Relativistic centripetal force

[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'=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} .

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'=\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'} 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 &#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.

 

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

which reduces to:

a_x &#039; = \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 &#039; = \gamma \frac {d \vec{v}}{dt}

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

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} .

proper acceleration by definition is , a &#039; = \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 &#039; = \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 &#039; = \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.