When we put a diamagnetic material in the presence of an external magnetic field $\vec B_0$, the magnetic field inside the material decreases to $$\vec B=(1+\chi_m)\vec B_0,$$ where the magnetic susceptibility $\chi_m$ is a small negative number. I am assuming the material is linear and isotropic.

On the other hand the diamagnetism is explained in terms of Lenz law. When we change the magnetic flux of the external field over the material, atomic currents generate an induced magnetic field trying to restore the flux. But then the induced field could have any sign (in the appropriated direction), depending if we are increasing or decreasing the flux of the external field. It seems diamagnetism is a dynamic effect. How come the sign of $\chi_m$ is always negative? Moreover, the magnitude of $\chi_m$ should depend on how large is the variation of the flux but I do not see any suggestion of this when looking at tables of $\chi_m$.


In general your relation is $$ \vec{B}(\omega) = (1 + \chi_m(\omega))\vec{B}_0(\omega) $$ or in the time domain $$ \vec{B}(t) =\vec{B}_0(t) + \int\limits_{-\infty}^\infty \chi_m(t,t') \vec{B}_0(t') \;\rm{d}t' $$ Only in the case of instantanous material response, i.e. $\chi_m(t,t') = \chi_{m,0} \cdot \delta(t-t') $, your equation is correct. This already tells us that in the usual approximation of constant susceptibility $\chi_m(\omega) = \chi_{m,0}$ the material response is much faster than the applied magnetic field. In free induction decay on the other hand one applies a very short magnetic pulse and can observe the behaviour of $\chi_m(t,t')$. The observed magnetic field for $\vec{B}_0(t) \propto \delta(t-t_0)$ is $$ \vec{B}(t) =\vec{B}_0(t) + \text{const}\cdot \chi_m(t,t_0) $$ The properties of $\chi_m(\omega)$ can normally only be understood with quantum mechanics. Furthermore I will also assume the static limit $\chi_m(\omega\rightarrow 0) = \chi_{m,0}$.


Diamagnetism is present in basically all matter and leads to a negative $\chi_{m,0}$. The simplest example is Helium. If one applies a magnetic field to a quantum system the electronic wavefunction will change due to this perturbation. This leads to an increase in the total energy and consequently to a counteracting force. The counteracting magnetic field is generated by a change in the orbital momentum of the electrons in the material. These are what one would classically interpret as induced currents but as the wavefunction is not time dependent in this case I would not call that a dynamic effect.

If the material has unpaired electron spins it will also show paramagnetism or ferromagnetism. Those are typically much stronger and overshadow the diamagnetism.

Switching of fields

Consider a Field that is turned on at time zero and constant afterwards with $B_0(t) = B_0\Theta(t)$ and a simple example for the susceptibility with $$\chi_m(t,t') = \left(\sin[w_0(t-t')]+\frac{\chi_{m,0}}{T_1}\right) \exp\left[-\frac{(t-t')}{T_1}\right]$$ The magnetization for $t>0$ is then given by $$M(t) = \frac{B_0}{\mu_0} \int\limits_{0}^t\chi_m(t,t')\text{d}t' = \\\chi_{m,0} (1-\exp(-t/T_1))-T_1\exp(-t/T_1) \frac{ -\exp(t/T_1) T_1 w_0+T_1 w_0 \cos[ w_0 t]+\sin[w_0 t])}{(1+T_1^2 w_0^2)} $$ and looks like this enter image description here

You can see that in the beginning the magnetization is oscillating and takes positive and negative values. For $t\rightarrow\infty$ it however approaches a negative value in the case of diamagnetic materials. Your confusion comes from the fact that you consider $\chi_m$ to be a number instead of a function. When we say a material is diamagnetic with $\chi_{m,0} = -1$ what we really mean is $\chi_m(\omega\rightarrow 0) = -1$

  • $\begingroup$ Even if $\chi_m=\chi_m(\omega)$, how can it never be positive? Consider a diamagnetic material in the presence of a external field. Then you turn down the external field. The induced current would generate a magnetic field in the same direction the external field. The resultant field would be greater than the external, i.e. $\chi_m>0$. I would like to understand where this reasoning is wrong. $\endgroup$ – Diracology May 19 '16 at 21:29
  • $\begingroup$ Your reasoning is not wrong at all. You are just mixing up frequency and time domain. The sign of$\chi_m(\omega)$ does not imply any sign condition for $M(t)$ $\endgroup$ – Jannick May 20 '16 at 9:51

The diamagnetism is definitely a static effect in the sense it happens even for static magnetic fields.

My mistake in the original post was to assume that the origin of the diamagnetism is the Faraday-Lenz law. In fact, the Bohr–van Leeuwen theorem says that magnetic phenomena such as diamagnetism, paramagnetism and ferromagnetism are strictly quantum effects.

The Hamiltonian for a charged particle, of charge $-e$, in a magnetic field $\vec B$ can be written as $$H=\frac{(\vec p+e\vec A)^2}{2m}+g\mu_B\vec B\cdot\vec\sigma+V(\vec r),$$ where $\vec\sigma$ is the electron spin. For a uniform magnetic field this can be rewritten as $$H=H_0+\mu_B\vec B\cdot(\vec l+g\vec\sigma)+\frac{e^2}{8m}|\vec B\times\vec r|^2,$$ where $\vec l$ is the orbital angular momentum of the electron. The first term on the rhs is just the Hamiltonian of the particle in the absence of magnetic field, the second term gives the paramagnetism and the third term originates diamagnetism.

Considering a $z$ oriented magnetic field, the expectation value of the diamagnetic field is $$E=\frac{e^2B^2}{12m}\langle r^2\rangle.$$ The magnetic moment per electron is $$-\frac{dE}{dB}=-\frac{e^2B}{6m}\langle r^2\rangle,$$ from which we obtain the magnetic susceptibility $$\chi=-\frac{ne^2\mu_0\langle r^2\rangle}{6m}.$$

This formula, obtained purely from quantum mechanics and using static fields exactly matches the classical Langevin expression for the magnetic susceptibility which can only be obtained for time varying fields.


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