Showing Four Acceleration and Four Velocity are Perpendicular I want to show that in general, $\vec{a}\cdot\vec{u}=0$, where $\vec{u}$ is the four-velocity and $\vec{a}$ is the four-acceleration. The four acceleration is defined as $\vec{a}=\nabla_{\vec{u}}\vec{u}$, or in component form as $a^{\alpha}=u^{\beta}u^{\alpha}_{\ \ \ ;\beta}$, that is the covariant derivative of $\vec{u}$ along the direction of $\vec{u}$.
Note also that the covariant derivative is defined component wise as;
$$(\nabla\vec{V})^{\alpha}_{\beta}=V^{\alpha}_{\ \ \ ;\beta}=\frac{\partial V^{\alpha}}{\partial x^{\beta}}+\Gamma^{\alpha}_{\ \ \ \lambda\beta}V^{\lambda}$$
To do so, I want to take the covariant derivative in the direction of $\vec{u}$, of both sides of the defining expression of the four-velocity: $\vec{u}\cdot\vec{u}=-1$
I know that the covariant derivative of a scalar is simply the partial derivative, so $\nabla(-1)=0$, but I'm not sure how to take the covariant derivative of the left hand side. I tried to apply it component wise as follows, but got very confused:
$$\left(\nabla_{\vec{u}}(\vec{u}\cdot\vec{u})\right)^{\alpha}=u^{\alpha}\left(\nabla(u_{\beta}u^{\beta})\right)$$
This is obviously wrong because the expression shouldnt have a free index, but I can't see any other way to apply the definition of covariant derivative.
I also feel like this will involve some kind product rule for covariant derivatives, but I was unable to derive or find such a rule.
I know this is really simple but the notation is just really giving me a hard time, and I'd really appreciate some clarification.
 A: I think I've proven the product rule I was hoping for.
Consider arbitrary four vectors $\vec{a}$, and $\vec{b}$, abd let $\phi=\vec{a}\cdot\vec{b}=a_{\alpha}b^{\alpha}$. Then, as $\phi$ ios a scalar, it's covariant derivative with respect to $x^{\beta}$, is simply the partial derivative:
\begin{align*}
\nabla(\vec{a}\cdot\vec{b})_{\beta}&=\frac{\partial\phi}{\partial x^{\beta}}=a_{\alpha}\frac{\partial b^{\alpha}}{\partial x^{\beta}}+b^{\alpha}\frac{\partial 
a_{\alpha}}{\partial x^{\beta}}\\
&=a_{\alpha}\left( \frac{\partial b^{\alpha}}{\partial x^{\beta}} +\Gamma^{\alpha}_{\lambda\beta}b^{\lambda}\right)+b^{\alpha}\left( \frac{\partial a_{\alpha}}{\partial x^{\beta}}-\Gamma^{\lambda}_{\alpha\beta}a_{\lambda}\right)+\left(a_{\lambda}b^{\alpha}\Gamma^{\lambda}_{\alpha\beta}-a_{\alpha}v^{\lambda}\Gamma^{\alpha}_{\lambda\beta}\right)\\
&=a_{\alpha}b^{\alpha}_{;\beta}+b^{\alpha}a_{\alpha ; \beta}\\
&=a_{\alpha}b^{\alpha}_{;\beta}+b_{\alpha}a^{\alpha}_{; \beta}\\
&=\vec{a}\cdot \nabla(\vec{b})_{\beta}+\vec{b}\cdot \nabla(\vec{a})_{\beta}\\
\end{align*}
Where we got rid of the $\left(a_{\lambda}b^{\alpha}\Gamma^{\lambda}_{\alpha\beta}-a_{\alpha}v^{\lambda}\Gamma^{\alpha}_{\lambda\beta}\right)$ term by cyclically permuting $\lambda$ and $\alpha$.
With this product rule in place, I think I've solved my initial problem, but would still really appreciate verification. I proceeded as follows:
$$\nabla(\vec{u}\cdot\vec{u})_{\alpha}=2\vec{u}\cdot\nabla(\vec{u})_{\alpha}=2u_{\beta}u^{\beta}_{\ \ \ ;\alpha}$$
Then to take this covariant derivative in the $\vec{u}$ direction, we simply contract with the components of $\vec{u}$ to give:
$$\nabla_{\vec{u}}(\vec{u}\cdot\vec{u})=u^{\alpha}\nabla(\vec{u}\cdot\vec{u})_{\alpha}=2u_{\beta}u^{\alpha}u^{\beta}_{\ \ \ ;\alpha}=2u_{\beta}a^{\beta}=2\vec{a}\cdot\vec{u}$$
So, $2\vec{a}\cdot\vec{b}=\nabla_{\vec{u}}(\vec{u}\cdot\vec{u})=\nabla_{\vec{u}}(-1)=0$, as required.
A: I find it more simpler written:
Let $ a_\mu = u^\sigma \nabla_\sigma u_\mu$
So: 
$$ u^\mu a_\mu = u^\mu u^\sigma \nabla_\sigma u_\mu = u^\sigma ( \nabla_\sigma(u^\mu u_\mu) - u_\mu \nabla_\sigma u^\mu )$$
We want $u^\mu u_\mu = -1$ and $u^\mu = g^{\mu\alpha}u_\alpha$ and $[\nabla_\sigma, g^{\alpha\beta}] = 0$
Thus $\nabla_\sigma(u^\mu u_\mu) = 0$, $u_\mu \nabla_\sigma u^\mu = u^\mu \nabla_\sigma u_\mu $, thus
$$ u^\mu a_\mu = -u^\mu a_\mu \Rightarrow 2 u^\mu a_\mu = 0 $$
