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The guiding center is the center of a charged particle's circular cyclotron orbit. In a nonuniform field it drifts slowly compared with the rapid cyclotron frequency.

In a uniform magnetic field, the Lorentz force leaves the component of velocity parallel to the field constant and rotates the perpendicular component at the nonrelativistic cyclotron frequency $|q|B/m$. The resulting path is a helix around a field line.

In classical charged particle in a uniform magnetic field motion, the Lorentz force and Newton's second law giveAssume $\mathbf{B}\ne 0$ and choose an orthonormal coordinate system with $\mathbf{B}=B{\mathbf{e}}_z$, $B=|\mathbf{B}|$. Write $\Omega =qB/m$, retaining the sign of the electric charge. The velocity equations areThus the parallel velocity is constant, and the perpendicular velocity rotates at constant angular frequency. Choosing the time origin and perpendicular axes conveniently givesIntegrating the velocity, one obtainsThe first two coordinates describe a circle of radius $m{v}_{\perp }/(|q|B)$, while the third advances uniformly. This is helical motion in a uniform magnetic field: the helix has axis parallel to $\mathbf{B}$ through the guiding centre. Its cyclotron frequency isand its signed angular velocity vector about that axis is $-q\mathbf{B}/m$. The minus sign follows from the ordering $\mathbf{v}\times \mathbf{B}$ in the Lorentz force. The helix degenerates to a circle when $v_{\parallel }=0$ and to a straight line when $v_{\perp }=0$. If $q=0$ or $B=0$, there is only straight-line motion. In a relativistic extension with fixed Lorentz factor $\gamma$, the frequency becomes $|q|B/(\gamma m)$; the classical result above uses $m\mathbf{v}$ as momentum.

Use a Lorentz boost along positive $x$ with $v=E/B$. Since $E<cB$, its dimensionless speed $\beta =E/(cB)$ is less than one. The given transformation of the electric field and magnetic field yields $E_y^{\prime }=\gamma (\beta )(E-vB)=0$ andAll other electric components vanish. In this frame a magnetic field does no work, so the particle's speed $u<c$ and Lorentz factor ${\gamma }_u=(1-u^2/c^2{)}^{-1/2}$ are constant. In proper time, the Lorentz force equations are $d({\gamma }_u{u}_x^{\prime })/d\tau =(q{B}^{\prime }/m){\gamma }_u{u}_y^{\prime }$ and $d({\gamma }_u{u}_y^{\prime })/d\tau =-(q{B}^{\prime }/m){\gamma }_u{u}_x^{\prime }$. Thus, choosing the phase and origin,This is the signed proper-time cyclotron frequency; the frequency with respect to $t^{\prime }$ is $\omega /{\gamma }_u$. The inverse Lorentz boost givesIn particular the average drift speed is $\beta c=E/B$ along $\mathbf{E}\times \mathbf{B}$.

For $u>0$, $q\ne 0$, set $s=\omega \tau$ and $d=\beta c/u$. The dimensionless spatial curve isThe drift between successive cycles is $2\pi \gamma (\beta )d$. When $2\pi \beta \ll u/c$, this is small compared with the oscillation size: the curve consists of nearly closed loops with slow rightward drift. When $2\pi \beta \gg u/c$, the longitudinal drift is large, producing a progressing oscillatory curve. More precisely $d\tilde{x}/ds=\gamma (\beta )(d+\cos s)$: loops occur for $d<1$, cusps at $d=1$, and monotone progression for $d>1$. The stated large-drift inequality compares the drift over a full cycle, so an example of that regime with $d>1$ is shown; that comparison alone does not replace the exact loop criterion.

Reversing $q$ reverses the cyclotron orientation; the drift remains along positive $x$. If $u=0$ the trajectory is pure straight drift and the dimensionless coordinates are undefined. If $q=0$ the particle is inertial and the displayed $1/\omega$ parametrization must be replaced by its inertial limit.

These potentials give the uniform magnetic field $\mathbf{B}=B\hat{\mathbf{z}}$ and zero electric field. With the signed frequency ${\omega }_c=qB/m$, the Lorentz force law becomesThusand explicitlyThe transverse trajectory is a circle about the fixed guiding centre $(x+\dot{y}/{\omega }_c,y-\dot{x}/{\omega }_c)$. The positive cyclotron frequency is

Let $\mathbf{E}=E\hat{\mathbf{x}}$ and $\mathbf{B}=B\hat{\mathbf{y}}$, and choose $\hat{\mathbf{z}}=\hat{\mathbf{x}}\times \hat{\mathbf{y}}$. Part (a) gives $y=0$. With the origin at ${\mathbf{x}}_0$, the remaining Cartesian components areThe initial conditions are $x=z=\dot{x}=\dot{z}=0$. Integrating the second equation gives $\dot{z}=\omega x$, soSolving this forced harmonic oscillator and then integrating $\dot{z}=\omega x$ yieldsThus the position vector is ${\mathbf{x}}_0+x(t)\hat{\mathbf{x}}+z(t)\hat{\mathbf{z}}$. The oscillation occurs at the signed cyclotron frequency, superposed on the usual E-cross-B drift.

Take $B>0$ and use the Landau gauge $\mathbf{A}=(0,Bx,0)$. With zero $z$-momentum, the Hamiltonian becomesFor a state $e^{i{k}_{y}y}\chi (x)$, this is a quantum harmonic oscillator in $x$, centred atwith cyclotron frequencyThe Landau levels are therefore

Periodicity in $y$ quantizes $k_y=2\pi \ell /R_y$. Requiring the corresponding oscillator centre to lie in an $x$ interval of length $R_x$ leavesallowed centres for each $n$. This is the degeneracy of a Landau level, equal to the magnetic flux through the rectangle divided by the magnetic flux quantum $h/|e|$. Consistent magnetic periodic boundary conditions make $D$ an integer.