## Introduction to covariant differentiation In this chapter, we define new operations, $\nabla_{i}$, *the covariant derivative* and, $\nabla^{i}$, *the contravariant derivative*. They satisfy $\nabla^{i} = Z^{ij} \nabla_{j}$ With this newly defined derivative, we can perform operations like the Laplacian on tensor fields and ensure the result remains a tensor. (recall that partial derivatives of tensors are not necessarily tensors, [[General idea on tensor properties#^78f2e3]]) For more detailed discussion on properties and proof, see [[Properties of covariant differentiation]]. ### Covariant derivative of an invariant vector field Here we illustrate the reasoning behind why the covariant derivative takes the form $ \nabla_j V^i = \frac{\partial V^i}{\partial Z^j} + \Gamma^i_{jm} V^m. $ When differentiating a vector field $\mathbf{V}$ in curvilinear coordinates $Z^i$, one might be tempted to treat the partial derivatives of its components, $\partial V^i / \partial Z^j$, as tensor components. However, this is misleading, because it neglects the variation of the basis vectors themselves. We start by expressing the vector field in terms of the covariant basis: $ \mathbf{V} = V^i \mathbf{Z}_i. $ Taking the partial derivative: $ \frac{\partial \mathbf{V}}{\partial Z^j} = \frac{\partial V^i}{\partial Z^j} \mathbf{Z}_i + V^i \frac{\partial \mathbf{Z}_i}{\partial Z^j}. $ The second term, involving the derivative of the basis, can be written using [[The Christoffel symbol|Christoffel symbols]]: $ \frac{\partial \mathbf{Z}_i}{\partial Z^j} = \Gamma^k_{ij} \mathbf{Z}_k. $ Substituting: $ \frac{\partial \mathbf{V}}{\partial Z^j} = \frac{\partial V^i}{\partial Z^j}\mathbf{Z}_i + \Gamma^k_{ij} V^i \mathbf{Z}_k. $ rename index, $ \frac{\partial \mathbf{V}}{\partial Z^j} = \left( \frac{\partial V^i}{\partial Z^j} + \Gamma^i_{jm} V^m \right) \mathbf{Z}_i. $ The key insight is that neither $\partial V^i / \partial Z^j$ nor $\Gamma^i_{jm} V^m$ is a tensor, but their sum **is**. This motivates defining a new operation—the covariant derivative—that corrects the non-tensorial behavior of the partial derivative. For a contravariant component $V^i$, we define: $ \nabla_j V^i = \frac{\partial V^i}{\partial Z^j} + \Gamma^i_{jm} V^m. $ This expression gives the correct tensorial rate of change of $V^i$ with respect to the coordinate $Z^j$. Similarly, for a covariant component $V_i$, the decomposition is: $ \mathbf{V} = V_i \mathbf{Z}^i. $ Taking the partial derivative and transforming the basis yields: $ \frac{\partial \mathbf{V}}{\partial Z^j} = \left( \frac{\partial V_i}{\partial Z^j} - \Gamma^m_{ij} V_m \right) \mathbf{Z}^i, $ which defines the covariant derivative of $V_i$ as: $ \nabla_j V_i = \frac{\partial V_i}{\partial Z^j} - \Gamma^m_{ij} V_m. $ The sign difference arises from whether the basis vectors appear covariantly or contravariantly. Thus, the covariant derivative provides a way to differentiate tensors in a manner that respects their geometric structure and **ensures that the result is also a tensor**. (proof in [[Properties of covariant differentiation#A direct proof of the tensor property]]) > [!Note] > An important application of this concept appears in the expression for the acceleration of a particle moving along a curve $Z^i(t)$. (This has been shown in [[The Christoffel symbol#^876ee2]]) If $V^i = dZ^i/dt$ is the velocity, then the acceleration is given by: > $ > A^i = \frac{d V^i}{dt} + \Gamma^i_{jk} V^j