Relativistic centripetal force

[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 $$\vec {F'}$$ 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:

$$a_x = \frac{a_x '}{\gamma^3 (1+u_x ' V/c^2)}$$

which reduces to:

$$a_x ' = \gamma^3 a_x$$

when $u_x '$ 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:

$$a_y = \frac{a_y '}{\gamma^2 (1+u_x ' V/c^2)}- \frac{a_x ' u_y ' V/c^2}{\gamma^2 (1+u_x ' V/c^2)}$$

which when $u_x ' = 0$ and $u_y ' = 0$ reduces to:

$$a_y ' = \gamma^2 a_y$$

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:

$$a_x ' = \frac{dv_x '}{dt} = \gamma^3 \frac{dv_x}{dt}$$

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:

$$a_y ' = \frac{dv_y/dt + v_y v (dv/dt)}{(1-v^2/c^2)}$$

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

$$a_y ' = \frac{dv_y/dt}{(1-v^2/c^2)} = \gamma^2 d\frac{v_y}{dt}$$

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.

 

[starthaus]

The expression you derived in attachment 1 of your blog starts with the statement "dv/dt and v have the same direction and sense"

I don't know what attachment you have been reading but your claim is clearly false. We are talking about circular motion so, acceleration (dv/dt) and velocity (v) certainly do not have the same direction and sense. Which attachment are you reading?

 

[starthaus]

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.

You are reading the wrong attachment. I have always pointed you towards the "Relativistic Lorentz Force" (attachment 1). For some reason, you decided to read a totally different attachment that has nothig to do with the subject of this thread.
I explained to you that this type of calculation is the foundation of cyclotron design, so why do you keep insisting that it is incorrect? Do you want me to get you a reference book on accelerator design? I

 

[starthaus]

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:

$$a_x = \frac{a_x '}{\gamma^3 (1+u_x ' V/c^2)}$$

which reduces to:

$$a_x ' = \gamma^3 a_x$$

Yes, this is textbook stuff for LINEAR motion, derived from Lorentz transforms for linear motion. Has nothing to do with the subject of circular motion. You are talking about LINEAR acceleration in the X direction, $$a_x$$. Why do you keep insisting on using concepts that are irrelevant for the subject being discussed?

 

[starthaus]

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:

I am not ignoring anything, I have just been telling you for a few days now that you have no right to use Lorentz transforms derived for linear motion in deriving the way force transforms for circular motion. I am getting tired of your continued bickering, I put up an attachment that deals with the Lorentz transforms for circular motion that show how to calculate the force transformation correctly. If you want to learn, I suggest you read it.

 

[starthaus]

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 $$d^2 \frac{y}{dt^2}$$

Thank you, it is an obvious typo. The equation is correct and the previous one should clue you in that it is about $$d^2 \frac{y}{dt^2}$$

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

No, they are left as an exercise for you to complete. If you manage to do it, you will have an interesting surprise as a result

 

[starthaus]

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.

I checked, pervect DOES NOT calculate the acceleration transformation in rotating frames (what we want to find out), he applies the well-known standard formula in calculating four-acceleration (what pervect wanted to find out).
Do you even understand the difference between the two concepts? If you did, you would not have brought this up as a relevant argument in the discussion.

 

[Jorrie]

I checked, pervect DOES NOT calculate the acceleration transformation in rotating frames (what we want to find out), he applies the well-known standard formula in calculating four-acceleration (what pervect wanted to find out).

But, I read pervect saying: "Note that I'm not the one calling this the "centripetal force". What I'd call what I'm computing is the proper acceleration of the body, i.e. the acceleration that someone on the body would measure with an accelerometer."

Read with kev and pervect's prior posts, it seems to me exactly the 'proper centripetal acceleration' that kev and I want. The only contentious issue that I spot is how to transform that proper acceleration to a force in the inertial frame of the center of the circle.

 

[starthaus]

But, I read pervect saying: "Note that I'm not the one calling this the "centripetal force". What I'd call what I'm computing is the proper acceleration of the body, i.e. the acceleration that someone on the body would measure with an accelerometer."

Read with kev and pervect's prior posts, it seems to me exactly the 'proper centripetal acceleration' ($d^2r/d\tau^2$) that kev and I want.

1. This is not what kev calculates in post #1. He calculates a transform for the force between frames by using the Lorentz transform for translation instead of rotation.
2. It is precisely this attempt of using the inappropiate transform that I have objected to throughout this thread, starting with post #3
3. Both you and kev have insisted that the Lorentz transform for translation is appropiate treatment for deriving the force in the observer's frame throughout this thread.
4. I do not think that this is correct and I provided the appropiate treatment using the transform for rotating frames.

