Thermal noise in an electrical resistor There is a close analogy between the paradigmatic Brownian particle discussed above and
Johnson noise, the electric voltage generated by thermal fluctuations in a resistor. The diagram at the right shows an electric circuit consisting of a
resistance R and a
capacitance C. The slow variable is the voltage
U between the ends of the resistor. The Hamiltonian reads \mathcal{H} = E / k_\text{B}T = CU^2 / (2k_\text{B}T), and the Langevin equation becomes \frac{\mathrm{d}U}{\mathrm{d}t} =-\frac{U}{RC} + \eta \left( t\right),\;\;\left\langle \eta \left( t\right) \eta \left( t'\right)\right\rangle = \frac{2k_\text{B}T}{RC^{2}}\delta \left(t-t'\right). This equation may be used to determine the correlation function \left\langle U\left(t\right) U\left(t'\right) \right\rangle = \frac{k_\text{B}T}{C} \exp \left(-\frac{\left| t - t'\right| } {RC}\right) \approx 2Rk_\text{B}T \delta \left( t - t'\right), which becomes white noise (Johnson noise) when the capacitance becomes negligibly small.
Critical dynamics The dynamics of the
order parameter \varphi of a second order phase transition slows down near the
critical point and can be described with a Langevin equation. \begin{align} R_{vv}(t_1,t_2) & \equiv \langle \mathbf{v}(t_1) \cdot \mathbf{v}(t_2) \rangle \\ & = v^2(0) e^{-(t_1+t_2)/\tau} + \int_0^{t_1} \int_0^{t_2} R_{aa}(t_1',t_2') e^{-(t_1+t_2-t_1'-t_2')/\tau} dt_1' dt_2' \\ & \simeq v^2(0) e^{-|t_2-t_1|/\tau} + \left[\frac{3k_\text{B}T}{m} - v^2(0)\right] \Big[e^{-|t_2-t_1|/\tau} - e^{-(t_1+t_2)/\tau}\Big], \end{align} where we have used the property that the variables \mathbf{a}(t_1') and \mathbf{a}(t_2') become uncorrelated for time separations t_2'-t_1' \gg t_c. Besides, the value of \lim_{t \to \infty} \langle v^2 (t) \rangle = \lim_{t \to \infty} R_{vv}(t,t) is set to be equal to 3k_\text{B}T/m such that it obeys the
equipartition theorem. If the system is initially at thermal equilibrium already with v^2(0) = 3 k_\text{B} T/m, then \langle v^2(t) \rangle = 3 k_\text{B} T/m for all t, meaning that the system remains at equilibrium at all times. The velocity \mathbf{v}(t) of the Brownian particle can be integrated to yield its trajectory \mathbf{r}(t). If it is initially located at the origin with probability 1, then the result is \mathbf{r}(t) = \mathbf{v}(0) \tau \left(1-e^{-t/\tau}\right) + \tau \int_0^t \mathbf{a}(t') \left[1 - e^{-(t-t') / \tau}\right] dt'. Hence, the average displacement \langle \mathbf{r}(t) \rangle = \mathbf{v}(0) \tau \left(1-e^{-t/\tau}\right) asymptotes to \mathbf{v}(0) \tau as the system relaxes. The
mean squared displacement can be determined similarly: \langle r^2(t) \rangle = v^2(0) \tau^2 \left(1 - e^{-t/\tau}\right)^2 - \frac{3k_\text{B}T}{m} \tau^2 \left(1 - e^{-t/\tau}\right) \left(3 - e^{-t/\tau}\right) + \frac{6k_\text{B}T}{m} \tau t. This expression implies that \langle r^2(t \ll \tau) \rangle \simeq v^2(0) t^2, indicating that the motion of Brownian particles at timescales much shorter than the relaxation time \tau of the system is (approximately)
time-reversal invariant. On the other hand, \langle r^2(t \gg \tau) \rangle \simeq 6 k_\text{B} T \tau t/m = 6 \mu k_\text{B} T t = 6Dt, which indicates an
irreversible,
dissipative process. . The left panel shows the time evolution of the phase portrait at different temperatures. The right panel captures the corresponding equilibrium probability distributions. At zero temperature, the velocity slowly decays from its initial value (the red dot) to zero, over the course of a handful of oscillations, due to damping. For nonzero temperatures, the velocity can be kicked to values higher than the initial value due to thermal fluctuations. At long times, the velocity remains nonzero, and the position and velocity distributions correspond to that of thermal equilibrium.
Recovering Boltzmann statistics If the external potential is conservative and the noise term derives from a reservoir in thermal equilibrium, then the long-time solution to the Langevin equation must reduce to the
Boltzmann distribution, which is the probability distribution function for particles in thermal equilibrium. In the special case of
overdamped dynamics, the inertia of the particle is negligible in comparison to the damping force, and the trajectory x(t) is described by the overdamped Langevin equation \lambda \frac{dx}{dt} = - \frac{\partial V(x)}{\partial x} + \eta(t)\equiv - \frac{\partial V(x)}{\partial x}+\sqrt{2 \lambda k_\text{B} T} \frac{dB_t}{dt}, where \lambda is the damping constant. The term \eta(t) is white noise, characterized by \left\langle\eta(t) \eta(t')\right\rangle = 2 k_\text{B} T \lambda \delta(t-t') (formally, the
Wiener process). One way to solve this equation is to introduce a test function f and calculate its average. The average of f(x(t)) should be time-independent for finite x(t), leading to \frac{d}{dt} \left\langle f(x(t))\right\rangle = 0,
Itô's lemma for the
Itô drift-diffusion process dX_t = \mu_t \, dt + \sigma_t \, dB_t says that the differential of a twice-differentiable function is given by df = \left(\frac{\partial f}{\partial t} + \mu_t\frac{\partial f}{\partial x} + \frac{\sigma_t^2}{2}\frac{\partial^2 f}{\partial x^2}\right)dt + \sigma_t\frac{\partial f}{\partial x}\,dB_t. Applying this to the calculation of \langle f(x(t)) \rangle gives \left\langle -f'(x)\frac{\partial V}{\partial x} + k_\text{B} T f''(x) \right\rangle=0. This average can be written using the probability density function p(x); \begin{align} & \int \left( -f'(x)\frac{\partial V}{\partial x}p(x) + {k_\text{B} T} f''(x)p(x) \right) dx \\ = &\int \left( -f'(x)\frac{\partial V}{\partial x} p(x) - {k_\text{B} T} f'(x)p'(x) \right) dx \\ = & \; 0 \end{align} where the second term was integrated by parts (hence the negative sign). Since this is true for arbitrary functions f, it follows that \frac{\partial V}{\partial x} p(x) + {k_\text{B} T} p'(x) = 0, thus recovering the Boltzmann distribution p(x) \propto \exp \left( {-\frac{ V(x)}{k_\text{B} T}}\right). == Equivalent techniques ==