Euclidean Tensor Analysis

The key ideas of tensor calculus on $\mathbb{R}^3$ are now outlined. For the sake of conceptual clarity, the presentation is first carried out in the Cartesian coordinate setting. The generalization to general coordinate systems is considered in the next chapter.

Coordinate systems

A coordinate system on $\mathbb{R}^3$ is an open subset $U \subseteq \mathbb{R}^3$, together with a smooth injective map $\mathsf\phi:\mathbb{R}^3 \to \mathbb{R}^3$ such that $\mathsf\phi^{-1}:\mathsf\phi(U) \to U$ is also smooth; such a map is called a diffeomorphism. The map $\mathsf\phi$ is called a coordinate chart and the open set $U$ is called a coordinate patch on $\mathbb{R}^3$. Given any $\mathsf{x} \in U$, the triple of real numbers $\mathsf\phi(\mathsf{x}) \in \mathbb{R}^3$ are called the curvilinear coordinates of $\mathsf{x}$. What is special about $\mathbb{R}^3$ (this is true for $\mathbb{R}^n$ too) is that it admits a global coordinate system $(\mathbb{R}^3, \mathsf\iota)$, also called the Cartesian coordinate system, where $\mathsf\iota:\mathbb{R}^3 \to \mathbb{R}^3$ is the identity map defined as follows: for any $\mathsf{x} \in \mathbb{R}^3$, $\mathsf\iota(\mathsf{x}) = \mathsf{x}$. The coordinates of $\mathsf{x}$ with respect to this global coordinate system are called the Cartesian coordinates of $\mathsf{x}$.

Remark

A few terminologies inspired by the more general case of a manifold are introduced here. A coordinate system $(U,\mathsf\phi)$ on $\mathbb{R}^3$ is said to be compatible with the Cartesian coordinate system on $\mathbb{R}^3$ if the maps $\mathsf\phi:U \to \mathsf\phi(U)$ and $\mathsf\phi^{-1}:\mathsf\phi(U) \to U$ are smooth and invertible. Note that all coordinate systems considered hereafter are considered to be compatible with the Cartesian coordinate system on $\mathbb{R}^3$. An atlas on $\mathbb{R}^3$ is a collection of compatible coordinate systems $((U_\alpha,\mathsf\phi_\alpha))_{\alpha \in A}$, where $A$ is some index set, such that $\cup_{\alpha \in A} U_\alpha = \mathbb{R}^3$. Note that adding a compatible coordinate system to an atlas yields a larger atlas. Consider now the Cartesian coordinate system on $\mathbb{R}^3$ and construct the maximal atlas compatible with it; this amounts to adding every compatible coordinate system to the atlas consisting of the global Cartesian coordinate system. This maximal atlas is called the standard differentiable structure on $\mathbb{R}^3$.

Remark

The fact that $\mathbb{R}^3$ admits a global coordinate system significantly simplifies the development of calculus on $\mathbb{R}^3$. There, however, exist many sets of practical importance that cannot be covered using a global coordinate system. The modern theory of differentiable manifolds was developed precisely to address these issues, and develop an appropriate extension of the tools of calculus to these sets.

Suppose that $(U,\mathsf\phi)$ is a coordinate system on $\mathbb{R}^3$ that is compatible with the Cartesian coordinate system on $\mathbb{R}^3$. The curvilinear coordinates of $\mathsf{x} \in U$ with respect to the coordinate system $(U,\mathsf\phi)$ is given by the triple of real numbers $\mathsf{y} = (y^1, y^2, y^3) \in \mathbb{R}^3$, where $$\mathsf{y} = \mathsf\phi(\mathsf{x}).$$ The inverse map $\mathsf\phi^{-1}:\mathsf\phi(U) \to U$ permits the computation of $\mathsf{x}$ given $\mathsf{y}$: $$\mathsf{x} = \mathsf\phi^{-1}(\mathsf{y}).$$ A useful simplified notation that will be adopted in the following development is the following: the foregoing relations will often be written as $$\mathsf{y} = \mathsf{y}(\mathsf{x}), \qquad \mathsf{x} = \mathsf{x}(\mathsf{y}).$$ Notice that this is a deliberate abuse of notation: $\mathsf{x}$ stands for both a point in $U$ and a function that takes a point in $\mathsf\phi(U)$ and returns a real number. Despite the ambiguity this introduces, this notation is often adopted in practice since it significantly simplifies the appearance of many equations. The ambiguity will not cause any problems as long as the appropriate meaning of the symbols are inferred from the context.

To quantify precisely the notion that the coordinate system $(U,\mathsf\phi)$ is compatible with the Cartesian coordinate system, note that the map $\mathsf\phi:U \to \mathbb{R}^3$ is smooth, and hence, its Fréchet derivative exists at each $\mathsf{x} \in U$. This implies, in particular that the determinant of the Jacobian matrix is non-zero: $$\text{det} \begin{bmatrix} \partial_1 y^1(\mathsf{x}) & \partial_2 y^1(\mathsf{x}) & \partial_3 y^1(\mathsf{x})\\ \partial_1 y^2(\mathsf{x}) & \partial_2 y^2(\mathsf{x}) & \partial_3 y^2(\mathsf{x})\\ \partial_1 y^3(\mathsf{x}) & \partial_2 y^3(\mathsf{x}) & \partial_3 y^3(\mathsf{x})\end{bmatrix} \neq 0.$$ Here, $\mathsf{y}(\mathsf{x}) = (y^1(\mathsf{x}), y^2(\mathsf{x}), y^3(\mathsf{x}))$. A similar condition holds for the inverse map $\mathsf{x}:\mathsf\phi(U) \to \mathbb{R}^3$.

Tangent spaces

A concept that will prove to be very useful later on is that of a tangent space. Rather than defining this precisely, it suffices for the purposes of this course to present a qualitative definition that provides some intuition about what tangent spaces are.

