MTH3008 Weekly Problems 3

Original Documents: Problem Sheet / Provided Solutions

Vibes: Builds directly on suffix notation for operators and coordinate transformation rules; proofs are mostly mechanical index substitution; medium difficulty, since recalling everything can be tricky.

Used Techniques:

• Vector transformation law: $V^{\prime i}=L_j^i{V}^j$, where $L_j^i=\partial x^{\prime i}/\partial x^j$.
• Chain rule: $\partial /\partial x^{\prime i}=L_i^j\partial /\partial x^j$.
• Orthogonality: $L_j^i{L}_k^i={\delta }_{jk}$.
• ${ϵ}^{ijk}$ vanishes on any symmetric pair of indices.
• Mixed partials commute; $\nabla \cdot (\nabla \times v)=0$; ${\nabla }^{2}u=\nabla (\nabla \cdot u)-\nabla \times (\nabla \times u)$.

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3.1. Showing ∇ Transforms as a Vector

Question

Use the tensor transformation law to show that the gradient operator $\nabla$ transforms as a vector under a change of Cartesian coordinates, and hence is itself a vector.

We need to show ${\partial }_i^{\prime }=L_i^j{\partial }_j$. The $i$th component of $\nabla$ is ${\partial }_i=\partial /\partial x^i$. Applying the chain rule under a coordinate change gives ${\partial }_i^{\prime }=\partial /\partial x^{\prime i}=(\partial x^j/\partial x^{\prime i}){\partial }_j=L_i^j{\partial }_j$. This is exactly the vector transformation law, so $\nabla$ is a vector.

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3.2. Laplacian of a Scalar is a Scalar

Question

Let $f$ be a scalar field. Use the tensor transformation law to show that $\nabla \cdot (\nabla f)$ transforms as a scalar under a change of Cartesian coordinates, and therefore $\nabla \cdot (\nabla f)$ is a scalar.

In suffix notation, $\nabla \cdot (\nabla f)={\partial }_i{\partial }_{i}f$. Since $f$ is a scalar, $f^{\prime }=f$, and since $\nabla$ is a vector (3.1), ${\partial }_i^{\prime }=L_i^j{\partial }_j$. So ${\partial }_i^{\prime }{\partial }_i^{\prime }f=L_i^j{\partial }_j{L}_i^k{\partial }_{k}f=L_i^j{L}_i^k{\partial }_j{\partial }_{k}f={\delta }_{jk}{\partial }_j{\partial }_{k}f={\partial }_k{\partial }_{k}f$, using orthogonality $L_i^j{L}_i^k={\delta }_{jk}$. The primed quantity equals the unprimed quantity, so it is a scalar.

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3.3. Directional Derivative as a Scalar

Question

Let $\nabla$ and $w$ be vectors. Using the tensor transformation law and the fact that both $\nabla$ and $w$ transform as vectors, show that the quantity $w\cdot \nabla$ is a scalar under a change of Cartesian coordinates.

In suffix notation, $w\cdot \nabla =w^i{\partial }_i$. Since both $w$ and $\nabla$ are vectors, their components transform as $w^{\prime i}=L_j^i{w}^j$ and ${\partial }_i^{\prime }=L_i^j{\partial }_j$. Then $w^{\prime i}{\partial }_i^{\prime }=L_j^i{w}^j{L}_i^k{\partial }_k=L_j^i{L}_i^k{w}^j{\partial }_k={\delta }_{jk}{w}^j{\partial }_k=w^j{\partial }_j$. Invariant under the transformation, so $w\cdot \nabla$ is a scalar.

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3.4. Constancy of Jacobian Entries in Cartesian Rotations

Question

Consider the matrix $L_{ij}=\frac{\partial x^{\prime i}}{\partial x^j}$ representing the components of the Jacobian for a (Cartesian) coordinate transformation. Show that $\frac{\partial L_{ij}}{\partial x^{\prime i}}=0$.

Differentiating directly, $\partial L_{ij}/\partial x^{\prime i}={\partial }^2{x}^{\prime i}/(\partial x^{\prime i}\partial x^j)$. Since we may swap the order of partial derivatives, this equals ${\partial }^2{x}^{\prime i}/(\partial x^j\partial x^{\prime i})=(\partial /\partial x^j)(\partial x^{\prime i}/\partial x^{\prime i})$. Now $\partial x^{\prime i}/\partial x^{\prime i}={\delta }_{ii}=3$, which is a constant, so its derivative with respect to $x^j$ is zero.

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3.5. Rotation Coefficients from Inner Products

Question

Suppose $e_1,e_2,e_3$ are basis vectors for a Cartesian coordinate system, and let $e_1^{\prime },e_2^{\prime },e_3^{\prime }$ be the images of $e_1,e_2,e_3$ under a rotation.

For each $i$, the original basis vector $e_i$ has an expansion in the rotated basis $e_i=a_{i1}{e}_1^{\prime }+a_{i2}{e}_2^{\prime }+a_{i3}{e}_3^{\prime }$. Find expressions for the coefficients $a_{ij}$ in terms of the inner products of $e_i$ and $e_j^{\prime }$.

