What is the difference between Larmor frequency and cyclotron frequency?

The Larmor frequency and cyclotron frequency differs by a factor of two. The later being twice the former. So when a magnetic field is applied when we can say that electron has Larmor frequency and when it will have cyclotron frequency? For delocalised electrons moving in the benzene ring, what frequency will they have if magnetic field is applied?

• Commenting just because the question and the most upvoted answer are currently incorrect, and this is the first google result for "larmor vs cyclotron frequency." For a free electron, the Larmor frequency and the cyclotron frequency DO NOT differ by a factor of two. In SI units, the larmor frequency is $\omega_L=g_e e B/2m$, and the cyclotron frequnecy is $\omega_c=eB/m$. $g_e=2.002318...\approx 2+\alpha/2\pi$. So the electron g factor nearly cancels the $2$ in the denominator of the larmor frequency formula (from the spin of the electron being 1/2). So the two frequencies are nearly the same. Apr 12, 2023 at 13:34

Cyclotron frequency is the frequency of electron rotation in magnetic field, due to the Lorentz force: $$m\frac{d\mathbf{v}}{dt}=\mathbf{F}=\frac{e}{c}\mathbf{v}\times\mathbf{B}\\\Rightarrow \frac{d\mathbf{v}}{dt}=-\frac{|e|B}{mc}\mathbf{v}\times\frac{\mathbf{B}}{B}= -\omega_c \mathbf{v}\times\mathbf{n}$$ where the cyclotron frequency is $$\omega_c=\frac{|e|B}{mc}$$.

Larmor frequency is the frequency of precession of a magnetic moment (see, e.g., Bloch equations) $$\frac{d\mathbf{M}}{dt}=\gamma \mathbf{M}\times\mathbf{B}=\gamma B \mathbf{M}\times\mathbf{n}=-\omega_L\mathbf{M}\times\mathbf{n}$$ where $$\gamma = -\frac{e}{2mc}$$ is the gyromagnetic ratio, and $$\omega_L=|\gamma|B=\frac{|e|B}{2mc}$$ is the Larmor frequency.

Amperian loop model
In order to relate intuitively gyromagnetic ration to one electron, let us consider an Amperian loop, carrying a current due to one electron charge. The magnetic moment of the loop is then $$M=JS/c$$, where $$S$$ is the cross-sectional area of the loop and $$J=-e\Omega/(2\pi)$$ with $$\Omega$$ being the frequency of electron rotation within the loop. The angular moment of the loop is $$\mathbf{L}=I\boldsymbol{\Omega}$$, where $$I=m R^2=mS/\pi$$ is the moment of inertia (for simplicity taking the loop to be circular.) We thus have $$\frac{d\mathbf{L}}{dt}=I\frac{d\boldsymbol{\Omega}}{dt}= \frac{mS}{\pi}\frac{d\boldsymbol{\Omega}}{dt}= \mathbf{M}\times\mathbf{B}=-\frac{e\boldsymbol{\Omega}S}{2\pi c}\times\mathbf{B}$$ which gives us $$\frac{d\boldsymbol{\Omega}}{dt}= \mathbf{M}\times\mathbf{B}=-\frac{e}{2mc}\boldsymbol{\Omega}\times\mathbf{B}$$

Larmor precession refers to precession of a magnetic moment - which in practice often means the spin of an electron, $$\mathbf{S}$$. Cyclotron frequency is the frequency of a charged particle rotating in a magnetic field, which in atomic physics is also seen as a precession - but of the orbital magnetic moment, $$\mathbf{L}$$. The two moments have different Landé g-factors (see also gyromagnetic ratio), that is different coefficients of proportionality between the mechanical moment and the magnetic moment, characterizing the response to the magnetic field: $$\boldsymbol{\mu}_L=-\mathbf{L}g_L\mu_B/\hbar,\\ \boldsymbol{\mu}_S=-\mathbf{S}g_S\mu_B/\hbar$$ where $$g_L=1$$, but $$g_S\approx 2$$. Hence the twofold difference in the precession frequencies: $$\omega_L=g_L\frac{eB}{mc} \text{ (cyclotron frequency)},\\ \omega_S=g_S\frac{eB}{mc} \text{ (Larmor frequency)}$$
• @Igris actually, what I wrote is wrong: for spin $g_S=2$ is cancelled by a factor of $2$ due to $S_z$ taking half integer values. The difference is really between the frequency of a free electron rotating in magnetic field and precession of the magnetic moment (two times smaller). I will fix my answer. Nov 30, 2022 at 12:53
• I still see an incorrect equation for $\omega_L$. The Larmor frequency ONLY differs from the cyclotron frequency by a small factor $(g-2)/2$, where $g\approx 2+\alpha/(2\pi)$. You seem to acknowledge some kind of mistake agout missing $g_s$ but the answer is still wrong as far as I can tell (and accepted with 8 upvotes). Apr 12, 2023 at 9:07