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In the book "relativity" by Vincenzo Barone, the relativistic angular momentum tensor is defined as:

$L^{ij}= x^ip^j - x^jp^i$

(With the indexes $i,j$ going from 0 to 3)

And the relativistic torque tensor as:

$N^{ij} = x^iF^j - x^jF^i$

Where $F^\mu$ is the four-force vector.

Afterwards it states the equations of motion for $\mathbf L$ as:

$\frac{dL^{ij}}{ds} = N^{ij}$

But if you derive the momentum with respect to the path element d$s$ you should end up with four terms (two for each term of $\mathbf L$) of which two are those of the torque. So how do you prove such a statement.

Mauricio
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Andrea
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1 Answers1

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I don't know if I understand your last sentence correctly. $ds$ is a scalar and $L^{ij}$ is a rank 2 tensor (or rather the spatial components of one). So if you take the derivative of $L^{ij}$ with respect to $ds$, you will get another rank 2 tensor (or rather its spatial components). The proof that that rank 2 tensor is $N^{ij}$ is rather trivial (basically just the product rule): $d/ds (L^{ij}) = (dx^i/ds) p^j + x^i (dp^j/ds) - (dx^j/ds) p^i - x^j (dp^i/ds) $ $= u^i p^j + x^i F^j - u^j p^i - x^j F^i = x^i F^j - x^j F^i = N^{ij}$, since $u^i p^j - u^j p^i = u^i m u^j - u^j m u^i = 0$ by definition of $(p^i)$.

Tarik
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