## Properties of covariant differentiation ### The tensor property A fundamental characteristic of the covariant derivative is that it maps tensors to tensors. Namely, $\nabla_i V_j$ is a tensor. This property is essential for ensuring that differentiation preserves the tensorial nature of quantities, thereby allowing geometric and physical laws to be expressed in a coordinate-independent manner. A formal proof of this result is provided in [[#A direct proof of the tensor property]]. With this property, then contraction with another tensor such as $Z^{ij}$ yields the invariant quantity $ Z^{ij} \nabla_i V_j = \nabla^j V_j, $ which serves as a definition for the divergence of the tensor field $V_j$. Similarly, Laplacian of a scalar field can be defined as $Z^{ij}\nabla_i \nabla_j F=\nabla_{i}\nabla^{i}F=\Delta F$ The tensor property holds for the covariant derivatives of scalar functions, vectors, and higher-rank tensors. In short, if $T^{ijk\ldots}_{rst\ldots}$ is a tensor, then its covariant derivative $ \nabla_p T^{ijk\ldots}_{rst\ldots} $ is also a tensor. > [!Example] > A vector field example: > > Let $T_{ij}$ be a rank-$(0,2)$ tensor. Define $\mathbf{T}_j = T_{ij} \mathbf{Z}^i$, which is a vector-valued field. Then the covariant derivative $\nabla_k \mathbf{T}_j$ is a tensor, and extracting the components yields > $ > \nabla_k \mathbf{T}_j = \nabla_k T_{ij} \, \mathbf{Z}^i \quad \Rightarrow \quad \nabla_k T_{ij} \text{ is a tensor}. > $ > This also applies for higher-rank tensors. > >[!Warning]- > >Here we see $\nabla_k \mathbf{T}_j$ is a tensor. This is proper defined, while in the reference [[Introduction to Tensor Analysis and the Calculus of Moving Surfaces]], equation 8.61 uses $\frac{\partial \mathbf{V}}{\partial Z^{j}}$, this is not a well defined quantity (at least when beyond Euclidean space) and is only correct under Cartesian system, Euclidean space. A connection is required to remove ambiguity. This completes the conceptual overview of the tensorial character of the covariant derivative. A more direct demonstration, based on its behavior under coordinate transformations, is presented in the following section. ### Covariant derivative applied to invariants For an invariant scalar (or any tensor of rank zero), the covariant derivative reduces to the ordinary partial derivative: $ \nabla_k F = \frac{\partial F}{\partial Z^k}. $ > [!Notee] > This is not so much a *property* as it is a consistent extension of the general definition of the covariant derivative to scalars. Since scalars have no indices, the Christoffel terms vanish by default — there are simply no terms to which the Christoffel symbols could apply. ### Covariant derivative in affine coordinates In affine coordinates (e.g., Cartesian coordinates in Euclidean space), the Christoffel symbols vanish. As a result, the covariant derivative reduces to the partial derivative: $ \nabla_k T^{i\ldots}_{j\ldots} = \frac{\partial T^{i\ldots}_{j\ldots}}{\partial Z^k}. $ > [!Note]- > This fact is often used to simplify proofs: one demonstrates a result in affine coordinates, where calculations are easier, and then extends it to arbitrary coordinate systems using the tensor property. ### Commutativity Covariant derivatives commute in **Euclidean space**: $ \nabla_k \nabla_l