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Energy for moving mass up a centrifugal gravity barrier

  1. Feb 28, 2014 #1
    Hi,

    I have been pondering on this -

    what is the energy required to move a point mass m0 from a radial position R to the center of a rotating frame, if the linear velocity v of the point mass at position R is relativistic (close to c).

    For an observer in the rotating frame, it should seem as if m is climbing against the centrifugal gravity. In the classical limit, the answer is [itex]\frac{mv^2}{2}[/itex]

    m0 is the rest mass


    Thanks!
     
    Last edited: Feb 28, 2014
  2. jcsd
  3. Feb 28, 2014 #2
    It would be the equivalent of stopping the point mass.
    If you know the apparent mass of the particle as it is moving (ma), then the energy to decelerate it would be (m0-ma)[itex]c^2[/itex].
    That would be a negative value, indicating that you are removing energy from the system.
     
  4. Feb 28, 2014 #3

    WannabeNewton

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    Just use the relativistic centrifugal force that appears in the rotating frame to calculate the work done relative to this frame in order to move a point mass from some radius to the origin of the frame along a desired path e.g. a radial line.
     
  5. Feb 28, 2014 #4

    PeterDonis

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    Don't you have to also include work done against the Coriolis force to keep the mass on the radial line?
     
  6. Feb 28, 2014 #5

    WannabeNewton

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    I don't think so because the Coriolis force is always perpendicular to the 3-velocity vector ##\vec{v}## of the path taken between two end points in the rotating frame so ##\vec{F}_{\text{Coriolis}}\cdot \vec{dr} = 0## i.e. it does no work.

    But now that you mention it, there is a relativistic term that appears in non-inertial frames to second order in ##\vec{v}## of the form ##\vec{F}_{\text{rel}} = 2m \vec{v} (\vec{a}\cdot \vec{v})## which becomes negligible for ##||\vec{v}|| << c## but will have to be taken into account for ##||\vec{v}|| \sim c##; here ##\vec{a}## corresponds to the acceleration of the frame itself so ##\vec{F}_{\text{rel}}## is a correction to the inertial acceleration. It won't matter in this case because the rotating frame is non-accelerating so ##\vec{F}_{\text{rel}} = 0## but for more general settings it would be relevant.

    See MTW exercise 6.8.
     
    Last edited: Feb 28, 2014
  7. Feb 28, 2014 #6
    I gather you meant the rotating frame moves with constant angular velocity?
     
  8. Feb 28, 2014 #7

    WannabeNewton

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    I'm assuming you're talking about starting with an inertial frame and rotating it about a fixed axis with some constant angular velocity ##\vec{\omega}## or equivalently performing the transformation ##\phi \rightarrow \phi - \omega t## of the cylindrical coordinates in the inertial frame.

    In such a case the rotating frame so achieved is non-accelerating in the sense that an observer fixed at its origin has vanishing acceleration. This can be made more mathematically precise but it's not really important at the moment.

    The main point is you still have a centrifugal potential in the rotating frame that gives rise to a centrifugal acceleration ##a_c = \gamma^2 \omega^2 r## where ##\gamma^2 = (1 - \omega^2 r^2)^{-1}##.
     
  9. Mar 1, 2014 #8
    Yes, the rotation is uniform (makes things easier :biggrin:).

    This is interesting, that you can actually think of the observer as rotating only, but not experiencing any acceleration because he is at the center. But he can still measure the centrifugal gravity, correct?
    If I understand you correctly, additional relativistic effects are avoided by having the rotating observer at the center of the frame.

    Then I guess the work to bring ##m## from ##R## to the center will be

    ##\int_R^0 m(r) a_c(r) dr= \int_R^0 \frac{m_0}{\sqrt(1-\frac{\omega^2r^2}{c^2})}\frac{\omega^2 r}{(1-\frac{\omega^2r^2}{c^2})}dr=\int_R^0 \frac{m_0 \omega^2 r dr}{(1-\frac{\omega^2r^2}{c^2})^{3/2}}=\frac{m_0c^2(\sqrt{c^2 - \omega^2 R^2}-c)}{\sqrt{c^2 - \omega^2 R^2}}##

    which diverges, because ##\omega R=c##. This disagrees with post #2. Either ##a_c## is wrong or I did something wrong in the derivation.
     
  10. Mar 1, 2014 #9

    WannabeNewton

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    If I'm floating in free space with my arms extended out and a pellet traveling at a constant speed hits one of my palms in a completely elastic collision then from conservation of momentum and angular momentum I will gain some constant speed and I will also start rotating. My acceleration is still zero but my rest frame is now rotating.

    Yes.

    Yes.

    I don't see why you're using ##\omega R = c##. You can't have the particle start out at the light barrier that's physically impossible.

    The expression can be rewritten as ##m_0 c^2 (1 - \gamma)## which is exactly what Scott had. Note that the expression is negative, as we would expect.
     
  11. Mar 1, 2014 #10
    I meant ##\omega R \rightarrow c##
     
  12. Mar 1, 2014 #11

    WannabeNewton

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    Well that clearly diverges because it takes an infinite amount of energy to bring a particle to the light barrier.
     
  13. Mar 2, 2014 #12
    Makes sense... Ah, and in the classical limit ##\omega R = v << c##, then

    ##\gamma = \frac{1}{\sqrt{1-\frac{\omega^2 r^2}{c^2}}} \approx 1+\frac{1}{2}\frac{\omega^2 r^2}{c^2}## then

    ##m_0c^2(\gamma-1) \approx m_0 c^2 (1+\frac{1}{2}\frac{\omega^2 r^2}{c^2} -1) = \frac{m_0 \omega^2 r^2}{2}##

    This settles it then :smile:
    Thank you to all who contributed!
     
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