Divergence of (A x B): Proving the Equation for Vector Calculus

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In summary, the conversation discusses the difficulty in proving the equation \nabla \cdot (\textbf{A} \times \textbf{B}) = \textbf{B}\cdot (\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times \textbf{B}) and provides a step-by-step explanation of the process followed. The mistake made in the process is identified as writing B_x\partial_yA_z instead of \partial_yA_zB_x. The importance of the product rule is also mentioned.
  • #1
r4nd0m
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Hi, I'm having trouble proving that:
[tex] \nabla \cdot (\textbf{A} \times \textbf{B}) = \textbf{B}\cdot
(\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times
\textbf{B}) [/tex]

This is how I proceeded:
[tex]\textbf{A} \times \textbf{B} = \overrightarrow{i}(A_y B_z - A_z
B_y) + \overrightarrow{j} (A_z B_x - A_x B_z) + \overrightarrow{k}
(A_x B_y - A_y B_x)[/tex]

[tex]\nabla \cdot (\textbf{A} \times \textbf{B}) = \frac{\partial A_y B_z - A_z
B_y}{\partial x} + \frac{\partial A_z B_x - A_x B_z}{\partial y} +
\frac {\partial A_x B_y - A_y B_x}{\partial z}[/tex]

[tex]\nabla \times \textbf{A} = \overrightarrow{i}(\frac{\partial A_z}{\partial y} - \frac{\partial A_y}{\partial z})
+ \overrightarrow{j}(\frac {\partial A_x}{\partial z} - \frac
{\partial A_z}{\partial x}) + \overrightarrow{k}(\frac {\partial
A_y}{\partial x} - \frac {\partial A_x}{\partial y}) [/tex]

[tex] \textbf{B} \cdot (\nabla \times \textbf{A}) = (\frac{\partial A_z B_x}{\partial y} - \frac{\partial A_y B_x}{\partial z})
+ (\frac {\partial A_x B_y}{\partial z} - \frac {\partial A_z
B_y}{\partial x}) + (\frac {\partial A_y B_z}{\partial x} - \frac
{\partial A_x B_z}{\partial y}) = [/tex]

[tex] = \frac{\partial A_y B_z - A_z
B_y}{\partial x} + \frac{\partial A_z B_x - A_x B_z}{\partial y} +
\frac {\partial A_x B_y - A_y B_x}{\partial z} = \nabla \cdot
(\textbf{A} \times \textbf{B}) [/tex]

[tex] \textbf{A} \cdot (\nabla \times \textbf{B}) = (\frac{\partial B_z A_x}{\partial y} - \frac{\partial B_y A_x}{\partial z})
+ (\frac {\partial B_x A_y}{\partial z} - \frac {\partial B_z
A_y}{\partial x}) + (\frac {\partial B_y A_z}{\partial x} - \frac
{\partial B_x A_z}{\partial y}) = -\nabla \cdot (\textbf{A} \times
\textbf{B}) [/tex]

Which finally yields:
[tex] \textbf{B}\cdot (\nabla \times \textbf{A}) - \textbf{A}\cdot (\nabla \times \textbf{B}) = 2\nabla \cdot (\textbf{A} \times \textbf{B}) [/tex]

Where did I make a mistake?
 
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  • #2
The obvious problem is that you've written B.(\/xA) and somehow got the components of the B's past the differential symbols, which is worrying

[tex]B_x\partial_yA_z[/tex]

is not the same as

[tex]\partial_yA_zB_x[/tex]

though it is the same as

[tex] (\partial_yA_z)B_x[/tex]

that might be one problem.
 
  • #3
Thanks a lot, I didn't realize that.
 
  • #4
It's just the product rule. if you differentiate f(x)g(x) the answer is not f'(x)g'(x), which I'm sure you do know.
 

FAQ: Divergence of (A x B): Proving the Equation for Vector Calculus

1. What is the concept of divergence in vector calculus?

Divergence is a mathematical operation that measures the flux or flow of a vector field at a given point in space. It is represented by the symbol ∇ ⋅ and is used to determine the extent to which a vector field originates or converges at a particular point.

2. How is divergence calculated for a vector field?

The divergence of a vector field is calculated by taking the dot product of the del operator (∇) and the vector field. This dot product is also known as the directional derivative and it represents the rate of change of the vector field in the direction of the del operator.

3. What is the significance of divergence in vector calculus?

Divergence is an important concept in vector calculus as it allows us to understand the behavior of vector fields in a given space. It helps us identify the regions where the vector field is expanding or contracting, and can be used to solve problems related to fluid flow, electromagnetism, and other physical phenomena.

4. How is the equation for divergence of (A x B) derived?

The equation for divergence of (A x B) is derived using vector calculus identities and properties such as the product rule, chain rule, and the identity that the divergence of a cross product is equal to the dot product of the original vector field and the curl of the vector field. This derivation involves multiple steps and can be found in most vector calculus textbooks.

5. Can the equation for divergence of (A x B) be proven mathematically?

Yes, the equation for divergence of (A x B) can be proven mathematically using the definition of divergence, the properties of vector calculus, and the properties of cross products. This proof is often included in vector calculus courses and can be found in mathematical journals and textbooks.

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