T^i_{\ j} = \nabla_l \nabla_k T^i_{\ j}. $ That is, switching the order of covariant derivatives leaves the result unchanged. This holds for any tensor with arbitrary indices in such spaces. > [!Note] > In Cartesian coordinates, the Christoffel symbols vanish, and this reduces to the commutativity of partial derivatives: > $ > \frac{\partial^2 T^i_{\ j}}{\partial Z^k \partial Z^l} - \frac{\partial^2 T^i_{\ j}}{\partial Z^l \partial Z^k} = 0. > $ > Since the difference is a tensor and vanishes in Cartesian coordinates, it must vanish in all coordinate systems (in Euclidean space). In general curved spaces (i.e., non-Euclidean), the difference $ \nabla_k \nabla_l T^i_{\ j} - \nabla_l \nabla_k T^i_{\ j} $ is nonzero and defines a new tensor. This commutator is often written as $(\nabla_k \nabla_l - \nabla_l \nabla_k) T^i_{\ j}$, and reflects the intrinsic curvature of the manifold. > [!Notice] > In curved spaces, covariant derivatives generally **do not commute**. The failure to commute is captured by the **Riemann curvature tensor**: > $ > R^k_{\cdot mij} = \frac{\partial \Gamma^k_{im}}{\partial Z^j} - \frac{\partial \Gamma^k_{jm}}{\partial Z^i} + \Gamma^k_{in} \Gamma^n_{jm} - \Gamma^k_{jn} \Gamma^n_{im}. > $ > This tensor measures the noncommutativity of the connection and is central to Riemannian geometry and general relativity. ### The sum rule The covariant derivative is linear: $ \nabla_k (T^i_{\ j} + S^i_{\ j}) = \nabla_k T^i_{\ j} + \nabla_k S^i_{\ j}, $ and more generally, $ \nabla_k (a T^i_{\ j} + b S^i_{\ j}) = a \nabla_k T^i_{\ j} + b \nabla_k S^i_{\ j}. $ > [!Proof]- > **Proof of the Sum Rule** > Let $T^i_{\ j}$ and $S^i_{\ j}$ be tensors of the same type. By the definition of the covariant derivative: > $ > \nabla_k (T^i_{\ j} + S^i_{\ j}) = \frac{\partial}{\partial Z^k} (T^i_{\ j} + S^i_{\ j}) + \Gamma^i_{km} (T^m_{\ j} + S^m_{\ j}) - \Gamma^m_{kj} (T^i_{\ m} + S^i_{\ m}). > $ > Distributing each term: > $ > = \left( \frac{\partial T^i_{\ j}}{\partial Z^k} + \Gamma^i_{km} T^m_{\ j} - \Gamma^m_{kj} T^i_{\ m} \right) > + \left( \frac{\partial S^i_{\ j}}{\partial Z^k} + \Gamma^i_{km} S^m_{\ j} - \Gamma^m_{kj} S^i_{\ m} \right) > $ > $ > = \nabla_k T^i_{\ j} + \nabla_k S^i_{\ j}. > $ > Hence, the covariant derivative is linear: > $ > \nabla_k (T^i_{\ j} + S^i_{\ j}) = \nabla_k T^i_{\ j} + \nabla_k S^i_{\ j}. > $ ### The product rule The covariant derivative satisfies the product rule. For tensors $T^i$ and $U_j$: $ \nabla_k (T^i U_j) = (\nabla_k T^i) U_j + T^i (\nabla_k U_j). $ > [!Proof]- > start with the definition: > $ > \nabla_k (T^i U_j) = \frac{\partial (T^i U_j)}{\partial Z^k} + \Gamma^i_{km} T^m U_j - \Gamma^m_{kj} T^i U_m. > $ > Apply the ordinary product rule: > $ > \nabla_k (T^i U_j)= \left( \frac{\partial T^i}{\partial Z^k} + \Gamma^i_{km} T^m \right) U_j + T^i \left( \frac{\partial U_j}{\partial Z^k} - \Gamma^m_{kj} U_m \right), > $ > which simplifies to: > $ > \nabla_k (T^i U_j)= (\nabla_k T^i) U_j + T^i (\nabla_k U_j). > $ This rule generalizes to arbitrary tensor products and ensures that the covariant derivative preserves tensor structure under multiplication. > [!Example] > Let $A^{ij}$, $B_k$, and $C^l_{\ m}$ be tensors. Define their product: > $ > S = A^{ij} B_k C^l_{\ m}. > $ > Then the covariant