A Review of Brownian Motion Based Solely on the Langevin Equation with White Noise

Abstract

We give a historical and mathematical review of Brownian motion based solely on the Langevin equation. We derive the main statistical properties without bringing in external and subsidiary issues, such as temperature, Focker-Planck equations, the Maxwell–Boltzmann distribution, spectral analysis, the fluctuation-dissipation theorem, among many other topics that are typically introduced in discussions of the Langevin equation. The method we use is the formal solution approach, which was the standard method devised by the founders of the field. In addition, we give some relevant historical comments.

Appendices

Appendix 1: Formal Solutions

Treating the Langevin equation as an ordinary differential equation with a driving force F(t), the solution, with initial condition v(0) is

$$\begin{aligned} v(t)=e^{-\beta t}v(0)+\int _{0}^{t}e^{\beta (t^{\prime }-t)}F(t^{\prime })dt^{\prime } \end{aligned}$$

(159)

This can be verified by direct substitution. Since

$$\begin{aligned} x(t)=x(0)+\int _{0}^{t}v(t^{\prime })\,dt^{\prime } \end{aligned}$$

(160)

we have

$$\begin{aligned} x(t)&=x(0)+\int _{0}^{t}\left[ e^{-\beta t^{\prime }}v(0)+\int _{0}^{t^{\prime }}e^{\beta (t^{\prime \prime }-t^{\prime })}F(t^{\prime \prime })dt^{\prime \prime }\right] dt^{\prime }\end{aligned}$$

(161)

$$\begin{aligned}&=x(0)+\frac{v(0)}{\beta }(1-e^{-\beta t})+\int _{0}^{t}\int _{0}^{t^{\prime } }e^{\beta (t^{\prime \prime }-t^{\prime })}F(t^{\prime \prime })dt^{\prime \prime }dt^{\prime } \end{aligned}$$

(162)

But

$$\begin{aligned} \int _{0}^{t}\int _{0}^{t^{\prime }}e^{\beta (t^{\prime \prime }-t^{\prime } )}F(t^{\prime \prime })dt^{\prime \prime }dt^{\prime }&=\int _{0}^{t}e^{-\beta t^{\prime }}\int _{0}^{t^{\prime }}e^{\beta t^{\prime \prime }}F(t^{\prime \prime })dt^{\prime \prime }\,dt^{\prime }\end{aligned}$$

(163)

$$\begin{aligned}&=-\frac{1}{\beta }e^{-\beta t}\int _{0}^{t}e^{\beta s}F(t^{\prime } )dt^{\prime }+\frac{1}{\beta }\int _{0}^{t}F(t^{\prime })dt^{\prime }\end{aligned}$$

(164)

$$\begin{aligned}&=-\frac{1}{\beta }\int _{0}^{t}\left( e^{\beta (t^{\prime }-t)}-1\right) F(t^{\prime })dt^{\prime } \end{aligned}$$

(165)

Therefore

$$\begin{aligned} x(t)=x(0)+\frac{v(0)}{\beta }(1-e^{-\beta t})+\frac{1}{\beta }\int _{0} ^{t}\left( 1-e^{\beta (t^{\prime }-t)}\right) F(t^{\prime })dt^{\prime } \end{aligned}$$

(166)

Also, rewrite Eq. (159) as

$$\begin{aligned} \int _{0}^{t}e^{\beta (t^{\prime }-t)}F(t^{\prime })dt^{\prime }=v(t)-e^{-\beta t}v(0) \end{aligned}$$

(167)

and substitute it into Eq. (166) to obtain

$$\begin{aligned} x(t)=x(0)+\frac{v(0)-v(t)}{\beta }+\frac{1}{\beta }\int _{0}^{t}F(t^{\prime })dt^{\prime } \end{aligned}$$

(168)

This gives a direct relation between x and v.

