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I have edited this question because I don't think that the related post answers my question fully. It refers to Noether's theorem but I would like an explicit illustration in an easier fashion: The angular momentum tensor is defined: $$L^{\mu\nu}~=~x^\mu p^\nu-x^\nu p^\mu$$

I would like to show that if a particle is not acted on by an external force that the angular momentum is conserved. Despite it being a direct consequence of the theorem I want to compute it in an inferior way, would this be sufficient?

Without loss of generality take the rest frame of the particle

$$\frac{dx^\mu}{d\tau}=(c,\underline{0})$$

Since the particle is not acted on by an external force, we have

$$\frac{dp^\mu}{d\tau}=\left(\frac{1}{c}\frac{dE}{ d\tau},\underline{0}\right)$$

$$L^{\mu\nu}L_{\mu\nu}=\left( 2 (x\cdot x) (p\cdot p) - 2 (x \cdot p)^2 \right)$$

Can I show the conservation by $$\frac{d(L^2)}{d\tau}=0$$

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shilov
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    In 3d, angular momentum is also 'really' an antisymmetric tensor, but in 3d you have the special property that vectors and antisymm. tensors are the same, i.e. you can encode a vector in such a tensor and vice versa through the relations $v^i = \epsilon^{ijk}L_{jk}$ or $L_{ij} = \epsilon_{ijk} p^k.$ – Vibert Apr 20 '13 at 12:10
  • Seems to me that in your chosen frame $L^{\mu\nu}\equiv 0$. – firtree May 11 '13 at 12:51
  • I can see that - I think I have messed that up – shilov May 11 '13 at 12:56

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