# How many independent components does the spin-tensor have?

In Loomis 2017, the equations of motion for a spinning particle with 4-momentum $P^{\alpha}$ and spin-tensor $S^{\alpha \beta}$ are considered.

It is stated in the paper that these equations of motion

give the time evolution of the four components of momentum, $P^{\alpha}$, and the six components of spin, $S^{\alpha \beta}$

Clearly in a 4D spacetime, $P^{\alpha}$ does have 4 components. My question is why does $S^{\alpha \beta}$ only have six components? I can see that the symmetry $S^{\alpha \beta} = S^{\beta \alpha}$ will reduce the number of independent components to 10, but I cannot see how to reduce this further to 6?

Thanks

• $S_{\alpha\beta}$ is antisymmetric. – Prahar Mitra Oct 9 '17 at 16:18
• Just write out the most general antisymmetric $4\times4$ matrix and count the number of independent components (I presume you already did this for a symmetric matrix since you reached the correct number 10 for that) – Prahar Mitra Oct 9 '17 at 16:52

The spin tensor is actually antisymmetric. For example, it can be related (at least in flat spacetime) to the spin vector $S^\mu$ as
$$S_{\mu\nu} = \frac12 \epsilon_{\mu\nu\alpha\beta}u^\alpha S^\beta.$$
If you doubt this, take a look at equation I.1a, where they contract the spin tensor with the Riemann tensor as $R_{\alpha\beta\mu\nu}S^{\mu\nu}$. Due to the antisymmetry of the Riemann tensor, this would vanish if the spin tensor were symmetric.