V^k, > $ > which is again a tensor due to the transformation behavior of each term. > This derivative is with respect to time and looks different. We will revisit this in [[Properties of covariant differentiation#A particle moving along a trajectory]] and discuss it as *intrinsic derivative*. ### Covariant derivative of an invariant scalar field and Laplacian For scalar fields (invariants) $F$, the covariant derivative reduces to the ordinary partial derivative: $ \nabla_i F = \frac{\partial F}{\partial Z^i} $ This follows from the fact that scalar fields have no components to transform, and thus no correction terms (like Christoffel symbols) are needed. This simple identity underlies many key expressions involving invariants. (This result is a tensor and has been shown in [[General idea on tensor properties#^08d177]]) The Laplacian operator is a fundamental second-order differential operator. In Cartesian coordinates, it is typically defined as: $ \Delta F = \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} $ However, this form lacks generality outside Cartesian systems. In tensor calculus, we define the Laplacian in a coordinate-independent way, which ensures invariance under general transformations. The tensor definition is: $ \Delta F = Z^{ij} \nabla_i \nabla_j F $ Substituting $\nabla_i F = \partial F / \partial Z^i$, the Laplacian becomes: $ \Delta F = Z^{ij} \frac{\partial^2 F}{\partial Z^i \partial Z^j} $ This shows that $\Delta F$ is itself an invariant scalar, since it is the result of fully contracting a rank-(0,2) tensor. > [!notice] > The definition $\Delta F = Z^{ij} \nabla_i \nabla_j F$ is valid in all coordinate systems and can be evaluated using the metric and Christoffel symbols. It ensures consistency and interpretability across curvilinear coordinates. > [!note] > **Special Case: Cartesian Coordinates** > > In Cartesian coordinates, the Christoffel symbols vanish and the metric tensor is the identity. Thus, the Laplacian simplifies to: > $ > \Delta F = \frac{\partial^2 F}{\partial Z^1 \partial Z^1} + \frac{\partial^2 F}{\partial Z^2 \partial Z^2} > $ > Relabeling $Z^1 = x$, $Z^2 = y$, we recover: > $ > \Delta F = \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2} > $ > [!example]- > **Polar Coordinates** (2D) > > Let $Z^1 = r$, $Z^2 = \theta$. The relevant Christoffel symbols are: > $ > \Gamma^1_{22} = -r, \quad \Gamma^2_{12} = \Gamma^2_{21} = \frac{1}{r} > $ > > For a scalar $F$, we again have: > $ > \nabla_1 F = \frac{\partial F}{\partial r}, \quad \nabla_2 F = \frac{\partial F}{\partial \theta} > $ > > The second covariant derivatives include correction terms: > $ > \nabla_1 \nabla_1 F = \frac{\partial^2 F}{\partial r^2}, \quad > \nabla_1 \nabla_2 F = \frac{\partial^2 F}{\partial r \partial \theta} - \frac{1}{r} \frac{\partial F}{\partial \theta} > $ > $ > \nabla_2 \nabla_1 F = \frac{\partial^2 F}{\partial \theta \partial r} - \frac{1}{r} \frac{\partial F}{\partial \theta}, \quad > \nabla_2 \nabla_2 F = \frac{\partial^2 F}{\partial \theta^2} + r \frac{\partial F}{\partial r} > $ > > The metric is $Z^{11} = 1$, $Z^{22} = \frac{1}{r^2}$, so the Laplacian becomes: > $ > \Delta F = \frac{\partial^2 F}{\partial r^2} + \frac{1}{r} \frac{\partial F}{\partial r} + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} > $ > [!example]- > **Cylindrical Coordinates** > > Let $Z^1 = r$, $Z^2 = \theta$, $Z^3 = z$. The [[The Christoffel symbol#Cylindrical