Is Angular Momentum also conserved in 3-dimensions.

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Angular momentum is conserved in three dimensions, similar to linear momentum, as long as no external torques are applied. The conservation can be understood through the angular momentum vector, which has components that are conserved in a closed system. The Lagrangian formulation demonstrates this conservation, as it remains invariant under 3D rotations, indicating that the physics does not change regardless of the perspective. The discussion emphasizes that the invariance of the metric under rotations supports the conservation of angular momentum. Thus, in a three-dimensional universe, angular momentum conservation is a fundamental principle.
Michio Cuckoo
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We know that a 3D universe like ours, linear momentum is conserved in 3 dimesions,

Up-Down
Left-Right
Forward-Backward


But is Angular Momentum also conserved in three dimensions?

Clockwise-Anticlockwise, but repeated 3 times.

Imagine three circles, all mutually perpendicular.
 
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Yes, there is an angular momentum vector whose magnitude is ##I \omega## for a rigid body or mvr for a point-mass going along a circular path, and whose direction is along the axis of rotation. Its three components are each conserved, in a system with no external torques, just like the three components of linear momentum are conserved, in a system with no external forces.
 
Yet another way, is to look at the Lagrangian, a warehouse of information for the system. The Lagrangian for a free particle is L=\frac{1}{2}m\, g_{ij}u_i u_j. Now remember that the metric is invariant under rotations, and transforms as g_{ij}=R_{ik}R_{jl}g_{kl} and vectors as a_{i}'=R_{ij}a_j. You can see that the Lagrangian is invariant under 3D rotations
<br /> L&#039;=\frac{1}{2}m g_{ij} u&#039;_i u&#039;_j=\frac{1}{2}mg_{ij}R_{ik}u_k R_{jl}u_l=\frac{1}{2}m(g_{ij}R_{ik}R_{jl})u_k u_l=\frac{1}{2}m\, g_{kl}u_k u_l<br />
which is the same thing we started with in a different frame of reference. Then if the system is invariant under 3D rotations, angular momentum is conserved, since what ever happens that this Lagrangian describes, could have happened from any rotated perspective, so rotating the system won't change the physics. But angular momentum is all about what happens when "going around". It's got to be conserved.
 
jfy4 said:
Yet another way, is to look at the Lagrangian, a warehouse of information for the system. The Lagrangian for a free particle is L=\frac{1}{2}m\, g_{ij}u_i u_j. Now remember that the metric is invariant under rotations, and transforms as g_{ij}=R_{ik}R_{jl}g_{kl} and vectors as a_{i}&#039;=R_{ij}a_j. You can see that the Lagrangian is invariant under 3D rotations
<br /> L&#039;=\frac{1}{2}m g_{ij} u&#039;_i u&#039;_j=\frac{1}{2}mg_{ij}R_{ik}u_k R_{jl}u_l=\frac{1}{2}m(g_{ij}R_{ik}R_{jl})u_k u_l=\frac{1}{2}m\, g_{kl}u_k u_l<br />
which is the same thing we started with in a different frame of reference. Then if the system is invariant under 3D rotations, angular momentum is conserved, since what ever happens that this Lagrangian describes, could have happened from any rotated perspective, so rotating the system won't change the physics. But angular momentum is all about what happens when "going around". It's got to be conserved.

What metric are you referring to here, the space-time metric as in GR?
 
just flat space, no time. so \delta_{ij}=g_{ij}.
 
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