Relativistic centripetal force

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  • #1
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Looking on the internet, it seems to very difficult to find a simple straightforward statement of the equations for relativistic centripetal force. This is my point of view and hopefully with some feedback we can come to a consensus.

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

[tex]a = \frac{v^2}{r} = r \omega^2 = r \frac{d\theta^2}{dt^2} [/tex].

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

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

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

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

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

[tex]F' = ma' = m \frac{v^2}{r}\gamma^2 = ma \gamma^2[/tex]

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

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

[tex] F = \frac{F'}{\gamma} = \frac{ma'}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]

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

Given that the Newtonian equation for angular momentum is [itex]L = mr^2\omega = mrv[/itex] it is not difficult to work out that the relativistic angular momentum of the test mass in the non rotating frame is [itex]m\gamma r v[/itex] from the above equations.
 
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  • #2
Jorrie
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To an observer co-moving with the test mass, the acceleration of the test mass is:

[tex]a' = r \frac{d\theta^2}{d\tau^2} \gamma^2 = r \frac{d\theta^2}{dt^2} \gamma^2 = \frac{v^2}{r}\gamma^2 = a \gamma^2[/tex]
I think you are correct and I'm trying to reconcile it with the equation that I arrived at in https://www.physicsforums.com/showpost.php?p=2678507&postcount=17" for 'centripetal acceleration' in a circular orbit:

[tex]
a_{cp} = \frac{v_o^2}{r} = \frac{GM}{g_{tt} r^2} = \frac{GM}{r^2}(1+2v_o^2)
[/tex]

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

[tex]
a'_{cp} = \frac{v_o^2}{r}\gamma^2 = \frac{GM}{r^2}\left(\frac{1+2v_o^2}{1-v_o^2}\right)
[/tex]

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

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

This is surprising, because the tension at one end of the rope is different from the tension at the other, which is not the case in Newtonian mechanics. It is also a measure of the centripetal force that would be provided by a non rotating electromagnetic field to keep the test mass moving in a circle. In both these cases, it is the force provided from the non-rotating frame to the moving test mass in the rotating frame.
Interesting observation, kev!
 
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  • #3
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It is known that the Lorentz transformation of transverse force is [itex]F = F ' \gamma^{-1}[/itex], so it follows that the force measured by the non rotating inertial observer at rest with the centre of rotation is:

[tex] F = \frac{F'}{\gamma} = \frac{ma'}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]
I don't think that you can extrapolate the Lorentz transform (derived for inertial frames) to the transform between the observer frame and the proper frame. The proper frame is rotating wrt the observer frame.
 
  • #4
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I don't think that you can extrapolate the Lorentz transform (derived for inertial frames) to the transform between the observer frame and the proper frame. The proper frame is rotating wrt the observer frame.
Agreed that the proper frame is rotating and therefore an accelerating or non-inertial frame, but we can use the concept of the momentarily comoving inertial frame and the clock hypothesis to treat calculations in the accelerating frame exactly like an instantaneous inertial frame.

This is a quote from the http://www.edu-observatory.org/physics-faq/Relativity/SR/experiments.html" [Broken]:
The clock hypothesis states that the tick rate of a clock when measured in an inertial frame depends only upon its velocity relative to that frame, and is independent of its acceleration or higher derivatives. The experiment of Bailey et al. referenced above stored muons in a magnetic storage ring and measured their lifetime. While being stored in the ring they were subject to a proper acceleration of approximately 10^18 g (1 g = 9.8 m/s2). The observed agreement between the lifetime of the stored muons with that of muons with the same energy moving inertially confirms the clock hypothesis for accelerations of that magnitude.
In other words the time dilation experienced by the muons at 10,000,000,000,000,000,000 g is exactly in agreement with the Lorentz transformation for muons moving inertially in a straight line at a constant velocity equal to the instantaneous velocity of the accelerating muons.

