Quoting from the text

[...] Evaluating (20.46) in such coordinates is just like evaluating the derivative of a function (20.1): $$\left( \mathbf{\nabla_t v}\right)^\alpha = t^\beta \frac{\partial v^\alpha}{\partial x^\beta} \quad \text{(LIF)}. \tag{20.47}$$ For the tensor $\mathbf{\nabla v}$ we therefore have $$\nabla_\beta v^\alpha = \frac{\partial v^\alpha}{\partial x^\beta} \quad \text{(LIF)}. \tag{20.48}$$ (The LIFs have been added as a reminder that the formulas hold only in a local inertial frame at the point $x^\alpha$ where the formula is evaluated.)

I have a silly doubt here: on equation 20.47 we have the compact notation for a covariant derivative on Local inertial Frame. Despite the context of GR, I would like to know why in equation 20.48 Hartle say "For the tensor...."(I mean, why and were is the Tensor?) and why in the phrase above, Hartle wrote \begin{equation} \nabla \textbf{v} \end{equation}

insted of

\begin{equation} \nabla _\textbf{t} \textbf{v} \end{equation}

and who can I reach on expression 20.48 (from notions constructed on 20.46 (definition of the covariant derivative) and 20.47 +linear algebra and tensor algebra notions)? What suppose to mean the beta on nabla?

I have serious doubt on notation.

  • $\begingroup$ A vector is first order tensor, and the covarient derivative of s tensor is another tensor , (which is the whole point of covarience, to preserve tensor properties). $\endgroup$
    – user154420
    Jul 9, 2017 at 19:39
  • $\begingroup$ I replaced the image with a typeset block quote of the text immediately around line 20.7 using a combination of markdown (for the blockquote) and MathJax (for LateX mathmode). You should check that and also determine if more should be quoted $\endgroup$ Jul 9, 2017 at 20:06
  • $\begingroup$ There is also this site, onlineocr.net which is 90% plus accurate. $\endgroup$
    – user154420
    Jul 9, 2017 at 20:13

1 Answer 1


Given a vector $\mathbf{t}$, $\nabla_{\mathbf{t}}\mathbf{v}$ is the covariant derivative of $\mathbf{v}$ in the direction of $\mathbf{t}$, which is another vector (which is why there is only one upper index). $\nabla \mathbf{v}$, however, is not a vector. It's a tensor, because if you give it a vector $\mathbf{t}$ it returns the vector $\nabla_\mathbf{t} \mathbf{v}$. A function which depends linearly on a vector and returns another vector is a $\binom 1 1$ tensor.

If you prefer an argument with indices, you can think of it like this. In an inertial frame the components of $(\nabla_{\mathbf{t}}\mathbf{v})^\alpha$ are

$$t^\beta \partial_\beta v^\alpha.$$

So the obvious definition of $\nabla \mathbf{v}$ is

$$(\nabla \mathbf{v})_\beta^\alpha = \partial_\beta v^\alpha$$

so that

$$ (\nabla_\mathbf{t} \mathbf{v})^\alpha = t^\beta (\nabla \mathbf{v})_\beta^\alpha = t^\beta \partial_\beta v^\alpha.$$


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