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Rotational energy and angular momentum in special relativity?



 
 
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  #1  
Old August 16th 03 posted to sci.physics.relativity
Tom Helmond
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Posts: 26
Default Rotational energy and angular momentum in special relativity?

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  
Old August 16th 03 posted to sci.physics.relativity
WaiteDavid137
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Posts: 574
Default Rotational energy and angular momentum in special relativity?

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

  #3  
Old August 16th 03 posted to sci.physics.relativity
WaiteDavid137
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Posts: 574
Default Rotational energy and angular momentum in special relativity?

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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