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| Tags: angular, energy, momentum, relativity, rotational, special |
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#1
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At high school we learn that there is a correspondence:
- linear momentum - angular momentum - mass - moment of inertia - velocity - angular velocity etc. For relativistic motion we have E^2 - p^2c^2 = m^2 c^4. and p = v E/c^2 What are the corresponding expressions using angular momentum, moment of inertia, and rotational energy? Do they exist at all? Thanks for any help! Tom Helmond |
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#2
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Subject: Rotational energy and angular momentum in special relativity?
From: (Tom Helmond) Date: 8/16/03 12:54 AM US Mountain Standard Time Message-id: At high school we learn that there is a correspondence: - linear momentum - angular momentum - mass - moment of inertia - velocity - angular velocity etc. For relativistic motion we have E^2 - p^2c^2 = m^2 c^4. and p = v E/c^2 What are the corresponding expressions using angular momentum, moment of inertia, and rotational energy? Do they exist at all? Consider cylindrical coordinates ( ct, z, r, theta ) and flat spacetime. The conserved angular momentum is L = (E/c^2)(r^2)(dtheta/dt) = r x p which you can say is L = I*omega if you take I to be I = (E/c^2)(r^2) You can't break up the energy itself into translational and rotational parts like you can in Newtonian mechanics where you can do so because the kinetic energy is proportional to v^2, but equations of motion are often written in terms of the conserved "kinetic energy parameter" T instead of the energy E. T = (p^2)/2m T = gamma*[(1/2)(E/c^2)(dz/dt)^2 + (1/2)(E/c^2)(dr/dt)^2 + (1/2)(E/c^2)(r^2)(dtheta/dt)^2] If you say I is (E/c^2)(r^2) as above then the rotational kinetic energy parameter is T = (1/2)gamma*I*omega^2 |
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#3
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L = (E/c^2)(r^2)(dtheta/dt) = r x p
should be L = (E/c^2)(r^2)(dtheta/dt) = |r x p| with mag bars as I'm only discussing a magnitude. |
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#4
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