Consider a dilute solution composed of a polar molecule solute in a nonpolar solvent (polar molecules are molecules with a permanent electric dipole). The electric polarization \(P(t)\) of this fluid is driven out of equilibrium by an electric field \(E(t)\) and obeys the Langevin equation of motion $$ \frac{d P(t)}{\mathrm{d} t}+5 P(t)=\xi(t)+5 a E(t) $$ where \(\mathcal{E}(t)\) is a delta-correlated white noise (due to background magnetic fluctuations) with zero mean \((\mathcal{\xi}(t)\rangle=0\), and \(a\) is a constant. (a) Compute the linear response function \(K(t) .\) (b) Compute the dynamic susceptibility \(X(\omega)\) and find its limit as \(\omega \rightarrow 0\), where \(1 / 2 C\left(Q(0)^{2}\right\rangle_{T}=\frac{1}{2} k_{B T}\) from the equipartition theorem.

Short Answer

Expert verified
Linear response function \(K(t) = 5a e^{-5t} \theta(t)\) and dynamic susceptibility \(X(\omega) = \frac{5a}{i\omega + 5}\). As \(\omega \rightarrow 0, X(\omega) = a\).

Step by step solution

01

Understanding the Langevin Equation

The given Langevin equation is: \(\frac{d P(t)}{\mathrm{d} t}+5 P(t)=\xi(t)+5 a E(t)\). Here, \(P(t)\) is the electric polarization, \(E(t)\) is the electric field, and \(\xi(t)\) represents delta-correlated white noise.
02

Solve for the Linear Response Function

The linear response function \(K(t)\) characterizes how the polarization \(P(t)\) responds to the electric field \(E(t)\). Looking at the deterministic part of the equation (ignoring noise), this is a first-order linear differential equation with a driving term. We take the Fourier transform of the equation to solve it.
03

Fourier Transform the Langevin Equation

Applying Fourier transform to both sides of the differential equation gives: \[i\omega \tilde{P}(\tilde{\omega}) + 5 \tilde{P}(\tilde{\omega}) = 5a \tilde{E}(\tilde{\omega})\] Solving for \(\tilde{P}(\tilde{\omega})\): \[\tilde{P}(\tilde{\omega}) = \frac{5a\tilde{E}(\tilde{\omega})}{i\omega + 5}\]
04

Inverse Fourier Transform to Find \(K(t)\)

To find \(K(t)\), we consider that \(\tilde{P}(\tilde{\omega}) = \tilde{K}(\tilde{\omega}) \tilde{E}(\tilde{\omega})\), which gives: \[\tilde{K}(\tilde{\omega}) = \frac{5a}{i\omega + 5}\].We can then take the inverse Fourier transform to find \(K(t)\): \[ K(t) = 5a e^{-5t} \theta(t) \] where \(\theta(t)\) is the Heaviside step function.
05

Define Dynamic Susceptibility, \(X(\tilde{\omega})\)

The dynamic susceptibility \(X(\tilde{\omega})\) is defined as the Fourier transform of the linear response function \(K(t)\). Thus, \(\tilde{K}(\tilde{\omega}) = \tilde{X}(\tilde{\omega})\): \[X(\omega) = \frac{5a}{i\omega + 5}\].
06

Find the Limit of \(X(\tilde{\omega})\) as \(\omega \rightarrow 0\)

To find the limit of \(X(\tilde{\omega})\) as \(\omega \rightarrow 0\), we simply evaluate: \[ \lim_{\omega \to 0} X(\tilde{\omega}) = \lim_{\omega \to 0} \frac{5a}{i\omega + 5} = \frac{5a}{5} = a \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Polarization
In the context of physics, **polarization** refers to the orientation of the dipole moments within a material. Specifically, for molecules with a permanent electric dipole, polarization quantifies the extent to which the dipoles align under an external electric field.

Polar molecules, such as water, inherently possess an electric dipole moment. When an external electric field is applied, these dipoles tend to align themselves with the field, contributing to the overall **electric polarization**. This process can be mathematically described using the Langevin equation. In our specific scenario, the polarization of a dilute solution of polar molecules in a nonpolar solvent is given by the equation:

\[\frac{d P(t)}{\text{d} t} + 5 P(t) = \xi(t) + 5 a E(t)\].