Let $U \subseteq \mathbb{R}^3$ be an open subset of $\mathbb{R}^3$. At each $\mathsf{x} \in U$, consider the set $T_{\mathsf{x}}U$ defined as follows: $$T_{\mathsf{x}}U = \{\mathsf{v} \in \mathbb{R}^3 \,|\, \text{there exists a (smooth) curve passing through $\mathsf{x}$ with tangent $\mathsf{v}$ at $\mathsf{x}$}\}.$$ This set is called the tangent space at $\mathsf{x}$ to $\mathbb{R}^3$. The point $\mathsf{x}$ is called the base point of the tangent space. In the special case of $\mathbb{R}^3$, given any $\mathsf{v} \in \mathbb{R}^3$, the straight line $\{\mathsf{x} + t\mathsf{v}\,|\,t \in \mathbb{R}\}$ is a smooth curve that passes through $\mathsf{x}$ with tangent $\mathsf{v}$ at $\mathsf{x}$. This shows that $T_{\mathsf{x}}\mathbb{R}^3 \cong \mathbb{R}^3$, and is hence a linear space. Thus the tangent space at any point in $\mathsf{x} \in \mathbb{R}^3$ is a local copy of the whole of $\mathbb{R}^3$ attached to $\mathsf{x}$, and hence need not be distinguished from $\mathbb{R}^3$. For conceptual clarity, however, the tangent space will be explicitly indicated whenever appropriate. Further, each tangent space is assumed to be an inner product space, with the inner product being identical to the standard inner product on $\mathbb{R}^3$. The union of all tangent spaces to $U$ is called the tangent bundle of $U$, and is written as $TU$: $$TU = \cup_{\mathsf{x} \in U} T_{\mathsf{x}}U.$$ Notice that this is a disjoint union.

Remark

For the special case of $U \subseteq \mathbb{R}^3$, where $U$ is open, the tangent space $T_{\mathsf{x}}U$ to $U$ at $\mathsf{x} \in U$ can be equivalently characterized as follows: $$T_{\mathsf{x}}U = \{(\mathsf{x},\mathsf{v}) \,|\, \mathsf{v} \in \mathbb{R}^3\}.$$ The set $T_{\mathsf{x}}U$ defined above acquires a linear structure with addition $+:T_{\mathsf{x}}U \times T_{\mathsf{x}} \to T_{\mathsf{x}}U$ and scalar multiplication $\cdot:\mathbb{R} \times T_{\mathsf{x}} \to T_{\mathsf{x}}U$ defined as follows: for any $\mathsf{v}, \mathsf{w} \in \mathbb{R}^3$ and $a \in \mathbb{R}$, $$(\mathsf{x},\mathsf{v}) + (\mathsf{x},\mathsf{w}) = (\mathsf{x}, \mathsf{v} + \mathsf{w}), \qquad a(\mathsf{x}, \mathsf{v}) = (\mathsf{x}, a\mathsf{v}).$$ Notice now this definition does not affect the base point $\mathsf{x}$: this is necessary to ensure that vector addition and scalar multiplication satisfy the closure property. With this definition of the tangent space, it is easy to check that the tangent bundle is given by $TU = U \times \mathbb{R}^3$.

It follows from the the foregoing discussion that each $T_{\mathsf{x}}U$ can be thought as a local copy of $\mathbb{R}^3$. Thus, the tangent bundle $TU$ can be envisaged as a copy of $\mathbb{R}^3$ attached to each point $\mathsf{x} \in U$. Each of the tangent spaces $T_{\mathsf{x}}U$, where $\mathsf{x} \in U$, can be equipped with a basis of $\mathbb{R}^3$ that could, in general, vary as $\mathsf{x}$ varies over $U$. The simplest choice is to use the same standard basis $(\mathsf{e}_i)$ of $\mathbb{R}^3$ as the basis for every tangent space $T_{\mathsf{x}}U$. With this choice, any $\mathsf{h} \in T_{\mathsf{x}}\mathbb{R}^3$ admits a representation of the form $\mathsf{h} = \sum h_i \mathsf{e}_i$. Note that other choices of bases are possible for each tangent space $T_{\mathsf{x}}\mathbb{R}^3$; this will in fact be the case when curvilinear coordinate systems are considered in a later section.

Tensor fields

Having defined the basic concepts of coordinate systems and tangent spaces, the stage is set to introduce the most important quantities from the point of view of continuum mechanics: tensor fields. In a loose sense, the term field is used here in the sense of a quantity that varies smoothly over a region of $\mathbb{R}^3$. The basic idea behind a tensor field is to associate a tensor to every point in some region of $\mathbb{R}^3$, and to do so in a manner that is smooth. These ideas are developed precisely in this section. Scalar fields, which are tensor fields of order $0$ are discussed first, followed by vector fields, which are tensor fields of order $1$, and finally tensor fields of order $k$ are discussed. Throughout this section $U \subseteq \mathbb{R}^3$ denotes an open subset of $\mathbb{R}^3$. For the coordinate representation of various fields, the global Cartesian coordinate system is used for $\mathbb{R}^3$ and the standard basis of $\mathbb{R}^3$ is used for each $T_{\mathsf{x}}U$ for every $\mathsf{x} \in U$.

Remark

The fact that $\mathbb{R}^3$ has a linear structure will turn out be very convenient for the ensuing development and simplify many of the calculations. Some of the terminology associated with the more general case will however be used when appropriate.

A scalar field is a function of the form $f:U\subseteq\mathbb{R}^3 \to \mathbb{R}$. Thus, the scalar field $f$ associates with each $\mathsf{x} \in U$ the real number $f(\mathsf{x})$. Note that the continuity, differentiability and smoothness of $f$ can be defined according to the definitions given in the previous section in the context of nonlinear maps between vector spaces. A vector field on $U \subseteq \mathbb{R}^3$ is a map of the form $\mathsf{v}:U \to TU$ such that $\mathsf{v}(\mathsf{x}) \in T_{\mathsf{x}}U$ for every $\mathsf{x} \in U$. The vector field $\mathsf{v}$ can be expressed with respect to the Cartesian coordinate system as follows: $$\mathsf{v}(\mathsf{x}) = \sum v_i(\mathsf{x}) \mathsf{e}_i,$$ where each $v_i:U \to \mathbb{R}$ is a scalar field. The continuity/differentiability/smoothness of the vector field $\mathsf{v}$ is defined in terms of the corresponding property of the scalar fields $v_i$. A tensor field of order $k$ on $U \subseteq \mathbb{R}^3$ is defined similarly as a map of the form $$\mathsf{A}:U \to \otimes^k \, TU,$$ such that $\mathsf{A}(\mathsf{x}) \in \otimes^k \, T_{\mathsf{x}}U$ for every $\mathsf{x} \in U$. In writing these expressions, the following notations have been introduced: $$\begin{split} \otimes^k T_{\mathsf{x}}U &= \{ \mathsf{T}:\times^k \, T_{\mathsf{x}}U \to \mathbb{R} \,|\, \mathsf{T}\text{ is multilinear }\} = \mathcal{T}^k(T_{\mathsf{x}}U),\\ \otimes^k TU &= \cup_{\mathsf{x} \in U} \left( \otimes^k T_{\mathsf{x}}U \right) = \cup_{\mathsf{x} \in U} \mathcal{T}^k(T_{\mathsf{x}}U). \end{split}$$ Thus, a tensor field of order $k$ associates with each point in $U$ a tensor of order $k$ over the tangent space at that point. In terms of the Cartesian coordinate system, it is evident that $$\mathsf{A}(\mathsf{x}) = \sum A_{i_1\ldots i_k}(\mathsf{x}) \, \mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k},$$ where each $A_{i_1 \ldots i_k}:U \to \mathbb{R}$ is a scalar field. As in the case of vector fields, the continuity/differentiability/smoothness of tensor fields are naturally defined in terms of the corresponding properties of their component scalar fields.

