# Calculating the proper acceleration of an observer along a scalar field

I'm trying to go through the derivation of various quantities from this paper by Madsen in 1988 (here, but unfortunately you have to log in), and I've been having some problems replicating the quantities the author found.

The paper considers a scalar field $\phi$ and an observer velocity (which is timelike), given by $u^a = \frac{\partial^a \phi}{\sqrt{-\partial^e \phi \partial_e \phi}}$ (I'm working with the convention that $u^au_a = -1$ and is timelike, unlike the author). Then, the author defines an acceleration, given by $\dot{u}^a = u^b \nabla_b u_a$. I'm having difficulty calculating this quantity.

When I take the covariant derivative, I have to apply the chain rule on $u^a$, correct? If so, I need to calculate a term that looks like $\nabla_b (-\partial^e \phi \partial_e \phi)^{-1/2}$. I'm a bit confused as to how I calculate this. On the one hand, it's a scalar, so this should reduce to a partial derivative. On the other hand, if I break it up with the chain rule, than they become pieces that aren't scalars. I think I'm missing just a small aspect to this calculation, and I'm hoping someone could point this out to me.

For reference, the author lists the solution as:

$$\dot{u}_a = ((-\partial^e \phi \partial_e \phi)^{-2} \left((-\partial^u \phi \partial_u \phi) \partial^b \phi \nabla_b \nabla_a \phi - \partial^b \phi \partial^n \phi \nabla_b \nabla_n \phi \partial_a \phi \right).$$

Consider a scalar as a product of two tensors $$S = V^\mu W_\mu$$ If we take the covariant derivative of S, we get \begin{align} \nabla_\alpha S &= \nabla_\alpha (V^\mu W_\mu) \\ &= (\nabla_\alpha V^\mu) W_\mu + V^\mu (\nabla_\alpha W_\mu) \\ &= (\partial_\alpha V^\mu + \Gamma_{\alpha \beta}^{\mu} V^\beta) W_\mu + V^\mu (\partial_\alpha W_\mu - \Gamma_{\alpha \mu}^{\beta} W_\beta) \\ &= W_\mu \partial_\alpha V^\mu + \Gamma_{\alpha \beta}^{\mu} V^\beta W_\mu + V^\mu \partial_\alpha W_\mu - \Gamma_{\alpha \mu}^{\beta} V^\mu W_\beta \\ &= W_\mu \partial_\alpha V^\mu + V^\mu \partial_\alpha W_\mu \end{align} as you can see that the second and the last terms are just the same up to dummy indices ($\mu$ and $\beta$). Therefore, $$\partial_\alpha (V^\mu W_\mu) = W_\mu \partial_\alpha V^\mu + V^\mu \partial_\alpha W_\mu$$