Skip to content

Appendix A Solutions

← Back to problems　|　Back to chapter

Table of Contents

Basic

Medium

Advanced

Basic

B-1. Component Calculation of the Cross Product

→ Back to problem

Problem: Calculate the cross product \(\boldsymbol{a} \times \boldsymbol{b}\) of \(\boldsymbol{a} = (2, 1, -1)\) and \(\boldsymbol{b} = (0, 3, 2)\).

Solution strategy: Expand the formal determinant by cofactors along the first row.

\[ \boldsymbol{a} \times \boldsymbol{b} = \begin{vmatrix} \boldsymbol{e}_1 & \boldsymbol{e}_2 & \boldsymbol{e}_3 \\ 2 & 1 & -1 \\ 0 & 3 & 2 \end{vmatrix} \]

Calculation:

\[ (\boldsymbol{a} \times \boldsymbol{b})_x = 1 \cdot 2 - (-1) \cdot 3 = 2 + 3 = 5 \]
\[ (\boldsymbol{a} \times \boldsymbol{b})_y = -(2 \cdot 2 - (-1) \cdot 0) = -(4 - 0) = -4 \]
\[ (\boldsymbol{a} \times \boldsymbol{b})_z = 2 \cdot 3 - 1 \cdot 0 = 6 \]
\[ \boxed{\boldsymbol{a} \times \boldsymbol{b} = (5,\, -4,\, 6)} \]

Verification: By the anticommutativity property, \(\boldsymbol{b} \times \boldsymbol{a} = (-5, 4, -6)\) should hold. Indeed, \((\boldsymbol{b} \times \boldsymbol{a})_x = 3 \cdot (-1) - 2 \cdot 1 = -5\) ✓

B-2. Orthogonality of the Cross Product

→ Back to problem

Problem: Verify that \(\boldsymbol{a} \times \boldsymbol{b} = (5, -4, 6)\) is orthogonal to \(\boldsymbol{a} = (2, 1, -1)\).

Calculation:

\[ (\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{a} = 5 \cdot 2 + (-4) \cdot 1 + 6 \cdot (-1) = 10 - 4 - 6 = 0 \]
\[ \boxed{(\boldsymbol{a} \times \boldsymbol{b}) \cdot \boldsymbol{a} = 0} \]

Verification: Checking the dot product with \(\boldsymbol{b}\) as well: \(5 \cdot 0 + (-4) \cdot 3 + 6 \cdot 2 = 0 - 12 + 12 = 0\) ✓

B-3. Scalar Triple Product

→ Back to problem

Problem: Compute the scalar triple product of \(\boldsymbol{a} = (1, 0, 2)\), \(\boldsymbol{b} = (3, 1, 0)\), \(\boldsymbol{c} = (0, -1, 1)\) as a determinant.

Calculation:

\[ \boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c}) = \begin{vmatrix} 1 & 0 & 2 \\ 3 & 1 & 0 \\ 0 & -1 & 1 \end{vmatrix} \]

Cofactor expansion along the first row:

\[ = 1 \cdot \begin{vmatrix} 1 & 0 \\ -1 & 1 \end{vmatrix} - 0 \cdot \begin{vmatrix} 3 & 0 \\ 0 & 1 \end{vmatrix} + 2 \cdot \begin{vmatrix} 3 & 1 \\ 0 & -1 \end{vmatrix} \]
\[ = 1 \cdot (1 \cdot 1 - 0 \cdot (-1)) - 0 + 2 \cdot (3 \cdot (-1) - 1 \cdot 0) \]
\[ = 1 \cdot 1 + 2 \cdot (-3) = 1 - 6 = -5 \]
\[ \boxed{\boldsymbol{a} \cdot (\boldsymbol{b} \times \boldsymbol{c}) = -5} \]

Verification (Sarrus' method):

Sum of downward-right diagonals: \(a_{11}a_{22}a_{33} + a_{12}a_{23}a_{31} + a_{13}a_{21}a_{32} = 1 \cdot 1 \cdot 1 + 0 \cdot 0 \cdot 0 + 2 \cdot 3 \cdot (-1) = 1 + 0 - 6 = -5\)

Sum of downward-left diagonals: \(a_{13}a_{22}a_{31} + a_{12}a_{21}a_{33} + a_{11}a_{23}a_{32} = 2 \cdot 1 \cdot 0 + 0 \cdot 3 \cdot 1 + 1 \cdot 0 \cdot (-1) = 0 + 0 + 0 = 0\)

Determinant \(= -5 - 0 = -5\) ✓

B-4. Gradient of a Scalar Field

→ Back to problem

Problem: Find the gradient of \(\varphi(x, y, z) = x^2 y + y^2 z + z^2 x\).

Calculation:

\[ \frac{\partial \varphi}{\partial x} = 2xy + z^2, \quad \frac{\partial \varphi}{\partial y} = x^2 + 2yz, \quad \frac{\partial \varphi}{\partial z} = y^2 + 2zx \]
\[ \boxed{\nabla\varphi = (2xy + z^2,\; x^2 + 2yz,\; y^2 + 2zx)} \]

Verification: Check the dimensions (partial derivatives of each monomial in \(\varphi\) with respect to each variable). \(\partial(x^2 y)/\partial x = 2xy\) ✓, \(\partial(y^2 z)/\partial x = 0\) ✓, \(\partial(z^2 x)/\partial x = z^2\) ✓. The other components follow similarly.

B-5. Divergence of a Vector Field

→ Back to problem

Problem: Compute the divergence of \(\boldsymbol{F} = (x^2 y,\; y^2 z,\; z^2 x)\).

Calculation:

\[ \nabla \cdot \boldsymbol{F} = \frac{\partial(x^2 y)}{\partial x} + \frac{\partial(y^2 z)}{\partial y} + \frac{\partial(z^2 x)}{\partial z} \]
\[ = 2xy + 2yz + 2zx \]
\[ \boxed{\nabla \cdot \boldsymbol{F} = 2(xy + yz + zx)} \]

Verification: Checking each partial derivative individually. \(\partial(x^2 y)/\partial x = 2xy\) (treating \(y\) as a constant) ✓

B-6. Curl of a Vector Field

→ Back to problem

Problem: Compute the curl of \(\boldsymbol{F} = (yz,\; xz,\; xy)\).

