Intuitive way to think about covariant derivative of a scalar made from contracting tensors

If a scalar quantity is formed from the contraction of multiple indexed quantities (e.g. let's suppose as a simple example $\phi = v^\lambda w_\lambda$), is it always correct to take the covariant derivative of that scalar quantity $\phi$ as \begin{eqnarray} \nabla_\mu \phi &=& \nabla_\mu (v^\lambda w_\lambda) \\ &=& (\nabla_\mu v^\lambda )w_\lambda + v^\lambda(\nabla_\mu w_\lambda ) \\ &=& (\partial_\mu v^\lambda + \Gamma^\lambda_{\mu\sigma}v^\sigma)w_\lambda + v^\lambda (\partial_\mu w_\lambda - \Gamma^\sigma_{\mu\lambda}w_\sigma) \end{eqnarray}

1. I want to check that this is indeed the case.
2. If this is the case, I am not completely comfortable with this idea. I feel that this is a statement we are asserting $v^\lambda$ and $w_\lambda$ are "more fundamental" than $\phi$. (I think that If $\phi$ is truly most fundamental, I should be able to perform all my calculations in terms of $\phi$ only. But before I make any claim as to what $\phi$, $v^\lambda$, and $w_\lambda$ represent, we have no knowledge as to "which quantity(s) are fundamental.) I suspect that "more fundamental" is not the best way to think about this question.....is there a natural/better way to think about this question?
• 1. It is indeed. Note also that since $\phi$ is a scalar, $\nabla_\mu \phi = \partial_\mu \phi$. You should check that the RHS of your above equation reduces to this. – gj255 Jan 10 '18 at 21:01
• Is this always the case? Part of the thing that bothers me is that often the covariant derivative of the Ricci Scalar is written as $\nabla_\mu R = 2\nabla^\rho R_{\rho\mu}$, but I never see $\nabla_\mu R =\partial_\mu R$. Or is it true that $\nabla_\mu R =\partial_\mu R= 2\nabla^\rho R_{\rho\mu}$, but $\nabla_\mu R = 2\nabla^\rho R_{\rho\mu}$ is just more useful in practice? – Bob Jan 10 '18 at 21:13
• It is certainly true that $\nabla_\mu R = \partial_\mu R$. I personally have not seen your other identity. Where have you seen it? – gj255 Jan 10 '18 at 23:51
• – Bob Jan 10 '18 at 23:58
• OK, note that (3.94) is a very particular formula which holds because the Riemann tensor has certain symmetries. In a sense it is separate from the question of how the covariant derivative acts on scalars and tensors. Since $R$ is a scalar, the covariant derivative of $R$ is the partial derivative. Which of these two relations is more useful in practice will depend on context. – gj255 Jan 11 '18 at 0:02

If you're worried that $\nabla_\mu\phi=\partial_\mu\phi$ should be "more fundamental" than how vectors transform, you'll be pleased to know this fact about scalar fields is one of the axioms that defines the covariant derivative. You can then take the differentiation of either covariant or contravariant vectors as a further axiom (both choices are equivalent), and also as the definition of Christoffel symbols. Your calculation can be rewritten to derive how one kind of vector transforms from how the other does. For example, $$0=(\nabla_\mu-\partial_\mu)(v^\lambda w_\lambda)=\Gamma_{\mu\sigma}^\lambda v^\sigma w_\lambda+v^\lambda(\nabla_\mu-\partial_\mu) w_\lambda.$$