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:


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.

  • 2
    $\begingroup$ I don't think it needs to be this complicated. You can just differentiate with respect to proper time, $(d/d\tau)(u\cdot u)=0$. You can just consider $u\cdot u$ to be a function of $\tau$, and none of the covariant derivative stuff is necessary. All you need is the product rule, and the product rule holds for any type of derivative -- that's essentially what we mean by a derivative operator, something that obeys the product rule. $\endgroup$
    – user4552
    Mar 31, 2018 at 2:04

2 Answers 2


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.

  • 1
    $\begingroup$ You could also note that the covariant derivative for higher rank tensors is defined so that the product rule holds, then this becomes much easier to do. $\endgroup$ Jun 28, 2018 at 13:17

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

$$ u^\mu a_\mu = -u^\mu a_\mu \Rightarrow 2 u^\mu a_\mu = 0 $$

  • $\begingroup$ Can you come up with a similar proof for the orthogonality of the position vector and the acceleration vector? (e.g. $x^\alpha a_\alpha$, where $u^\alpha = dx^\alpha/d\tau$) $\endgroup$
    – aRockStr
    Jan 16, 2020 at 4:50
  • $\begingroup$ @aRockStr position and acceleration don't have to be orthogonal. $\endgroup$ Apr 10, 2021 at 22:31

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