The clock hypothesis can be extended to the transformation of all physical measurements in an accelerating frame, so that the accelerating frame obeys Lorentz transformations exactly as if it was a measurement in an instantaneous comoving inertial frame. This is another quote from ahttp://www.edu-observatory.org/physics-faq/Relativity/SR/clock.html" [Broken]:
On a more advanced note, the clock postulate can be generalised to say something about measurements we make in a noninertial frame. First, it tells us that noninertial objects only age and contract by the same gamma factor as that of their MCIF. So, any measurements we make in a noninertial frame that use clocks and rods will be identical to measurements made in our MCIF--at least, if these measurements are made over a small enough region. This is because, in general, different regions of the noninertial frame have different MCIFs, a complicating factor that makes the construction of noninertial frames very difficult. In fact, it can only readily be done for constant-velocity and constant-acceleration frames; and for the last, that is rather difficult.

But we now choose to extend the clock postulate to include all measurements (though perhaps it can be argued that all measurements only ever use clocks and rods anyway). This idea leads onto "covariance", which is a way of using tensors to write the language of physics in a way that applies to all frames, noninertial as well as inertial.
If a local observer observes an accelerating particle, all he needs to know is its instantaneous velocity and he can work out all the proper measurements of the particle, by using the Lorentz transformations and treating the particle as if it was moving in a straight line in flat spcae. This is a powerful concept and very convenient!
 
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  • #5
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Agreed that the proper frame is rotating and therefore an accelerating or non-inertial frame, but we can use the concept of the momentarily comoving inertial frame and the clock hypothesis to treat calculations in the accelerating frame exactly like an instantaneous inertial frame.
The point was that you can't use results that were derived for frames in relative linear
uniform motion.
 
  • #6
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The point was that you can't use results that were derived for frames in relative linear uniform motion.
The point is that the clock postulate says you can.

The result I obtained by using that postulate:

[tex] F = \frac{F '}{\gamma} = \frac{ma '}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]

is supported by experimental measurements of the force required to deflect an electron from a straight path in a cathode ray tube. They confirm that the deflection force required is greater than the Newtonian prediction by a factor of [itex]\gamma = 1/ \sqrt{1-v^2/c^2}[/itex]. It is also confirmed in hundreds of cyclotrons and particle accelerators all over the world.
 
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  • #7
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The point is that the clock postulate says you can.

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


[tex] F = \frac{F '}{\gamma} = \frac{ma '}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]

is supported by experimental measurements of the force required to deflect an electron from a straight path in a cathode ray tube.
No, it isn't. The electron deflection is obtained from solving a totally different equation:

[tex]\frac{d}{dt}(\gamma m_0 v)=[/tex]qvxB

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

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


[tex]\frac{d}{dt}(m_0 v)=q v x B[/tex]

Note that neither equation has nothing to do with any centrifugal force nor does it have anything to do with the way forces (centripetal or not) transform, so, I don't even see how your quote is relevant to the subject.
 
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  • #8
general relativity is needed for this problem. accelerated frame.
 
  • #9
Jorrie
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The clock postulate says nothing about attempting to use a formula derived for one case (linear relative motion between two frames with aligned axes) to a totally different situation (motion with a variable direction velocity between two frames that don't have aligned axes).
But surely one can set up an inertial frame where the center of the circle is static have an inertial observer momentarily comoving with the circling mass, as kev used. The clock postulate and all Lorentz transformations between the static an the comoving observer should hold. Why would what happens after that moment (variable direction) influence the momentary observations?
 
  • #10
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But surely one can set up an inertial frame where the center of the circle is static have an inertial observer momentarily comoving with the circling mass, as kev used.
This is not good enough. The transformations were derived for the case when the axes of the two inertial frames are parallel and when the relative motion between frames coincides with the common x-axis. Neither of these two conditions is satisfied.
You need to do the derivation correctly, by using the extension of the Lorentz transforms to rotating frames.


The clock postulate and all Lorentz transformations between the static an the comoving observer should hold.
The clock postulate has nothing to do with this exercise.
The Lorentz transforms for linear motion do not hold in the case of rotation, a different set is used.
 