coordinates|nonzero Christoffel symbols]] are: > $ > \Gamma^1_{22} = -r, \quad \Gamma^2_{12} = \Gamma^2_{21} = \frac{1}{r} > $ > > For a scalar field $F(r, \theta, z)$, we again have: > $ > \nabla_i F = \frac{\partial F}{\partial Z^i} > $ > > The second covariant derivatives include: > $ > \nabla_1 \nabla_1 F = \frac{\partial^2 F}{\partial r^2}, \quad > \nabla_1 \nabla_2 F = \frac{\partial^2 F}{\partial r \partial \theta} - \frac{1}{r} \frac{\partial F}{\partial \theta} > $ > $ > \nabla_2 \nabla_2 F = \frac{\partial^2 F}{\partial \theta^2} + r \frac{\partial F}{\partial r}, \quad > \nabla_3 \nabla_3 F = \frac{\partial^2 F}{\partial z^2} > $ > > The inverse metric is: > $ > Z^{11} = 1, \quad Z^{22} = \frac{1}{r^2}, \quad Z^{33} = 1 > $ > > Therefore, the Laplacian is: > $ > \Delta F = \frac{\partial^2 F}{\partial r^2} > + \frac{1}{r} \frac{\partial F}{\partial r} > + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} > + \frac{\partial^2 F}{\partial z^2} > $ > [!example]- > **Spherical Coordinates** > > Let $Z^1 = r$, $Z^2 = \theta$, $Z^3 = \phi$. The [[The Christoffel symbol#Spherical coordinates|nonzero Christoffel symbols]] are: > $ > \Gamma^1_{22} = -r, \quad \Gamma^1_{33} = -r \sin^2 \theta, \quad > \Gamma^2_{12} = \Gamma^2_{21} = \frac{1}{r}, \quad > \Gamma^2_{33} = -\sin \theta \cos \theta > $ > $ > \Gamma^3_{13} = \Gamma^3_{31} = \frac{1}{r}, \quad > \Gamma^3_{23} = \Gamma^3_{32} = \cot \theta > $ > > For scalar field $F(r, \theta, \phi)$, we still have: > $ > \nabla_i F = \frac{\partial F}{\partial Z^i} > $ > > The second covariant derivatives include: > $ > \nabla_1 \nabla_1 F = \frac{\partial^2 F}{\partial r^2}, \quad > \nabla_2 \nabla_2 F = \frac{\partial^2 F}{\partial \theta^2} + r \frac{\partial F}{\partial r}, \quad > \nabla_3 \nabla_3 F = \frac{\partial^2 F}{\partial \phi^2} + r \sin^2 \theta \frac{\partial F}{\partial r} + \sin \theta \cos \theta \frac{\partial F}{\partial \theta} > $ > > The inverse metric is: > $ > Z^{11} = 1, \quad Z^{22} = \frac{1}{r^2}, \quad Z^{33} = \frac{1}{r^2 \sin^2 \theta} > $ > > Contracting these terms gives the Laplacian: > $ > \Delta F = > \frac{\partial^2 F}{\partial r^2} > + \frac{2}{r} \frac{\partial F}{\partial r} > + \frac{1}{r^2} \frac{\partial^2 F}{\partial \theta^2} > + \frac{\cot \theta}{r^2} \frac{\partial F}{\partial \theta} > + \frac{1}{r^2 \sin^2 \theta} \frac{\partial^2 F}{\partial \phi^2} > $ This tensor formulation ensures that regardless of the coordinate system, we always recover a consistent expression for the Laplacian (and even higher order derivatives), highlighting the power and necessity of covariant derivatives in general geometry. ### Covariant derivative of covariant/contravariant basis We now consider how the covariant derivative acts on the basis vectors themselves. For a covariant basis vector $\mathbf{Z}_j$, we define: $ \nabla_i \mathbf{Z}_j = \frac{\partial \mathbf{Z}_j}{\partial Z^i} - \Gamma^k_{ij} \mathbf{Z}_k. $ But recall that the Christoffel symbols are [[The Christoffel symbol#Expressions of Christoffel symbol|defined through the derivatives of basis vectors]]: $ \frac{\partial \mathbf{Z}_j}{\partial Z^i} = \Gamma^k_{ij} \mathbf{Z}_k. $ Substituting this into the expression for the covariant derivative: $ \nabla_i \mathbf{Z}_j = \Gamma^k_{ij} \mathbf{Z}_k - \Gamma^k_{ij} \mathbf{Z}_k = 0. $ Thus, we find that: $ \nabla_i \mathbf{Z}_j = 0. $ This result expresses the **[[Properties of covariant differentiation#The