Remark

Note that a tensor field of order $0$ is a scalar field since, for any $\mathsf{x} \in U$, $\mathcal{T}^0(T_{\mathsf{x}}U) \cong \mathbb{R}$. Similarly, a tensor field of order $1$ is a vector field since, for any $\mathsf{x} \in U$, $\mathcal{T}^1(T_{\mathsf{x}}U) \cong T_{\mathsf{x}}U$. Notice that this is true since each tangent space $T_{\mathsf{x}}U$ is a finite dimensional inner product space. Thus, it suffices to study tensor fields of order $k$ on $U$, where $k \ge 0$, since this includes scalar and vector fields too. For pedagogical reasons, however, scalar, vector and tensor fields will be considered successively in these notes.

Covariant derivative and Gradient

How does a tensor field vary across its domain of definition, and, in particular, how can this change be quantified precisely? This is the fundamental question that is addressed in this and the next few subsections. The fundamental quantity, towards this end, is what is called the covariant derivative of the tensor field. This is a generalization of the familiar notion of the directional derivative encountered earlier.

Let $f:U \subseteq \mathbb{R}^3 \to \mathbb{R}$, where $U$ is an open subset of $\mathbb{R}^3$ be a smooth scalar field on $U$. The directional derivative, also called the covariant derivative, of $f$ at $\mathsf{x} \in U$ along any $\mathsf{v} \in T_{\mathsf{x}}U$ is defined as the scalar $\nabla_{\mathsf{v}}f(\mathsf{x}) \in \mathbb{R}$ such that $$\nabla_{\mathsf{v}}f(\mathsf{x}) = \frac{d}{dt}\bigg\vert_{t=0} f(\mathsf{x} + t\mathsf{v}).$$ Note that in the term $f(\mathsf{x} + t\mathsf{v})$ in the definition above, $\mathsf{x} \in U$ and $\mathsf{v} \in T_{\mathsf{x}}U$. Since $U$ and $T_{\mathsf{x}}U$ are not the same spaces, the quantity $\mathsf{x} + t\mathsf{v}$ is ill-defined, strictly speaking. The saving grace, in this occasion, is the fact that $U$ is an open subset of $\mathbb{R}^3$, which as a linear structure, and since $T_{\mathsf{x}}U \cong \mathbb{R}^3$. Thus, $\mathsf{x} + t\mathsf{v}$ is to be understood as the point in $\mathbb{R}^3$ whose coordinates are obtained by summing the triples of real numbers $\mathsf{x}$ and $t\mathsf{v}$. Note that this point lies inside $U$ when $t$ is sufficiently small, as noted earlier.

Remark

In the more general case when $U$ is not a linear set, the foregoing definition breaks down. The modern theory of differentiable manifolds provides a more rigorous definition of the covariant derivative in more general cases where the argument just given breaks down by introducing an additional structure called a connection on a manifold.

The gradient of $f$ is defined as the vector field $\nabla f:U \to TU$ such that, at every $\mathsf{x} \in U$, $$\nabla f(\mathsf{x}) \cdot \mathsf{v} = \nabla_{\mathsf{v}}f(\mathsf{x}),$$ for any $\mathsf{v} \in T_{\mathsf{x}}U$. Note that $\nabla f(\mathsf{x}) \in T_{\mathsf{x}}U$ by definition. Since $\mathsf{v} \in T_{\mathsf{x}}U$, the dot product $\nabla f(\mathsf{x}) \cdot \mathsf{v}$ is indeed well-defined.

Remark

In this course, the alternative notation $\text{grad }f(\mathsf{x})$ will often be used to indicate $\nabla f(\mathsf{x})$.

The directional derivative of $f$ at $\mathsf{x} \in U$ along $\mathsf{v} \in T_{\mathsf{x}}U$ can be written in terms of the Cartesian coordinate system on $\mathbb{R}^3$ as follows: $$\begin{split} \nabla_{\mathsf{v}}f(\mathsf{x}) &= \frac{d}{dt}\bigg\vert_{t=0} f(x_1 + tv_1, x_2 + tv_2, x_3 + tv_3)\\ &= \sum \partial_i f(\mathsf{x}) v_i. \end{split}$$ Noting that $$\sum \partial_i f(\mathsf{x})v_i = \left(\sum \partial_i f(\mathsf{x}) \, \mathsf{e}_i\right)\cdot\left(\sum v_j \mathsf{e}_j\right),$$ it follows that the Cartesian coordinate representation of the gradient of $f$ is given by $\nabla f(\mathsf{x}) = \sum \partial_i f(\mathsf{x})\,\mathsf{e}_i$.

The covariant derivative and gradient of a smooth vector field $\mathsf{v}:U \to TU$ are defined similarly. The covariant derivative of $\mathsf{v}$ at $\mathsf{x} \in U$ along $\mathsf{w} \in T_{\mathsf{x}}U$ is defined as the vector $\nabla_{\mathsf{w}}\mathsf{v}(\mathsf{x}) \in T_{\mathsf{x}}U$ such that $$\nabla_{\mathsf{w}}\mathsf{v}(\mathsf{x}) = \frac{d}{dt}\bigg\vert_{t=0} \mathsf{v}(\mathsf{x} + t\mathsf{w}).$$ Note that $\mathsf{v}(\mathsf{x}) \in T_{\mathsf{x}}U$, whereas $\mathsf{v}(\mathsf{x} + t\mathsf{w}) \in T_{\mathsf{x} + t\mathsf{w}}U$. Strictly speaking, the vectors $\mathsf{v}(\mathsf{x})$ and $\mathsf{v}(\mathsf{x} + t\mathsf{w})$ cannot be compared since they belong to different vector spaces. However, the fact that both $T_{\mathsf{x}}U$ and $T_{\mathsf{x} + t\mathsf{w}}U$ are isomorphic to $\mathbb{R}^3$ comes to the rescue again. What is actually compared are the corresponding images of these vectors in $\mathbb{R}^3$, which is a valid algebraic operation.

