• rsammas
In summary, the homework statement is that the equation for the stress tensor is (∇xB)xB = (B∇)B - ∇ (B²/2).
rsammas

## Homework Statement

Show that:

∇x(∇xB) = (B∇)B - ∇ (1/2B2)

## Homework Equations

r = (x,y,z) = xiei

∂xi/∂xj = δij

r2 = xkxk

δij = 1 if i=j, 0 otherwise (kronecker delta)
εijk is the alternating stress tensor and summn convn is assumed.

## The Attempt at a Solution

On the LHS I simplified to get:

εijk2/∂xj∂xk

but was unsure what to do next because the RHS contains only first order derivatives

On the RHS I was able to get to:

(B∇)B - ∇ (1/2B2) = B(∂Bi/∂i)-B
= B(∂Bi/∂i-1)

I feel like I'm just not seeing some simple trick, or there is a rule that I don't remember/haven't learned. This is for my Classical Mechanics class BTW.

There must be something wrong in your problem statement or how can you get an expression which is quadratic in $\vec{B}$ taking derivatives of an expression that contains only one $\vec{B}$? The correct equation to prove is
$$\vec{\nabla} \times (\vec{\nabla} \times \vec{B})=\vec{\nabla} (\vec{\nabla} \cdot \vec{B}) - \vec{\nabla}^2 \vec{B},$$
which holds, however, only in Cartesian coordinates!

that's what I was thinking, but the assignment is what I wrote above

(∇xB)xB = (B∇)B - ∇ (B²/2)

is famous in MHD

that's still not what I'm asking. but maybe showing a proof might help me out a bit

Showing a proof of what?
The original identity is obviously wrong (∇x(∇xB) = (B∇)B - ∇ (1/2B2) is wrong).

The proof of the second identity, (∇xB)xB = (B∇)B - ∇ (B²/2), is straightforward by using components representation.

Using the notation "eik" for the Levi-Civita tensor,
using 'F,l" to denote the derivative of F with respect to xl,
(∇xB)xB can be developed as follows:

((∇xB)xB)i
= eijk (ejlm Bm,l) Bk
= - eikj elmj Bm,l Bk
= -(eil ekm - eim ekl) Bm,l Bk
= - Bk,i Bk + Bi,k Bk
= - (Bk²/2),i + Bi,k Bk

which ends the proof.

"εijk is the alternating stress tensor ..."
"On the LHS I simplified to get: εijk∂2/∂xj∂xk"

I have the feeling you lack some basic understanding, since it makes almost no meaning.
I don't know if your question is part of a math course or a physics course (electrodynamics).
In any case, you need to go back to the basics.
The strange thing is that the identity "(∇xB)xB = (B∇)B - ∇ (B²/2)" is indeed related to the Maxwell stress tensor in electrodynamics (if B is the magnetic field). The second term is then called the magnetic pressure.
As you posted in the "Calculus & Beyond Homework" section I wonder how you could have mixed that "math exercise" with electrodynamics. Is Google the reason?

Last edited:

## 1. What does the "grad x" symbol mean in this equation?

The "grad x" symbol, also known as the gradient operator, represents the vector of partial derivatives in the x, y, and z directions. It is commonly used in vector calculus to represent the rate and direction of change in a given function.

## 2. How do you solve for grad x (grad x B)?

To solve for grad x (grad x B), you will need to use the vector calculus identity known as the Laplacian operator (∇²). This operator is applied to the function B to find the second-order partial derivatives, which are then multiplied by the gradient operator once again. This will result in a vector equation with components in the x, y, and z directions.

## 3. What is the purpose of solving for grad x (grad x B)?

The purpose of solving for grad x (grad x B) is to find the vector field representing the rate and direction of change of the function B. This information is useful in many scientific and engineering applications, such as analyzing fluid flow, electromagnetic fields, and heat transfer.

## 4. Can this equation be solved analytically?

Yes, the equation grad x (grad x B) can be solved analytically using vector calculus techniques. However, in some cases, it may be more practical to use numerical methods to approximate the solution.

## 5. Are there any real-world applications of this equation?

Yes, there are many real-world applications of this equation. Some examples include modeling fluid flow in pipes and channels, analyzing magnetic fields in electrical systems, and predicting temperature distribution in heat transfer problems. It can also be used in computer graphics to create realistic 3D images.

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