Calculation:

\[ (\nabla \times \boldsymbol{F})_x = \frac{\partial(xy)}{\partial y} - \frac{\partial(xz)}{\partial z} = x - x = 0 \]
\[ (\nabla \times \boldsymbol{F})_y = \frac{\partial(yz)}{\partial z} - \frac{\partial(xy)}{\partial x} = y - y = 0 \]
\[ (\nabla \times \boldsymbol{F})_z = \frac{\partial(xz)}{\partial x} - \frac{\partial(yz)}{\partial y} = z - z = 0 \]
\[ \boxed{\nabla \times \boldsymbol{F} = \boldsymbol{0}} \]

Verification: Since \(\boldsymbol{F} = (yz, xz, xy) = \nabla(xyz)\), the curl of a gradient field is identically zero. Indeed, \(\partial(xyz)/\partial x = yz\), \(\partial(xyz)/\partial y = xz\), \(\partial(xyz)/\partial z = xy\) ✓

B-7. Laplacian of a Scalar Field

→ Back to problem

Problem: Compute the Laplacian of \(\varphi = e^x \sin y + z^3\).

Calculation:

\[ \frac{\partial \varphi}{\partial x} = e^x \sin y, \quad \frac{\partial^2 \varphi}{\partial x^2} = e^x \sin y \]
\[ \frac{\partial \varphi}{\partial y} = e^x \cos y, \quad \frac{\partial^2 \varphi}{\partial y^2} = -e^x \sin y \]
\[ \frac{\partial \varphi}{\partial z} = 3z^2, \quad \frac{\partial^2 \varphi}{\partial z^2} = 6z \]
\[ \Delta\varphi = e^x \sin y - e^x \sin y + 6z = 6z \]
\[ \boxed{\Delta\varphi = 6z} \]

Verification: The \(e^x \sin y\) part is a two-dimensional harmonic function (\(\Delta_{2D}(e^x \sin y) = 0\)), so its contribution to the Laplacian is zero. The Laplacian of \(z^3\) is \(6z\). Consistent ✓

B-8. Verification of the Lagrange Identity

→ Back to problem

Problem: For \(\boldsymbol{a} = (1, 2, 0)\) and \(\boldsymbol{b} = (0, 1, 3)\), verify the Lagrange identity.

Calculation (right-hand side):

\[ |\boldsymbol{a}|^2 = 1 + 4 + 0 = 5, \quad |\boldsymbol{b}|^2 = 0 + 1 + 9 = 10 \]
\[ \boldsymbol{a} \cdot \boldsymbol{b} = 0 + 2 + 0 = 2 \]
\[ |\boldsymbol{a}|^2|\boldsymbol{b}|^2 - (\boldsymbol{a} \cdot \boldsymbol{b})^2 = 50 - 4 = 46 \]

Calculation (left-hand side):

\[ \boldsymbol{a} \times \boldsymbol{b} = (2 \cdot 3 - 0 \cdot 1,\; 0 \cdot 0 - 1 \cdot 3,\; 1 \cdot 1 - 2 \cdot 0) = (6, -3, 1) \]
\[ |\boldsymbol{a} \times \boldsymbol{b}|^2 = 36 + 9 + 1 = 46 \]
\[ \boxed{|\boldsymbol{a} \times \boldsymbol{b}|^2 = |\boldsymbol{a}|^2|\boldsymbol{b}|^2 - (\boldsymbol{a} \cdot \boldsymbol{b})^2 = 46} \]

Both sides agree ✓

B-9. Verification of the BAC-CAB Formula

→ Back to problem

Problem: Verify the BAC-CAB formula with \(\boldsymbol{a} = (1,0,0)\), \(\boldsymbol{b} = (0,1,0)\), \(\boldsymbol{c} = (0,0,1)\).

Calculating the left-hand side:

\[ \boldsymbol{b} \times \boldsymbol{c} = (0,1,0) \times (0,0,1) = (1 \cdot 1 - 0 \cdot 0,\; 0 \cdot 0 - 0 \cdot 1,\; 0 \cdot 0 - 1 \cdot 0) = (1, 0, 0) \]
\[ \boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c}) = (1,0,0) \times (1,0,0) = (0, 0, 0) = \boldsymbol{0} \]

Calculating the right-hand side:

\[ \boldsymbol{a} \cdot \boldsymbol{c} = 1 \cdot 0 + 0 \cdot 0 + 0 \cdot 1 = 0 \]
\[ \boldsymbol{a} \cdot \boldsymbol{b} = 1 \cdot 0 + 0 \cdot 1 + 0 \cdot 0 = 0 \]
\[ \boldsymbol{b}(\boldsymbol{a} \cdot \boldsymbol{c}) - \boldsymbol{c}(\boldsymbol{a} \cdot \boldsymbol{b}) = (0,1,0) \cdot 0 - (0,0,1) \cdot 0 = \boldsymbol{0} \]
\[ \boxed{\text{LHS} = \text{RHS} = \boldsymbol{0}} \]

Verification: Since \(\boldsymbol{a} = \boldsymbol{e}_1\), \(\boldsymbol{b} = \boldsymbol{e}_2\), \(\boldsymbol{c} = \boldsymbol{e}_3\) are mutually orthogonal standard basis vectors, \(\boldsymbol{a} \cdot \boldsymbol{b} = \boldsymbol{a} \cdot \boldsymbol{c} = 0\) is natural. Also, since \(\boldsymbol{b} \times \boldsymbol{c} = \boldsymbol{e}_1 = \boldsymbol{a}\), the result \(\boldsymbol{a} \times \boldsymbol{a} = \boldsymbol{0}\) is also natural ✓

B-10. Verifying rot(grad) = 0

→ Back to problem

Problem: Verify that \(\nabla \times (\nabla\varphi) = \boldsymbol{0}\) for \(\varphi = x^2 + y^2 + z^2\).

