How is $\nabla (u\cdot A) =u\cdot \nabla A+ u\times (\nabla \times A) $?

Question

This was used in the answer here, in the derivation of the Lorentz force law from the Lagrangian. $u$ and $A$ are vectors, the velocity of the particle and the spacetime dependent Magnetic field

As part of the Euler Lagrange equation, we had to calculate $$\frac{\partial L}{\partial x}=\frac{\partial {(u\cdot A) }}{\partial x}$$

Since the Lagrangian treats $x$ and $u$, i.e. position and velocity, as independent variables, I think this partial derivative should treat $u$ as a constant to give:

$$\frac{\partial L}{\partial x}=\frac{\partial {(u\cdot A) }}{\partial x}$$

$$=\frac{\partial {(u_x A_x + u_yA_y +u_z A_z) }}{\partial x}$$

$$=u\cdot \frac{\partial A}{\partial x}$$

The vector form of this would be :

$$\frac{\partial L}{\partial r}=u\cdot \nabla A$$

I only got the first term. I don't understand where $u\times ({\nabla \times A})$ came from. Please help

Answer 1

Recall that for any vectors $a, b, c$ $$ a\times(b\times c) = (a\cdot c)b - (a\cdot b)c. $$ It is valid to manipulate $\nabla$ as a vector so long as you are aware of what you are differentiating. In the expression $$ u\times(\nabla\times A), $$ we are differentiating $A$. Let us keep track of this by putting a dot over $\nabla$ and $A$. We then see $$ u\times(\dot\nabla\times\dot A) = (u\cdot\dot A)\dot\nabla - (u\cdot\dot\nabla)\dot A = \dot\nabla(u\cdot\dot A) - (u\cdot\dot\nabla)\dot A. $$ The last equality is just from moving around the scalar quantity $u\cdot\dot A$. Since $u$ is to be treated as constant, we can now drop the dots and write in standard notation $$ u\times(\nabla\times A) = \nabla(u\cdot A) - (u\cdot\nabla) A. $$ Rearranging, we finally get $$ \nabla(u\cdot A) = (u\cdot\nabla) A + u\times(\nabla\times A). $$

Answer 2

OMG this is such ambiguous notation.

The thing is:

$$\vec{a} \cdot (\nabla \vec{b}) \neq (\vec{a}\cdot \nabla) \vec{b}$$

The answer that I linked derived a formula involving $ (\vec{a}\cdot \nabla) \vec{b}$ . This is why they got an extra cross product term.

I derived a formula involving $ \vec{a} \cdot (\nabla \vec{b}) $, which is why I didn't get the cross product term.

$(\vec{a}\cdot \nabla) \vec{b}$ can unambiguously be interpreted as $$a^i \partial ^i b^j$$

$\vec{a}\cdot (\nabla\vec{b})$ has to be interpreted carefully because the $\nabla \vec{b}$ is a matrix. When we matrix multiply this with $\vec{a}$, we have to contract the right indices. It is to be interpreted as the matrix multiplication :

$$a^j \partial ^i b^j$$

To see this, let's look at the derivation of this formula:

$$\partial ^i (v^jw^j) =w^j\partial ^i v^j +v^j \partial ^i w^j$$

In my post $v^j$ is a constant, so the first term vanishes.