A variant of Einstein’s partial differential equation for Brownian motion, applied to quantum mechanics

In one of his famous papers from 1905 (his “annus mirabilis”) Albert Einstein analysed Brownian motion and derived the following partial differential equation from physical principles to describe the process:

$\displaystyle D \frac{\partial^2 f}{\partial x^2} = \frac{\partial f}{\partial t}$

He showed that this equation has as a solution the probability density function

$\displaystyle f(x, t) = \frac{1}{\sqrt{4 \pi Dt}} \exp\bigg(-\frac{x^2}{4Dt}\bigg)$

(see Einstein, A, 1905, Investigations on the theory of the Brownian movement, pp. 15-16). Writing $\displaystyle D = \frac{\sigma^2}{2}$ this formula can be rewritten as

$\displaystyle f(x, t) = \frac{1}{\sqrt{2 \pi \sigma^2 t}} \exp\bigg(-\frac{x^2}{2 \sigma^2 t}\bigg)$

This is the density of the $\displaystyle N(0, \sigma^2 t)$ distribution of the increments of a Wiener process $\displaystyle \{X(t): t \geq 0, X(0) = 0\}$, which is more commonly used today to refer to a standard Brownian motion. This continuous-time stochastic process is symmetric about zero, continuous, and has stationary independent increments, i.e., the change from time $\displaystyle u$ to time $\displaystyle u + t$, given by the random variable $\displaystyle X(u, u + t) = X(u + t) - X(u)$, has the same $\displaystyle N(0, \sigma^2 t)$ probability distribution as the change from time $\displaystyle 0$ to time $\displaystyle t$, given by the random variable $\displaystyle X(0, t) = X(t)$, and the change is also independent of the history of the process before time $\displaystyle u$.
For certain reasons, I was exploring a continuous-time stochastic process which is also symmetric about zero and continuous like the Wiener process, but which has non-stationary increments. The probability density function of this alternative process I was investigating has the form

$\displaystyle f(x, t) = \frac{1}{\sqrt{2 \pi \sigma^2 t^2}} \exp\bigg(-\frac{x^2}{2 \sigma^2 t^2}\bigg)$

The fact that time appears as a squared term in this formula rather than linearly is enough to destroy the increment-stationarity property of the Wiener process. This can be demonstrated by observing that increment-stationarity requires

$\displaystyle X(u, u + t) \,{\buildrel d \over =}\, X(t)$

and therefore

$\displaystyle X(u, u + t) \sim N(0, \sigma^2 t^2)$

but by definition of $\displaystyle X(u, u + t)$ we must also have

$\displaystyle X(u, u + t) + X(u) \,{\buildrel d \over =}\, X(u + t)$

which is not true here since

$\displaystyle X(u, u + t) + X(u) \sim N(0, \sigma^2(u^2 + t^2))$

whereas

$\displaystyle X(u + t) \sim N(0, \sigma^2(u + t)^2)$

We have a contradiction and must therefore conclude that $\displaystyle X(u, u + t) \nsim N(0, \sigma^2 t^2)$, which means that this alternative continuous-time process does not have the increment-stationarity property of the Wiener process. (Note that this kind of contradiction does not arise with the Wiener process: we have $\displaystyle X(u, u + t) + X(u) \sim N(0, \sigma^2(u + t))$ which is the same as the distribution of $\displaystyle X(u + t)$).
As an aside, it occurred to me to wonder if there might be a partial differential equation describing this alternative continuous-time process analogous to the partial differential equation derived by Einstein for the standard Brownian motion process. I did indeed find such a partial differential equation for the alternative process, as follows. Taking partial derivatives of the probability density function

$\displaystyle f(x, t) = \frac{1}{\sqrt{2 \pi \sigma^2 t^2}} \exp\bigg(-\frac{x^2}{2 \sigma^2 t^2}\bigg)$

we obtain

$\displaystyle \frac{\partial f}{\partial t} = \frac{1}{\sqrt{2 \pi \sigma^2 t^2}} \exp\bigg(-\frac{x^2}{2 \sigma^2 t^2}\bigg)\bigg\{\frac{x^2}{\sigma^2 t^3} - \frac{1}{t}\bigg\}$

$\displaystyle \frac{\partial f}{\partial x} = \frac{1}{\sqrt{2 \pi \sigma^2 t^2}} \exp\bigg(-\frac{x^2}{2 \sigma^2 t^2}\bigg)\bigg\{-\frac{x}{\sigma^2 t^2}\bigg\}$

$\displaystyle \frac{\partial^2 f}{\partial x^2} = \frac{1}{\sqrt{2 \pi \sigma^2 t^2}} \exp\bigg(-\frac{x^2}{2 \sigma^2 t^2}\bigg)\bigg\{\frac{x^2}{(\sigma^2 t^2)^2} - \frac{1}{\sigma^2 t^2}\bigg\}$

Comparing the expressions for $\displaystyle \frac{\partial f}{\partial t}$ and $\displaystyle \frac{\partial^2 f}{\partial x^2}$ we see that

$\displaystyle \sigma^2 \frac{\partial^2 f}{\partial x^2} = \frac{1}{t}\frac{\partial f}{\partial t}$

and this is the required variant of Einstein’s partial differential equation for the alternative continuous-time process I was investigating.

