# Can Einstein relation be used to calculate mobility under equilibrium dynamics?

In weak field non-equilibrium dynamics, mobility can be calculated by Einstein relation $$\mu=\frac{eD}{K_BT}$$, where $$D$$ is diffusion constant.

Mobility can also be calculated by the definition $$\mu=\frac{v}{F}$$, where $$v$$ is drift velocity and $$F$$ is external field.

Question:

1. What if there is no drift (equilibrium dynamics), does the mobility $$\mu=0$$? Does this result contradict with Einstein relation?

2. what if external field $$F$$ becomes large enough and it's beyond linear response theory? Einstein relation collapses for far non-equilibrium area. How to calculate the diffusion constant and mobility?

• Yes, indeed you can. – Jon Custer Jan 9 at 22:03
• I edited the question, pls take a look. – kinder chen Jan 9 at 22:31
• In equilibrium, things still move. So, yes, there is mobility. – Jon Custer Jan 9 at 22:42

This is the topic of linear response theory. A typical static example, which makes a good analogy, is when you imagine measuring the response of a polar molecule in a liquid to an applied electric field. At equilibrium, the average dipole moment in any direction, say $$x$$, is $$\langle p_x\rangle_0=0$$. The subscript $$0$$ means that we are averaging in an unperturbed ensemble. But if we apply a small external electric field $$E_x$$, which can be thought of as introducing a perturbation term $$-p_x E_x$$ in the energy, the induced dipole will be proportional to $$E_x$$: $$\langle p_x\rangle = \alpha E_x \quad\Rightarrow\quad \alpha = \frac{\langle p_x\rangle}{E_x} \tag{1}$$ where now we are averaging in a perturbed ensemble. Linear response theory (I won't reproduce the whole derivation) tells us that we can also express the response coefficient $$\alpha$$ as an equilibrium ensemble average of dipole fluctuations $$\alpha = \frac{\langle p_x^2\rangle_0}{k_BT} \tag{2}$$ where $$k_B$$ is Boltzmann's constant and $$T$$ is the temperature. In the case of eqn (1), we can allow $$E_x\rightarrow 0$$ and $$\langle p_x\rangle\rightarrow 0$$ while their ratio $$\alpha$$ remains well defined. And eqn (2) makes it clear that one can still define $$\alpha$$ even in the absence of an external field. (I've used a classical statistical mechanics version of this, but very similar ideas turn up when considering the polarizability of an atom in quantum mechanics, relating it to dipolar fluctuations of the atom's electron cloud, using perturbation theory).
Turning to the mobility, dynamical linear response theory gives us an expression for the steady state drift velocity $$v_x$$ of a molecule in a liquid which has been subjected to a constant external force $$F_x$$, as \begin{align*} \langle v_x\rangle &= \frac{F_x}{k_B T} \int_{0}^\infty \langle v_x(0) v_x(t)\rangle_0 \, d t \\ \Rightarrow\quad \mu \equiv \frac{\langle v_x\rangle}{F_x} &= \frac{1}{k_B T} \int_{0}^\infty \langle v_x(0) v_x(t)\rangle_0 \, d t = \frac{D}{k_BT}. \end{align*} On the right we have the time integral of an equilibrium quantity, the velocity autocorrelation function. The link with the diffusion coefficient comes from noting that the displacement can be expressed as $$\Delta x(t) = x(t)-x(0) = \int_0^t v_x(t') dt'$$ and working out the mean-squared displacement at long times. Again I've omitted the details (you can easily look them up). So we get the Einstein relation (your version has an extra factor $$e$$ which probably means that you are thinking of the electrical mobility, defined in terms of the electric field, rather than the mobility, defined in terms of the applied force). But once again, it is always possible to allow $$F_x\rightarrow 0$$ and $$\langle v_x\rangle\rightarrow 0$$ while their ratio remains well defined, and the equilibrium time correlation function formula emphasizes that the mobility itself is still defined even in the absence of an external field.
• Thanks, considering both positive and negative direction the molecule or electron can move to, the ensemble average displacement $\Delta x(t)$ should be 0, correct? And then $⟨v_x⟩$=0? – kinder chen Jan 10 at 18:57
• But by Einstein relation $\mu$ is nonzero, isn't this contradictory？ – kinder chen Jan 10 at 20:12
• Thanks a lot, Prof. I want to numerically calculate the mobility by $\mu=v/F$ but failed, maybe that's why I thought it would be 0. – kinder chen Jan 10 at 21:53