Spinning Moving Objects: Angular and Linear Momentum

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In summary: This is the Schwarzschild mass, which is the maximum mass an object can have. If an object has L/m=1, Q/m=1, then it has the Schwarzschild mass which is 3*M.
  • #1
alvarogz
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If an "Object A" spins at nearly "c", and this object also is moving at any given posible speed. What happen with an "Object B" on the surface of "Object A" .I asume that there's a mechanism that "fix" this relation between angular and linear momentum to conserve the speed limit of light, but I can't realize what is the mechanism to get this explanation.

Best regards big thinkers,

AGZ
 
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  • #2
alvarogz said:
If an "Object A" spins at nearly "c", This makes no sense. An object spins with angular velocity which may be measured in "radians per second", not a velocity which may be measured in "meters per second". You can mark any single point on object A, at distance r from the axis of rotation and calculate that it is moving at speed [itex]r\omega[/itex] where [itex]\omega[/itex] is the angular velocity.

and this object also is moving at any given posible speed. What happen with an "Object B" on the surface of "Object A" .I asume that there's a mechanism that "fix" this relation between angular and linear momentum to conserve the speed limit of light, but I can't realize what is the mechanism to get this explanation.
The sum of two speed, u and v, is given by
[tex]\dfrac{u+ v}{1+ \frac{uv}{c^2}}[/tex]

If you look at that closely, you will see that will never be larger than c. For example, if u= v= .9c, the "sum" of the speeds is not any where near .9c+ .9c= 1.8c, it is
[tex]\dfrac{.9c+ .9c}{1+ \frac{(.9c)(.9c)}{c^2}}= \dfrac{1.8c}{1+ .81}= \dfrac{1.8}{1.81}c= 0.9945c[/tex]

Best regards big thinkers,

AGZ
And us small thinkers, too!
 
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  • #3
HallsofIvy said:
alvarogz said:
If an "Object A" spins at nearly "c", This makes no sense. An object spins with angular velocity which may be measured in "radians per second", not a velocity which may be measured in "meters per second". You can mark any single point on object A, at distance r from the axis of rotation and calculate that it is moving at speed [itex]r\omega[/itex] where [itex]\omega[/itex] is the angular velocity.

I haven't considered this. Thank you, now this has more sense to me.
 
  • #4
alvarogz said:
If an "Object A" spins at nearly "c", and this object also is moving at any given posible speed. What happen with an "Object B" on the surface of "Object A" .I asume that there's a mechanism that "fix" this relation between angular and linear momentum to conserve the speed limit of light, but I can't realize what is the mechanism to get this explanation.

Best regards big thinkers,

AGZ

In Relativity theory energy per unit mass [E/m], momentum per unit mass [p/m], and angular momentum per unit mass [L/m] are constants of the motion over an objects natural path [geodesic]. In the Kerr metric the angular momentum per unit mass is the rotation parameter [a]. In the appropriate metrics the angular momentum per unit mass and charge per unit mass sum to the Schwarzschild mass [what the mass would be if the object was spherically symmetric, non rotating, no charge]. When L/m=1, Q/m=1 [the extremal case] the mass of the system is 3 times the Schwarzschild mass.
 
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  • #5
Brucep said:
In Relativity theory energy per unit mass [E/m], momentum per unit mass [p/m], and angular momentum per unit mass [L/m] are constants of the motion over an objects natural path [geodesic]. In the Kerr metric the angular momentum per unit mass is the rotation parameter [a]. In the appropriate metrics the angular momentum per unit mass and charge per unit mass sum to the Schwarzschild mass [what the mass would be if the object was spherically symmetric, non rotating, no charge]. When L/m=1, Q/m=1 [the extremal case] the mass of the system is 3 times the Schwarzschild mass.

You are confusing a lot of different things here and this is not at all relevant to the OP's original question which has already been answered.
 
  • #6
WannabeNewton said:
You are confusing a lot of different things here and this is not at all relevant to the OP's original question which has already been answered.

Most likely true. I just wanted to add some information about how those quantities are conserved since he seemed confused about that in his original post. Maybe you learned something?
 
  • #7
The angular momentum parameter ##a## of the Kerr metric is the conserved angular momentum of the space-time, not the conserved angular momentum per unit mass along a geodesic in Kerr space-time. This is the main thing you confused in your post.
 
  • #8
WannabeNewton said:
The angular momentum parameter ##a## of the Kerr metric is the conserved angular momentum of the space-time, not the conserved angular momentum per unit mass along a geodesic in Kerr space-time. This is the main thing you confused in your post.

I didn't confuse anything. To start with this is the constant of motion for all geodesics in the Kerr geometry.

L/m=R^2 (dphi/dTau) - (2M^2/r)dt/dTau.

There's only one. So I didn't confuse it with anything else. All I did was give a simple explanation for components in the metric. It's interesting and worth knowing when you're discussing a rotation parameter going to the limit 1. Whatever.
 
  • #9
Your exact statement was "In Relativity theory...angular momentum per unit mass [L/m] are constants of the motion over an objects natural path [geodesic]. In the Kerr metric the angular momentum per unit mass is the rotation parameter [a]." which is wrong for the reason stated in post #7. The rotation parameter ##a = \frac{J}{M}## of Kerr space-time is the reduced angular momentum of Kerr space-time obtained from ##J = \int _{S^{2}_{\infty}}\epsilon_{abcd}\nabla^c \psi^d## where ##\psi## is the axial killing field of Kerr space-time. The reduced angular momentum of an observer with 4-velocity ##u^{a}## on the other hand is gotten from ##L = u_{a}\psi^{a}##. You are confusing these two things, simple as that.

Anyways like I noted this has absolutely nothing to do with the OP's original question.
 

1. What is the difference between angular momentum and linear momentum?

Angular momentum is a measure of an object's rotational motion, while linear momentum is a measure of an object's straight-line motion. Angular momentum is dependent on an object's moment of inertia and its angular velocity, while linear momentum is dependent on an object's mass and velocity.

2. How is angular momentum conserved in spinning objects?

Angular momentum is conserved in spinning objects because of the law of conservation of angular momentum. This law states that the total angular momentum of a system remains constant, unless acted upon by an external torque.

3. Can an object have both angular and linear momentum?

Yes, an object can have both angular and linear momentum. For example, a spinning top has both angular momentum due to its rotation and linear momentum due to its translation as it moves across a surface.

4. How does the distribution of mass affect the angular momentum of an object?

The distribution of mass affects the angular momentum of an object by influencing its moment of inertia. Objects with a larger moment of inertia will have a greater resistance to changes in their rotational motion, resulting in a higher angular momentum.

5. Is there a relationship between angular and linear momentum?

Yes, there is a relationship between angular and linear momentum. This relationship is described by the equation L = Iω, where L is angular momentum, I is moment of inertia, and ω is angular velocity. This equation shows that as angular momentum increases, so does linear momentum, and vice versa.

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