Calculation:

\[ \nabla\varphi = (2x,\; 2y,\; 2z) \]

Let \(\boldsymbol{G} = \nabla\varphi = (2x, 2y, 2z)\):

\[ (\nabla \times \boldsymbol{G})_x = \frac{\partial(2z)}{\partial y} - \frac{\partial(2y)}{\partial z} = 0 - 0 = 0 \]
\[ (\nabla \times \boldsymbol{G})_y = \frac{\partial(2x)}{\partial z} - \frac{\partial(2z)}{\partial x} = 0 - 0 = 0 \]
\[ (\nabla \times \boldsymbol{G})_z = \frac{\partial(2y)}{\partial x} - \frac{\partial(2x)}{\partial y} = 0 - 0 = 0 \]
\[ \boxed{\nabla \times (\nabla\varphi) = \boldsymbol{0}} \]

Verification: This is a concrete example of the identity \(\operatorname{curl}(\operatorname{grad}\varphi) = \boldsymbol{0}\). Each component automatically vanishes due to the symmetry of mixed partial derivatives (since \(\varphi\) is of class \(C^2\)), e.g., \(\frac{\partial^2 \varphi}{\partial y \partial z} - \frac{\partial^2 \varphi}{\partial z \partial y} = 0\) ✓

Medium

M-1. Proof of the BAC-CAB Formula

→ Back to problem

Problem: Prove the BAC-CAB formula by comparing the \(x\) components.

Solution Strategy

Expand the \(x\) component of the left-hand side \(\boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c})\) by applying the definition of the cross product twice. Also expand the \(x\) component of the right-hand side \(\boldsymbol{b}(\boldsymbol{a} \cdot \boldsymbol{c}) - \boldsymbol{c}(\boldsymbol{a} \cdot \boldsymbol{b})\), and show that both are equal.

Detailed Calculation

\(x\) component of the left-hand side:

First, write out the components of \(\boldsymbol{b} \times \boldsymbol{c}\):

\[ (\boldsymbol{b} \times \boldsymbol{c})_1 = b_2 c_3 - b_3 c_2, \quad (\boldsymbol{b} \times \boldsymbol{c})_2 = b_3 c_1 - b_1 c_3, \quad (\boldsymbol{b} \times \boldsymbol{c})_3 = b_1 c_2 - b_2 c_1 \]

The \(x\) component of \(\boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c})\) is:

\[ [\boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c})]_1 = a_2 (\boldsymbol{b} \times \boldsymbol{c})_3 - a_3 (\boldsymbol{b} \times \boldsymbol{c})_2 \]
\[ = a_2(b_1 c_2 - b_2 c_1) - a_3(b_3 c_1 - b_1 c_3) \]
\[ = a_2 b_1 c_2 - a_2 b_2 c_1 - a_3 b_3 c_1 + a_3 b_1 c_3 \]
\[ = b_1(a_2 c_2 + a_3 c_3) - c_1(a_2 b_2 + a_3 b_3) \]

Now add and subtract \(a_1 c_1\) and \(a_1 b_1\) (since \(b_1 a_1 c_1 - c_1 a_1 b_1 = 0\), the value is unchanged):

\[ = b_1(a_1 c_1 + a_2 c_2 + a_3 c_3) - c_1(a_1 b_1 + a_2 b_2 + a_3 b_3) \]
\[ = b_1(\boldsymbol{a} \cdot \boldsymbol{c}) - c_1(\boldsymbol{a} \cdot \boldsymbol{b}) \]

\(x\) component of the right-hand side:

\[ [\boldsymbol{b}(\boldsymbol{a} \cdot \boldsymbol{c}) - \boldsymbol{c}(\boldsymbol{a} \cdot \boldsymbol{b})]_1 = b_1(\boldsymbol{a} \cdot \boldsymbol{c}) - c_1(\boldsymbol{a} \cdot \boldsymbol{b}) \]

Final Answer

The \(x\) components of the left-hand side and right-hand side are equal:

\[ [\boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c})]_1 = b_1(\boldsymbol{a} \cdot \boldsymbol{c}) - c_1(\boldsymbol{a} \cdot \boldsymbol{b}) \]

Regarding the \(y\) and \(z\) components: The calculation above uses only the cyclic structure of the indices \(1, 2, 3\). Since the definition of the cross product has the same form under cyclic permutations of the indices \((1 \to 2 \to 3 \to 1)\), the proof for the \(y\) component is obtained by replacing \(1 \to 2\), \(2 \to 3\), \(3 \to 1\) in the \(x\) component proof, and the proof for the \(z\) component is obtained by replacing \(1 \to 3\), \(2 \to 1\), \(3 \to 2\).

Therefore, the equality holds for all three components, and the BAC-CAB formula is proved. \(\blacksquare\)

Verification

Confirmed with a specific numerical example (standard basis vectors) in D9 ✓

M-2. Proof that div(curl) = 0

→ Back to problem

Problem: Prove that \(\operatorname{div}(\operatorname{rot}\boldsymbol{F}) = 0\) and state its physical meaning.

Solution Strategy

Write out each component of \(\operatorname{rot}\boldsymbol{F}\), then take the \(\operatorname{div}\). By exchanging the order of partial derivatives, all terms pair up and cancel.

Detailed Calculation

Let \(\boldsymbol{F} = (F_1, F_2, F_3)\). The components of the curl are:

\[ (\nabla \times \boldsymbol{F})_1 = \frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}, \quad (\nabla \times \boldsymbol{F})_2 = \frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}, \quad (\nabla \times \boldsymbol{F})_3 = \frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y} \]

Taking the divergence:

\[ \nabla \cdot (\nabla \times \boldsymbol{F}) = \frac{\partial}{\partial x}\left(\frac{\partial F_3}{\partial y} - \frac{\partial F_2}{\partial z}\right) + \frac{\partial}{\partial y}\left(\frac{\partial F_1}{\partial z} - \frac{\partial F_3}{\partial x}\right) + \frac{\partial}{\partial z}\left(\frac{\partial F_2}{\partial x} - \frac{\partial F_1}{\partial y}\right) \]
\[ = \frac{\partial^2 F_3}{\partial x \partial y} - \frac{\partial^2 F_2}{\partial x \partial z} + \frac{\partial^2 F_1}{\partial y \partial z} - \frac{\partial^2 F_3}{\partial y \partial x} + \frac{\partial^2 F_2}{\partial z \partial x} - \frac{\partial^2 F_1}{\partial z \partial y} \]