Remark

Using the characterization of $\mathsf{v}(\mathsf{x}) \in T_{\mathsf{x}}U$ as the pair $(\mathsf{x}, \mathsf{v}(\mathsf{x}))$, the derivative in the definition of the covariant derivative of $\mathsf{v}$ can be understood in the following sense: $$\nabla_{\mathsf{w}}\mathsf{v}(\mathsf{x}) = \lim_{t \to 0} \frac{(\mathsf{x},\mathsf{v}(\mathsf{x} + t\mathsf{w})) - (\mathsf{x}, \mathsf{v}(\mathsf{x}))}{t} = \left(\mathsf{x}, \lim_{t \to 0} \frac{\mathsf{v}(\mathsf{x} + t\mathsf{w}) - \mathsf{v}(\mathsf{x})}{t}\right) \in T_{\mathsf{x}}U.$$ Notice how the vector $\mathsf{v}(\mathsf{x} + t\mathsf{w})$ has been parallely shifted from the tangent space $T_{\mathsf{x} + t\mathsf{w}}U$ to $T_{\mathsf{x}}U$. Such a shift is possible because of the fact that the Euclidean space $\mathbb{R}^3$ is flat.

The gradient of the vector field $\mathsf{v}$ is defined as the second order tensor field $\nabla \mathsf{v}:U \to \otimes^2 TU$ such that, at every $\mathsf{x} \in U$, $$\nabla \mathsf{v}(\mathsf{x}) \cdot \mathsf{w} = \nabla_{\mathsf{w}} \mathsf{v}(\mathsf{x}),$$ for every $\mathsf{w} \in T_{\mathsf{x}}U$. Recall that the term $\nabla \mathsf{v}(\mathsf{x}) \cdot \mathsf{w}$ in the definition of the gradient above stands for $\mathcal{C}_{2,3}(\nabla \mathsf{v}(\mathsf{x}) \otimes \mathsf{w})$.

The Cartesian coordinate representation of the covariant derivative of the vector field $\mathsf{v}:U \to TU$ at $\mathsf{x} \in U$ along $\mathsf{w} \in T_{\mathsf{x}}U$ is computed as follows: $$\begin{split} \nabla_{\mathsf{w}}\mathsf{v}(\mathsf{x}) &= \frac{d}{dt}\bigg\vert_{t=0} \sum v_i(x_1 + tw_1, x_2 + tw_2, x_3 + tw_3) \mathsf{e}_i\\ &= \sum \partial_j v_i(\mathsf{x}) w_j \mathsf{e}_i\\ &= \left(\sum \partial_j v_i(\mathsf{x}) \, \mathsf{e}_i \otimes \mathsf{e}_j\right)\cdot\left(\sum w_k \mathsf{e}_k\right). \end{split}$$ In deriving this expression, use has been made of the fact that the standard basis $\mathsf{e}_i$ of $T_{\mathsf{x}}U$ does not change as $\mathsf{x}$ varies over $U$. This calculation thus shows that the gradient of the vector field $\mathsf{v}$ has the following Cartesian coordinate representation: $$\nabla \mathsf{v}(\mathsf{x}) = \sum \partial_j v_i(\mathsf{x}) \, \mathsf{e}_i \otimes \mathsf{e}_j.$$ This calculation also shows that the components of the gradient of a vector field with respect to the Cartesian coordinate system can be represented in matrix form as follows: $$[\nabla \mathsf{v}(\mathsf{x})] = \begin{bmatrix} \partial_1 v_1(\mathsf{x}) & \partial_2 v_1(\mathsf{x}) & \partial_3 v_1(\mathsf{x})\\ \partial_1 v_2(\mathsf{x}) & \partial_2 v_2(\mathsf{x}) & \partial_3 v_2(\mathsf{x})\\ \partial_1 v_3(\mathsf{x}) & \partial_2 v_3(\mathsf{x}) & \partial_3 v_3(\mathsf{x})\end{bmatrix}.$$ As remarked earlier, the alternative notation $\text{grad }\mathsf{v}(\mathsf{x})$ will often be used for $\nabla \mathsf{v}(\mathsf{x})$.

The definition of the covariant derivative of a vector field can be readily extended to the covariant derivative of tensor fields. Suppose that $\mathsf{A}:U \to \otimes^k TU$ is a smooth tensor field of order $k$ on $U$. The covariant derivative of $\mathsf{A}$ at $\mathsf{x} \in U$ along $\mathsf{v} \in T_{\mathsf{x}}U$ is defined as the $k^{\text{th}}$ order tensor $\nabla_{\mathsf{v}} \mathsf{A}(\mathsf{x}) \in \otimes^k T_{\mathsf{x}}U$ such that $$\nabla_{\mathsf{v}} \mathsf{A}(\mathsf{x}) = \frac{d}{dt}\bigg\vert_{t=0} \mathsf{A}(\mathsf{x} + t\mathsf{v}).$$ The gradient of the tensor field $\mathsf{A}$ is defined as the tensor field $\nabla \mathsf{A}:U \to \otimes^{k + 1} TU$ of order $k + 1$ on $U$ such that, at every $\mathsf{x} \in U$, $$\nabla \mathsf{A}(\mathsf{x}) \cdot \mathsf{v} = \nabla_{\mathsf{v}} \mathsf{A}(\mathsf{x}),$$ for every $\mathsf{v} \in T_{\mathsf{x}}U$. As before, the Cartesian coordinate representation of $\nabla \mathsf{A}(\mathsf{x})$ can be worked out to be $$\nabla \mathsf{A}(\mathsf{x}) = \sum \partial_j A_{i_1 \ldots i_k}(\mathsf{x}) \, \mathsf{e}_{i_1} \otimes \ldots \mathsf{e}_{i_k} \otimes \mathsf{e}_j.$$ Notice how the special structure of $\mathbb{R}^3$, and the special choice of the same basis for every tangent space of $U$ have resulted in the simple expressions for gradient of tensor fields.

