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.

  • 1
    $\begingroup$ How is $J=qv$ ? $\endgroup$
    – GeeJay
    Commented Mar 12, 2017 at 7:15

1 Answer 1


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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.