I was intrigued to find that a slight generalisation of this framework makes it applicable to quantum wave-packet dispersion. To see this, let $\displaystyle \sigma(t^2) = \sqrt{a + bt^2}> 0$ where $a$ and $b$ are some parameters. Then the partial differential equation

$\displaystyle b \frac{\partial^2 f}{\partial x^2} = \frac{1}{t}\frac{\partial f}{\partial t}$

has as a solution the probability density function

$\displaystyle f(x, t) = \frac{1}{\sqrt{2 \pi \sigma^2(t^2)}} \exp\bigg(-\frac{x^2}{2 \sigma^2(t^2)}\bigg)$

as can be verified by comparing the partial derivatives

$\displaystyle \frac{\partial f}{\partial t} = \frac{1}{\sqrt{2 \pi \sigma^2 (t^2)}} \exp\bigg(-\frac{x^2}{2 \sigma^2 (t^2)}\bigg)\bigg\{\frac{bx^2 t}{\sigma^4(t^2)} - \frac{bt}{\sigma^2(t^2)}\bigg\}$

and

$\displaystyle \frac{\partial^2 f}{\partial x^2} = \frac{1}{\sqrt{2 \pi \sigma^2(t^2)}} \exp\bigg(-\frac{x^2}{2 \sigma^2(t^2)}\bigg)\bigg\{\frac{x^2}{\sigma^4(t^2)} - \frac{1}{\sigma^2(t^2)}\bigg\}$

(As a check, note that all of this reduces to the previously obtained differential equation and probability density function when $\displaystyle a = 0$ and $\displaystyle b = \sigma^2$). Now, in quantum mechanics a wave representation of a moving body is obtained as a wave-packet consisting of a superposition of individual plane waves of different wavelengths (or equivalently, different wave numbers $k = \frac{2 \pi}{\lambda}$) in the form

$\displaystyle \psi(x, t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} \Phi(k) e^{i(kx - i\hbar k^2 t/2m)}dk$

where $\displaystyle \Phi(k)$ is the Fourier transform of the $x$-space wavefunction $\displaystyle \psi(x, t)$ at $\displaystyle t = 0$, i.e.,

$\displaystyle \Phi(k) = \int_{-\infty}^{\infty} \psi(x', 0)e^{-ikx'}dx'$

The wave-packet $\displaystyle \psi(x, t)$ disperses over time and it has been shown that the probability density as a function of time of such a moving body as the wave-packet disperses, given by $\displaystyle |\psi(x, t)|^2$, always becomes Gaussian (irrespective of the original shape of the wave-packet) and has the form of the probability density function above, i.e.,

$\displaystyle |\psi(x, t)|^2 = \frac{1}{\sqrt{2 \pi \sigma^2(t^2)}} \exp\bigg(-\frac{x^2}{2 \sigma^2(t^2)}\bigg)$

See, for example, Mita, K, 2007, Dispersion of non-Gaussian free particle wave packets, Am. J. Phys. 75 (10), who derives an expression for $\displaystyle |\psi(x, t)|^2$ like the one above with $\displaystyle x^2$ replaced by $\displaystyle (x - \gamma_1)^2$ and

$\displaystyle \sigma^2(t^2) = \frac{\gamma_2^2 + (\hbar t/m)^2}{2\gamma_2}$

(see equations (15) to (18) on page 952 of the paper). Therefore the partial differential equation

$\displaystyle b \frac{\partial^2 f}{\partial x^2} = \frac{1}{t}\frac{\partial f}{\partial t}$

has as a solution the probability density of a moving body undergoing quantum wave-packet dispersion as time progresses (with $b = \frac{(\hbar/m)^2}{2\gamma_2}$ in the above set-up).

Gauss’s Flux Theorem for Gravity and Newton’s Law

Gauss’s flux theorem for gravity (also known as Gauss’s law for gravity) in differential form says that

$\nabla \cdot \mathbf{g} = -4 \pi G \rho$

In this note I want to show that one can get quite far towards deriving Gauss’s law for gravity without knowing Newton’s law of universal gravitation, but not all the way. To explore this, suppose that all we know is that the gravitational force depends on mass and radial distance:

$\mathbf{g} = k(M, r) \cdot \mathbf{e_r}$

Here, $k$ is an unspecified function, $M$ is a mass which can be taken as being located at the origin, $r$ is the radial distance from the origin, and $\mathbf{e_r}$ is a radial unit vector.

Now we imagine a closed spherical surface $\delta V$ of radius $r$ centered at the origin. The total flux of the gravitational field $\mathbf{g}$ over the closed surface $\delta V$ is

$\oint_{\delta V} \mathbf{g} \cdot d\mathbf{A} = \oint_{\delta V} k(M, r) \cdot \mathbf{e_r} \cdot d\mathbf{A}$

= $k(M, r) \oint_{\delta V} \mathbf{e_r} \cdot \mathbf{e_r} \cdot dA$

= $k(M, r) \oint_{\delta V} dA$

= $k(M, r) \cdot 4 \pi r^2$ (this explains where the $4 \pi$ comes from).

The total flux is independent of $r$ so to eliminate $r^2$ we must have $k(M, r) = \frac{k^*(M)}{r^2}$ where $k^*(M)$ is some unspecified function of $M$, and therefore

$\oint_{\delta V} \mathbf{g} \cdot d\mathbf{A} = k^*(M) \cdot 4 \pi$

By the divergence theorem we can write this as

$\oint_{V} \nabla \cdot \mathbf{g} dV = k^*(M) \cdot 4 \pi$

and therefore differentiating both sides with respect to $V$ we get

$\nabla \cdot \mathbf{g} = k^{* \prime}(M)\frac{dM}{dV} 4 \pi$

If we set $\frac{dM}{dV} = \rho$ we see that this is *nearly* Gauss’s law:

$\nabla \cdot \mathbf{g} = k^{* \prime}(M) 4 \pi \rho$

We only need Newton’s law to tell us that $k^{* \prime}(M) = -G$ at this final stage.

The link with the scalar potential comes through $\mathbf{g} = -\nabla \phi$ which gives us

$\nabla^2 \phi = 4 \pi G \rho$

(a well known type of partial differential equation known as Poisson’s equation).