Is (B.\nabla)A the same as B(\nabla.A)?

In summary, the problem asks to prove the vector identity \nabla\times(A\times B) = (B\cdot\nabla)A - (A\cdot\nabla)B + A(\nabla\cdot B) - B(\nabla\cdot A), where A and B are vector fields. To solve this, the LHS and RHS must be expanded and shown to be equal. The product rule must be used when taking the gradient of components with more than one term multiplied together. The confusion lies in the expression (B\cdot\nabla)A, which is not equal to B(\nabla\cdot A) due to the different components. The components of the two
  • #1
featheredteap
4
0

Homework Statement



Prove the following vector identity:

[tex]\nabla[/tex]x(AxB) = (B.[tex]\nabla[/tex])A - (A.[tex]\nabla[/tex])B + A([tex]\nabla[/tex].B) - B([tex]\nabla[/tex].A)

Where A and B are vector fields.

Homework Equations



Curl, divergence, gradient

The Attempt at a Solution



I think I know how to do this: I have to expand out the LHS and the RHS and show that they equal one another. To do this I need to use the product rule when taking the gradient of components with more than one term multiplied together.

What I don't understand is what's going on on the RHS: doesn't (B.[tex]\nabla[/tex])A = B([tex]\nabla[/tex].A) ? (Obviously this can't be the case since then all the components would cancel to zero.) So how does this really work?
 
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  • #2
No, since if
[tex]
\begin{displaymath}
\begin{array}{rcl}
\mathbf{A} & = & A_{x}\mathbf{i}+A_{x}\mathbf{j}+A_{z}\mathbf{k} \\
\mathbf{B} & = & B_{x}\mathbf{i}+B_{x}\mathbf{j}+B_{z}\mathbf{k}
\end{array}
\end{displaymath}
[/tex]
Then:
[tex]
\begin{displaymath}
\begin{array}{rcl}
\mathbf{B}\cdot\nabla & = & B_{x}\partial_{x}+B_{y}\partial_{y}+B_{z}\partial_{z} \\
\nabla\cdot\mathbf{A} & = & \partial_{x}A_{x}+\partial_{y}A_{y}+\partial_{z}A_{z}
\end{array}
\end{displaymath}
[/tex]
Using the above compare one component of the two vector expressions and see if they're the same.
 

1. What is the purpose of proving vector identities?

The purpose of proving vector identities is to mathematically show the relationship between different vector quantities. This helps to better understand and manipulate vectors in various applications, such as physics and engineering.

2. How do you prove a vector identity?

To prove a vector identity, you must use mathematical operations and properties to manipulate the given vectors and show that both sides of the equation are equal. This typically involves using properties like commutativity, associativity, and distributivity, as well as vector algebra rules such as the dot product and cross product.

3. What are some common techniques used to prove vector identities?

Common techniques used to prove vector identities include expanding both sides of the equation using the definition of vector operations, using algebraic manipulations to rearrange terms, and using geometric interpretations of vectors to visualize and understand the relationship between different quantities.

4. Are there any special cases or exceptions when proving vector identities?

Yes, there can be special cases or exceptions when proving vector identities. For example, some identities may only hold for certain types of vectors (e.g. unit vectors) or may have additional conditions that need to be met. It is important to carefully consider the given vectors and any restrictions or assumptions when proving an identity.

5. How can proving vector identities be applied in real-world situations?

Proving vector identities has various real-world applications, such as in physics and engineering. For example, it can be used to derive equations for motion, analyze forces and moments in structures, and solve problems involving electric and magnetic fields. It can also be used to simplify and manipulate equations in order to make them more usable in practical situations.

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