From $e_i=a_{ik}{e}_k^{\prime }$, take the inner product of both sides with $e_j^{\prime }$: $e_i\cdot e_j^{\prime }=a_{ik}(e_k^{\prime }\cdot e_j^{\prime })=a_{ik}{\delta }_{kj}=a_{ij}$. Hence, $a_{ij}=e_i\cdot e_j^{\prime }$.

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3.6. Curl of a Radial Vector Field

Question

Let $u$ be the vector field defined by $u=h(r)r$, where $h(r)$ is an arbitrary differentiable scalar function of $r$, and $r$ is the position vector $r=(x_1,x_2,x_3)$ with magnitude $r=|r|$.

Using suffix (index) notation, show that $\nabla \times u=0$. Hint: You may use the result from Exercise 2.8 that $\nabla h(r)=h^{\prime }(r)r/r$.

The $i$th component of $\nabla \times u$ is ${ϵ}^{ijk}{\partial }_j{u}_k={ϵ}^{ijk}{\partial }_j(h{x}_k)$. By the product rule and the hint, ${ϵ}^{ijk}{\partial }_j(h{x}_k)={ϵ}^{ijk}[(h^{\prime }/r)x_j{x}_k+h{\delta }_{jk}]$. The second term gives $h{ϵ}^{ijk}{\delta }_{jk}=h{ϵ}^{ijj}=0$. For the first term, $x_j{x}_k$ is symmetric in $j,k$ while ${ϵ}^{ijk}$ is antisymmetric in $j,k$, so ${ϵ}^{ijk}{x}_j{x}_k=0$. Hence, $\nabla \times u=0$.

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3.7. Commuting Divergence and Laplacian on a Vector Field

Question

Show that, for a sufficiently smooth vector field $u$, the identity $$\begin{gathered} \nabla \cdot {\nabla }^{2}u={\nabla }^2( \\ nabla \\ cdotu) \end{gathered}$$ holds in two different ways:

1. Prove the identity directly using suffix (index) notation.
2. First use the vector identity from the lectures $$\begin{gathered} nabl{a}^{2}u= \\ nabla( \\ nabla \\ cdotu)- \\ nabla \\ times( \\ nabla \\ timesu) \end{gathered}$$, and then rewrite the resulting expressions using suffix notation to verify the equality.

1. In suffix notation, $\nabla \cdot {\nabla }^{2}u={\partial }_j({\partial }_k{\partial }_k{u}_j)$. Since partial derivatives commute for smooth fields, we may reorder: ${\partial }_j{\partial }_k{\partial }_k{u}_j={\partial }_k{\partial }_k({\partial }_j{u}_j)={\nabla }^2(\nabla \cdot u)$.

2. Substituting the identity: $\nabla \cdot {\nabla }^{2}u=\nabla \cdot [\nabla (\nabla \cdot u)-\nabla \times (\nabla \times u)]=\nabla \cdot \nabla (\nabla \cdot u)-\nabla \cdot [\nabla \times (\nabla \times u)]$. The second term is $\nabla \cdot (\nabla \times v)=0$ for any smooth $v$ (since ${ϵ}^{ijk}{\partial }_i{\partial }_j{v}_k=0$ by antisymmetry), leaving $\nabla \cdot \nabla (\nabla \cdot u)={\nabla }^2(\nabla \cdot u)$.

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3.8. Curl and Divergence of Nonlinear Scalar Combinations

Question

Let $f$ and $g$ be scalar fields.

1. Show, using suffix (index) notation, that $$\begin{gathered} \nabla \\ times(f \\ nablaf)=0 \end{gathered}$$.
2. Simplify the expression $$\begin{gathered} nabla \\ cdot(g \\ nablag) \end{gathered}$$ to an equivalent expression involving just a single differential operator acting on a single scalar field.

1. The $i$th component is ${ϵ}^{ijk}{\partial }_j(f{\partial }_{k}f)$. Applying the product rule: ${ϵ}^{ijk}[({\partial }_{j}f)({\partial }_{k}f)+f{\partial }_j{\partial }_{k}f]$. The second term vanishes since ${\partial }_j{\partial }_{k}f$ is symmetric in $j,k$. For the first term, $({\partial }_{j}f)({\partial }_{k}f)$ is also symmetric in $j,k$, so contraction with ${ϵ}^{ijk}$ gives zero. Hence, $\nabla \times (f\nabla f)=0$.

2. $\nabla \cdot (g\nabla g)={\partial }_i(g{\partial }_{i}g)=({\partial }_{i}g)({\partial }_{i}g)+g{\partial }_i{\partial }_{i}g=|\nabla g{|}^2+g{\nabla }^{2}g$. Recognising that $({\partial }_{i}g)({\partial }_{i}g)=|\nabla g{|}^2=\frac{1}{2}{\nabla }^2(g^2)-g{\nabla }^{2}g$ is not simpler, the cleanest form is $\nabla \cdot (g\nabla g)=\frac{1}{2}{\nabla }^2(g^2)$, since ${\partial }_i{\partial }_i(g^2)={\partial }_i(2g{\partial }_{i}g)=2({\partial }_{i}g{)}^2+2g{\partial }_i{\partial }_{i}g=2|\nabla g{|}^2+2g{\nabla }^{2}g$.