With this, the guiding center's equation of motion
is reduced to the following:
$$\begin{aligned}
m \dv{\vb{u}_{gc}}{t}
= q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
\end{aligned}$$
Let us now split $$\vb{u}_{gc}$$ into
components $$\vb{u}_{gc\perp}$$ and $$u_{gc\parallel} \vu{b}$$,
which are respectively perpendicular and parallel
to the magnetic unit vector $$\vu{b}$$,
such that $$\vb{u}_{gc} = \vb{u}_{gc\perp} \!+\! u_{gc\parallel} \vu{b}$$.
Consequently:
$$\begin{aligned}
\dv{\vb{u}_{gc}}{t}
= \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} + u_{gc\parallel} \dv{\vu{b}}{t}
\end{aligned}$$
Inserting this into the guiding center's equation of motion,
we now have:
$$\begin{aligned}
\dv{\vb{u}_{gc}}{t}
= m \bigg( \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} + u_{gc\parallel} \dv{\vu{b}}{t} \bigg)
= q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
\end{aligned}$$
The derivative of $$\vu{b}$$ can be rewritten as follows,
where $$R_c$$ is the radius of the field's [curvature](/know/concept/curvature/),
and $$\vb{R}_c$$ is the corresponding vector from the center of curvature:
$$\begin{aligned}
\dv{\vu{b}}{t}
\approx - u_{gc\parallel} \frac{\vb{R}_c}{R_c^2}
\end{aligned}$$
With this, we arrive at the following equation of motion
for the guiding center:
$$\begin{aligned}
m \bigg( \dv{\vb{u}_{gc\perp}}{t} + \dv{u_{gc\parallel}}{t} \vu{b} - u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} \bigg)
= q \bigg( \vb{E} + \vb{u}_{gc} \cross \vb{B} - \frac{u_L^2}{2 \omega_c} \nabla B \bigg)
\end{aligned}$$
Since both $$\vb{R}_c$$ and any cross product with $$\vb{B}$$
will always be perpendicular to $$\vb{B}$$,
we can split this equation into perpendicular and parallel components like so:
$$\begin{aligned}
m \dv{\vb{u}_{gc\perp}}{t}
&= q \vb{E}_{\perp} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\perp} B + m u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} + q \vb{u}_{gc} \cross \vb{B}
\\
m \dv{u_{gc\parallel}}{t}
&= q E_{\parallel} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\parallel} B
\end{aligned}$$
The parallel part simply describes an acceleration.
The perpendicular part is more interesting:
we rewrite it as follows, defining an effective force $$\vb{F}_{\!\perp}$$:
$$\begin{aligned}
m \dv{\vb{u}_{gc\perp}}{t}
= \vb{F}_{\!\perp} + q \vb{u}_{gc} \cross \vb{B}
\qquad \quad
\vb{F}_{\!\perp}
\equiv q \vb{E}_\perp + m u_{gc\parallel}^2 \frac{\vb{R}_c}{R_c} - \frac{q u_L^2}{2 \omega_c} \nabla_{\!\perp} B
\end{aligned}$$
To solve this, we make a crude approximation now, and improve it later.
We thus assume that $$\vb{u}_{gc\perp}$$ is constant in time,
such that the equation reduces to:
$$\begin{aligned}
0
\approx \vb{F}_{\!\perp} + q \vb{u}_{gc} \cross \vb{B}
= \vb{F}_{\!\perp} + q \vb{u}_{gc\perp} \cross \vb{B}
\end{aligned}$$
This is analogous to the previous case of a uniform electric field,
with $$q \vb{E}$$ replaced by $$\vb{F}_{\!\perp}$$,
so it is also solved by crossing with $$\vb{B}$$ in front,
yielding a drift:
$$\begin{aligned}
\vb{u}_{gc\perp}
\approx \vb{v}_F
\equiv \frac{\vb{F}_{\!\perp} \cross \vb{B}}{q B^2}
\end{aligned}$$
From the definition of $$\vb{F}_{\!\perp}$$,
this total $$\vb{v}_F$$ can be split into three drifts:
the previously seen electric field drift $$\vb{v}_E$$,
the **curvature drift** $$\vb{v}_c$$,
and the **grad-$$\vb{B}$$ drift** $$\vb{v}_{\nabla B}$$:
$$\begin{aligned}
\boxed{
\vb{v}_c
= \frac{m u_{gc\parallel}^2}{q} \frac{\vb{R}_c \cross \vb{B}}{R_c^2 B^2}
}
\qquad \quad
\boxed{
\vb{v}_{\nabla B}
= \frac{u_L^2}{2 \omega_c} \frac{\vb{B} \cross \nabla B}{B^2}
}
\end{aligned}$$
Such that $$\vb{v}_F = \vb{v}_E + \vb{v}_c + \vb{v}_{\nabla B}$$.
We are still missing a correction,
since we neglected the time dependence of $$\vb{u}_{gc\perp}$$ earlier.
This correction is called $$\vb{v}_p$$,
where $$\vb{u}_{gc\perp} \approx \vb{v}_F + \vb{v}_p$$.
We revisit the perpendicular equation, which now reads:
$$\begin{aligned}
m \dv{}{t}\big( \vb{v}_F + \vb{v}_p \big)
= \vb{F}_{\!\perp} + q \big( \vb{v}_F + \vb{v}_p \big) \cross \vb{B}
\end{aligned}$$
We assume that $$\vb{v}_F$$ varies much faster than $$\vb{v}_p$$,
such that $$\idv{}{\vb{v}p}{t}$$ is negligible.
In addition, from the derivation of $$\vb{v}_F$$,
we know that $$\vb{F}_{\!\perp} + q \vb{v}_F \cross \vb{B} = 0$$,
leaving only:
$$\begin{aligned}
m \dv{\vb{v}_F}{t}
= q \vb{v}_p \cross \vb{B}
\end{aligned}$$
To isolate this for $$\vb{v}_p$$,
we take the cross product with $$\vb{B}$$ in front,
like earlier.
We thus arrive at the following correction,
known as the **polarization drift** $$\vb{v}_p$$:
$$\begin{aligned}
\boxed{
\vb{v}_p
= - \frac{m}{q B^2} \dv{\vb{v}_F}{t} \cross \vb{B}
}
\end{aligned}$$
In many cases $$\vb{v}_E$$ dominates $$\vb{v}_F$$,
so in some literature $$\vb{v}_p$$ is approximated as follows:
$$\begin{aligned}
\vb{v}_p
\approx - \frac{m}{q B^2} \dv{\vb{v}_E}{t} \cross \vb{B}
= - \frac{m}{q B^2} \Big( \dv{}{t}(\vb{E}_\perp \cross \vb{B}) \Big) \cross \vb{B}
= - \frac{m}{q B^2} \dv{\vb{E}_\perp}{t}
\end{aligned}$$
The polarization drift stands out from the others:
it has the opposite sign,
it is proportional to $$m$$,
and it is often only temporary.
Therefore, it is also called the **inertia drift**.
## References
1. F.F. Chen,
*Introduction to plasma physics and controlled fusion*,
3rd edition, Springer.
2. M. Salewski, A.H. Nielsen,
*Plasma physics: lecture notes*,
2021, unpublished.