Divergence

Suppose that $\mathsf{A}:U \to \otimes^k TU$ is a smooth tensor field of order $k$, where $k \ge 1$, over an open subset $U \subseteq \mathbb{R}^3$. The divergence of $\mathsf{A}$ is defined as the smooth tensor field $\nabla \cdot \mathsf{A}:U \to \otimes^{k-1}TU$ of order $k - 1$ on $U$ such that, for any $\mathsf{x} \in U$, $$\nabla \cdot \mathsf{A} (\mathsf{x}) = \mathcal{C}_{k,k+1} \nabla \mathsf{A}(\mathsf{x}).$$ Note that the gradient $\nabla \mathsf{A}:U \to \otimes^{k+1} TU$ of $\mathsf{A}$ is a tensor field of order $k + 1$ on $U$, whence its contraction is a tensor field of order $k - 1$, as required. Note also since $\mathsf{A}$ is a tensor field of order $k$, there are $k$ possible contractions of $\nabla \mathsf{A}$. Each of these defines a distinct notion of divergence, and the convention adopted here to call the $(k,k+1)$ contraction of $\nabla \mathsf{A}$ as the divergence of $\mathsf{A}$ is just a matter of convention that will turn out to be helpful later on.

Remark

The notation $\text{div }\mathsf{A}$ will also be used in this course to denote the divergence $\nabla \cdot \mathsf{A}$ of the tensor field $\mathsf{A}$.

In terms of the Cartesian coordinate system on $\mathbb{R}^3$, it is straightforward to check that $$\nabla \cdot \mathsf{A}(\mathsf{x}) = \sum \partial_a A_{i_1 \ldots i_{k - 1}a}(\mathsf{x}) \, \mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_{k-1}}.$$ It is helpful to work out a few examples to illustrate the definition of the divergence of a tensor field. Suppose that $\mathsf{v}:U \to TU$ is a smooth vector field on $U$. The divergence of $\mathsf{v}$ is a scalar field on $U$ that is easily computed as follows: for any $\mathsf{x} \in U$, $$\nabla \cdot \mathsf{v}(\mathsf{x}) = \mathcal{C}_{1,2} \nabla \mathsf{v}(\mathsf{x}) = \sum \partial_i v_i(\mathsf{x}).$$ Similarly, given a smooth second order tensor field $\mathsf{A}:U \to \otimes^2 TU$ on $U$, its divergence is a vector field on $U$ that is obtained as follows: for any $\mathsf{x} \in U$, $$\nabla \cdot \mathsf{A}(\mathsf{x}) = \mathcal{C}_{2,3} \nabla \mathsf{A}(\mathsf{x}) = \sum \partial_j A_{ij}(\mathsf{x}) \mathsf{e}_i.$$ These expressions agree with the familiar expressions for divergence from elementary calculus.

Curl

The curl operation, specific to tensor fields over $\mathbb{R}^3$, is introduced now. The curl of a tensor field of order $k$, where $k > 1$ is defined recursively in terms of the curl of a tensor field of order $k - 1$. Hence, the curl of a vector field is discussed first, followed by the general definition and an illustration of how to compute the curl of a second order tensor field.

Suppose that $\mathsf{v}:U \to \otimes^k TU$ is a smooth vector field over an open subset $U \subseteq \mathbb{R}^3$. The curl of $\mathsf{v}$ is defined as the smooth vector field $\nabla \times \mathsf{v}:U \to \otimes TU$ such that, for any $\mathsf{x} \in U$, $$\mathsf{w} \cdot \nabla \times \mathsf{v}(\mathsf{x}) = \nabla \cdot (\mathsf{v} \times \mathsf{w})(\mathsf{x}).$$ In the definition above, $\mathsf{w}:U \to TU$ is a constant vector field on $U$: this can be thought of as the set $\cup_{\mathsf{x} \in U} (\mathsf{x}, \mathsf{w})$. Since $\mathsf{w}$ is a constant vector field, its dependence on $\mathsf{x}$ will be suppressed. The vector field $\mathsf{v} \times \mathsf{w}:U \to \otimes TU$ is defined as follows: for any $\mathsf{x} \in U$, $\mathsf{v} \times \mathsf{w}(\mathsf{x}) = \mathsf{v}(\mathsf{x}) \times \mathsf{w}$.

Remark

The alternative notation $\text{curl }\mathsf{v}$ will often be used to denote $\nabla \times \mathsf{v}$.

To see how this definition of the curl of a vector field matches up with the familiar expression from undergraduate calculus, it is helpful to work out the Cartesian coordinate represention of $\nabla \times \mathsf{v}$. Since the curl of $\mathsf{v}$ is computed from the divergence of $\mathsf{v}$, which, in turn, is computed from the gradient of $\mathsf{v}$, the gradient of the vector field $\mathsf{v} \times \mathsf{w}:U \to TU$ is computed first: for any $\mathsf{x} \in U$ and $\mathsf{z} \in T_{\mathsf{x}}U$, $$\begin{split} \nabla (\mathsf{v} \times \mathsf{w})(\mathsf{x}) \cdot \mathsf{z} &= \frac{d}{dt}\bigg\vert_{t=0} \mathsf{v}(\mathsf{x} + t\mathsf{z}) \times \mathsf{w}\\ &= \frac{d}{dt}\bigg\vert_{t=0} \sum \epsilon_{ijk} v_j(x_1 + tz_1, x_2 + tz_2, x_3 + tz_3) \, w_k \mathsf{e}_i\\ &= \left(\sum \epsilon_{ijk} \partial_a v_j(\mathsf{x}) w_k \mathsf{e}_i \otimes \mathsf{e}_a\right) \cdot \left(\sum z_b \mathsf{e}_b\right). \end{split}$$ This calculation thus yields the Cartesian coordinate representation of $\nabla (\mathsf{v} \times \mathsf{w})(\mathsf{x})$ as $$\nabla (\mathsf{v} \times \mathsf{w})(\mathsf{x}) = \sum \epsilon_{ijk} \partial_a v_j(\mathsf{x}) w_k \mathsf{e}_i \otimes \mathsf{e}_a.$$ The Cartesian coordinate representation of the divergence of the vector field $\mathsf{v} \times \mathsf{w}$ immediately follows as $$\begin{split} \nabla \cdot (\mathsf{v} \times \mathsf{w})(\mathsf{x}) &= \sum \epsilon_{ijk} \partial_i v_j(\mathsf{x}) w_k\\ &= \left(\sum \epsilon_{kij} \partial_i v_j(\mathsf{x}) \, \mathsf{e}_k\right) \cdot \left(\sum w_l \mathsf{e}_l\right). \end{split}$$ Comparing the final expression just obtained with $\mathsf{w} \cdot \nabla \times \mathsf{v}(\mathsf{x})$ yields the following expression for the Cartesian coordinate representation of the curl of the vector field $\mathsf{v}$: $$\nabla \times \mathsf{v}(\mathsf{x}) = \sum \epsilon_{ijk} \partial_j v_k(\mathsf{x}) \, \mathsf{e}_i,$$ which is the familiar expression for the curl of a vector field from undergraduate calculus.

