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Relativistic centripetal force

  1. Apr 22, 2010 #1
    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.
     
    Last edited: Apr 23, 2010
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  3. Apr 22, 2010 #2

    Jorrie

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    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.

    Interesting observation, kev!
     
    Last edited by a moderator: Apr 25, 2017
  4. Apr 22, 2010 #3
    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.
     
  5. Apr 22, 2010 #4
    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]:
    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]:
    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!
     
    Last edited by a moderator: May 4, 2017
  6. Apr 22, 2010 #5
    The point was that you can't use results that were derived for frames in relative linear
    uniform motion.
     
  7. Apr 22, 2010 #6
    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.
     
    Last edited: Apr 23, 2010
  8. Apr 23, 2010 #7
    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.


    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.
     
    Last edited: Apr 23, 2010
  9. Apr 23, 2010 #8
    general relativity is needed for this problem. accelerated frame.
     
  10. Apr 23, 2010 #9

    Jorrie

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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. 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?
     
  11. Apr 23, 2010 #10
    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 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.
     
  12. Apr 23, 2010 #11
    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.


    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]
     
    Last edited by a moderator: Apr 25, 2017
  13. Apr 23, 2010 #12
    The above has nothing to do with transformation of force between frames.



    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.



    Nope, there is nothing wrong, this is basic relativistic electrodynamics. There is no [tex]\gamma[/tex] in the RHS.
     
    Last edited by a moderator: Apr 25, 2017
  14. Apr 23, 2010 #13
    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.
     
  15. Apr 23, 2010 #14
    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).
     
  16. Apr 23, 2010 #15
    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.
     
  17. Apr 23, 2010 #16
    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]

    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.
     
  18. Apr 23, 2010 #17
    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.
     
  19. Apr 23, 2010 #18
    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.
     
  20. Apr 23, 2010 #19
    Nope. I am using coordinate time as well.
     
  21. Apr 23, 2010 #20
    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:
    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.
     
    Last edited: Apr 23, 2010
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