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I'm working through Stancil and Prabhakar's 'Spin Waves', and am stuck with a vector identity which I am not sure how the authors have justified.

On page 34, we adopt the use of a scalar potential $\phi$, and a vector potential, $\vec{A}$. Then we use these to recast the electric and magnetic field in terms of the Coulomb gauge:

$$ \tag{1} B = \nabla \times \vec{A} $$ $$ \tag{2} E = - \frac{\partial \vec{A}}{\partial t} - \nabla \phi$$

Next we attempt to rewrite the Lorentz force in terms of the scalar and vector potentials. This looks like:

$$ \tag{3} m \frac{d \vec{v}}{dt} = q(-\frac{\partial \vec{A}}{dt} - \nabla \phi + \vec{v} \times (\nabla \times \vec{A})) $$

Here, we see a particular term, $\vec{v} \times (\nabla \times \vec{A})$.

If we use equation 3, and then use the following expression of $\frac{d A_x}{dt}$:

$$ \tag{4} \frac{d A_x}{dt} = \frac{\partial A_x}{\partial t} + (v_x \frac{\partial A_x}{\partial x} + v_y \frac{\partial A_x}{\partial y} + v_z \frac{\partial A_x}{\partial z}) $$

If we then take the x-component of $\vec{v} \times (\nabla \times \vec{A})$, we get:

$$ \tag{5} (\vec{v} \times (\nabla \times \vec{A}))_x = v_y (\frac{d A_y}{dx} - \frac{d A_x}{dy}) - v_z(\frac{d A_x}{dz} - \frac{d A_z}{dx}) $$

The next line is where my issue arises. The book states that then, this follows:

$$ \tag{6} (\vec{v} \times (\nabla \times \vec{A}))_x = \partial_x (\vec{v} \cdot \vec{A}) - \frac{d A_x}{dt} + \frac{\partial A_x}{\partial t} $$

However, working backwards from the RHS of equation 6's first term, I find that:

$$ \tag{7} \partial_x (\vec{v} \cdot \vec{A}) = (A_x \frac{\partial v_x}{\partial x} + A_y \frac{\partial v_y}{\partial x} + A_z \frac{\partial v_z}{\partial x}) + (v_x \frac{\partial A_x}{\partial x} + v_y \frac{\partial A_y}{\partial x} + v_z \frac{\partial A_z}{\partial x}) $$

And, rearranging equation 4:

$$ \tag{8} \frac{d A_x}{dt} - \frac{\partial A_x}{\partial t} = (v_x \frac{\partial A_x}{\partial x} + v_y \frac{\partial A_x}{\partial y} + v_z \frac{\partial A_x}{\partial z}) $$

Then, equation 6 becomes:

$$ \tag{9} \partial_x (\vec{v} \cdot \vec{A}) - \frac{d A_x}{dt} + \frac{\partial A_x}{\partial t} = (A_x \frac{\partial v_x}{\partial x} + A_y \frac{\partial v_y}{\partial x} + A_z \frac{\partial v_z}{\partial x})\\ + v_y \frac{\partial A_y}{\partial x} + v_z \frac{\partial A_z}{\partial x} - v_y \frac{\partial A_x}{\partial y} - v_z \frac{\partial A_x}{\partial z} $$

Which is the original statement made in equation 5, plus the bracketed terms $(A_x \frac{\partial v_x}{\partial x} + A_y \frac{\partial v_x}{\partial y} + A_z \frac{\partial v_x}{\partial y})$.

This means that, for equation 6 to be true, it would be necessary to state that

$$ \tag{10} A_x \frac{\partial v_x}{\partial x} + A_y \frac{\partial v_x}{\partial y} + A_z \frac{\partial v_x}{\partial z} = 0 $$

Is this justified? And if so, how? I'm not sure how I could prove this. The only way I can see this being possible, is by requiring $\vec{v}$ to be unchanging, but this is not an assumed detail in the textbook.

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I did not go through the details but your Eq. (10) is satisfied as $\vec{v}$ is not a function of position. It is the time derivative of the position of a particle.

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  • $\begingroup$ Surely it is though? Velocity is not uniform from position to position inherently. $\endgroup$ – Andrew Mar 31 '18 at 12:01
  • $\begingroup$ The position of a particle is not a field that depends on position, it is the position. It and its time derivative only depend on time. $\endgroup$ – my2cts Mar 31 '18 at 19:07

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