The only contentious issue that I spot is how to transform that proper acceleration to a force in the inertial frame of the center of the circle.

Using the Lorentz transforms for translation is not justified. I gave you both the correct method based on the Lorentz transforms for rotation. If you or kev finish the computations, you are in for a big surprise.

 

[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 $$\vec{a}=\gamma \frac {d \vec{v}}{dt}$$.

What is even more amusing is that the assertion you make above in #34

proper acceleration by definition is , $$a ' = \gamma \frac {d \vec{v}}{dt}$$

directly contradicts what you said in #29 (below):

$$F'=m_0 a'$$ (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: $$\frac {d}{dt} (\gamma v)=\gamma^3 \frac{dv}{dt}$$ .

proper acceleration by definition is , $$a ' = \gamma^3 \frac {d \vec{v}}{dt}$$

What is really funny, is that having come up with two contradicting definitions of proper acceleration, neither of them is correct. The correct solution for proper acceleration in the case of centripetal acceleration is:

$$a ' = \gamma^2 \frac {d \vec{v}}{dt}$$

I checked, pervect DOES NOT calculate the acceleration transformation in rotating frames (what we want to find out), he applies the well-known standard formula in calculating four-acceleration (what pervect wanted to find out).
Do you even understand the difference between the two concepts? If you did, you would not have brought this up as a relevant argument in the discussion.

I do understand the difference. Four acceleration is the acceleration in terms of proper time so the equation that pervect gives is the proper centripetal acceleration so:

$$a ' = \gamma^2 \frac {d \vec{v}}{dt}$$

Note that the proper time is all we need because transverse distances are not subject to length contraction. It is also clear that pervect is talking about acceleration in a rotation context because the title of that thread is "relativistic centripetal force" (very like the title of this thread).

However I do concede that pervect, jorrie and I are talking about the proper acceleration as measured by an instantaneously comoving inertial observer, while a comoving rotating non-inertial observer might measure some things differently. One obvious difference is that to a non-inertial non-instantaneous comoving observer the orbiting particle has no acceleration or movement at all in his rest frame. In that sense the proper acceleration measured by an observer in the comoving rotating reference frame is always zero, no matter what the accleration according to the coordinate lab frame is measured to be.

So it seems we have to come to a consensus at to what exactly you mean by "proper acceleration" in the context of your blog article which is very sparce on explanatory or supporting text. One definition that comes to mind, is what acceleration an particle would be measured to have by a non-inertial comoving observer, if the particle was released and shot off on a straight tangential trajectory as seen in the non-rotating lab frame.

So is proper acceleration that which is measured by an accelerometer (which is the usual definition of proper acceleration in relativity) or is it change in spatial location per unit time per unit time as you seem to be using in your blog?

 

[yuiop]

You are reading the wrong attachment. I have always pointed you towards the "Relativistic Lorentz Force" (attachment 1). For some reason, you decided to read a totally different attachment that has nothig to do with the subject of this thread.

I read the "wrong" attachment because earlier you said:

... especially in the light of explaining (see also attachment 1) that the proper acceleration is, by definition, $$\frac {d}{dt}(\gamma \vec{v})$$
...

so naturally I assumed you said something about proper acceleration in attachment 1 but not a single equation there relates directly to acceleration or even to to a transformation between frames. Seeing nothing there, I looked in the other attachments where you do talk about acceleration, to see how you come to the conclusion that proper acceleration has a factor of gamma^3 and find that the source of your confusion is that the only kind of acceleration that you analyse is linear acceleration, rather than the transverse acceleration that we require here.

 

[yuiop]

Using the Lorentz transforms for translation is not justified. I gave you both the correct method based on the Lorentz transforms for rotation. If you or kev finish the computations, you are in for a big surprise.

I have finished your computations and the final result is just as messy and ugly as the equations that precede it. If your computations are correct then they have to resolve in the simplest situation to either $a' = \gamma a$ or $a' =\gamma^3 a$ because that is what proper acceleration is by (your) definition. It does not seem to do that but maybe I am doing it wrong. Can you demonstrate that your definition is correct and in agreement with your rotation transforms?

In your blog attachment you state:

$$t = \gamma \left(t ' + \frac{\omega R y '}{c^2} \right)$$

It seems odd in that in that expression there is no x'. Is the transformation between coordinate time and proper time really completely independent of movement in the x' direction? Somehow I doubt it because these are supposed to generalised transformations with no preferential direction.

Can you elaborate on how you arrive at the equations for dx and dy in (4)?