  • #11
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[tex] F = \frac{F '}{\gamma} = \frac{ma '}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]

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

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

[tex]\frac{d}{dt}(\gamma m_0 v)=q v x B[/tex]

The LHS represents the force described as [tex]\frac {dp}{dt}[/tex] and the RHS is the Lorentz force.
The LHS of your relativistic equation is not "a totally different equation". I can obtain it from my equation showing it is the same. Starting with my equation:

[tex] F = \frac{F '}{\gamma} = \frac{ma '}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]

[tex] F = ma \gamma = \gamma m \frac{v^2}{r}[/tex]

Since [itex]v^2/r = dv/dt[/itex]

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

it follows that:

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

Q.E.D.


[tex]\frac{d}{dt}(\gamma m_0 v)=q v x B[/tex]

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

[tex]\frac{d}{dt}( m_0 v)=q v x B[/tex]
There is something wrong with RHS of your two eqautions:

If

[tex]\frac{d}{dt}( m_0 v)=q v x B[/tex]

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

[tex]\frac{d}{dt}(\gamma m_0 v)= (q v x B) \gamma [/tex]
 
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  • #12
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The LHS of your relativistic equation is not "a totally different equation". I can obtain it from my equation showing it is the same. Starting with my equation:

[tex] F = \frac{F '}{\gamma} = \frac{ma '}{\gamma} = m \frac{v^2}{r} \gamma = ma \gamma [/tex]

[tex] F = ma \gamma = \gamma m \frac{v^2}{r}[/tex]

Since [itex]v^2/r = dv/dt[/itex]

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

it follows that:

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

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



There is something wrong with RHS of your two eqautions:

If

[tex]\frac{d}{dt}( m_0 v)=q v x B[/tex]

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



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

[tex]\frac{d}{dt}(\gamma m_0 v)= (q v x B) \gamma [/tex]
Nope, there is nothing wrong, this is basic relativistic electrodynamics. There is no [tex]\gamma[/tex] in the RHS.
 
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  • #13
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The above has nothing to do with transformation of force between frames.
I obtained my final equation by using the clock postulate and doing a Lorentz transformation from the accelerating frame as it was an instantaneous inertial frame (i.e using the concept of a Momentarily Comoving Inertial Frame) and the result I get agrees with the equation you gave, so I must be doing something right.
 
  • #14
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Nope, there is nothing wrong, this is basic relativistic electrodynamics. There is no [tex]\gamma[/tex] in the RHS.
The rules of mathematics say that one of your two equations is wrong (unless you are saying [itex]\gamma[/itex] is a constant of numerical value 1, which does not agree with relativity).
 
  • #15
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I obtained my final equation by using the clock postulate and doing a Lorentz transformation from the accelerating frame as it was an instantaneous inertial frame (i.e using the concept of a Momentarily Comoving Inertial Frame) and the result I get agrees with the equation you gave, so I must be doing something right.
You are getting things "right" by accident. To do things truly right you need to have a rigorous derivation. The fact that your derivation lacks rigor is illustrated by your confusion in solving the cyclotron equations, yours are not even stated correctly (see above).
The subject of SR in rotating frames is quite interesting and deserves a lot of attention.
 
  • #16
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There is something wrong with RHS of your two eqautions:

If

[tex]\frac{d}{dt}( m_0 v)=q v x B[/tex]

is true...
It is true only for non-relativistic speeds, it isn't true for relativistic speeds. This is why, the LHS is different from what you wrote above.
It follows that [itex] q v x B[/itex] is also only true for non-relativistic speeds and that for relativistic speeeds you must use [itex] \gamma(q v x B)[/itex]

The subject of SR in rotating frames is quite interesting and deserves a lot of attention.
I am sure other members here will agree that I given the subject of SR in rotating frame a LOT of attention in other threads over the years.
 
  • #17
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It follows that [itex] q v x B[/itex] is also only true for non-relativistic speeds and that for relativistic speeeds you must use [itex] \gamma(q v x B)[/itex]
Nope, this is definitely not true. For a correct treatment, please see attachment #1 (LorentzForce) in my blog. Or, you can consult Griffith , I can't give you the exact page off the top of my head.
 
  • #18
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Nope, this is definitely not true. For a correct treatment, please see attachment #3 in my blog. Or, you can consult Griffith , I can't give you the exact page off the top of my head.
I think you will find that the root of our differences is that I am using coordinate time as measured in the lab in which the cathode ray tube is at res,t while you and Grffith are probably using the proper time of the electron.
 