derivative of $S$ satisfies the Leibniz rule: > $ > \nabla_n S = (\nabla_n A^{ij}) B_k C^l_{\ m} + A^{ij} (\nabla_n B_k) C^l_{\ m} + A^{ij} B_k (\nabla_n C^l_{\ m}). > $ > This property generalizes to any finite tensor product. The result remains a tensor of the appropriate rank and type, with the covariant derivative acting linearly on each factor. The product rule also applies when taking the covariant derivative of a scalar formed by a dot product of tensors: $ \nabla_k \left( \mathbf{T}^i \cdot \mathbf{U}_j \right) = \nabla_k \mathbf{T}^i \cdot \mathbf{U}_j + \mathbf{T}^i \cdot \nabla_k \mathbf{U}_j. $ > [!Proof]- > Let $\mathbf{T}^i = T^i \mathbf{Z}^i$ and $\mathbf{U}_j = U_j \mathbf{Z}_j$. Their dot product is a scalar: > $ > \mathbf{T}^i \cdot \mathbf{U}_j = T^i U_j \, \delta^j_i. > $ > Taking the covariant derivative: > $ > \nabla_k (T^i U_j) = (\nabla_k T^i) U_j + T^i (\nabla_k U_j), > $ > since $\delta^j_i$ is invariant and $\nabla_k \delta^j_i = 0$, > $ > \nabla_k \left( \mathbf{T}^i \cdot \mathbf{U}_j \right) > = \nabla_k T^i \cdot U_j + T^i \cdot \nabla_k U_j. > $ > The result is consistent with the usual product rule for scalars derived from vector contractions. ### The metrinilic property The metrinilic property states that certain metric-related tensors vanish under covariant differentiation: $ \nabla_j \mathbf{Z}_i = \nabla_j \mathbf{Z}^i = 0, \quad \nabla_k Z_{ij} = \nabla_k Z^{ij} = 0, \quad \nabla_p \delta^i_j = \nabla_p \delta^{ij}_{rs} = \nabla_p \delta^{ijk}_{rst} = 0, $ and it extends to the [[Relative tensor and Levi-Civita symbol#The Levi-Civita symbols|Levi-Civita symbols]] as well: $ \nabla_p \varepsilon^{ijk} = \nabla_p \varepsilon_{ijk} = 0. $ This property guarantees that raising and lowering indices can be performed freely across a covariant derivative. It also allows for tensor identities involving contractions and index rearrangements to remain valid when covariant derivatives are present. > [!Proof]- > For the basis vectors, by construction of the covariant derivative, > $ > \nabla_j \mathbf{Z}_i = \nabla_j \mathbf{Z}^i = 0. > $ > For the metric tensor, since $Z_{ij} = \mathbf{Z}_i \cdot \mathbf{Z}_j$, we apply the product rule: > $ > \nabla_k Z_{ij} = \nabla_k (\mathbf{Z}_i \cdot \mathbf{Z}_j) = (\nabla_k \mathbf{Z}_i) \cdot \mathbf{Z}_j + \mathbf{Z}_i \cdot (\nabla_k \mathbf{Z}_j) = 0. > $ > Similarly, for the inverse metric $Z^{ij} = \mathbf{Z}^i \cdot \mathbf{Z}^j$, we have: > $ > \nabla_k Z^{ij} = (\nabla_k \mathbf{Z}^i) \cdot \mathbf{Z}^j + \mathbf{Z}^i \cdot (\nabla_k \mathbf{Z}^j) = 0. > $ > For the Kronecker delta $\delta^i_j = \mathbf{Z}^i \cdot \mathbf{Z}_j$, again applying the product rule: > $ > \nabla_k \delta^i_j = (\nabla_k \mathbf{Z}^i) \cdot \mathbf{Z}_j + \mathbf{Z}^i \cdot (\nabla_k \mathbf{Z}_j) = 0. > $ > The same logic extends to higher-order delta symbols $\delta^{ij}_{rs} = \delta^i_r \delta^j_s$ and $\delta^{ijk}_{rst} = \delta^i_r \delta^j_s \delta^k_t$, as well as to the Levi-Civita symbols: > $ > \nabla_p \varepsilon^{ijk} = 0, \quad \nabla_p \varepsilon_{ijk} = 0. > $ > The proof is given [[Relative tensor and Levi-Civita symbol#The metrinilic property of Levi-Civita symbol|later]]. The metrinilic property may help to do contraction and [[Index juggling|index juggling]]. More