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?
 A: 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.