metrinilic property|metrinilic property]]** of the covariant derivative: the basis vectors are preserved under parallel transport. That is, the covariant derivative is compatible with the metric and the basis. Similarly, one can show: $ \nabla_i \mathbf{Z}^j = 0. $ > [!example]- > **Polar coordinates verification** > > In polar coordinates, the covariant basis vectors are: > $ > \mathbf{Z}_1 = \mathbf{e}_r, \quad \mathbf{Z}_2 = r \mathbf{e}_\theta, > $ > where $\mathbf{e}_r$ and $\mathbf{e}_\theta$ are unit vectors in the radial and angular directions. > > Let us compute $\nabla_1 \mathbf{Z}_2$ explicitly: > > - $\partial \mathbf{Z}_2 / \partial r = \mathbf{e}_\theta = \frac{1}{r} \mathbf{Z}_2$, > - $\Gamma^2_{12} = \frac{1}{r}$ is the only relevant nonzero Christoffel symbol, > > Then: > $ > \nabla_1 \mathbf{Z}_2 = \frac{\partial \mathbf{Z}_2}{\partial r} - \Gamma^k_{12} \mathbf{Z}_k = \frac{1}{r} \mathbf{Z}_2 - \frac{1}{r} \mathbf{Z}_2 = 0. > $ > > Similarly, all other combinations vanish as well: > $ > \nabla_1 \mathbf{Z}_1 = \nabla_2 \mathbf{Z}_1 = \nabla_2 \mathbf{Z}_2 = 0. > $ > Therefore: > $ > \nabla_i \mathbf{Z}_j = 0. > $ ### Covariant derivative of general tensors The definition of the covariant derivative naturally extends to tensors of arbitrary type—tensors with any number and combination of covariant and contravariant indices. The rule is simple: for each index, we include a Christoffel symbol term that corrects the partial derivative to ensure tensorial transformation properties. For a tensor $T^i_{\ j}$ with one upper and one lower index, we define: $ \nabla_k T^i_{\ j} = \frac{\partial T^i_{\ j}}{\partial Z^k} + \Gamma^i_{km} T^m_{\ j} - \Gamma^m_{kj} T^i_{\ m}. $ Each term corresponds to a correction for each index: - For the contravariant index $i$, we add $\Gamma^i_{km} T^m_{\ j}$, - For the covariant index $j$, we subtract $\Gamma^m_{kj} T^i_{\ m}$. > [!Example] > This pattern generalizes to higher-order tensors. For example, a triply covariant tensor $T_{rst}$ has: > $ > \nabla_k T_{rst} = \frac{\partial T_{rst}}{\partial Z^k} - \Gamma^m_{kr} T_{mst} - \Gamma^m_{ks} T_{rmt} - \Gamma^m_{kt} T_{rsm}. > $ > The sign of each correction term depends on whether the index is covariant ($-$) or contravariant ($+$), and the Christoffel symbol always appears with the lower two indices matching the derivative and tensor index, and the upper index as a dummy summed over. This construction ensures that the covariant derivative of any tensor is again a tensor. > [!NOTE] > A simple but powerful identity arising from this structure is: > $ > \nabla_k \delta^i_{\ j} = 0, > $ > which reflects the metric compatibility of the covariant derivative and will be discussed further under the [[Properties of covariant differentiation#The metrinilic property|metrinilic property]]. > > [!Proof]- > > The Kronecker delta $\delta^i_{\ j}$ is a mixed tensor. Applying the definition of the covariant derivative for such a tensor, we have: > > $ > > \nabla_k \delta^i_{\ j} = \frac{\partial \delta^i_{\ j}}{\partial Z^k} + \Gamma^i_{km} \delta^m_{\ j} - \Gamma^m_{kj} \delta^i_{\ m}. > > $ > > Since $\delta^i_{\ j}$ is constant, the partial derivative vanishes. The remaining terms simplify via contraction: > > $ > > \Gamma^i_{km} \delta^m_{\ j} = \Gamma^i_{kj}, \quad \Gamma^m_{kj} \delta^i_{\ m} = \Gamma^i_{kj}. > > $ > > Thus, > > $ > > \nabla_k \delta^i_{\ j} = \Gamma^i_{kj} - \Gamma^i_{kj} = 0. > > $