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Suppose we have some charged particle above the ionosphere. Then in a simplified model, it will experience a curvature and gradient drift. Assume further that a particle is in the equatorial plane. Since the particle is in the equatorial plane, using spherical coordinates, the magnetic field will yield the form:

$$ \vec{B}(r)=B_\theta(r) \hat{\theta} $$

In many textbooks a formula for the curvature drift comes as:

$$ \vec{v}_c=\frac{2U_{||}}{qB^4}[(\vec{B}\cdot\nabla)\vec{B}]\times\vec{B} $$

However, by applying this expression I get:

$$ \vec{B}\cdot\nabla=\frac{B_\theta}{r}\frac{\partial}{\partial\theta} $$

$$ (\vec{B}\cdot\nabla)\vec{B}=\frac{B_\theta}{r}\frac{\partial B_\theta (r)}{\partial\theta}=0 $$

Then we have $\vec{v_c}=0$. But there is another definition that states:

$$ \vec{v}_c=\frac{2U_{||}}{qR_c^2}\frac{\vec{R_c}\times \vec{B}}{B^2} $$

And since $R_c$ is defined as $R_c=\frac{B_\theta}{\frac{dB_\theta}{dr}}$ then $\vec{R_c}$ pointing in the radial direction which will clearly give a non zero drift curvature. I know that the last one is the correct one, but where is the mistake on the first try?

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The gradient drift velocity of a particle is given by: $$ V_{\nabla B} = \frac{ w_{\perp} }{ q_{s} \ B } \ \left[ \frac{ \mathbf{B} \times \nabla B }{ B^{2} } \right] \tag{0} $$ where $\mathbf{B}$ is the magnetic field vector, $B$ is its magnitude, $q_{s}$ is the charge (including sign) of the particle species $s$, and $w_{\perp}$ is given by: $$ w_{\perp} = \frac{ m_{s} }{ 2 } \Omega_{cs}^{2} \ \rho_{cs}^{2} \tag{1} $$ where $m_{s}$ is the mass of species $s$, $\Omega_{cs} = \tfrac{ q_{s} \ B }{ m_{s} }$ is the cyclotron frequency of species $s$, $\rho_{cs} = \tfrac{ m_{s} \ v_{\perp} }{ q_{s} \ B }$ is the gyroradius of species $s$, and $v_{\perp}$ is the perpendicular velocity of the particle with respect to $\mathbf{B}$.

The curvature drift velocity of a particle is given by: $$ V_{curv} = \frac{ 2 \ w_{\parallel} }{ q_{s} \ B^{2} } \ \left[ \frac{ \mathbf{B} \times \mathbf{R}_{c} }{ R_{c}^{2} } \right] \tag{2} $$ where $\mathbf{R}_{c}$ is the radius of curvature vector and $R_{c}$ its magnitude, $w_{\parallel} = \tfrac{ 1 }{ 2 } m_{s} \ v_{\parallel}^{2}$ is the parallel kinetic energy, and $v_{\parallel}$ is parallel velocity of the particle with respect to $\mathbf{B}$. Note that Equation 2 can be rewritten as: $$ V_{curv} = \frac{ 2 \ w_{\parallel} }{ q_{s} \ B } \ \left[ \frac{ \mathbf{B} \times \nabla B }{ B^{2} } \right] \tag{3} $$ which means we can combine these into one form called the gradient curvature drift given by: $$ V_{gc} = \frac{ m_{s} }{ 2 \ q_{s} \ B } \left( v_{\perp}^{2} + 2 \ v_{\parallel}^{2} \right) \left[ \frac{ \mathbf{B} \times \nabla B }{ B^{2} } \right] \tag{4} $$

I know that the last one is the correct one, but where is the mistake on the first try?

Well, technically the proper version of the one you present should have a term like: $$ \mathbf{B} \times \left( \mathbf{B} \cdot \nabla \mathbf{B} \right) $$ where the last term is rank-2 tensor, not the gradient of a scalar. Then there's an extra term using your field in spherical coordinates. That is, the result would be something like: $$ \mathbf{B} \cdot \nabla \mathbf{B} = - \hat{r} \frac{ B_{\theta}^{2} }{ r } + \hat{\theta} \frac{ B_{\theta} }{ r } \frac{ \partial \ B_{\theta} }{ \partial \ \theta } \tag{5} $$ Then the $\mathbf{B}$ crossed into this vector will not be zero.

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  • $\begingroup$ Could you tell me a bit more about where that extra term from the rank two tensor comes from? I've never used the formula before, so in my head it was as in the question $\endgroup$
    – Bidon
    Jun 2, 2020 at 18:16
  • $\begingroup$ @Bidon - It's just a rule from vector calculus with tensors. You can see the $\mathbf{B} \cdot \nabla B$ version at en.wikipedia.org/wiki/…. The full tensor version takes a little bit of tedious algebra but it's not terribly difficult to work out by hand. $\endgroup$ Jun 2, 2020 at 18:22
  • $\begingroup$ oh the material derivative, thank you! $\endgroup$
    – Bidon
    Jun 2, 2020 at 18:25

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