# Help me to understand this conversion (4-vectors)

$u^{\mu}$ - 4-velocity

$b^{\mu}$ - 4-vector of magnetic field

$u_{\mu}u^{\mu}=-1, \qquad u_{\mu}b^{\mu}=0$

$$u_{\beta}u^{\alpha}\nabla_{\alpha}b^{\beta}-u_{\beta}b^{\alpha}\nabla_{\alpha}u^{\beta}+\nabla_{\alpha}b^{\alpha}=0$$ I don't understand why this equation gives this $$u^{\alpha}u^{\beta}\nabla_{\alpha}b^{\beta}+\nabla_{\alpha}b^{\alpha}=0$$

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## migrated from math.stackexchange.comApr 6 '13 at 14:17

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This is because $u_\alpha \nabla_\beta u^\alpha =0$. To show this, just act with $\nabla_\beta$ on both sides of the equality $u_\alpha u^\alpha = -1$. You get $$u_\alpha \nabla_\beta u^\alpha + u^\alpha \nabla_\beta u_\alpha = 0$$ and thus $u_\alpha \nabla_\beta u^\alpha =0$ as promised.
You are correct, but perhaps use different indices (for consistency with the original question) and note the term $u_{\beta}b^{\alpha}\nabla_{\alpha}u^{\beta}$ vanishes because $u_{\beta}\nabla_{\alpha}u^{\beta}=0$? –  Alex Nelson Apr 6 '13 at 17:55
Exactly. Actually the names for the indicies does not really matter as long as your are consistent with yourself. Moreover, it is obvious that if $u_\alpha \nabla_\beta u^\alpha=0$, then of course $u_\beta \nabla_\alpha u^\beta=0$. –  Bru Apr 6 '13 at 18:11