar.

You volunteered to study the chapter on rotating frames in Gron's book. Why don't we reprise this discussion in a different thread, once you have studied the chapter?

 

[Dale]

You volunteered to study the chapter on rotating frames in Gron's book. Why don't we reprise this discussion in a different thread, once you have studied the chapter?

Do you or do you not agree that the worldline of a particle undergoing uniform circular motion in some inertial reference frame is given by:
(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

 

[starthaus]

Do you or do you not agree that the worldline of a particle undergoing uniform circular motion in some inertial reference frame is given by:
(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

Why don't you spend some time understanding the Gron chapter on the subject? Once you do that, we can talk in a separate thread about how to calculate the proper acceleration. The value you calculated is wrong, there is no point in muddling this thread.

 

[Dale]

You are really starting to irritate me with your repeated assertions that it is wrong followed by a completely evasive non-answer every time your assertion is challenged. You have stated in this thread that my derivation is wrong so defend your statement in this thread and stop trying to weasel your way out of it.

If that expression does not represent the worldline of a particle undergoing uniform circular motion in some inertial reference frame then what expression does?

 

[starthaus]

You are really starting to irritate me with your repeated assertions that it is wrong followed by a completely evasive non-answer every time your assertion is challenged. You have stated in this thread that my derivation is wrong so defend your statement in this thread and stop trying to weasel your way out of it.

Since you don't want to read the Gron chapter, I'll help you out:

The transformation between IF and RF is:

t=T
\theta=\Theta-\omega T

If you calculate the proper acceleration \frac{d^2\Theta}{dT^2} you find out that it is equal to the coordinate acceleration in IF \frac{d^2\theta}{dt^2} . Earlier in this thread (post 120), the coordinate acceleration in IF has been found to be R\omega^2. You can read the same exact value straight of Gron's line element. What does this tell you?

If that expression does not represent the worldline of a particle undergoing uniform circular motion in some inertial reference frame then what expression does?

The answer is found in the Nikolic paper. I provided a link to it several times.

 

[Dale]

OK, so according to Gron the worldline of a particle undergoing uniform circular motion:
In the rotating frame is given by:
(ct,r_0,\theta_0,0)

And in an inertial frame using cylindrical coordinates is given by:
(cT,r_0,\theta_0+\omega T,0)

Transforming this into the inertial frame using Cartesian coordinates we obtain:
(ct,\; r_0 \; cos(\theta_0 + \omega t),\; r_0 \; sin(\theta_0 + \omega t),\; 0)

Which is the same form as:
(ct,\; r \; cos(\phi + \omega t),\; r \; sin(\phi + \omega t),\; 0)

So Gron agrees with me wrt the form of the worldline.

 

[starthaus]

So Gron agrees with me wrt the form of the worldline.

...but not on your derivation for the acceleration. This is very simple stuff, why is so difficult for you to admit that you are wrong?