If \(\boldsymbol{F}\) is sufficiently smooth (class \(C^2\)), the order of partial differentiation is interchangeable: \(\frac{\partial^2 F_i}{\partial x_j \partial x_k} = \frac{\partial^2 F_i}{\partial x_k \partial x_j}\). Therefore:

  • \(\frac{\partial^2 F_3}{\partial x \partial y}\) and \(-\frac{\partial^2 F_3}{\partial y \partial x}\) cancel
  • \(-\frac{\partial^2 F_2}{\partial x \partial z}\) and \(\frac{\partial^2 F_2}{\partial z \partial x}\) cancel
  • \(\frac{\partial^2 F_1}{\partial y \partial z}\) and \(-\frac{\partial^2 F_1}{\partial z \partial y}\) cancel
\[ \boxed{\nabla \cdot (\nabla \times \boldsymbol{F}) = 0} \]

Physical Meaning

In electromagnetism, the magnetic field \(\boldsymbol{B}\) can be written using a vector potential \(\boldsymbol{A}\) as \(\boldsymbol{B} = \nabla \times \boldsymbol{A}\). Applying the identity above:

\[ \nabla \cdot \boldsymbol{B} = \nabla \cdot (\nabla \times \boldsymbol{A}) = 0 \]

This corresponds to one of Maxwell's equations, \(\nabla \cdot \boldsymbol{B} = 0\) (the non-existence of magnetic monopoles). In other words, the fact that the magnetic field can be expressed as the curl of a vector potential is automatically consistent with the non-existence of magnetic monopoles, by virtue of the mathematical identity \(\operatorname{div}(\operatorname{rot}) = 0\). Conversely, if magnetic monopoles were to exist, then \(\nabla \cdot \boldsymbol{B} \neq 0\), and \(\boldsymbol{B}\) could no longer be globally expressed as the curl of a single vector potential. \(\blacksquare\)

M-3. Inner Product of Two Cross Products Formula

→ Back to problem

Problem: Derive the Lagrange identity from the formula for the inner product of two cross products, and show that \(|\boldsymbol{a} \times \boldsymbol{b}| = |\boldsymbol{a}||\boldsymbol{b}|\sin\theta\).

Detailed Calculation

Derivation of the Lagrange identity:

The formula for the inner product of two cross products:

\[ (\boldsymbol{a} \times \boldsymbol{b}) \cdot (\boldsymbol{c} \times \boldsymbol{d}) = (\boldsymbol{a} \cdot \boldsymbol{c})(\boldsymbol{b} \cdot \boldsymbol{d}) - (\boldsymbol{a} \cdot \boldsymbol{d})(\boldsymbol{b} \cdot \boldsymbol{c}) \]

Substituting \(\boldsymbol{c} = \boldsymbol{a}\), \(\boldsymbol{d} = \boldsymbol{b}\):

\[ (\boldsymbol{a} \times \boldsymbol{b}) \cdot (\boldsymbol{a} \times \boldsymbol{b}) = (\boldsymbol{a} \cdot \boldsymbol{a})(\boldsymbol{b} \cdot \boldsymbol{b}) - (\boldsymbol{a} \cdot \boldsymbol{b})(\boldsymbol{b} \cdot \boldsymbol{a}) \]
\[ \boxed{|\boldsymbol{a} \times \boldsymbol{b}|^2 = |\boldsymbol{a}|^2|\boldsymbol{b}|^2 - (\boldsymbol{a} \cdot \boldsymbol{b})^2} \]

Derivation of \(|\boldsymbol{a} \times \boldsymbol{b}| = |\boldsymbol{a}||\boldsymbol{b}|\sin\theta\):

Let \(\theta\) (\(0 \leq \theta \leq \pi\)) be the angle between \(\boldsymbol{a}\) and \(\boldsymbol{b}\). Since \(\boldsymbol{a} \cdot \boldsymbol{b} = |\boldsymbol{a}||\boldsymbol{b}|\cos\theta\):

\[ |\boldsymbol{a} \times \boldsymbol{b}|^2 = |\boldsymbol{a}|^2|\boldsymbol{b}|^2 - |\boldsymbol{a}|^2|\boldsymbol{b}|^2\cos^2\theta = |\boldsymbol{a}|^2|\boldsymbol{b}|^2(1 - \cos^2\theta) = |\boldsymbol{a}|^2|\boldsymbol{b}|^2\sin^2\theta \]

Since \(\sin\theta \geq 0\) for \(0 \leq \theta \leq \pi\), taking the positive square root of both sides:

\[ \boxed{|\boldsymbol{a} \times \boldsymbol{b}| = |\boldsymbol{a}||\boldsymbol{b}|\sin\theta} \]

Verification

When \(\theta = 0\) (parallel): \(|\boldsymbol{a} \times \boldsymbol{b}| = 0\) ✓ (the cross product of parallel vectors is zero)

When \(\theta = \pi/2\) (orthogonal): \(|\boldsymbol{a} \times \boldsymbol{b}| = |\boldsymbol{a}||\boldsymbol{b}|\) ✓ (for example, \(\boldsymbol{e}_1 \times \boldsymbol{e}_2 = \boldsymbol{e}_3\) with \(|\boldsymbol{e}_3| = 1 = 1 \cdot 1\))

When \(\theta = \pi\) (antiparallel): \(|\boldsymbol{a} \times \boldsymbol{b}| = 0\) ✓ (\(\sin\pi = 0\), and the cross product of antiparallel vectors is also zero)

Also confirmed with the numerical example from D8 ✓

M-4. Product Rule for Divergence

→ Back to problem

Problem: Prove that \(\operatorname{div}(\varphi\,\boldsymbol{F}) = (\nabla\varphi) \cdot \boldsymbol{F} + \varphi\,(\operatorname{div}\boldsymbol{F})\).

