- #51
yuiop
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Analysing the relativistic angular velocity and momentum of a individual particle at any given instant on its cycle can be very complex. The gory details involving a second order antisymmetric tensor and a time dependent quantity called the dynamic mass moment are given in this Wikipedia article. However, if we look at the bigger picture in terms of complete revolutions, then considerable simplifications can be made.
For a flywheel with angular velocity ##\omega_0## as measured in a reference frame in which the rotating flywheel is linearly at rest, then the angular velocity as measured in any other frame with relative motion v is ##\omega = \omega_0/\gamma(v)##. This is true no matter what the orientation of the flywheel is, because ##\omega## is proportional to 1/dt and dt is a scalar. Put another way angular velocity can be expressed in terms of rotations per unit time, so that ##\omega =2\pi r/dt##. Under a Lorentz transformation ##\omega' = 2\pi r/dt'##. More generally, if the flywheel as a whole has linear motion with a velocity component u parallel to the x axis, then after a Lorentz boost of v in the x direction, the transformed angular velocity is given by:
##\omega ' = \omega \frac{\sqrt{1-v^2}}{(1-vu)} = \frac{2 \pi r}{dt}\frac{\sqrt{1-v^2}}{(1-vu)}##
Any component of the linear motion of the flywheel that is not parallel to the direction of the boost has no effect on how the angular velocity transforms. For completeness we can represent the above in the form of a four-position vector as ##(dt, dx, dy, dz) = (2\pi r/\omega, dx, dy, dz)## where the linear 3 velocity of the flywheel is ##(u_x,u_y,u_z) = (dx/dt, dy/dt, dz/dt)##.
Now for the angular momentum. The linear version of momentum is given by ##p = m_0 v \gamma(v)##. The angular momentum analogue is ##L = I\omega_0\gamma(v)## where I is the moment of inertia and ##\omega_0## is the angular velocity in the irf where the flywheel has no linear motion. For a simple flywheel with the bulk of its rest mass ##m_0## at the rim, ##I=m_0 r^2##. When the flywheel is rotating, ##I = m_0 \gamma(\omega_0) r^2 = m_{\omega}r^2## where ##m_\omega## represents the total effective mass of the rotating flywheel in an inertial reference frame where it is linearly at rest. This means that ##m_{\omega}## is the new effective rest mass of the rotating flywheel and so ##m_{\omega}## is invariant under a Lorentz transformation. Putting this together we have ##L= m_{\omega}r^2\omega \gamma(v)##. As mentioned above ##\omega =\omega_0/\gamma(v)## so we end up with ##L = m_{\omega}r^2\omega_0## and since all the components are Lorentz invariant, angular momentum must be Lorentz invariant and L=L' for any orientation of the flywheel rotation axis.
I am not totally sure of the above analysis so any criticisms and corrections are welcome.
For a flywheel with angular velocity ##\omega_0## as measured in a reference frame in which the rotating flywheel is linearly at rest, then the angular velocity as measured in any other frame with relative motion v is ##\omega = \omega_0/\gamma(v)##. This is true no matter what the orientation of the flywheel is, because ##\omega## is proportional to 1/dt and dt is a scalar. Put another way angular velocity can be expressed in terms of rotations per unit time, so that ##\omega =2\pi r/dt##. Under a Lorentz transformation ##\omega' = 2\pi r/dt'##. More generally, if the flywheel as a whole has linear motion with a velocity component u parallel to the x axis, then after a Lorentz boost of v in the x direction, the transformed angular velocity is given by:
##\omega ' = \omega \frac{\sqrt{1-v^2}}{(1-vu)} = \frac{2 \pi r}{dt}\frac{\sqrt{1-v^2}}{(1-vu)}##
Any component of the linear motion of the flywheel that is not parallel to the direction of the boost has no effect on how the angular velocity transforms. For completeness we can represent the above in the form of a four-position vector as ##(dt, dx, dy, dz) = (2\pi r/\omega, dx, dy, dz)## where the linear 3 velocity of the flywheel is ##(u_x,u_y,u_z) = (dx/dt, dy/dt, dz/dt)##.
Now for the angular momentum. The linear version of momentum is given by ##p = m_0 v \gamma(v)##. The angular momentum analogue is ##L = I\omega_0\gamma(v)## where I is the moment of inertia and ##\omega_0## is the angular velocity in the irf where the flywheel has no linear motion. For a simple flywheel with the bulk of its rest mass ##m_0## at the rim, ##I=m_0 r^2##. When the flywheel is rotating, ##I = m_0 \gamma(\omega_0) r^2 = m_{\omega}r^2## where ##m_\omega## represents the total effective mass of the rotating flywheel in an inertial reference frame where it is linearly at rest. This means that ##m_{\omega}## is the new effective rest mass of the rotating flywheel and so ##m_{\omega}## is invariant under a Lorentz transformation. Putting this together we have ##L= m_{\omega}r^2\omega \gamma(v)##. As mentioned above ##\omega =\omega_0/\gamma(v)## so we end up with ##L = m_{\omega}r^2\omega_0## and since all the components are Lorentz invariant, angular momentum must be Lorentz invariant and L=L' for any orientation of the flywheel rotation axis.
I am not totally sure of the above analysis so any criticisms and corrections are welcome.
We have a paradox here! Compression wave analysis says one thing, common sense about transitory momentum change says another thing.