In addition

$$\begin{aligned} (x(t)-x(0))\beta +(v(t)-v(0))=\int _{0}^{t}F(t^{\prime })dt^{\prime } \end{aligned}$$

(169)

and squaring gives

$$\begin{aligned} (x(t)-x(0))^{2}\beta ^{2}+(v(t)-v(0))^{2}+(x(t)-x(0))(v(t)-v(0))2\beta =\left( \int _{0}^{t}F(t^{\prime })dt^{\prime }\right) ^{2} \end{aligned}$$

(170)

and using

$$\begin{aligned} \beta (x(t)-x(0))=-(v(t)-v(0)+\int _{0}^{t}F(t^{\prime })dt^{\prime } \end{aligned}$$

(171)

gives

$$\begin{aligned} (x(t)-x(0))^{2}\beta ^{2}=(v(t)-v(0))^{2}-2(v(t)-v(0))\int _{0}^{t}F(t^{\prime })dt^{\prime }+\left( \int _{0}^{t}F(t^{\prime })dt^{\prime }\right) ^{2} \end{aligned}$$

(172)

This gives a direct relation between $x^{2}(t)$ and $v^{2}(t)$.

Appendix 2: Evaluation of $\int _{0}^{t}\int _{0}^{t} e^{\beta (t^{\prime }+t^{\prime \prime })}\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime }dt^{\prime \prime }$

Consider the evaluation of

$$\begin{aligned} \int _{0}^{t}\int _{0}^{t}e^{\beta (t^{\prime }+t^{\prime \prime })}\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime }dt^{\prime \prime } \end{aligned}$$

(173)

Using

$$\begin{aligned} \left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle =2D\delta (t^{\prime }-t^{\prime \prime }) \end{aligned}$$

(174)

we have

$$\begin{aligned} \int _{0}^{t}\int _{0}^{t}e^{\beta (t^{\prime }+t^{\prime \prime })}\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime \prime }dt^{\prime }=2D\int _{0}^{t}\int _{0}^{t}e^{\beta (t^{\prime }+t^{\prime \prime })} \delta (t^{\prime }-t^{\prime \prime })dt^{\prime \prime }dt^{\prime } \end{aligned}$$

(175)

The inner integral gives

$$\begin{aligned} \int _{0}^{t}e^{\beta (t^{\prime }+t^{\prime \prime })}\delta (t^{\prime } -t^{\prime \prime })dt^{\prime \prime }=\left\{ \begin{array} [c]{cc} 2De^{2\beta t^{\prime }} &{} 0<t^{\prime }<t\\ 0 &{} \text {otherwise} \end{array} \right. \end{aligned}$$

(176)

Hence

$$\begin{aligned} \int _{0}^{t}\int _{0}^{t}e^{\beta (t^{\prime }+t^{\prime \prime })}\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime }dt^{\prime \prime }=2D\int _{0}^{t}e^{2\beta t^{\prime }}dt^{\prime }=\frac{D}{\beta }(e^{-2\beta t}-1) \end{aligned}$$

(177)

Multiplying both sides by $e^{-2\beta t}$ we have

$$\begin{aligned} e^{-2\beta t}\int _{0}^{t}\int _{0}^{t}e^{\beta (t^{\prime }+t^{\prime \prime } )}\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime }dt^{\prime \prime }=\frac{D}{\beta }(1-e^{-2\beta t}) \end{aligned}$$

(178)

which is Eq. (31) of the main text.

Appendix 3: Evaluation of $\int _{0}^{t}\int _{0}^{s}e^{\beta t^{\prime }}e^{\beta t^{\prime \prime }}\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime }dt^{\prime \prime }$

We evaluate

$$\begin{aligned} \int _{0}^{t}\int _{0}^{s}e^{\beta t^{\prime }}e^{\beta t^{\prime \prime } }\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime }dt^{\prime \prime } \end{aligned}$$

(179)

which enters in the valuation of $\left\langle v(t)v(s)\right\rangle$ as per Eq. (51). We have

$$\begin{aligned} \int _{0}^{t}\int _{0}^{s}e^{\beta t^{\prime }}e^{\beta t^{\prime \prime } }\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime }dt^{\prime \prime }&=\int _{0}^{t}\int _{0}^{s}e^{\beta t^{\prime }}e^{\beta t^{\prime \prime }}\delta (t^{\prime }-t^{\prime \prime })dt^{\prime }dt^{\prime \prime }\end{aligned}$$

(180)

$$\begin{aligned}&=2D\int _{0}^{t}e^{2\beta t^{\prime }}dt^{\prime }\qquad 0<t^{\prime }<s \end{aligned}$$

(181)

