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So I'm trying to understand this derivation of magnetic pressure:

starting with force $ f = J \times B$ where $f$ is the force per unit volume. In the low frequency limit, $\frac{dE}{dt}$ goes to 0. So we can use Ampere's law with $\nabla \times B = \mu_0J$. Inserting this into our force equation, we get: $$f = \frac{1}{\mu_0}[(\nabla \times B) \times B]$$ which becomes $$ f = -\nabla(\frac{B^2}{2\mu_0}) + \frac{1}{\mu_0}(B.\nabla)B$$ The vector calculus equality $A \times (B \times C) = (A.C)B - (A.B)C$ has obviously been used, but where does the 2 on the denominator in the first term on the RHS come from? Thank you for any answer. I know I'm missing something basic.

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    $\begingroup$ How is $J=qv$ ? $\endgroup$
    – GeeJay
    Commented Mar 12, 2017 at 7:15

1 Answer 1

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While $\nabla = \mathbf{\hat x}\frac{\partial}{\partial x} + \mathbf{\hat y}\frac{\partial}{\partial y} + \mathbf{\hat z}\frac{\partial}{\partial z}$ acts like a vector when applying to the gradient $\nabla f$, divergence $\nabla\cdot\mathbf F$ and curl $\nabla \times\mathbf F$, such correspondence is only formal, $\nabla$ is not really a vector, so identities like $$\mathbf A\times(\mathbf B\times \mathbf C) = \mathbf B(\mathbf A\cdot\mathbf C) - (\mathbf A\cdot \mathbf B)\mathbf C$$ won't always work when blindly inserting a $\nabla$.


If we expand $\mathbf A\times(\nabla \times \mathbf B)$ abusing the identity above, we get $$ \mathbf A\times(\nabla \times \mathbf B) \stackrel{\text{wrong}}{=} \color{red}{\nabla}(\mathbf A\cdot \mathbf B) - (\mathbf A\cdot \nabla)\mathbf B $$ The problem is that the $\nabla$ on the LHS applies derivatives only on the field $\mathbf B$, but the $\color{red}{\nabla}$ on the RHS applies derivatives that involve both $\mathbf A$ and $\mathbf B$. This causes the identity to break down. It is correct if we only apply the $\color{red}{\nabla}$ to $\mathbf B$, and treats $\mathbf A$ as a constant: $$ \mathbf A\times(\nabla \times \mathbf B) = \nabla_{\mathbf{B}} (\mathbf A\cdot\mathbf B) - (\mathbf A\cdot \nabla)\mathbf B $$ This is known as the Feymann subscript notation. When you set $\mathbf A$ to $\mathbf B$, this identity becomes \begin{align} (\nabla \times \color{green}{\mathbf B})\times \color{blue}{\mathbf B} &= -\nabla_{\color{green}{\mathbf B}}(\underbrace{\quad \color{blue}{\mathbf B} \quad}_{\text{treat as constant}} \cdot \color{green}{\mathbf B}) + (\color{blue}{\mathbf B}\cdot\nabla)\color{green}{\mathbf B} \\ &= -\frac12 \nabla(B^2) + (\mathbf B\cdot\nabla)\mathbf B \end{align} The situation is actually similar to: \begin{align} \frac{\partial(f(\color{blue}y)f(\color{green}x))}{\partial \color{green}x} = f(\color{blue}y)f'(\color{green}x) \xrightarrow{\text{put $\color{blue}y$ as $\color{blue}x$}}\frac{\partial(\overbrace{f(\color{blue}x)}^{\text{const}}f(\color{green}x))}{\partial \color{green}x} &= f(\color{blue}x)f'(\color{green}x) \\ \text{but}\quad \frac{\partial(f(\color{green}x)^2)}{\partial \color{green}x} &= 2f(\color{green}x)f'(\color{green}x) \end{align} which illustrates why we need to insert the $\frac12$ to cancel out the extra factor of $2$.


For sure, this is just some heuristic arguments, you may want to mathematically expand both sides to partial derivatives in Cartesian coordinates to verify this is indeed true.

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