I have managed to locate a copy of the article you reference in your blog (Generalized Lorentz transformation for an accelerated, rotating frame of reference [J. Math. Phys. 28, 2379-2383 (1987)] Robert A. Nelson) and none of the equations in that paper match the equations in your blog. I guess that is a credit to you that you are not just copying other people's work, but since it is your work perhaps you could clarify what you are thinking. One advantage of a forum over books is supposed to be that you can ask the author what he means or to elaborate on something. All this "surprise" stuff is not very helpful. Relativity is complicated enough with plenty of opportunities for error and misunderstanding, without playing silly games.

 

[yuiop]

Let's have a look at the method you use in your generalised transformation for rotating reference frames. You start with:

$$\frac{dx}{dt} = \frac{dx}{dt '} \frac{dt '}{dt} = \gamma^{-1}\frac{dx}{dt '}$$

Fine. In the next step you obtain an equation for acceleration by:

$$\frac{d^2x}{dt ^2} = \frac{d^2 x}{dt '^2} \frac{dt '^2}{dt^2} = \gamma^{-2}\frac{d^2x}{dt '^2}$$

Rearrange:

$$a ' = \gamma^2 \frac{d^2x}{dt ^2}$$

Now by definition:

$$a ' = \gamma^2 \frac{d^2x}{dt ^2} = \gamma^2 \frac{dv}{dt}$$

which is the result obtained by pervect and me (which you say is wrong) using your method.

 

[starthaus]

I have finished your computations and the final result is just as messy and ugly as the equations that precede it.

These are not "my" computations. These are the "correct" computations for the application of the appropiate transforms. If you complete the calculations (left to you as an exercise), you will have the pleasant surprise to find the correct transformation of accelerations, which, by the way, looks very ellegant. That is, if you want to learn.

$$t = \gamma \left(t ' + \frac{\omega R y '}{c^2} \right)$$

It seems odd in that in that expression there is no x'. Is the transformation between coordinate time and proper time really completely independent of movement in the x' direction? Somehow I doubt it because these are supposed to generalised transformations with no preferential direction.

Why don't you read the reference?

Can you elaborate on how you arrive at the equations for dx and dy in (4)?

Through simple differentiation.

I have managed to locate a copy of the article you reference in your blog (Generalized Lorentz transformation for an accelerated, rotating frame of reference [J. Math. Phys. 28, 2379-2383 (1987)] Robert A. Nelson) and none of the equations in that paper match the equations in your blog. I guess that is a credit to you that you are not just copying other people's work, but since it is your work perhaps you could clarify what you are thinking.

If you have difficulty with simple math, HERE is another reference where the calculations are all done for you.

One advantage of a forum over books is supposed to be that you can ask the author what he means or to elaborate on something. All this "surprise" stuff is not very helpful. Relativity is complicated enough with plenty of opportunities for error and misunderstanding, without playing silly games.

Hey, you need to learn how to calculate for yourself, not to cherry pick from formulas derived by others.

 

[starthaus]

I read the "wrong" attachment because earlier you said:

...which is the definition of proper acceleration. This is the 4-th time I'm explaining this to you.

so naturally I assumed you said something about proper acceleration in attachment 1 but not a single equation there relates directly to acceleration or even to to a transformation between frames. Seeing nothing there,

I guess everything needs to be spelled out to you, you did not see the $$\frac{d}{dt}(m_0 \gamma \vec {v})=q \vec{v} X \vec{B}$$ equation. One more time, the term $$\frac{d}{dt} (\gamma \vec {v})$$ is the proper acceleration. By defintion.

I looked in the other attachments where you do talk about acceleration, to see how you come to the conclusion that proper acceleration has a factor of gamma^3

The other attachments talk about translation motion. It is not my fault that you seem unable to tell the difference between rotation and translation.
If you were able to do that maybe you could stop ascribing me all the errors that you keep making.

 

[starthaus]

Let's have a look at the method you use in your generalised transformation for rotating reference frames. You start with:

$$\frac{dx}{dt} = \frac{dx}{dt '} \frac{dt '}{dt} = \gamma^{-1}\frac{dx}{dt '}$$

Fine.

I am glad that it met with your approval.

In the next step you obtain an equation for acceleration by:

$$\frac{d^2x}{dt ^2} = \frac{d^2 x}{dt '^2} \frac{dt '^2}{dt^2}$$

You sure about this? This is basic calculus. What you wrote is wrong, I am doing none of the stuff you are claiming I am doing. Here is what I am really doing:

$$a= \frac{d^2x}{dt ^2} =\frac{d}{dt} \frac{dx}{dt}=\frac {d^2x}{dtdt'} \frac{dt'}{dt}$$

Do you see the difference? The object is to get the correct expression for $$a$$.