Solution Strategy

Write \(\operatorname{div}(\varphi\,\boldsymbol{F})\) in component form and apply the Leibniz product rule to expand each term.

Detailed Calculation

Let \(\boldsymbol{F} = (F_1, F_2, F_3)\).

\[ \operatorname{div}(\varphi\,\boldsymbol{F}) = \frac{\partial(\varphi F_1)}{\partial x} + \frac{\partial(\varphi F_2)}{\partial y} + \frac{\partial(\varphi F_3)}{\partial z} \]

Applying the Leibniz product rule \(\frac{\partial(\varphi F_i)}{\partial x_i} = \frac{\partial\varphi}{\partial x_i}F_i + \varphi\frac{\partial F_i}{\partial x_i}\) to each term:

\[ \frac{\partial(\varphi F_1)}{\partial x} = \frac{\partial\varphi}{\partial x}F_1 + \varphi\frac{\partial F_1}{\partial x} \]
\[ \frac{\partial(\varphi F_2)}{\partial y} = \frac{\partial\varphi}{\partial y}F_2 + \varphi\frac{\partial F_2}{\partial y} \]
\[ \frac{\partial(\varphi F_3)}{\partial z} = \frac{\partial\varphi}{\partial z}F_3 + \varphi\frac{\partial F_3}{\partial z} \]

Summing all terms:

\[ \operatorname{div}(\varphi\,\boldsymbol{F}) = \left(\frac{\partial\varphi}{\partial x}F_1 + \frac{\partial\varphi}{\partial y}F_2 + \frac{\partial\varphi}{\partial z}F_3\right) + \varphi\left(\frac{\partial F_1}{\partial x} + \frac{\partial F_2}{\partial y} + \frac{\partial F_3}{\partial z}\right) \]

The first bracket is the dot product of the gradient with the vector field \((\nabla\varphi) \cdot \boldsymbol{F}\), and the second bracket is the divergence of the vector field \(\operatorname{div}\boldsymbol{F}\), so:

\[ \boxed{\operatorname{div}(\varphi\,\boldsymbol{F}) = (\nabla\varphi) \cdot \boldsymbol{F} + \varphi\,(\operatorname{div}\boldsymbol{F})} \]

\(\blacksquare\)

Verification

We verify with concrete examples. Let \(\varphi = x\), \(\boldsymbol{F} = (y, 0, 0)\):

  • Left-hand side: \(\operatorname{div}(xy, 0, 0) = \frac{\partial(xy)}{\partial x} + 0 + 0 = y\)
  • Right-hand side: \((\nabla x) \cdot (y, 0, 0) + x \cdot \operatorname{div}(y, 0, 0) = (1, 0, 0) \cdot (y, 0, 0) + x \cdot 0 = y + 0 = y\)

Match ✓

Another example. Let \(\varphi = x^2 + y^2 + z^2\), \(\boldsymbol{F} = (x, y, z)\):

  • \(\varphi\boldsymbol{F} = ((x^2+y^2+z^2)x,\; (x^2+y^2+z^2)y,\; (x^2+y^2+z^2)z)\)
  • Partial derivative of the \(x\)-component on the left-hand side: \(\frac{\partial}{\partial x}[(x^2+y^2+z^2)x] = (2x)x + (x^2+y^2+z^2) = 2x^2 + x^2+y^2+z^2\). Computing the \(y\) and \(z\) components similarly and summing: \(2(x^2+y^2+z^2) + 3(x^2+y^2+z^2) = 5(x^2+y^2+z^2)\)
  • Right-hand side: \((\nabla\varphi) \cdot \boldsymbol{F} + \varphi(\operatorname{div}\boldsymbol{F}) = (2x,2y,2z)\cdot(x,y,z) + (x^2+y^2+z^2) \cdot 3 = 2(x^2+y^2+z^2) + 3(x^2+y^2+z^2) = 5(x^2+y^2+z^2)\)

Match ✓

M-5. Gauss's Theorem and the Coulomb Field

→ Back to problem

Problem: Using Gauss's theorem, show that the electric flux of a Coulomb field is \(q/\varepsilon_0\).

Solution Strategy

Inside an arbitrary closed surface \(S\) containing the origin, take a sphere \(S_\epsilon\) of sufficiently small radius \(\epsilon\) centered at the origin. In the region \(V'\) between \(S\) and \(S_\epsilon\), since \(r \neq 0\), we have \(\operatorname{div}\boldsymbol{E} = 0\). Applying Gauss's theorem to this region, we reduce the surface integral over \(S\) to a surface integral over \(S_\epsilon\).

Detailed Calculation

Step 1: Verifying \(\operatorname{div}\boldsymbol{E} = 0\) for \(r \neq 0\)

For \(\boldsymbol{E} = \frac{q}{4\pi\varepsilon_0}\frac{\boldsymbol{r}}{r^3}\), we compute the divergence of \(\frac{\boldsymbol{r}}{r^3} = \left(\frac{x}{(x^2+y^2+z^2)^{3/2}},\; \frac{y}{(x^2+y^2+z^2)^{3/2}},\; \frac{z}{(x^2+y^2+z^2)^{3/2}}\right)\).

Partial derivative of the \(x\)-component:

\[ \frac{\partial}{\partial x}\frac{x}{r^3} = \frac{r^3 - x \cdot 3r^2 \cdot (x/r)}{r^6} = \frac{r^3 - 3x^2 r}{r^6} = \frac{1}{r^3} - \frac{3x^2}{r^5} \]

Similarly computing the \(y\) and \(z\) components and summing:

\[ \nabla \cdot \frac{\boldsymbol{r}}{r^3} = \left(\frac{1}{r^3} - \frac{3x^2}{r^5}\right) + \left(\frac{1}{r^3} - \frac{3y^2}{r^5}\right) + \left(\frac{1}{r^3} - \frac{3z^2}{r^5}\right) = \frac{3}{r^3} - \frac{3(x^2 + y^2 + z^2)}{r^5} = \frac{3}{r^3} - \frac{3}{r^3} = 0 \quad (r \neq 0) \]

Step 2: Applying Gauss's Theorem

Consider the region \(V'\) enclosed between \(S\) and \(S_\epsilon\). The boundary of \(V'\) consists of the outer surface \(S\) (with outward-pointing normal) and the inner surface \(S_\epsilon\) (whose outward normal with respect to \(V'\) points toward the origin, i.e., in the \(-\hat{\boldsymbol{r}}\) direction).