Here:
  • **P(t)** represents the electric polarization at time **t**.
  • **E(t)** is the applied electric field.
  • **\xi(t)** is a term representing background magnetic fluctuations treated as white noise.
  • **a** is a constant related to the material properties.
This equation tells us how the polarization, **P(t)**, evolves over time under the influence of an electric field **E(t)** and random fluctuations **\xi(t)**.

Linear Response Function
The **Linear Response Function (K(t))** is a measure of how a system responds to an external perturbation. In our case, it tells us how the electric polarization **P(t)** responds to the applied electric field **E(t)**.

To compute **K(t)**, we consider the deterministic part of the Langevin equation, ignoring the noise term **\xi(t)**:
\[ \frac{d P(t)}{\text{d} t} + 5 P(t) = 5 a E(t) \]
By taking the Fourier transform of both sides, we move from the time domain to the frequency domain. This helps simplify the equation, allowing us to solve for **K(t)**:

\[i\omega \tilde{P}(\tilde{\omega}) + 5 \tilde{P}(\tilde{\omega}) = 5a \tilde{E}(\tilde{\omega}) \].
Solving for \tilde{P}(\tilde{\omega}) gives us:
\[ \tilde{P}(\tilde{\omega}) = \frac{5a \tilde{E}(\tilde{\omega})}{i\omega + 5} \].
The Fourier transform of **K(t)**, denoted as \tilde{K}(\tilde{\omega}), relates to **P(t)** and **E(t)** via:
\[ \tilde{K}(\tilde{\omega}) = \frac{5a}{i\omega + 5} \].
To find **K(t)** in the time domain, we take the inverse Fourier transform:

\[ K(t) = 5a e^{-5t} \theta(t) \],
where **\theta(t)** is the Heaviside step function, which ensures the function is zero for **t < 0** and 1 for **t >= 0**.

Simply put, this function describes how quickly the polarization returns to equilibrium after a sudden change in the electric field.

Dynamic Susceptibility
The **Dynamic Susceptibility (X(\tilde{\omega}))** of a system characterizes how it reacts to changes in an external field over different frequencies. It’s essentially the frequency domain equivalent of the **Linear Response Function**.

In our case, the dynamic susceptibility is the Fourier transform of the linear response function **K(t)**. From the previous computation, we have:

\[X(\omega) = \tilde{K}(\tilde{\omega}) = \frac{5a}{i\omega + 5} \].
This tells us how the system's polarization responds to varying frequencies of the applied electric field. When you want to understand the response at a specific frequency, this function gives you that information.

To find the behavior as the frequency approaches zero (the static limit), we evaluate:
\[\text{Lim}_{\omega \to 0} X(\tilde{\omega}) = \frac{5a}{i\omega + 5} = a \].
This result means that at static conditions (very low frequencies), the response is simply proportional to the constant **a**, which describes the material's intrinsic properties.

Key points to remember about dynamic susceptibility:
  • It provides insights into how a system responds to varying frequencies.
  • At static equilibrium conditions (\omega \to 0), the susceptibility simplifies to a constant.
  • It is derived from the linear response function, bridging the time and frequency domains.
This concept is pivotal in understanding the behavior of materials under varying external influences.

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Most popular questions from this chapter

The magnetization \(M(t)\) of a spin system is driven out of equilibrium by a magnetic field \(H(t)\) and obeys the Langevin equation of motion $$ \frac{\mathrm{d} M(t)}{\mathrm{d} t}+\Gamma M(t)=\xi(t)-\Gamma_{X_{0}} H(t) $$ where \(\mathcal{\xi}(t)\) is a delta-correlated white noise with zero mean that is due to background magnetic field fluctuations and \(\Gamma\) is a positive constant. (a) Compute the linear response function, \(K(t)\), which is defined through the equation \(\langle M(t)\rangle_{\varepsilon}=\int_{-\infty}^{\infty} \mathrm{d} t^{\prime} K\left(t-t^{\prime}\right) H\left(t^{\prime}\right),(M(t))_{\xi}=\) (b) Compute the equilibrium dynamic correlation function \((M(t) M(0)\rangle_{e q}+(c)\) What is \(\chi_{0}\) ?