will be shown [[#Commutativity with contraction|later]]. > [!Example] > We know that $T_i = S^j_i U^k_j V_{kl}$ implies $T^i = T^{ij} U^k_j V_{kl}$. If we apply a covariant derivative: > $ > S_i = \nabla_k T^j_i U^k_j V_l, > $ > then contract with the metric to raise the index: > $ > S^r = Z^{ir} \nabla_k T^j_i U^k_j V_l = \nabla_k (Z^{ir} T^j_i) U^k_j V_l, > $ > where the last equality uses the product rule and the metrinilic property $\nabla_k Z^{ir} = 0$. Hence, the metric passes freely through the covariant derivative and we obtain: > $ > S^r = \nabla_k T^{rj} U^k_j V_l. > $ This justifies that **free indices can be juggled across the covariant derivative**. In a similar way, we can exchange flavors of contracted indices across derivatives: $ S^j_i T^k_j = S_{ij} T^{ik} \quad \Rightarrow \quad S^j_i \nabla_l T^k_j = S_{ij} \nabla_l T^{ik}. $ > [!Note]- > This property also implies that the components $\nabla_k V^i$ of a vector field $\mathbf{V} = V^i \mathbf{Z}_i$ fully describe its rate of change: > > $ > \frac{\partial \mathbf{V}}{\partial Z^k} = \nabla_k \mathbf{V} = \nabla_k V^i \, \mathbf{Z}_i, > $ > while the partial derivative $\partial V^i / \partial Z^k$ alone fails to capture this variation due to the nontrivial change of the basis vectors. > ($\frac{\partial \mathbf{V}}{\partial Z^k}$ only meaningful in Cartesian system) ### Commutativity with Contraction Covariant differentiation commutes with contraction. That is, for a tensor $T^i_{\ rj}$, we may take its covariant derivative and then contract on $i = r$, or first contract to obtain $T^i_{\ ij}$ and then differentiate — the result is the same: $ \nabla_k T^i_{\ ij} = \nabla_k (T^i_{\ rj} \delta^r_i) = (\nabla_k T^i_{\ rj}) \delta^r_i = \nabla_k T^i_{\ rj} \big|_{r = i}. $ > [!Notice] > **Two interpretations of $\nabla_k T^i_{\ ij}$:** > > **(1) Contract first.** Let $S_j = T^i_{\ ij}$ be a rank-1 tensor. Then, > $ > \nabla_k T^i_{\ ij} = \nabla_k S_j = \frac{\partial T^i_{\ ij}}{\partial Z^k} - \Gamma^m_{kj} T^i_{\ im}. > $ > > **(2) Differentiate first.** Begin with the rank-3 tensor $T^i_{\ rj}$, then take the covariant derivative: > $ > \nabla_k T^i_{\ rj} = \frac{\partial T^i_{\ rj}}{\partial Z^k} + \Gamma^i_{km} T^m_{\ rj} - \Gamma^m_{kr} T^i_{\ mj} - \Gamma^m_{kj} T^i_{\ rm}. > $ > Contract $i = r$ to obtain: > $ > \nabla_k T^i_{\ ij} = \frac{\partial T^i_{\ ij}}{\partial Z^k} + \Gamma^i_{km} T^m_{\ ij} - \Gamma^m_{ki} T^i_{\ mj} - \Gamma^m_{kj} T^i_{\ im}. > $ > Now note that the first and third correction terms are equal and cancel, so: > $ > \nabla_k T^i_{\ ij} = \frac{\partial T^i_{\ ij}}{\partial Z^k} - \Gamma^m_{kj} T^i_{\ im}. > $ > which matches the result from (1). This confirms that **the contraction operation and the covariant derivative commute**, even when contraction is implicit via Einstein notation. Had this not been the case, one would need to reintroduce summation symbols to distinguish between $\sum_{i=r} \nabla_k T^i_{\ rj} \quad \text{and} \quad \nabla_k \sum_{i=r} T^i_{\ rj}$ This commutativity is a cornerstone of tensor calculus. It ensures that covariant differentiation respects the structure of contracted indices, and preserves the consistency of index manipulation without ambiguity. ### A direct proof of the tensor property *This part is purely proof. One may skip them if not interested.