Imposing the constraint $0<t^{\prime }<s$ on the remaining integral we have that

$$\begin{aligned} \int _{0}^{t}e^{2\beta t^{\prime }}dt^{\prime }=\left\{ \begin{array} [c]{cc} \frac{1}{2\beta }\left( e^{2\beta s}-1\right) &{} t>s\\ \frac{1}{2\beta }\left( e^{2\beta t}-1\right) &{} t<s \end{array} \right. \end{aligned}$$

(182)

Therefore

$$\begin{aligned} \int _{0}^{t}\int _{0}^{s}e^{\beta t^{\prime }}e^{\beta t^{\prime \prime } }\left\langle F(t^{\prime })F(t^{\prime \prime })\right\rangle dt^{\prime }dt^{\prime \prime }=\frac{D}{\beta }\left\{ \begin{array} [c]{cc} \left( e^{2\beta s}-1\right) &{} t>s\\ \left( e^{2\beta t}-1\right) &{} t<s \end{array} \right. \end{aligned}$$

(183)

Appendix 4: Standard Deviation of x(t)

We obtain the standard deviation of position,

$$\begin{aligned} \sigma _{x}^{2}(t)=\left\langle \left( x(t)-\left\langle x\right\rangle _{t}\right) ^{2}\right\rangle =\sigma _{x}^{2}(0)+\frac{2D}{\beta ^{2}} t+\frac{D}{\beta ^{3}}\left( 4e^{-\beta t}-3-e^{-2\beta t}\right) \end{aligned}$$

(184)

Staring with

$$\begin{aligned} x(t)=x(0)+\int _{0}^{t}v(t^{\prime })\,dt^{\prime } \end{aligned}$$

(185)

and

$$\begin{aligned} \left\langle x\right\rangle _{t}=\left\langle x\right\rangle _{0}+\frac{1}{\beta }\left\langle v\right\rangle _{0}(1-e^{-\beta t}) \end{aligned}$$

(186)

and subtracting Eq. (186) from Eq. (185) we have

$$\begin{aligned} x(t)-\left\langle x\right\rangle _{t}&=x(0)-\left\langle x\right\rangle _{0}-\frac{1}{\beta }\left\langle v\right\rangle _{0}(1-e^{-\beta t} )+x(0)+\int _{0}^{t}v(t^{\prime })\,dt^{\prime }\end{aligned}$$

(187)

$$\begin{aligned}&=\int _{0}^{t}v(t^{\prime })\,dt^{\prime }-\frac{1}{\beta }\left\langle v\right\rangle _{0}(1-e^{-\beta t}) \end{aligned}$$

(188)

Squaring and taking expectation values yields

$$\begin{aligned} \sigma _{x}^{2}(t)=\sigma _{x}^{2}(0)+\int _{0}^{t}\int _{0}^{t}\left\langle v(t^{\prime })v(t^{\prime \prime })\right\rangle \,dt^{\prime }dt^{\prime \prime }+\frac{1}{\beta ^{2}}\left\langle v\right\rangle _{0}^{2}(1-e^{-\beta t} )^{2}-\frac{2}{\beta }\left\langle v\right\rangle _{0}(1-e^{-\beta t})\int _{0}^{t}\left\langle v(t^{\prime })\right\rangle \,dt^{\prime } \end{aligned}$$

(189)

Using Eq. (24), we have that

$$\begin{aligned} \int _{0}^{t}\left\langle v(t^{\prime })\right\rangle \,dt^{\prime }=\int _{0} ^{t}\left\langle v\right\rangle _{0}e^{-\beta t^{\prime }}\,dt^{\prime } =\frac{1}{\beta }\left\langle v\right\rangle _{0}(1-e^{-\beta t}) \end{aligned}$$

(190)

Therefore

$$\begin{aligned} \sigma _{x}^{2}(t)=\sigma _{x}^{2}(0)+\int _{0}^{t}\int _{0}^{t}\left\langle v(t^{\prime })v(t^{\prime \prime })\right\rangle \,dt^{\prime }dt^{\prime \prime }-\frac{1}{\beta ^{2}}\left\langle v\right\rangle _{0}^{2}(1-e^{-\beta t})^{2} \end{aligned}$$

(191)

But we know from Eq. (55) that

$$\begin{aligned} \left\langle v(t)v(s)\right\rangle =e^{-\beta (t+s)}\left\langle v^{2} \right\rangle _{0}+\frac{D}{\beta }\left( e^{-\beta (\left| t-s\right| }-e^{-\beta (t+s)}\right) \end{aligned}$$