Remark

The foregoing result can more easily be obtained by using the Cartesian coordinate expression for the divergence of a vector field. The reason for illustrating it in the longer version is to show the following dependence tree of the various differential quantities introduced in this section: $$\text{Covariant derivative} \to \text{Gradient} \to \text{Divergence} \to \text{Curl}.$$ The covariant derivative, which is a generalization of the directional derivative, is thus the fundamental starting point for all these computations.

Before introducing the definition of the curl of a tensor field on $U$, it is pertinent to point out two useful identities. First, given any smooth vector field $\mathsf{v}:U \to TU$, the following result holds: $\nabla \cdot \nabla \times \mathsf{v} = \mathsf{0}$, where $\mathsf{0}:U \to TU$ is the vector field that takes every $\mathsf{x} \in U$ to the zero vector $\mathsf{0} \in T_{\mathsf{x}}U$. This is most easily seen by using Cartesian coordinates: for any $\mathsf{x} \in U$, $$\nabla \cdot \nabla \times \mathsf{v}(\mathsf{x}) = \sum \partial_i \epsilon_{ijk} \partial_j v_k(\mathsf{x}) = \sum \epsilon_{ijk} \partial_i \partial_j v_k(\mathsf{x}) = 0.$$ The last step follows from the equality of mixed partial derivatives and the properties of the Levi-Civita symbols. A second identity that is useful in practice is the following: if $f:U \to \mathbb{R}$ is a smooth scalar field, then $\nabla \times \nabla f = \mathsf{0}$. This is also easily proved in a Cartesian coordinate setting: for any $\mathsf{x} \in U$, $$\nabla \times \nabla f(\mathsf{x}) = \sum \epsilon_{ijk} \partial_j \partial_k f(\mathsf{x}) \mathsf{e}_i = 0.$$

Remark

Despite the fact that a special coordinate system, namely the Cartesian coordinate system, was used to prove the identities $\text{div curl }\mathsf{v} = \mathsf{0}$ and $\text{curl grad }f = \mathsf{0}$, it should be noted that the final expression obtained in the coordinate setting, when cast back into the coordinate-free form, is still valid in any coordinate system. This trick will be used throughout this course to simplify a variety of calculations since it is significantly easier to work in a Cartesian coordinate setting.

The curl of a smooth tensor field $\mathsf{A}:U \to \otimes^k TU$ of order $k$ over $U$, is defined as the smooth tensor field $\nabla \times \mathsf{A}:U \to \otimes^k TU$ of order $k$ on $U$ such that, for any $\mathsf{x} \in U$, $$\mathsf{w} \cdot \nabla \times \mathsf{A}(\mathsf{x}) = \nabla \times (\mathsf{w} \cdot \mathsf{A})(\mathsf{x}).$$ Here, $\mathsf{w}:U \to TU$ is a constant vector field on $U$, as before. Notice that this is a recursive definition. The quantity $\mathsf{w} \cdot \mathsf{A}$ is a tensor field of order $k-1$ over $U$. To see this, note that $\mathsf{w}\cdot\mathsf{A}$ has the following Cartesian coordinate representation: for any $\mathsf{x} \in U$, $$\mathsf{w} \cdot \mathsf{A}(\mathsf{x}) = \sum w_{i_1} A_{i_1i_2 \ldots i_k}(\mathsf{x}) \, \mathsf{g}_{i_2} \otimes \ldots \otimes \mathsf{g}_{i_k} \in \otimes^{k-1} T_{\mathsf{x}}U.$$ This shows that $\mathsf{w}\cdot\mathsf{A}:U \to \otimes^{k-1}TU$ is indeed a smooth tensor field of order $k - 1$ on $U$. The curl of the tensor field $\mathsf{w} \cdot \mathsf{A}$ is, in turn, defined using the curl of a tensor field of order $k - 2$, and so on, until a vector field, whose curl is computable directly, is obtained.

To illustrate the recursive definition, it is instructive to work out the curl of a second order tensor field $\mathsf{A}:U \to \otimes^2 TU$ on $U$. Note that, in this case, $\mathsf{w} \cdot \mathsf{A}:U \to TU$, where $\mathsf{w}:U \to TU$ is a constant vector field on $U$, is a smooth vector field on $U$. Hence, using Cartesian coordinates, it is evident that for any $\mathsf{x} \in U$, $$\nabla \times (\mathsf{w} \cdot \mathsf{A})(\mathsf{x}) = \sum \epsilon_{ijk} \partial_j (w_a A_{ak}(\mathsf{x})) \mathsf{e}_i = \left(\sum w_b \mathsf{e}_b\right) \cdot \left(\sum \epsilon_{ijk} \partial_j A_{ak}(\mathsf{x}) \, \mathsf{e}_a \otimes \mathsf{e}_i\right).$$ Using the definition $\mathsf{w} \cdot \text{curl }\mathsf{A} = \text{curl }(\mathsf{w} \cdot \mathsf{A})$, it follows that $$\nabla \times \mathsf{A}(\mathsf{x}) = \sum \epsilon_{ijk} \partial_j A_{lk}(\mathsf{x}) \, \mathsf{e}_l \otimes \mathsf{e}_i.$$

Remark

Notice how the order of the terms in the dot product $\mathsf{w} \cdot \mathsf{A}$ used in the definition of the curl operation decides the final coordinate expression of the curl of a tensor field. The choice to use $\mathsf{w} \cdot \mathsf{A}$, instead of $\mathsf{A} \cdot \mathsf{w}$ is purely conventional, and is chosen with an eye towards certain applications.

Integration of tensor fields

The integration of scalar, vector and tensor fields on $\mathbb{R}^3$ is now outlined. In particular, attention is focused on line, surface and volume integrals of scalar fields on $\mathbb{R}^3$. Certain important integral theorems that are very useful for continuum mechanics are finally discussed. To keep the discussion simple, the Cartesian coordinate system is adopted throughout this section.