Rearrange:

$$a ' = \gamma^2 \frac{d^2x}{dt ^2}$$

Nope, you need to get your basic calculus straightened out. Only after that your physics will come out correct.

 

[yuiop]

You sure about this? This is basic calculus. What you wrote is wrong, I am doing none of the stuff you are claiming I am doing. Here is what I am really doing:

$$a= \frac{d^2x}{dt ^2} =\frac{d}{dt} \frac{dx}{dt}=\frac {d^2x}{dtdt'} \frac{dt'}{dt}$$

Do you see the difference? The object is to get the correct expression for $$a$$.

I can extend your expression, using the methods you use, like this:

$$a= \frac{d^2x}{dt ^2} =\frac{d}{dt} \frac{dx}{dt}= \frac {d^2x}{dtdt'} \frac{dt'}{dt} = \gamma^{-1} \frac {d^2x}{dtdt'} = \gamma^{-1} \frac {d^2x}{dt'dt'} \frac{dt'}{dt} = \gamma^{-2} \frac {d^2x}{dt'dt'} = \gamma^{-2} \frac {d^2x}{dt'^2}$$

which gives:

$$a' = \gamma^2 \frac{d^2x}{dt ^2}$$

which is what I have always claimed and which you keep insisting is wrong. I think that since it obvious we are never going to agree it might be helpful if some of the pf staff gave a second opinion on the disputed items in this thread.

 

[starthaus]

I can extend your expression, using the methods you use, like this:

$$a= \frac{d^2x}{dt ^2} =\frac{d}{dt} \frac{dx}{dt}= \frac {d^2x}{dtdt'} \frac{dt'}{dt} = \gamma^{-1} \frac {d^2x}{dtdt'} = \gamma^{-1} \frac {d^2x}{dt'dt'} \frac{dt'}{dt} = \gamma^{-2} \frac {d^2x}{dt'dt'} = \gamma^{-2} \frac {d^2x}{dt'^2}$$

which gives:

$$a' = \gamma^2 \frac{d^2x}{dt ^2}$$

which is what I have always claimed and which you keep insisting is wrong. I think that since it obvious we are never going to agree it might be helpful if some of the pf staff gave a second opinion on the disputed items in this thread.

Look at the attachment I wrote for you, $$\frac {d^2x}{dt'^2}$$ is not $$a'$$.
You should know better than that:

$$a'=\frac {d^2x'}{dt'^2}$$$$\frac {d^2x}{dt'^2}$$ means nothing.

Your exercise is to express $$\frac {d^2x}{dt^2}$$ as a function of $$\frac {d^2x'}{dt'^2}$$. I gave you all the tools to do that correctly.

 

[yuiop]

At this step:

$$a= \frac{d^2x}{dt dt} = \frac {d^2x}{dtdt'} \frac{dt'}{dt}$$

you are using the simple fact that:

$$\frac{1}{dt} = \frac{1}{dt'}\frac{dt'}{dt}$$

(Nothing wrong with that - basic algebra.) I am using exactly the same algebraic fact to complete this step:

$$\gamma^{-1} \frac {d^2x}{dtdt'} = \gamma^{-1} \frac {d^2x}{dt'dt'} \frac{dt'}{dt}$$

Are you really saying the above equality is invalid? If so, then it you who needs to think about it.

 

[yuiop]

Look at the attachment I wrote for you, $$\frac {d^2x}{dt'^2}$$ is not $$a'$$.

We are discussing centripetal acceleration which is orthogonal to the instantaneous velocity. Orthogonal distances do not length contract so dx = dx' and they are interchangeable.

$$\frac {d^2x}{dt'^2}$$ is the same thing as $$\frac {d^2x'}{dt'^2}$$

You should know better than that:

$$a'=\frac {d^2x'}{dt'^2}$$

See above.

$$\frac {d^2x}{dt'^2}$$ means nothing.

It means the same thing as

$$\frac {d^2x'}{dt'^2}$$

when dx' = dx, as it does in this case.

 

[starthaus]

At this step:

$$a= \frac{d^2x}{dt dt} = \frac {d^2x}{dtdt'} \frac{dt'}{dt}$$

you are using the simple fact that:

$$\frac{1}{dt} = \frac{1}{dt'}\frac{dt'}{dt}$$

(Nothing wrong with that - basic algebra.) I am using exactly the same algebraic fact to complete this step:

$$\gamma^{-1} \frac {d^2x}{dtdt'} = \gamma^{-1} \frac {d^2x}{dt'dt'} \frac{dt'}{dt}$$

Are you really saying the above equality is invalid? If so, then it you who needs to think about it.

$$\frac {d^2x}{dt'^2}$$ is physically a meaningless entity, you are mixing frames. Can you write down the correct definition for $$a'$$?