(192)

and therefore

$$\begin{aligned} \sigma _{x}^{2}(t)&=\sigma _{x}^{2}(0)+\left\langle v^{2}\right\rangle _{0}\int _{0}^{t}\int _{0}^{t}e^{-\beta (t^{\prime }+t^{\prime \prime })}dt^{\prime }dt^{\prime \prime }+\frac{D}{\beta }\int _{0}^{t}\int _{0}^{t}\left( e^{-\beta (\left| t^{\prime }-t^{\prime \prime }\right| }-e^{-\beta (t^{\prime }+t^{\prime \prime })}\right) \,dt^{\prime }dt^{\prime \prime }\end{aligned}$$

(193)

$$\begin{aligned}&-\frac{1}{\beta ^{2}}\left\langle v\right\rangle _{0}^{2}(1-e^{-\beta t})^{2} \end{aligned}$$

(194)

But clearly

$$\begin{aligned} \left\langle v^{2}\right\rangle _{0}\int _{0}^{t}\int _{0}^{t}e^{-\beta (t^{\prime }+t^{\prime \prime })}dt^{\prime }dt^{\prime \prime }=\frac{1}{\beta ^{2} }\left\langle v\right\rangle _{0}^{2}(1-e^{-\beta t})^{2} \end{aligned}$$

(195)

and therefore

$$\begin{aligned} \sigma _{x}^{2}(t)=\sigma _{x}^{2}(0)+\frac{D}{\beta }\int _{0}^{t}\int _{0} ^{t}\left( e^{-\beta (\left| t^{\prime }-t^{\prime \prime }\right| }-e^{-\beta (t^{\prime }+t^{\prime \prime })}\right) \,dt^{\prime }dt^{\prime \prime } \end{aligned}$$

(196)

We now evaluate the integral in Eq. (196). First consider

$$\begin{aligned} \int _{0}^{t}\left( e^{-\beta (\left| t^{\prime }-t^{\prime \prime }\right| }\right) \,dt^{\prime }&=\int _{0}^{t^{\prime \prime }}\left( e^{-\beta (t^{\prime \prime }-t^{\prime })}\right) \,dt^{\prime }+\int _{t^{\prime \prime }}^{t}\left( e^{-\beta (t^{\prime }-t^{\prime \prime })}\right) \,dt^{\prime }\end{aligned}$$

(197)

$$\begin{aligned}&=\frac{1}{\beta }e^{-\beta t^{\prime \prime }}(e^{\beta t^{\prime \prime } }-1)-e^{\beta t^{\prime \prime }}\frac{1}{\beta }(e^{-\beta t}-e^{-\beta t^{\prime \prime }})\end{aligned}$$

(198)

$$\begin{aligned}&=\frac{1}{\beta }\left( 1-e^{-\beta t^{\prime \prime }}-e^{-\beta t}e^{\beta t^{\prime \prime }}+1\right) \end{aligned}$$

(199)

or

$$\begin{aligned} \int _{0}^{t}\left( e^{-\beta (\left| t^{\prime }-t^{\prime \prime }\right| }\right) \,dt^{\prime }=\frac{1}{\beta }\left( 2-e^{-\beta t^{\prime \prime }}-e^{-\beta t}e^{\beta t^{\prime \prime }}\right) \end{aligned}$$

(200)

Integrating over $t^{\prime \prime }$ we obtain

$$\begin{aligned} \int _{0}^{t}\frac{1}{\beta }\left( 2-e^{-\beta t^{\prime \prime }}-e^{-\beta t}e^{\beta t^{\prime \prime }}\right) dt^{\prime \prime }&=\frac{1}{\beta }\frac{1}{\beta }\left( 2t+\frac{1}{\beta }(e^{-\beta t}-1)-\frac{1}{\beta }e^{-\beta t}(e^{\beta t}-1)\right) \end{aligned}$$

(201)

$$\begin{aligned}&=\frac{1}{\beta ^{2}}\left( 2t+\frac{2}{\beta }(e^{-\beta t}-1)\right) \end{aligned}$$

(202)

Giving

$$\begin{aligned} \int _{0}^{t}\int _{0}^{t}\left( e^{-\beta (\left| t^{\prime }-t^{\prime \prime }\right| }\right) dt^{\prime }dt^{\prime \prime }=\frac{1}{\beta ^{2} }\left( 2t+\frac{2}{\beta }(e^{-\beta t}-1)\right) \end{aligned}$$