* Following the same scheme, proof for higher-rank tensor could be established. #### Tensor property for $\nabla_{j}T_{i}$ We aim to show that $\nabla_j T_i$ is a tensor. In the primed coordinates $Z^{i'}$, the covariant derivative is defined as: $ \nabla_{j'} T_{i'} = \frac{\partial T_{i'}}{\partial Z^{j'}} - \Gamma^{k'}_{i' j'} T_{k'}. $ To verify tensorial transformation, we express each term on the right-hand side in terms of unprimed coordinates using the chain rule: - For the derivative: $ \frac{\partial T_{i'}}{\partial Z^{j'}} = \frac{\partial T_i}{\partial Z^j} J^i_{i'} J^j_{j'} + J^i_{i' j'} T_i, $ where $J^i_{i'} = \frac{\partial Z^i}{\partial Z^{i'}}$, and $J^i_{i' j'} = \frac{\partial^2 Z^i}{\partial Z^{i'} \partial Z^{j'}}$. - For the Christoffel symbol: $ \Gamma^{k'}_{i' j'} = \Gamma^k_{ij} J^i_{i'} J^j_{j'} J^{k'}_k + J^k_{i' j'} J^{k'}_k. $ - For the tensor component: $ T_{k'} = T_k J^k_{k'}. $ Substituting all into the definition of $\nabla_{j'} T_{i'}$, we obtain: $ \nabla_{j'} T_{i'} = \left( \frac{\partial T_i}{\partial Z^j} J^i_{i'} J^j_{j'} + J^i_{i' j'} T_i \right) - \left( \Gamma^k_{ij} J^i_{i'} J^j_{j'} J^{k'}_k + J^k_{i' j'} J^{k'}_k \right) T_k J^k_{k'}. $ Multiplying out and rearranging: $ \nabla_{j'} T_{i'} = \left( \frac{\partial T_i}{\partial Z^j} - \Gamma^k_{ij} T_k \right) J^i_{i'} J^j_{j'} = \nabla_j T_i \, J^i_{i'} J^j_{j'}. $ This matches the transformation rule for a rank-$(0,2)$ tensor, confirming that $\nabla_j T_i$ is a tensor. #### Tensor property for $\nabla_j T^i$ Similarly, to prove the tensor property of $\nabla_j T^i$, we begin in the primed coordinate system $Z^{i'}$, where the covariant derivative is given by $ \nabla_j T^{i'} = \frac{\partial T^{i'}}{\partial Z^{j'}} + \Gamma^{i'}_{j'k'} T^{k'}. $ Each term in this expression transforms as follows: - The partial derivative obeys the chain rule: $ \frac{\partial T^{i'}}{\partial Z^{j'}} = \frac{\partial T^i}{\partial Z^j} J^{j}_{j'} J^{i'}_i + T^i \frac{\partial J^{i'}_i}{\partial Z^{j'}}. $ - The Christoffel symbols transform as $ \Gamma^{i'}_{j'k'} = \Gamma^i_{jk} J^{j}_{j'} J^{k}_{k'} J^{i'}_i + J^{i'}_{j'k'} J^i_i. $ - The vector components transform as $ T^{k'} = T^l J^{k'}_l. $ Substituting all terms: $ \nabla_j T^{i'} = \frac{\partial T^i}{\partial Z^j} J^j_{j'} J^{i'}_i + T^i \frac{\partial J^{i'}_i}{\partial Z^{j'}} + \Gamma^i_{jk} J^j_{j'} J^k_{k'} J^{i'}_i T^k + J^{i'}_{j'k'} J^i_i T^l J^{k'}_l. $ The first and third terms combine to yield $\nabla_j T^i J^j_{j'} J^{i'}_i$. However, to simplify the remaining terms, we group and factor $T^i$: $ \nabla_j T^{i'} = \nabla_j T^i J^j_{j'} J^{i'}_i + \left( J^{i'}_{ij'} + J^{j}_{j'k'} J^{k'}_i \right) T^i. $ To show the expression in parentheses vanishes, we use the identity: $ J^{i'}_i J^i_{j'} = \delta^{i'}_{j'}, $ and differentiate with respect to $Z^i$: $ J^{i'}_{ij'} J^i_{j'} + J^{j}_{j'k'} J^{k'}_i = 0. $ This confirms the non-tensor term vanishes, and we obtain: $ \nabla_j T^{i'} = \nabla_j T^i Z^{i'}_i Z^j_{j'}, $ which shows $\nabla_j T^i$ is a tensor. #### Tensor property for $\nabla_{k}T_{j}^{i}$ We now prove the tensor property of the covariant derivative of a mixed tensor $T^i_{\ j}$. In the primed coordinates $Z^{i'}$, the covariant derivative is expressed as: $ \nabla_{k'} T^{i'}_{\ j'} = \frac{\partial T^{i'}_{\ j'}}{\partial Z^{k'}} + \Gamma^{i'}_{k'm'} T^{m'}_{\ j'} - \Gamma^{m'}_{j'k'} T^{i'}_{\ m'}. $ The transformation of $T^i_{\ j}$ under coordinate change is: $ T^{i'}_{\ j'} = T^i_{\ j} J^{i'}_i J^j_{\ j'}. $ By applying the product and sum rules, the partial derivative term expands to: $ \frac{\partial T^{i'}_{\ j'}}{\partial Z^{k'}} = \colorbox{#fff20066}{$\frac{\partial T^i_{\ j}}{\partial Z^k} J^{i'}_i J^j_{\ j'} J^k_{\ k'}$} + \colorbox{#a5d8ff66}{$T^j_{ik} J^{i'}_i J^j_{\ j'} J^k_{\ k'}$} + \colorbox{#ffa94d66}{$T^i_{\ j} J^{i'}_{ik} J^j_{\ j'} J^k_{\ k'}$}. $ For the Christoffel term $\Gamma^{i'}_{k'm'} T^{m'}_{\ j'}$, we use: $ \Gamma^{i'}_{k'm'} T^{m'}_{\ j'} = \left( \Gamma^i_{km} J^{i'}_i J^k_{\ k'} J^m_{\ m'} + J^{i'}_{k'm'} \right) T^m_{\ j} J^j_{\ j'} = \colorbox{#fff20066}{$\Gamma^i_{km} T^m_{\ j} J^{i'}_i J^k_{\ k'} J^j_{\ j'}$} + \colorbox{#ffa94d66}{$J^{i'}_{k'm'} T^m_{\ j} J^j_{\ j'}$}. $ For the final term $\Gamma^{m'}_{j'k'} T^{i'}_{\ m'}$, we expand: $ \Gamma^{m'}_{j'k'} T^{i'}_{\ m'} = \left( \Gamma^m_{jk} J^m_{\ m'} J^j_{\ j'} J^k_{\ k'} + J^m_{j'k'} \right) T^i_{\ m} J^{i'}_i = \colorbox{#fff20066}{$\Gamma^m_{jk} T^i_{\ m} J^{i'}_i J^j_{\ j'} J^k_{\ k'}$} + \colorbox{#a5d8ff66}{$T^i_{\ s} J^{i'}_i J^j_{\ j'} J^s_{\ m'} J^m_{j'k'}$}. $ Now combining all terms, we observe that: - The terms highlighted in $\colorbox{#fff20066}{yellow}$ reconstruct the expression $ \left( \frac{\partial T^i_{\ j}}{\partial Z^k} + \Gamma^i_{km} T^m_{\ j} - \Gamma^m_{kj} T^i_{\ m} \right) J^{i'}_i J^j_{\ j'} J^k_{\ k'} = \nabla_k T^i_{\ j} \,\, J^{i'}_i J^j_{\ j'} J^k_{\ k'} $ as expected from the tensor transformation law. - The $\colorbox{#a5d8ff66}{blue}$ terms from the derivative and Christoffel expansion cancel due to symmetry in dummy indices, a result consistent with the [[#Tensor property for $ nabla_j T i$|proof]] of the tensor property for $\nabla_j T^i$. - The $\colorbox{#ffa94d66}{orange}$ terms cancel due to the metric identity used in the previous sections (from applying the chain rule to second-order derivatives: $\partial^2 Z^{i'}/\partial Z^i \partial Z^k$): $ J^{i'}_{ik} J^k_{\ k'} + J^k_{k'm'} J^{i'}_i J^m_{\ m'} = 0. $ Thus, we conclude: $ \nabla_{k'} T^{i'}_{\ j'} = \nabla_k T^i_{\ j} \,\, J^{i'}_i J^j_{\ j'} J^k_{\ k'}, $ which proves that $\nabla_k T^i_{\ j}$ transforms as a tensor. ### Case study: A particle moving along a trajectory > [!Notice] > This problem setup has appeared in [[The Christoffel symbol#^876ee2]]. The exercises under this part is more important. Let a particle move along a curve parameterized by $Z^i(t)$, with velocity vector $ \mathbf{V}(t) = \frac{d \mathbf{R}}{dt}, $ where $\mathbf{R}(t)$ is the position vector. Then the components of the velocity in the basis $\mathbf{Z}_i$ are $ V^i = \frac{d Z^i}{dt}. $ From this, we define the acceleration vector $\mathbf{A}(t) = d\mathbf{V}/dt$, whose components are: $ A^i = \frac{d V^i}{dt} + \Gamma^i_{jk} V^j V^k. $ This motivates the definition of the **intrinsic derivative** along the trajectory: $ \frac{\delta U^i}{\delta t} = \frac{d U^i}{dt} + \Gamma^i_{jk} V^j U^k, $ which is the covariant derivative of $\mathbf{U}(t)$ along the