# 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.

• 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.
– user4552
Mar 31, 2018 at 2:04

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.

• 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. 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

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

• 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 $u^\alpha = dx^\alpha/d\tau$) Jan 16, 2020 at 4:50
• @aRockStr position and acceleration don't have to be orthogonal. Apr 10, 2021 at 22:31