Consider the "short time" relaxation of a free Brownian particle. The Langevin equation for the velocity is \(m \mathrm{~d} v / \mathrm{d} t=-\gamma v+\xi(t)\). (a) Find the Fokker-Planck equation for the probability, \(P(v, t) \mathrm{d} v\), to find the Brownian particle with velocity, \(v \rightarrow v+\mathrm{d} v\), at time \(t\). (b) Solve the FokkerPlanck equation, assuming that at time, \(t=0\), the velocity is \(v=v_{0}\). (Hint: Use the transform in Section Z.3.3.1 to write the Fokker-Planck in terms of a Hermitian operator. The eigenfunctions of that operator will be Hermite polynomials.)

Consider a 1D ballistic waveguide which is coupled to low-temperature electron reservoirs with Fermi distributions \(f_{\mathrm{L}}\left(E_{n}\right)\) and \(f_{\mathrm{R}}\left(E_{n}\right)\) on the left and right ends, respectively. Assume that the temperatures of the reservoirs are the same but that the left reservoir has an electric potential energy bias \(-e V\), where \(e\) is the electron charge. Assume the waveguide has a length \(L\) that is very large so you can take the limit \(L \rightarrow \infty\), and keep only the lowest-order term in \(V\). Compute the average current in the waveguide and show that the conductivity of the waveguide is given by \(G=e^{2} / h\), where \(h\) is Planck's constant.

Prove that $$ \langle\alpha(t)\rangle=2 \mathrm{i} \int_{-\infty}^{t} \mathrm{~d} t^{\prime} \bar{K}^{\prime \prime}\left(t-t^{\prime}\right) \cdot \bar{F}\left(t^{\prime}\right) $$ where \(\bar{K}^{\prime \prime}(t)\) is the Fourier transform of \(\bar{X}^{\prime \prime}(\omega) .\) This form of the linear response is commonly seen in the literature. [Note that the identity $$ \lim _{\eta \rightarrow 0} \frac{1}{\omega^{\prime}-\omega \mp \mathrm{i} \eta}=P \frac{1}{\omega^{\prime}-\omega} \pm \mathrm{i} \pi \delta\left(\omega^{\prime}-\omega\right) $$ and the spectral representation of the Heaviside function $$ \theta\left(t-t^{\prime}\right)=-\lim _{\eta \rightarrow 0} \int_{-\infty}^{\infty} \frac{\mathrm{d} \omega}{(2 \pi \mathrm{i})} \frac{\mathrm{e}^{-\mathrm{i} \omega\left(t-t^{\prime}\right)}}{\omega+i \eta} $$ are useful.]

A Brownian particle of mass \(m\) is attached to a harmonic spring with force constant, \(k\), and is driven by an external force, \(F(t)\). The particle is constrained to move in one dimension. The Langevin equation is $$ m \frac{\mathrm{d}^{2} x(t)}{\mathrm{d} t^{2}}+\gamma \frac{\mathrm{d} x(t)}{\mathrm{d} t}+m \omega_{0}^{2} x(t)=\xi(t)+F(t) $$ where \(\omega_{0}=k / m, \gamma\) is the friction constant, and \(\mathcal{\xi}(t)\) is a Gaussian white noise with zero mean, \((\xi(t)\rangle_{\varepsilon}=0 .\) Here ()\(\varepsilon\) denotes the average over values of the random force. Consider the overdamped case. (a) Compute the equilibrium correlation function, \((\langle x(t) x(0)\rangle \xi\rangle T\), starting from the Langevin equation above with \(F(t)=0\). Let (\rangle\(_{T}\) denote the thermal average over the initial position and velocity of the Brownian particle. Assume that \((x(0) v(0)\rangle_{T}=0\) and \(\left(x(0)^{2}\right\rangle_{T}=k_{\mathrm{B}} T /\left(m \omega_{0}^{2}\right)\), (b) Show that the dynamic susceptibility for the Brownian oscillator is \(X(\omega)=\left(-m \omega^{2}+m \omega_{n}^{2}-\mathrm{i} \gamma \omega\right)^{-1}\). Use the fluctuation- dissipation theorem, $$ (x(t) x(0)\rangle=\frac{k_{\mathrm{B}} T}{\mathrm{i} \pi} P \int_{-\infty}^{\infty} \frac{1}{\omega} X(\omega) \cos (\omega t) $$ to compute the equilibrium correlation function. Do your results in (a) and (b) agree?

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