Apparent disagreement in curvature drift formulas

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?

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: $$V_{gc} = \frac{ m_{s} }{ 2 \ q_{s} \ B } \left( v_{\perp}^{2} + 2 \ v_{\parallel}^{2} \right) \left[ \frac{ \mathbf{B} \times \nabla B }{ B^{2} } \right] \tag{4}$$
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.
• @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. Jun 2, 2020 at 18:22