Since \(\operatorname{div}\boldsymbol{E} = 0\) in \(V'\), by Gauss's theorem:

\[ 0 = \int_{V'} \nabla \cdot \boldsymbol{E}\, dV = \oint_S \boldsymbol{E} \cdot d\boldsymbol{S}_{\text{outward}} + \oint_{S_\epsilon} \boldsymbol{E} \cdot d\boldsymbol{S}_{V'\text{-outward}} \]

On \(S_\epsilon\), the outward normal with respect to \(V'\) is in the \(-\hat{\boldsymbol{r}}\) direction, so \(d\boldsymbol{S}_{V'\text{-outward}} = -\hat{\boldsymbol{r}}\, dA\). Therefore:

\[ 0 = \oint_S \boldsymbol{E} \cdot d\boldsymbol{S} - \oint_{S_\epsilon} \boldsymbol{E} \cdot \hat{\boldsymbol{r}}\, dA \]
\[ \oint_S \boldsymbol{E} \cdot d\boldsymbol{S} = \oint_{S_\epsilon} \boldsymbol{E} \cdot \hat{\boldsymbol{r}}\, dA \]

Step 3: Evaluating the Surface Integral over the Sphere

On \(S_\epsilon\), \(r = \epsilon\) (constant), and \(\boldsymbol{E}\) points in the radial direction with constant magnitude:

\[ \boldsymbol{E}\big|_{S_\epsilon} = \frac{q}{4\pi\varepsilon_0}\frac{\hat{\boldsymbol{r}}}{\epsilon^2} \]

Therefore:

\[ \boldsymbol{E} \cdot \hat{\boldsymbol{r}} = \frac{q}{4\pi\varepsilon_0\epsilon^2} \]

Integrating over the entire sphere:

\[ \oint_{S_\epsilon} \boldsymbol{E} \cdot \hat{\boldsymbol{r}}\, dA = \frac{q}{4\pi\varepsilon_0\epsilon^2} \cdot 4\pi\epsilon^2 = \frac{q}{\varepsilon_0} \]

Final Answer

\[ \boxed{\oint_S \boldsymbol{E} \cdot d\boldsymbol{S} = \frac{q}{\varepsilon_0}} \]

This holds for any closed surface \(S\) containing the origin. \(\blacksquare\)

Verification

  • Dimensional analysis: \([E] = \text{V/m}\), \([dS] = \text{m}^2\), so \([E \cdot dS] = \text{V·m}\). Simplifying \([q/\varepsilon_0] = \text{C}/(\text{C}^2\text{s}^2/(\text{kg·m}^3))\) gives \(\text{kg·m}^3/(\text{C·s}^2) = \text{V·m}\). Consistent ✓
  • Special case: When \(S\) itself is a sphere of radius \(R\), direct calculation gives \(\frac{q}{4\pi\varepsilon_0 R^2} \cdot 4\pi R^2 = q/\varepsilon_0\) ✓
  • Case \(q < 0\): The electric field points toward the origin (in the \(-\hat{\boldsymbol{r}}\) direction), so its inner product with the outward normal is negative. The flux \(q/\varepsilon_0 < 0\), which is consistent ✓

Advanced

A-1. Identities Using the Levi-Civita Symbol

→ Back to problem

Problem: Index calculations using the Levi-Civita symbol.

(a) Formula for the inner product of two cross products

Detailed calculation:

\[ (\boldsymbol{a} \times \boldsymbol{b}) \cdot (\boldsymbol{c} \times \boldsymbol{d}) = \sum_i (\boldsymbol{a} \times \boldsymbol{b})_i (\boldsymbol{c} \times \boldsymbol{d})_i = \sum_i \left(\sum_{j,k} \varepsilon_{ijk}\, a_j\, b_k\right)\left(\sum_{l,m} \varepsilon_{ilm}\, c_l\, d_m\right) \]
\[ = \sum_{j,k,l,m} \left(\sum_i \varepsilon_{ijk}\,\varepsilon_{ilm}\right) a_j\, b_k\, c_l\, d_m \]

Applying the contraction formula \(\sum_i \varepsilon_{ijk}\,\varepsilon_{ilm} = \delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl}\):

\[ = \sum_{j,k,l,m} (\delta_{jl}\delta_{km} - \delta_{jm}\delta_{kl})\, a_j\, b_k\, c_l\, d_m \]
\[ = \sum_{j,k} a_j\, b_k\, c_j\, d_k - \sum_{j,k} a_j\, b_k\, c_k\, d_j \]
\[ = \left(\sum_j a_j c_j\right)\left(\sum_k b_k d_k\right) - \left(\sum_j a_j d_j\right)\left(\sum_k b_k c_k\right) \]
\[ \boxed{= (\boldsymbol{a} \cdot \boldsymbol{c})(\boldsymbol{b} \cdot \boldsymbol{d}) - (\boldsymbol{a} \cdot \boldsymbol{d})(\boldsymbol{b} \cdot \boldsymbol{c})} \]

\(\blacksquare\)

(b) BAC-CAB formula

Detailed calculation:

\[ [\boldsymbol{a} \times (\boldsymbol{b} \times \boldsymbol{c})]_i = \sum_{j,k} \varepsilon_{ijk}\, a_j\, (\boldsymbol{b} \times \boldsymbol{c})_k = \sum_{j,k} \varepsilon_{ijk}\, a_j \sum_{l,m} \varepsilon_{klm}\, b_l\, c_m \]
\[ = \sum_{j,l,m} \left(\sum_k \varepsilon_{ijk}\,\varepsilon_{klm}\right) a_j\, b_l\, c_m \]

Since \(\varepsilon_{klm} = \varepsilon_{lmk}\) (invariant under cyclic permutations), we have \(\sum_k \varepsilon_{ijk}\,\varepsilon_{klm} = \sum_k \varepsilon_{ijk}\,\varepsilon_{lmk}\).

