geometric interpretation on covariant derivateve in curvilinear coordinates.

Question

I'm having trouble understanding what does the covariant derivative do in a coordinate system where we have changing basis vectors. I always thought it was giving us the change in coordinates while taking into account the change in basis vectors $\vec{a}_i(\xi^1,\xi^2,...,\xi^n)$. I tried to visualize it with the following scenario: $$$$ Let't say we have a constant vector field $\vec{u}$ in a curvilinear basis and we take the covariant derivative w.r.t to $\xi^i$. Then according to the sketch I have drawn to approximately plot how things would look geometrically I should get a vector that is opposite to what I get from the formula since if we look at the new (red) basis, according to it our vector component has decreased and is now less than 0. However the calculation say that the covariant derivative is positive.Why is that? A mistake in the calculations or a fundamental misunderstanding of the role of the covariant derivative in curvilinear basis? Can you help me build intuition?

Answer 1

Your notation is a bit hard to understand. Let's look at a vector field that is expressed in Cartesian and polar coordinates in the plane: \begin{align}\tag{1} \boldsymbol{V}&=V^x\hat{\boldsymbol{x}}+V^y\hat{\boldsymbol{y}}=V^r\hat{\boldsymbol{ r}}+V^\phi\hat{\boldsymbol{\phi}}\\[2mm]\tag{2} \hat{\boldsymbol{r}}&=(\cos\phi)\,\hat{\boldsymbol{x}}+(\sin\phi)\,\hat{\boldsymbol{y}}\,,\\[2mm]\tag{3} \hat{\boldsymbol{\phi}}&=-r(\sin\phi)\,\hat{\boldsymbol{x}}+r(\cos\phi)\,\hat{\boldsymbol{y}}\,,\\[2mm]\tag{4} V^r&=(\cos\phi)V^x+(\sin\phi)V^x\,,\\[2mm]\tag{5} V^\phi&=-\tfrac1r(\sin\phi)V^x+\tfrac1r(\cos\phi)V^x\,. \end{align} When $\boldsymbol{V}$ is constant then all Cartesian ingredients are constant: components $V^x,V^y$ are constnt and basis vectors $\hat{\boldsymbol{x}},\hat{\boldsymbol{y}}$ are constant anyway. When moving around $\mathbb R^2$ the polar basis vectors $\hat{\boldsymbol{r}},\hat{\boldsymbol{\phi}}$ hovewer rotate as we see from (2) and (3). This means that $(V^r,V^\phi)$ must rotate in the opposite direction (be "contravariant") which is ensured by (4) and (5).

• From the product rule \begin{align}\tag{6} \nabla_r\boldsymbol{V}&=(\partial_r V^r)\,\hat{\boldsymbol{r}}+V^r\partial_r\hat{\boldsymbol{r}}+(\partial_rV^\phi)\,\hat{\boldsymbol{\phi}}+V^\phi\partial_r\hat{\boldsymbol{\phi}}\end{align} you should be able to derive now the familiar expression of the covariant derivative that contains Christoffel symbols and figure out that the latter are nothing else than the components of the partial derivatives of the basis vectors $\hat{\boldsymbol{r}},\hat{\boldsymbol{\phi}}\,.$

• It seems to me that in OP you have instead taken the partial derivatives of the components of those basis vectors.

From (6):\begin{align}\nabla_r\boldsymbol{V}&=(\partial_r V^r)\,\hat{\boldsymbol{r}}+\left\{(\partial_rV^\phi)+V^\phi\frac{1}r\right\}\hat{\boldsymbol{\phi}}\\\partial_r\hat{\boldsymbol{r}}&=0\,,\quad\partial_r\hat{\boldsymbol{\phi}}=-(\sin\phi)\,\hat{\boldsymbol{x}}+\cos\phi)\,\hat{\boldsymbol{y}}=\frac{\hat{\boldsymbol{\phi}}}r=\Gamma^\phi_{\phi r}\hat{\boldsymbol{\phi}}\,,\\[2mm]\Gamma^r_{\phi\phi}&=-r\,,\quad\Gamma^\phi_{\phi r}=\frac1r\end{align}