  • #19
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I think you will find that the root of our differences is that I am using coordinate time as measured in the lab in which the cathode ray tube is at res,t while you and Grffith are probably using the proper time of the electron.
Nope. I am using coordinate time as well.
 
  • #20
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Nope. I am using coordinate time as well.
Actually, my suggestion that you are using proper time does not work. It seems that the Lorentz force devised by Lorentz in 1898 turns out to be relativistically correct even though it discovered before Einstein's published his paper on relativity in 1905. My statement that one of your two equations is incorrect still stands: Your statement would be more accurately stated as:
(Relativistic:)

[tex]\frac{d}{dt}(\gamma m_0 v) = q v x B[/tex]

(Newtonian)

[tex]\frac{d}{dt}( m_0 v) \ne q v x B [/tex]
although the Newtonian expression is approximately true for v << c. The Newtonian equation is only exact when v=0.

Anyway, none of this means that any of the equations I posted in #1 are wrong and all your equations demonstrate, is that relativistic equations approximate Newtonian equations at low velocities.
 
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  • #21
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Actually, my suggestion that you are using proper time does not work. It seems that the Lorentz force devised by Lorentz in 1898 turns out to be relativistically correct even though it discovered before Einstein's published his paper on relativity in 1905. My statement that one of your two equations is incorrect still stands:
This is very simple stuff, I don't understand why you have so much trouble with it.

Your statement would be more accurately stated as:
I suggest that you study Griffith. You can find this sort of stuff in a lot of books on electrodynamics, both classical and relativistic.

although the Newtonian expression is approximately true for v << c. The Newtonian equation is only exact when v=0.
The Newtonian expression always holds within the Newtonian domain. It is simply due to the fact that, in Newtonian physics [tex]F=d(m_0v)[/tex]. By definition.
 
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  • #22
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This is very simple stuff, I don't understand why you have so much trouble with it...
The reason I having trouble is that I do not play around with the equations of electromagnetics much. Here is one problem I am still having:

In your relativistic equation:

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

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

[tex]F = \frac{d}{dt} \gamma mv = q v B[/tex]

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

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

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

which when E = 0 reduces to:

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

The RHS of the Wikipedia equation differs from your equation by a factor of gamma. Maybe I am not the only one who is having trouble with this "simple stuff"?
 
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  • #23
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The reason I having trouble is that I do not play around with the equations of electromagnetics much. Here is one problem I am still having:

In your relativistic equation:

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

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

[tex]F = \frac{d}{dt} \gamma mv = q v B[/tex]

It can be seen that when v = c the LHS is infinite and the RHS remains finite indicating that your equation is wrong.
:lol: How can you set v=c when you are dealing with massive particles?
Have you made the effort to read the attachment in the blog?
 
  • #24
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:lol: How can you set v=c when you are dealing with massive particles?
Mathematically you can set v = c and the infinite result for the LHS indicates the impossibilty or uphysical nature of doing that. The RHS of your equation does not indicate that setting v = c is unphysical, indicating that the RHS of your equation is not a relativistically valid term.

Have you made the effort to read the attachment in the blog?
I have and it starts out with assuming the Lorentz force is relativistic with no proof or explanation why you think it is.

Have you read the Wikipedia article? Can you explain why their result differs from yours?
 
  • #25
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Mathematically you can set v = c and the infinite result for the LHS indicates the impossibilty or uphysical nature of doing that. The RHS of your equation does not indicate that setting v = c is unphysical, indicating that the RHS of your equation is not a relativistically valid term.
At this point I will recommend that you read pages 26-11 thru 26-13 in Feyman's "Lectures on Physics" Volume II.

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



I have and it starts out with assuming the Lorentz force is relativistic with no proof or explanation why you think it is.
Very simple, because this is the standard equation in the three-fource formalism in SR. Precisely equation 26.24 in Feynman.
You are mixing indiscriminately four-force formalism (from wiki) with the three-force formalism (from your and my posts) and you are confusing yourself in the process.


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

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

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

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