# Partial derivatives vs Covariant derivatives in polar coordinates

Covariant derivatives take into account for both component and basis changes, thereby applicable for curved spaces - where partial derivatives only take component changes into account - is this statement correct? In a polar coordinate system basis changes, even though the space is flat. Will that be considered in partial derivatives? Or we need covariant derivative for that?

Covariant derivatives take into account for both component and basis changes, thereby applicable for curved spaces - where partial derivatives only take component changes into account - is this statement correct?

Yes!

In a polar coordinate system basis changes, even though the space is flat. Will that be considered in partial derivatives? Or we need covariant derivative for that?

The Covariant Derivative is defined as $$D_\mu V_\nu := \partial_\mu V_\nu - \Gamma^\alpha_{\mu\nu} V_\alpha$$ where $$\Gamma^\alpha_{\mu\nu}$$ is the Christoffel Symbol. Note that it's called Christoffel Symbol, not Christoffel Tensor. That's because it's not a tensor. It can be zero in one coordinate system and nonzero in another.

Now consider flat space. In ordinary cartesian coordinates, the Christoffel symbol is zero, such that the Covariant derivative is equal to the ordinary partial derivative. $$D_\mu V_\nu = \partial_\mu V_\nu$$ However, in Polar coordinates, the Christoffel symbol is not zero, even though you are still describing flat space. Thus, in Polar Coordinates, the covariant derivative is NOT equal to the ordinary partial derivative.

Nonzero Christoffel symbols do not necessarily imply curved space.

• Your example is actually of a components of a covector (thus lowered component indices). Commented May 30, 2023 at 4:09
• That's true. I get them mixed up all the time :D For a contravariant vector, you get a different sign: $$D_\mu V^\nu = \partial_\mu V^\nu + \Gamma_{\mu\alpha}^\nu V^\alpha$$ Commented May 30, 2023 at 11:51
• Ok then, my question was actually about the covariant and partial derivatives of the total vector, meaning $D_\mu V$ and $d_\mu V$, not just its components. Commented May 30, 2023 at 11:56
1. As OP correctly points out connections introduce a concept of differentiation of tensor fields or more in general of sections of vector bundles that takes into account how the bases of the fibers change. When we have a (pseudo-)Riemannian manifold $$(M,g)$$ there is a unique torsion-free connection compatible with the metric, namely the Levi-Civita connection, whose components are locally of the form $$\Gamma^\mu_{\nu\sigma}=\frac{1}{2}g^{\mu\alpha}(\partial_\nu g_{\alpha\sigma}+\partial_\sigma g_{\alpha\nu}-\partial_\alpha g_{\nu\sigma})\tag{1}\label{1}$$ where the sum over the dummy indices is understood.
2. Even in the case of a flat space, i.e. zero curvature tensor, the components of the connection \eqref{1} don't have to be zero. In fact, \eqref{1} are not the components of a tensor and might be zero with a given choice of coordinates and non-zero with others. As an example, consider the Euclidean $$\mathbb{R}^n$$ $$(\mathbb{R}^3, g)$$ in cartesian coordinates $$(x,y,z)$$, i.e. $$g_{ij}=\delta_{ij}=\mathrm{diag}(1,1,1,1)\implies \Gamma^{i}_{jk}=0$$. On the other hand, the metric tensor in spherical polar coordinates $$(r,\theta,\phi)$$ is not constant $$g_{ij}=\mathrm{diag}(1,r^2, r^2\sin^2\theta)$$ leading to non-zero Christoffel symbols with this choice of coordinates. Note that this wouldn't be possible with tensors, whose components transform linearly.
3. Note that for a non-euclidean metric, there is a distinction between the components of the differential of a function $$f$$, namely $$df=(\partial_\mu f) dx^\mu$$ and its gradient $$\mathrm{grad} f=g^{\mu\nu}(\partial_\nu f) \partial_\mu$$. Actually, as long as we're not interested in differentiating vector fields or tensor fields, that's all we should care about.
4. Regarding OP's main question, we have to use covariant derivatives, being the most general construction we have in this case. Clearly, in the case of zero Christoffel symbols, we'd get back classical partial derivatives, but that won't happen for a generic choice of coordinates. In fact, acting with partial derivatives wouldn't yield something tensorial in the general case$$^1$$.

$$^1$$ This doesn't mean that we can't build tensorial object with partial derivatives of tensor components. In fact, using antisymmetrized combinations of partial derivatives of $$p$$-forms, we can build the components of their exterior derivatives, which are $$p+1$$ forms.