3-angular momentum : independent of pivot?

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SUMMARY

The discussion focuses on proving that the time derivative of angular momentum, dL/dt, equals zero, and establishing the relationship between the spatial components and the three-dimensional angular momentum vector. The solution to part (i) demonstrates that the derivative of the angular momentum tensor L^{ab} with respect to proper time τ results in zero, confirming its conservation. In part (ii), the angular momentum L_i is expressed in terms of position x_j and momentum p_k, leading to a successful resolution of the problem. The independence of the three-dimensional angular momentum vector from the pivot point is also confirmed.

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Homework Statement


[/B]
(i) Prove that dL/dt = 0
(ii) Find the relation between space part and 3-angular momentum vector
(iii)Show that 3-angularmomentum vector is independent of pivot
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Homework Equations

The Attempt at a Solution


[/B]
I'm not sure what part (iii) is trying to get at, but I think I've solved (i) and (ii):

Part(i)
[tex]\frac{\partial L^{ab}}{\partial \tau} = P^b \frac{\partial X^a}{\partial \tau} - P^a \frac{\partial X^b}{\partial \tau}[/tex]

[tex]= \frac{P^bP^a}{m_0} - \frac{P^aP^b}{m_0} =0[/tex]

Part(ii)
[tex]L_i = x_jp_k - x_kp_j[/tex]
 
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Solved it, thanks!
 

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