direction of motion $\mathbf{V}(t)$. This ensures that if $\mathbf{U}(t)$ is invariant along the trajectory (i.e., "parallel transported"), then: $ \frac{\delta U^i}{\delta t} = 0. $ >[!Notice] >**Parallel transport** is a very important concept in differential geometry. This may be discussed in the future. > [!Note] > The intrinsic derivative generalizes naturally to higher-order derivatives: > $ > \frac{\delta A^i}{\delta t} = \frac{d A^i}{dt} + \Gamma^i_{jk} V^j A^k, > $ > and in general, > $ > \frac{\delta T^i}{\delta t} = \frac{d T^i}{dt} + \Gamma^i_{jk} V^j T^k > $ > for any vector $T^i(t)$ evolving along the curve. This recursive structure defines a covariant time-derivative sequence for velocity, acceleration, and beyond, all preserving tensorial character. This framework connects the geometry of a curved space with the motion of a particle and gives a coordinate-free meaning to time derivatives in general curvilinear coordinates. > [!Example] > Extensions of intrinsic derivative > > 1. **Arbitrary index structure**: > For a general tensor field $T^{i_1 i_2 \cdots}_{\quad\;\; j_1 j_2 \cdots}$ transported along a curve with tangent vector $V^k = \frac{dZ^k}{dt}$, the intrinsic derivative is: > $ > \frac{\delta T^{i_1 i_2 \cdots}_{\quad\;\; j_1 j_2 \cdots}}{\delta t} > = > \frac{d T^{i_1 i_2 \cdots}_{\quad\;\; j_1 j_2 \cdots}}{dt} > + \sum_r \Gamma^{i_r}_{km} V^k T^{i_1 \cdots m \cdots}_{\quad\;\; j_1 \cdots} > - \sum_s \Gamma^{m}_{k j_s} V^k T^{i_1 \cdots}_{\quad\;\; j_1 \cdots m \cdots}. > $ > The Christoffel symbols correct for each upper and lower index. > > 2. **Higher-order derivatives (e.g. jolt)**: > Continuing from acceleration $A^i = \frac{\delta V^i}{\delta t}$, the **jolt** is defined as: > $ > J^i = \frac{\delta A^i}{\delta t} > = \frac{d A^i}{dt} + \Gamma^i_{jk} V^j A^k. > $ > Similarly, this process can be extended recursively to define higher derivatives (snap, etc.), always using: > $ > \frac{\delta X^i}{\delta t} = \frac{d X^i}{dt} + \Gamma^i_{jk} V^j X^k, > $ > where $X^i$ is the previous derivative term along the curve. > [!Notice] > **Intrinsic Derivative and the Role of a Path** > > In a curved space, we cannot directly subtract tensors at different points. However, by introducing a *path*—a parameterized curve $Z^i(t)$—we define a direction along which we can consistently compare tensor values. > > The *intrinsic derivative* along this path is a generalization of the [[The fundamental properties of tensors#The directional derivative|directional derivative]]. For a vector field $\mathbf{U}(t) = U^i(t) \mathbf{Z}_i$ evolving along the curve, the intrinsic derivative is: > $ > \mathbf{U}'(t) = \left( \frac{dU^i}{dt} + \Gamma^i_{jk} V^j U^k \right) \mathbf{Z}_i > $ > where $V^i = \frac{dZ^i}{dt}$ is the velocity along the path. > > This derivative accounts for both the variation of components $U^i$ and the change in the basis $\mathbf{Z}_i$ due to curvature, via the Christoffel symbols $\Gamma^i_{jk}$. > > Therefore, **the path encodes the direction of comparison**, and intrinsic derivative gives the *correct geometric rate of change* of a tensor field along that path. Higher derivatives (acceleration, jolt, etc.) follow the same structure recursively, and this framework naturally supports arbitrary tensor types.