Skip to main content
Physics

Return to Answer

added 47 characters in body
Source Link
Tom-Tom
Tom-Tom
  • 2k
  • 12
  • 23

Let us use the definition $\langle\eta(s)\eta(t)\rangle=\Gamma\gamma^2\delta(t-s)$. First of all, $C(s,t)$ depends on $t$ because $$C(s,t)=\Gamma\min(s,t).$$

It is clear, from causality, that $\frac{\delta x(t)}{\delta \eta(s)}=0$ if $t<s$. If $t>s$, compute the difference $\delta x(t)$ causecaused by two realisations of noise that differ only at time $s$ by an amount $\delta\eta$. This means that $\delta\eta(t)=\delta\eta\,\delta(t-s)$. Then you have $\frac{\partial \delta x}{\partial t}=0$ for $t>s$ and $\delta x(s)=\gamma^{-1}\delta\eta$,$$\frac{\partial \delta x}{\partial t}=\gamma^{-1}\delta\eta \,\delta(t-s)$$ and therefore $\delta x(t)=\gamma^{-1}\delta\eta(s)$$\delta x(t)=\gamma^{-1}\delta\eta$ for $t>s$, which means that $$\frac{\delta x(t)}{\delta\eta(s)}=\frac1\gamma\,\Theta(t-s).$$

Now we compute $$\frac{\partial C}{\partial s}(s,t)=\Gamma\Theta(t-s).$$ So this agrees with the fluctuation-dissipation theorem if $$\Gamma=\frac1\gamma\qquad\text{or}\qquad\langle\eta(s)\eta(t)\rangle=\gamma\delta(t-s).$$

The value of $\Gamma$ that you proposed, $2k_{\text B}T\gamma$, has no meaning here, since the particle has no mass, it has no kinetic energy and it is therefore impossible for it to be thermalized.

Let us use the definition $\langle\eta(s)\eta(t)\rangle=\Gamma\gamma^2\delta(t-s)$. First of all, $C(s,t)$ depends on $t$ because $$C(s,t)=\Gamma\min(s,t).$$

It is clear, from causality, that $\frac{\delta x(t)}{\delta \eta(s)}=0$ if $t<s$. If $t>s$, compute the difference $\delta x(t)$ cause by two realisations of noise that differ only at time $s$ by an amount $\delta\eta$. Then you have $\frac{\partial \delta x}{\partial t}=0$ for $t>s$ and $\delta x(s)=\gamma^{-1}\delta\eta$, therefore $\delta x(t)=\gamma^{-1}\delta\eta(s)$ for $t>s$, which means that $$\frac{\delta x(t)}{\delta\eta(s)}=\frac1\gamma\,\Theta(t-s).$$

Now we compute $$\frac{\partial C}{\partial s}(s,t)=\Gamma\Theta(t-s).$$ So this agrees with the fluctuation-dissipation theorem if $$\Gamma=\frac1\gamma\qquad\text{or}\qquad\langle\eta(s)\eta(t)\rangle=\gamma\delta(t-s).$$

The value of $\Gamma$ that you proposed, $2k_{\text B}T\gamma$, has no meaning here, since the particle has no mass, it has no kinetic energy and it is therefore impossible for it to be thermalized.

Let us use the definition $\langle\eta(s)\eta(t)\rangle=\Gamma\gamma^2\delta(t-s)$. First of all, $C(s,t)$ depends on $t$ because $$C(s,t)=\Gamma\min(s,t).$$

It is clear, from causality, that $\frac{\delta x(t)}{\delta \eta(s)}=0$ if $t<s$. If $t>s$, compute the difference $\delta x(t)$ caused by two realisations of noise that differ only at time $s$ by an amount $\delta\eta$. This means that $\delta\eta(t)=\delta\eta\,\delta(t-s)$. Then you have $$\frac{\partial \delta x}{\partial t}=\gamma^{-1}\delta\eta \,\delta(t-s)$$ and therefore $\delta x(t)=\gamma^{-1}\delta\eta$ for $t>s$, which means that $$\frac{\delta x(t)}{\delta\eta(s)}=\frac1\gamma\,\Theta(t-s).$$

Now we compute $$\frac{\partial C}{\partial s}(s,t)=\Gamma\Theta(t-s).$$ So this agrees with the fluctuation-dissipation theorem if $$\Gamma=\frac1\gamma\qquad\text{or}\qquad\langle\eta(s)\eta(t)\rangle=\gamma\delta(t-s).$$

The value of $\Gamma$ that you proposed, $2k_{\text B}T\gamma$, has no meaning here, since the particle has no mass, it has no kinetic energy and it is therefore impossible for it to be thermalized.

Source Link
Tom-Tom
Tom-Tom
  • 2k
  • 12
  • 23

Let us use the definition $\langle\eta(s)\eta(t)\rangle=\Gamma\gamma^2\delta(t-s)$. First of all, $C(s,t)$ depends on $t$ because $$C(s,t)=\Gamma\min(s,t).$$

It is clear, from causality, that $\frac{\delta x(t)}{\delta \eta(s)}=0$ if $t<s$. If $t>s$, compute the difference $\delta x(t)$ cause by two realisations of noise that differ only at time $s$ by an amount $\delta\eta$. Then you have $\frac{\partial \delta x}{\partial t}=0$ for $t>s$ and $\delta x(s)=\gamma^{-1}\delta\eta$, therefore $\delta x(t)=\gamma^{-1}\delta\eta(s)$ for $t>s$, which means that $$\frac{\delta x(t)}{\delta\eta(s)}=\frac1\gamma\,\Theta(t-s).$$

Now we compute $$\frac{\partial C}{\partial s}(s,t)=\Gamma\Theta(t-s).$$ So this agrees with the fluctuation-dissipation theorem if $$\Gamma=\frac1\gamma\qquad\text{or}\qquad\langle\eta(s)\eta(t)\rangle=\gamma\delta(t-s).$$

The value of $\Gamma$ that you proposed, $2k_{\text B}T\gamma$, has no meaning here, since the particle has no mass, it has no kinetic energy and it is therefore impossible for it to be thermalized.