Applying the contraction formula \(\sum_k \varepsilon_{ijk}\,\varepsilon_{lmk} = \delta_{il}\delta_{jm} - \delta_{im}\delta_{jl}\):

\[ = \sum_{j,l,m} (\delta_{il}\delta_{jm} - \delta_{im}\delta_{jl})\, a_j\, b_l\, c_m \]
\[ = \sum_{j} a_j\, b_i\, c_j - \sum_{j} a_j\, b_j\, c_i \]
\[ = b_i \sum_j a_j c_j - c_i \sum_j a_j b_j \]
\[ = b_i (\boldsymbol{a} \cdot \boldsymbol{c}) - c_i (\boldsymbol{a} \cdot \boldsymbol{b}) \]
\[ \boxed{= [\boldsymbol{b}(\boldsymbol{a} \cdot \boldsymbol{c}) - \boldsymbol{c}(\boldsymbol{a} \cdot \boldsymbol{b})]_i} \]

Since this holds for all \(i\), the BAC-CAB formula is proved. \(\blacksquare\)

(c) Proof of \(\operatorname{div}(\operatorname{curl}\boldsymbol{F}) = 0\) by index calculation

The \(i\)-th component of \(\operatorname{curl}\boldsymbol{F}\):

\[ (\nabla \times \boldsymbol{F})_i = \sum_{j,k} \varepsilon_{ijk}\, \partial_j F_k \]

where \(\partial_j = \frac{\partial}{\partial x_j}\).

Calculation of \(\operatorname{div}(\operatorname{curl}\boldsymbol{F})\):

\[ \nabla \cdot (\nabla \times \boldsymbol{F}) = \sum_i \partial_i (\nabla \times \boldsymbol{F})_i = \sum_i \partial_i \sum_{j,k} \varepsilon_{ijk}\, \partial_j F_k = \sum_{i,j,k} \varepsilon_{ijk}\, \partial_i \partial_j F_k \]

Here, \(\partial_i \partial_j F_k\) is symmetric in the indices \(i, j\) (when \(\boldsymbol{F}\) is of class \(C^2\), \(\partial_i \partial_j = \partial_j \partial_i\)), while \(\varepsilon_{ijk}\) is antisymmetric in the indices \(i, j\) (\(\varepsilon_{ijk} = -\varepsilon_{jik}\)).

The contraction of a symmetric tensor with an antisymmetric tensor vanishes. To show this explicitly:

\[ S \equiv \sum_{i,j,k} \varepsilon_{ijk}\, \partial_i \partial_j F_k \]

Swapping the dummy indices \(i \leftrightarrow j\):

\[ S = \sum_{j,i,k} \varepsilon_{jik}\, \partial_j \partial_i F_k = \sum_{i,j,k} (-\varepsilon_{ijk})\, \partial_i \partial_j F_k = -S \]

(In the first equality, we relabeled the dummy variables \(i \leftrightarrow j\); in the second equality, we used \(\varepsilon_{jik} = -\varepsilon_{ijk}\) (antisymmetry) and \(\partial_j \partial_i = \partial_i \partial_j\) (commutativity of partial derivatives).)

Therefore \(S = -S\), i.e., \(2S = 0\), giving:

\[ \boxed{\nabla \cdot (\nabla \times \boldsymbol{F}) = 0} \]

\(\blacksquare\)

Connection to four dimensions: In the main text on general relativity, the 4-dimensional Levi-Civita tensor \(\varepsilon^{\mu\nu\rho\sigma}\) is used to construct dual tensors. Just as the 3-dimensional \(\varepsilon_{ijk}\) defines the curl, the 4-dimensional \(\varepsilon^{\mu\nu\rho\sigma}\) defines the dual of the electromagnetic field tensor \(F_{\mu\nu}\): \({}^*F^{\mu\nu} = \frac{1}{2}\varepsilon^{\mu\nu\rho\sigma}F_{\rho\sigma}\). The 4-dimensional version of \(\operatorname{div}(\operatorname{curl}) = 0\) corresponds to the Bianchi identity \(\partial_{[\mu}F_{\nu\rho]} = 0\), which yields the homogeneous Maxwell equations (\(\nabla \cdot \boldsymbol{B} = 0\) and Faraday's law).

Verification

(a) Setting \(\boldsymbol{c} = \boldsymbol{a}\), \(\boldsymbol{d} = \boldsymbol{b}\) yields Lagrange's identity, consistent with S3 ✓

(b) Consistent with the numerical example in D9 ✓

(c) Consistent with the component-wise proof in S2 ✓

A-2. Applications of Stokes' Theorem

→ Back to problem

Problem: Arguments using Stokes' theorem.

(a) The closed curve integral of a gradient field is zero

Proof:

Let \(\boldsymbol{F} = \nabla\varphi\). For any closed curve \(C\), take a surface \(S\) whose boundary is \(C\). By Stokes' theorem:

\[ \oint_C \boldsymbol{F} \cdot d\boldsymbol{r} = \int_S (\nabla \times \boldsymbol{F}) \cdot d\boldsymbol{S} = \int_S (\nabla \times \nabla\varphi) \cdot d\boldsymbol{S} \]

From the identity \(\nabla \times (\nabla\varphi) = \boldsymbol{0}\) (verified in D10; in general this follows from the commutativity of partial derivatives):

\[ \boxed{\oint_C \nabla\varphi \cdot d\boldsymbol{r} = \int_S \boldsymbol{0} \cdot d\boldsymbol{S} = 0} \]

\(\blacksquare\)

Alternative proof (direct method without using Stokes' theorem): Parametrize \(C\) by \(\boldsymbol{r}(t)\) (\(t: a \to b\), \(\boldsymbol{r}(a) = \boldsymbol{r}(b)\)):

\[ \oint_C \nabla\varphi \cdot d\boldsymbol{r} = \int_a^b \nabla\varphi \cdot \frac{d\boldsymbol{r}}{dt}\, dt = \int_a^b \frac{d\varphi(\boldsymbol{r}(t))}{dt}\, dt = \varphi(\boldsymbol{r}(b)) - \varphi(\boldsymbol{r}(a)) = 0 \]

(Since the curve is closed, the start and end points coincide.) This alternative proof does not require Stokes' theorem. ✓

(b) An irrotational field possesses a scalar potential

Proof:

Assume that \(\nabla \times \boldsymbol{F} = \boldsymbol{0}\) holds in a simply connected region \(D\).

