When I am asked to evaluate, $\mathbf{U^{\alpha}_{~,~~\beta}}$ for all $\alpha$ and $\beta$, what does it mean? I have not able to understand this notation.

In case of $\mathbf{g(~~,~\bar{A})}$ I understand that the blank in the metric tensor means it acts like a one-form which takes in a vector and outputs a real number. For this case it makes sense to me because it is a

$\begin{bmatrix} 0 \\ 2 \end{bmatrix}$ tensor. How do we understand the same for a mixed tensor like the one above ?

Reference: Chapter 3, Problem 30: A First Course in General Relativity, Second Edition, B. Schutz, pg 82.


1 Answer 1


Let us assume $U^\alpha$ is a vector, i.e. described by its contravariant components. The subscript $_{, \beta}$ means the partial derivative, that is $_{, \beta} = \partial_\beta$. Therefore $U^\alpha _{, \beta} = \partial_\beta U^\alpha$.

However in general that is not a tensor, as the partial derivative is not a tensor in arbitrary coordinates. To have a tensorial expression, you should replace the partial derivative with the covariant derivative, that is $\partial_\mu$ replaced by $\nabla_\mu = _{; \mu}$.

So the expression $U^\alpha _{; \beta} = \nabla_\beta U^\alpha$ is a tensor, contravariant in $\alpha$ and covariant in $\beta$.

  • $\begingroup$ Thank you ! Thant makes sense. In Minkowski space-time, the partial derivative and the covariant derivative would be the same which was my case :-) $\endgroup$
    – Astronomer
    Commented Sep 24, 2018 at 22:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.