Question

Show that \(\nabla \cdot(\mathbf{U} \times \mathbf{r})=\mathbf{r} \cdot(\nabla \times \mathbf{U})\) where \(\mathbf{U}\) is a vector function of \(x, y, z,\) and \(\mathbf{r}=x \mathbf{i}+y \mathbf{j}+z \mathbf{k}\).

Short answer

\( abla \cdot ( \mathbf{U} \times \mathbf{r} ) = \mathbf{r} \cdot ( abla \times \mathbf{U} ) \)

Step-by-step solution

- Understand the Given Vectors

Identify the given vectors. Here, \( \mathbf{U} \) is a vector function of \( x, y, z \) and \( \mathbf{r} = x \mathbf{i} + y \mathbf{j} + z \mathbf{k} \) is the position vector.

- Use Vector Triple Product Identity

Recall the vector triple product identity: \( abla \cdot ( \mathbf{U} \times \mathbf{r} ) = \mathbf{r} \cdot ( abla \times \mathbf{U} ) \). We need to show that this identity holds in this particular case.

- Compute \( \mathbf{U} \times \mathbf{r} \)

Calculate \( \mathbf{U} \times \mathbf{r} \). Using the cross product formula, we get:\[ \mathbf{U} \times \mathbf{r} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \ U_x & U_y & U_z \ x & y & z \end{vmatrix} \]Expanding this determinant, we have:\[ \mathbf{U} \times \mathbf{r} = (U_y z - U_z y) \mathbf{i} + (U_z x - U_x z) \mathbf{j} + (U_x y - U_y x) \mathbf{k} \]

- Take the Divergence

Now take the divergence of \( \mathbf{U} \times \mathbf{r} \): \[ abla \cdot ( \mathbf{U} \times \mathbf{r} ) = \frac{\partial }{\partial x} (U_y z - U_z y) + \frac{\partial }{\partial y} (U_z x - U_x z) + \frac{\partial }{\partial z} (U_x y - U_y x) \]

- Compute Partial Derivatives

Evaluate each partial derivative term by term:\( \frac{\partial }{\partial x} (U_y z - U_z y) = z \frac{\partial U_y}{\partial x} - y \frac{\partial U_z}{\partial x} \)\( \frac{\partial }{\partial y} (U_z x - U_x z) = x \frac{\partial U_z}{\partial y} - z \frac{\partial U_x}{\partial y} \)\( \frac{\partial }{\partial z} (U_x y - U_y x) = y \frac{\partial U_x}{\partial z} - x \frac{\partial U_y}{\partial z} \)

- Combine Results

Sum the results from the partial derivatives: \[ abla \cdot ( \mathbf{U} \times \mathbf{r} ) = z \frac{\partial U_y}{\partial x} - y \frac{\partial U_z}{\partial x} + x \frac{\partial U_z}{\partial y} - z \frac{\partial U_x}{\partial y} + y \frac{\partial U_x}{\partial z} - x \frac{\partial U_y}{\partial z} \]Notice that these terms are exactly the components of the dot product \( \mathbf{r} \cdot ( abla \times \mathbf{U} ) \).

- Confirm Equivalent Expression for Curl

Compare this to the calculation of the curl:\( abla \times \mathbf{U} = \left( \frac{\partial U_z}{\partial y} - \frac{\partial U_y}{\partial z} \right) \mathbf{i} + \left( \frac{\partial U_x}{\partial z} - \frac{\partial U_z}{\partial x} \right) \mathbf{j} + \left( \frac{\partial U_y}{\partial x} - \frac{\partial U_x}{\partial y} \right) \mathbf{k} \)Thus, \( \mathbf{r} \cdot ( abla \times \mathbf{U} ) = x \left( \frac{\partial U_z}{\partial y} - \frac{\partial U_y}{\partial z} \right) + y \left( \frac{\partial U_x}{\partial z} - \frac{\partial U_z}{\partial x} \right) + z \left( \frac{\partial U_y}{\partial x} - \frac{\partial U_x}{\partial y} \right) \)

- Verify Equality

Observe that the expression for the divergence \( abla \cdot ( \mathbf{U} \times \mathbf{r} ) \) is exactly the same as the dot product \( \mathbf{r} \cdot ( abla \times \mathbf{U} ) \). Thus,\[ abla \cdot ( \mathbf{U} \times \mathbf{r} ) = \mathbf{r} \cdot ( abla \times \mathbf{U} ) \]