Step 1: Construction of \(\varphi\)

Choose a fixed point \(\boldsymbol{r}_0 \in D\), and for any point \(\boldsymbol{r} \in D\), define:

\[ \varphi(\boldsymbol{r}) = \int_{\boldsymbol{r}_0}^{\boldsymbol{r}} \boldsymbol{F} \cdot d\boldsymbol{r}' \]

where the integral is taken along any curve from \(\boldsymbol{r}_0\) to \(\boldsymbol{r}\) within \(D\).

Step 2: Path independence

Take two paths \(C_1\) and \(C_2\) from \(\boldsymbol{r}_0\) to \(\boldsymbol{r}\). Connecting \(C_1\) and \(C_2^{-1}\) (the reverse of \(C_2\)) yields a closed curve \(C = C_1 + C_2^{-1}\). Since \(D\) is simply connected, there exists a surface \(S \subset D\) with \(C\) as its boundary. By Stokes' theorem:

\[ \oint_C \boldsymbol{F} \cdot d\boldsymbol{r} = \int_S (\nabla \times \boldsymbol{F}) \cdot d\boldsymbol{S} = \int_S \boldsymbol{0} \cdot d\boldsymbol{S} = 0 \]

Therefore:

\[ \int_{C_1} \boldsymbol{F} \cdot d\boldsymbol{r}' - \int_{C_2} \boldsymbol{F} \cdot d\boldsymbol{r}' = 0 \]

Thus \(\varphi(\boldsymbol{r})\) is well-defined regardless of the choice of path.

Step 3: Verifying \(\nabla\varphi = \boldsymbol{F}\)

Consider the straight-line path from \(\boldsymbol{r}\) to \(\boldsymbol{r} + \delta\boldsymbol{r}\) (taking \(\delta\boldsymbol{r} = \delta x\, \boldsymbol{e}_1\)):

\[ \varphi(\boldsymbol{r} + \delta x\, \boldsymbol{e}_1) - \varphi(\boldsymbol{r}) = \int_{\boldsymbol{r}}^{\boldsymbol{r} + \delta x\, \boldsymbol{e}_1} \boldsymbol{F} \cdot d\boldsymbol{r}' \approx F_1(\boldsymbol{r})\, \delta x \]

Therefore:

\[ \frac{\partial\varphi}{\partial x} = \lim_{\delta x \to 0} \frac{\varphi(\boldsymbol{r} + \delta x\, \boldsymbol{e}_1) - \varphi(\boldsymbol{r})}{\delta x} = F_1(\boldsymbol{r}) \]

Similarly for the \(y\) and \(z\) components, we obtain \(\frac{\partial\varphi}{\partial y} = F_2\) and \(\frac{\partial\varphi}{\partial z} = F_3\), so:

\[ \boxed{\nabla\varphi = \boldsymbol{F}} \]

\(\blacksquare\)

Remark: The assumption of simple connectedness is essential. For example, in \(D = \mathbb{R}^3 \setminus \{z\text{-axis}\}\) (space with the \(z\)-axis removed), \(\boldsymbol{F} = \frac{1}{x^2+y^2}(-y, x, 0)\) satisfies \(\nabla \times \boldsymbol{F} = \boldsymbol{0}\), but the line integral along a closed curve encircling the \(z\)-axis is \(2\pi \neq 0\), and no global scalar potential exists.

(c) Relationship between Stokes' theorem and curvature

Qualitative discussion:

Stokes' theorem states that the circulation (line integral) of a vector field \(\boldsymbol{F}\) along a closed curve \(C\) equals the surface integral of \(\nabla \times \boldsymbol{F}\) over a surface \(S\) bounded by \(C\):

\[ \oint_C \boldsymbol{F} \cdot d\boldsymbol{r} = \int_S (\nabla \times \boldsymbol{F}) \cdot d\boldsymbol{S} \]

In particular, for an infinitesimal closed curve (area \(\delta A\), normal \(\hat{\boldsymbol{n}}\)):

\[ \oint_{\delta C} \boldsymbol{F} \cdot d\boldsymbol{r} \approx (\nabla \times \boldsymbol{F}) \cdot \hat{\boldsymbol{n}}\, \delta A \]

That is, \(\operatorname{rot}\boldsymbol{F}\) represents the "circulation per unit area."

In general relativity, the curvature tensor \(R^\mu{}_{\nu\rho\sigma}\) plays a completely analogous role. When a vector \(V^\mu\) is parallel transported along an infinitesimal closed parallelogram (with sides \(\delta x^\rho\) and \(\delta x^\sigma\)), the vector changes upon returning to the starting point. The change is given by

\[\delta V^\mu = R^\mu{}_{\nu\rho\sigma}\,V^\nu\,\delta x^\rho\,\delta x^\sigma \]

This can be regarded as the general relativistic version of Stokes' theorem:

Vector calculus (\(\mathbb{R}^3\)) General relativity (curved spacetime)
Circulation of vector field \(\boldsymbol{F}\) Change of vector \(V^\mu\) under parallel transport
Curl \(\nabla \times \boldsymbol{F}\) Riemann curvature tensor \(R^\mu{}_{\nu\rho\sigma}\)
Area of infinitesimal closed curve \(\delta A\) Area of infinitesimal parallelogram \(\delta x^\rho \wedge \delta x^\sigma\)
  • In flat space (\(R^\mu{}_{\nu\rho\sigma} = 0\)), parallel transport is path-independent (zero circulation).
  • In curved space (\(R^\mu{}_{\nu\rho\sigma} \neq 0\)), parallel transport is path-dependent (non-zero circulation).

This correspondence shows that the curvature tensor measures the "non-commutativity of parallel transport," which is consistent with the derivation of the Riemann tensor studied in Ch. 12. \(\square\)