(203)

The second integral is

$$\begin{aligned} \int _{0}^{t}\int _{0}^{t}e^{-\beta (t^{\prime }+t^{\prime \prime })}\,dt^{\prime }dt^{\prime \prime }=\frac{1}{\beta ^{2}}(e^{-\beta t}-1)^{2} \end{aligned}$$

(204)

Subtracting Eq. (203) from Eq. (204) and multiplying by $\frac{D}{\beta }$ we have

$$\begin{aligned} \frac{D}{\beta }\int _{0}^{t}\int _{0}^{t}\left( e^{-\beta (\left| t^{\prime }-t^{\prime \prime }\right| }-e^{-\beta (t^{\prime }+t^{\prime \prime } )}\right) \,dt^{\prime }dt^{\prime \prime }=\frac{D}{\beta }\left[ \frac{1}{\beta ^{2}}\left( 2t+\frac{2}{\beta }(e^{-\beta t}-1)\right) -\frac{1}{\beta ^{2}}(e^{-\beta t}-1)^{2}\right] \end{aligned}$$

(205)

which simplifies to

$$\begin{aligned} \frac{D}{\beta }\int _{0}^{t}\int _{0}^{t}\left( e^{-\beta (\left| t^{\prime }-t^{\prime \prime }\right| }-e^{-\beta (t^{\prime }+t^{\prime \prime } )}\right) \,dt^{\prime }dt^{\prime \prime }=\frac{2D}{\beta ^{2}}t+\frac{D}{\beta ^{3}}\left( 4e^{-\beta t}-3-e^{-2\beta t}\right) \end{aligned}$$

(206)

Appendix 5: Evaluation of Eq. (115)

We show Eq. (115) of the text. We have that

$$\begin{aligned}&\frac{1}{\beta ^{2}}\int _{0}^{t}\int _{0}^{t}\left( 1-e^{\beta (t^{\prime }-t)}\right) \left( 1-e^{\beta (t^{\prime \prime }-t)}\right) F(t^{\prime })F(t^{\prime \prime })dt^{\prime }dt^{\prime \prime }\end{aligned}$$

(207)

$$\begin{aligned}&=\frac{1}{\beta ^{2}}\int _{0}^{t}\int _{0}^{t}\left( 1-e^{\beta (t^{\prime }-t)}\right) \left( 1-e^{\beta (t^{\prime \prime }-t)}\right) 2D\delta (t^{\prime }-t^{\prime \prime })dt^{\prime }dt^{\prime \prime }\end{aligned}$$

(208)

$$\begin{aligned}&=\frac{2D}{\beta ^{2}}\int _{0}^{t}\left( 1-e^{\beta (t^{\prime }-t)}\right) ^{2}dt^{\prime }\end{aligned}$$

(209)

$$\begin{aligned}&=\frac{2D}{\beta ^{2}}\int _{0}^{t}\left[ 1-2e^{\beta (t^{\prime } -t)}+e^{2\beta (t^{\prime }-t)}\right] dt^{\prime }\end{aligned}$$

(210)

$$\begin{aligned}&=\frac{2D}{\beta ^{2}}\left[ t-\frac{2}{\beta }e^{\beta (t^{\prime }-t)} +\frac{1}{2\beta }e^{2\beta (t^{\prime }-t)}\right] |_{0}^{t}\end{aligned}$$

(211)

$$\begin{aligned}&=\frac{2D}{\beta ^{2}}\left[ t-\frac{2}{\beta }+\frac{1}{2\beta }+\frac{2}{\beta }e^{-\beta t}-\frac{1}{2\beta }e^{-2\beta t}\right] \end{aligned}$$

(212)

Therefore

$$\begin{aligned} \frac{1}{\beta ^{2}}\int _{0}^{t}\int _{0}^{t}\left( 1-e^{\beta (t^{\prime } -t)}\right) \left( 1-e^{\beta (t^{\prime \prime }-t)}\right) F(t^{\prime })F(t^{\prime \prime })dt^{\prime }dt^{\prime \prime }=\frac{D}{\beta ^{3}}\left[ 2\beta t-3+4e^{-\beta t}-e^{-2\beta t}\right] \end{aligned}$$

(213)