Key concepts

Vector Triple Product Identity

The vector triple product identity is a crucial tool in vector calculus. It simplifies the manipulation of cross products and dot products involving three vectors. The identity states that given vectors \(\textbf{A}\), \(\textbf{B}\), and \(\textbf{C}\), the expression \(\textbf{A} \times (\textbf{B} \times \textbf{C})\) can be rewritten as \((\textbf{A} \cdot \textbf{C}) \textbf{B} - (\textbf{A} \cdot \textbf{B}) \textbf{C}\). This identity helps in simplifying components and understanding vector fields.
To show that \(abla \cdot (\textbf{U} \times \textbf{r}) = \textbf{r} \cdot (abla \times \textbf{U})\), where \textbf{U}\ is a vector function and \(\textbf{r}= x \textbf{i} + y \textbf{j} + z \textbf{k}\), we recognize that both sides of the equation describe the same vector operation in different forms.
By applying the vector triple product identity, we can handle these cross and dot products in a more structured and concise way.

Cross Product

The cross product of two vectors results in a third vector that is perpendicular to the plane of the first two. For vectors \(\textbf{A} = a_1 \textbf{i} + a_2 \textbf{j} + a_3 \textbf{k}\) and \(\textbf{B} = b_1 \textbf{i} + b_2 \textbf{j} + b_3 \textbf{k}\), the cross product \(\textbf{A} \times \textbf{B}\) is given by the determinant: \(\textbf{A} \times \textbf{B} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ a_1 & a_2 & a_3 \ b_1 & b_2 & b_3 \end{vmatrix}\).
Expanding this determinant, we get \(\textbf{A} \times \textbf{B} = (a_2 b_3 - a_3 b_2) \textbf{i} + (a_3 b_1 - a_1 b_3) \textbf{j} + (a_1 b_2 - a_2 b_1) \textbf{k}\). In the given exercise, applying the cross product formula to \(\textbf{U} \times \textbf{r}\) yields: \((U_y z - U_z y) \textbf{i} + (U_z x - U_x z) \textbf{j} + (U_x y - U_y x) \textbf{k}\).
This operation is essential for constructing vector fields and capturing rotational aspects of the system in question.

Divergence

Divergence measures how much a vector field is spreading out from a given point. For a vector field \(\textbf{F} = P \textbf{i} + Q \textbf{j} + R \textbf{k}\), its divergence is given by \(abla \cdot \textbf{F} = \frac{\text{∂}P}{\text{∂}x} + \frac{\text{∂}Q}{\text{∂}y} + \frac{\text{∂}R}{\text{∂}z}\).
In the given exercise, we apply this to \(\textbf{U} \times \textbf{r}\) to find its divergence: \(abla \cdot (\textbf{U} \times \textbf{r}) = \frac{\text{∂}}{\text{∂}x}(U_y z - U_z y) + \frac{\text{∂}}{\text{∂}y}(U_z x - U_x z) + \frac{\text{∂}}{\text{∂}z}(U_x y - U_y x)\).
By computing these partial derivatives step-by-step, we demonstrate that the divergence of the cross product equates to the dot product of \(\textbf{r}\) and the curl of \(\textbf{U}\), providing insight into how vector fields interact and evolve.

Curl

Curl measures the rotational component of a vector field. It is a vector operation that takes a vector field and returns another vector field, which represents the rotation at each point. For a vector field \(\textbf{F} = P \textbf{i} + Q \textbf{j} + R \textbf{k}\), the curl is given by \(abla \times \textbf{F} = \begin{vmatrix} \textbf{i} & \textbf{j} & \textbf{k} \ \frac{\text{∂}}{\text{∂}x} & \frac{\text{∂}}{\text{∂}y} & \frac{\text{∂}}{\text{∂}z} \ P & Q & R \end{vmatrix}\).
This results in: \(abla \times \textbf{F} = \bigg(\frac{\text{∂}R}{\text{∂}y} - \frac{\text{∂}Q}{\text{∂}z}\bigg) \textbf{i} + \bigg(\frac{\text{∂}P}{\text{∂}z} - \frac{\text{∂}R}{\text{∂}x}\bigg) \textbf{j} + \bigg(\frac{\text{∂}Q}{\text{∂}x} - \frac{\text{∂}P}{\text{∂}y}\bigg) \textbf{k}\).
In the given exercise, evaluating \(abla \times \textbf{U}\) and taking the dot product with \(\textbf{r}\) lets us connect to the divergence of \(\textbf{U} \times \textbf{r}\). Summing these ways to explore vector fields reveals how they twist and turn, providing a deeper grasp of physical phenomena in fields like electromagnetism and fluid mechanics.