Volume integrals

Given a scalar field $f:U \subseteq \mathbb{R}^3 \to \mathbb{R}$ on an open subset $U$ of $\mathbb{R}^3$. The (volume) integral of $f$ over $U$ is written as $$\int_U f(\mathsf{x})\,dv = \int_U f(x_1,x_2,x_3) \, dx_1 dx_2 dx_3.$$ The integral in the expression above is a triple integral with limits defined such that they cover $U$. The integral is understood in the usual Riemannian sense.

Remark

It is more useful to interpret the integral $\int_U f(x_1, x_2, x_3)\,dx_1dx_2dx_3$ in the sense of Lebesgue. Since this course will not focus on the analytic aspects on continuum mechanics, it is sufficient for the present purposes to interpret the integral as the limit of a Riemannian sum.

Remark

The volume element $dv$ in $\int_U f(\mathsf{x}) \, dv$ can be given a precise meaning as a differential $3$-form on $\mathbb{R}^3$. Since this is outside the scope of these notes, it is sufficient to think of $dv$ as just a useful symbol.

If $\mathsf{A}:U \to \otimes^k TU$ is a tensor field of order $k$ on $U$, recall that the Cartesian coordinate representation of $\mathsf{A}$ is given by $$\mathsf{A}(\mathsf{x}) = \sum A_{i_1\ldots i_k}(x_1, x_2, x_3) \, \mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k},$$ for any $\mathsf{x} \in U$. Note that each of the components $A_{i_1\ldots i_k}:U \to \mathbb{R}$ is a scalar field on $U$. The (volume) integral of $\mathsf{A}$ over $U$ can be defined in terms of its Cartesian components as follows: $$\int_U \mathsf{A}(\mathsf{x}) \, dv = \sum \left(\int_U A_{i_1\ldots i_k}(x_1, x_2, x_3)\,dx_1dx_2dx_3\right) \mathsf{e}_{i_1} \otimes \ldots \otimes \mathsf{e}_{i_k}.$$ Despite the fact that the Cartesian coordinate system is used to define the integral above, the change of variables formula ensures that the value of the integral does not depend on the specific choice of coordinate system. The expression for the integral in a general coordinate system will be presented in a later section.

Remark

It is important to note that the definition of the integral of tensor fields is, in general, not defined over differentiable manifolds. What are integrable are special tensor fields called differential forms. In the special case of Euclidean spaces $\mathbb{R}^n$, and in particular, in $\mathbb{R}^3$, this problem does not arise, and the integral of tensor fields is defined in terms of the integral of its component fields, as done here.

Line integrals

A curve in $\mathbb{R}^3$ is a smooth and injective map of the form $\mathsf\gamma:I\subseteq\mathbb{R} \to \mathbb{R}^3$, where $I$ is an open subset of $\mathbb{R}$. Thus, for every $t \in I$, there is a point $(\gamma_1(t), \gamma_2(t), \gamma_3(t)) \in \mathbb{R}^3$ and the set of all such points constitute the curve $\mathsf\gamma$. Let the line $C = \mathsf\gamma(I) \subseteq \mathbb{R}^3$ be the image of the curve in $\mathbb{R}^3$. Note that a curve is not be identified with its image here. The tangent vector to the curve $\mathsf\gamma$ at $\mathsf\gamma(t)$ is the vector $\dot{\mathsf\gamma}(t) = (\dot{\gamma}_1(t), \dot{\gamma}_2(t), \dot{\gamma}_3(t)) \in T_{\mathsf\gamma(t)}C$, where $\dot{\gamma}_i(t)$ denotes the derivative of the function $\gamma_i:I \to \mathbb{R}$ at $t$. It is assumed that $\dot{\mathsf\gamma} \neq \mathsf{0}$ for every $t \in I$. A curve that has this property is called a regular curve.

Remark

The tangent space $T_{\mathsf{x}}C$ to the curve $C$ at $\mathsf{x} \in C$ is defined in a manner similar to the tangent space of an open subset of $\mathbb{R}^3$. Thus, the tangent space to $C$ at $\mathsf{x}$ consists of all possible tangent vectors to $C$ at $\mathsf{x}$. Note that $T_{\mathsf{x}}C \subset T_{\mathsf{x}}\mathbb{R}^3$.

Given a scalar field $f:U \subseteq \mathbb{R}^3 \to \mathbb{R}$ where $U$ is open and $C \subseteq U$, the line integral of $f$ over $C$ is defined as follows: $$\int_C f(\mathsf{x}) \, ds = \int_I (f \circ \mathsf\gamma)(t) \lVert \dot{\mathsf\gamma}(t)\rVert \, dt.$$

Remark

The line element $ds$ used in the line integral $\int_C f(\mathsf{x})\,ds$ can be given a precise definition as a differential $1$-form on $\mathbb{R}^3$. For the present purposes, though, it is sufficient to think of $ds$ as just a useful symbol.

Note that the definition of the line integral provided here depends on the choice of a curve $\mathsf\gamma$. Such a curve is called a parametrization of the line $C$. It can be shown that the value of the line integral is independent of the choice of parametrization, thereby establishing that the definition of the integral is well-defined.

The line integral of tensor fields over a regular curve is defined in terms of the line integral of its component scalar fields, as in the case of the volume integral.

Surface integrals

To define surface integrals, it is necessary to introduce a few basic concepts. A parametrized surface in $\mathbb{R}^3$ is a smooth and injective map of the form $\mathsf{a}:R \subseteq \mathbb{R}^2 \to \mathbb{R}^3$. The components of $\mathsf{a}$ are the maps $\mathsf{a}_i:R \to \mathbb{R}^3$, where $i = 1,2,3$. Let the surface $S = \mathsf{a}(R) \subseteq \mathbb{R}^3$ be the image of the parametrized surface $\mathsf{a}$ in $\mathbb{R}^3$. For any $(r,s) \in R$, the tangents to the surface $S$ at $\mathsf{a}(r,s) \in S$ are defined as the vectors $\mathsf{t}_1(r,s), \mathsf{t}_2(r,s) \in T_{\mathsf{a}(r,s)}S$ given by $$\mathsf{t}_i(r,s) = \partial_i \mathsf{a}(r,s) = \sum \partial_i a_j(r,s) \mathsf{e}_j, \qquad i = 1,2.$$ The map $\mathsf{a}:R \to \mathbb{R}^3$ is called a regular surface if $\mathsf{t}_1(r,s) \times \mathsf{t}_2(r,s) \neq \mathsf{0}$ for every $(r,s) \in R$.

Remark

The tangent space to the surface $S$ at $\mathsf{x} \in S$ is defined along the same lines as the tangent to a curve in $\mathbb{R}^3$ or an open subset of $\mathbb{R}^3$. Thus $T_{\mathsf{x}}S$ consists of all possible tangent vectors to $S$ at the point $\mathsf{x}$. Note that $T_{\mathsf{x}}S \subset T_{\mathsf{x}}\mathbb{R}^3$.

Let $\mathsf{t}_1(r,s),\mathsf{t}_2(r,s) \in T_{\mathsf{a}(r,s)}S$ be the tangent vectors at $\mathsf{a}(r,s) \in S$, where $(r,s) \in R$, to a regular surface $S$. The unit normal to the surface $S$ at $\mathsf{a}(r,s)$ is defined as the vector $\mathsf{n}(r,s) \in T_{\mathsf{a}(r,s)}\mathbb{R}^3$ given by $$\mathsf{n}(r,s) = \frac{\mathsf{t}_1(r,s) \times \mathsf{t}_2(r,s)}{\lVert\mathsf{t}_1(r,s) \times \mathsf{t}_2(r,s)\rVert}.$$ Note that it is necessary for the surface $S$ to be regular for the unit normal to be well-defined. It is noted, without proof, that different parametrizations of $S$ yield the same unit normal vector, up to a sign, at any point on $S$. One of the two possible unit normal vectors at point on $S$ is defined as the positive unit normal, and the set of all parametrizations that yield this positive unit normal are said to constitute an orientation on $S$.

If $f:U \subseteq \mathbb{R}^3 \to \mathbb{R}$, where $U$ is open in $\mathbb{R}^3$, is a smooth scalar field, and $\mathsf{a}:R \subseteq \mathbb{R}^2 \to \mathbb{R}^3$ is a regular surface such that $S = \mathsf{a}(R) \subseteq U$, then the surface integral of $f$ over $S$ is defined as follows: $$\int_S f(\mathsf{x}) \, da = \int_R (f \circ \mathsf{a})(r,s) \lVert \partial_1 \mathsf{a}(r,s) \times \partial_2 \mathsf{a}(r,s)\rVert \, dr ds.$$ As in the case of the line integral, it can be shown that this definition of the surface integral of the scalar field $f$ is independent of the choice of the parametrization $\mathsf{a}$.

Remark

The area element $da$ in $\int_S f(\mathsf{x}) \, da$ can be given a precise definition as a differential $2$-form on $\mathbb{R}^3$. As remarked before, it is sufficient to consider $da$ as just a useful symbol for this course.

The surface integral of tensor fields over a regular surface is defined in terms of the surface integral of its component scalar fields, as in the case of the volume integral.

Divergence and Gauss theorems

To conclude this section, two important theorems are presented without proof. The divergence theorem provides a means to convert volume integrals into an integral over the surface bounding the volume, and Stokes’ theorem provides a means to reduce a surface integral to a line integral over the bounding curve of the surface.

Suppose that $\mathsf{A}:U \to \otimes^k TU$ is a tensor field of order $k$ on an open subset $U \subseteq \mathbb{R}^3$. Suppose further that the boundary of $U$ is the set $\partial U$. Then, the divergence theorem states that $$\int_U \nabla \mathsf{A}(\mathsf{x}) \, dv = \int_{\partial U} \mathsf{A}(\mathsf{x}) \otimes \mathsf{n}(\mathsf{x}) \, da,$$ where $\mathsf{n}(\mathsf{x})$ is the outward unit normal at $\mathsf{x} \in \partial U$. It is to be noted that the statement of the divergence theorem presented here is valid only for tensor fields defined on Euclidean spaces.

A few special cases are now presented to illustrate the divergence theorem. First, if $f:U \to \mathbb{R}$, where $U \subseteq \mathbb{R}^3$ is open, is a scalar field, then the divergence theorem takes the form $$\int_U \nabla f(\mathsf{x}) \, dv = \int_{\partial U} f(\mathsf{x})\mathsf{n}(\mathsf{x})\,da.$$ If $\mathsf{v}:U \to TU$ is a vector field on $U$, the statement of the divergence theorem reads $$\int_U \nabla \mathsf{v}(\mathsf{x})\,dv = \int_{\partial U} \mathsf{v}(\mathsf{x}) \otimes \mathsf{n}(\mathsf{x})\,da.$$ Taking the trace this equation yields the familiar form of the divergence theorem: $$\int_U \nabla \cdot \mathsf{v}(\mathsf{x}) \, dv = \int_{\partial U} \mathsf{v}(\mathsf{x}) \cdot \mathsf{n}(\mathsf{x}) \, da.$$ Finally, if $\mathsf{A}:U \to \otimes^2 TU$ represents a second order tensor field on $U$, the foregoing argument can be extended to yield the following form of the divergence theorem: $$\int_U \nabla \cdot \mathsf{A}(\mathsf{x}) \, dv = \int_{\partial U} \mathsf{A}(\mathsf{x})\cdot\mathsf{n}(\mathsf{x})\,da.$$ This statement of the divergence theorem will also prove to be quite handy later on in the study of continuum mechanics.

A second theorem that is useful in applications relates surface and line integrals. Specifically, if $\mathsf{v}:U \to TU$ is a vector field on an open subset $U$ of $\mathbb{R}^3$, and $S$ is a regular surface in $U$ with boundary $\partial S$, then Stokes’ theorem states that $$\int_S \nabla \times \mathsf{v}(\mathsf{x}) \cdot \mathsf{n}(\mathsf{x}) \, da = \int_{\partial S} \mathsf{v}(\mathsf{x}) \cdot \mathsf{t}(\mathsf{x}) \, ds.$$ Here $\mathsf{t}(\mathsf{x}) \in T_{\mathsf{x}}\partial S$ is the tangent vector at the point $\mathsf{x} \in \partial S$ on the boundary curve $\partial S$ of $S$.

Remark

There is an elegant theory using the notion of differential forms that generalizes these theorems to the general case of differentiable manifolds. The generalized Stokes’ theorem unifies both the theorems presented above and takes a remarkably simple form in the language of differential forms.