Thomas–Fermi equation

Numerical solutions of the Thomas–Fermi equation

In mathematics, the Thomas–Fermi equation for the neutral atom is a second order non-linear ordinary differential equation, named after Llewellyn Thomas and Enrico Fermi, which can be derived by applying the Thomas–Fermi model to atoms. The equation reads

[math]\displaystyle{ \frac{d^2y}{dx^2} = \frac{1}{\sqrt x} y^{3/2} }[/math]

subject to the boundary conditions

[math]\displaystyle{ y(0)=1, \quad \quad \begin{cases}y(\infty)=0 \quad \text{for neutral atoms}\\ y(x_0)=0 \quad \text{for positive ions}\\ y(x_1)-x_1y'(x_1)=0\quad \text{for compressed neutral atoms}\end{cases} }[/math]

If [math]\displaystyle{ y }[/math] approaches zero as [math]\displaystyle{ x }[/math] becomes large, this equation models the charge distribution of a neutral atom as a function of radius [math]\displaystyle{ x }[/math]. Solutions where [math]\displaystyle{ y }[/math] becomes zero at finite [math]\displaystyle{ x }[/math] model positive ions. For solutions where [math]\displaystyle{ y }[/math] becomes large and positive as [math]\displaystyle{ x }[/math] becomes large, it can be interpreted as a model of a compressed atom, where the charge is squeezed into a smaller space. In this case the atom ends at the value of [math]\displaystyle{ x }[/math] for which [math]\displaystyle{ dy/dx=y/x }[/math].

Transformations

Introducing the transformation [math]\displaystyle{ z=y/x }[/math] converts the equation to

[math]\displaystyle{ \frac{1}{x^2}\frac{d}{dx}\left(x^2\frac{dz}{dx}\right) - z^{3/2}=0 }[/math]

This equation is similar to Lane–Emden equation with polytropic index [math]\displaystyle{ 3/2 }[/math] except the sign difference. The original equation is invariant under the transformation [math]\displaystyle{ x\rightarrow c x, \ y\rightarrow c^{-3} y }[/math]. Hence, the equation can be made equidimensional by introducing [math]\displaystyle{ y=x^{-3} u }[/math] into the equation, leading to

[math]\displaystyle{ x^2 \frac{d^2u}{dx^2} - 6x\frac{du}{dx} + 12 u = u^{3/2} }[/math]

so that the substitution [math]\displaystyle{ x=e^t }[/math] reduces the equation to

[math]\displaystyle{ \frac{d^2u}{dt^2} - 7\frac{du}{dt} +12 u = u^{3/2}. }[/math]

Treating [math]\displaystyle{ w =du/dt }[/math] as the dependent variable and [math]\displaystyle{ u }[/math] as the independent variable, we can reduce the above equation to

[math]\displaystyle{ w \frac{dw}{du} - 7w = u^{3/2}-12 u. }[/math]

But this first order equation has no known explicit solution, hence, the approach turns to either numerical or approximate methods.

Sommerfeld's approximation

The equation has a particular solution [math]\displaystyle{ y_p(x) }[/math], which satisfies the boundary condition that [math]\displaystyle{ y\rightarrow 0 }[/math] as [math]\displaystyle{ x\rightarrow\infty }[/math], but not the boundary condition y(0)=1. This particular solution is

[math]\displaystyle{ y_p(x) = \frac{144}{x^3}. }[/math]

Arnold Sommerfeld used this particular solution and provided an approximate solution which can satisfy the other boundary condition in 1932. If the transformation [math]\displaystyle{ x=1/t, \ w = yt }[/math] is introduced, the equation becomes

[math]\displaystyle{ t^4\frac{d^2w}{dt^2} = w^{3/2}, \quad w(0)=0, \ w(\infty)\sim t. }[/math]

The particular solution in the transformed variable is then [math]\displaystyle{ w_p(t)= 144 t^4 }[/math]. So one assumes a solution of the form [math]\displaystyle{ w=w_p(1+\alpha t^\lambda) }[/math] and if this is substituted in the above equation and the coefficients of [math]\displaystyle{ \alpha }[/math] are equated, one obtains the value for [math]\displaystyle{ \lambda }[/math], which is given by the roots of the equation [math]\displaystyle{ \lambda^2 + 7\lambda -6=0 }[/math]. The two roots are [math]\displaystyle{ \lambda_1 = 0.772, \ \lambda_2 = -7.772 }[/math], where we need to take the positive root to avoid the singularity at the origin. This solution already satisfies the first boundary condition ([math]\displaystyle{ w(0)=0 }[/math]), so, to satisfy the second boundary condition, one writes to the same level of accuracy for an arbitrary [math]\displaystyle{ n }[/math]

[math]\displaystyle{ W=w_p(1+\beta t^\lambda)^n = [144 t^3(1+\beta t^\lambda)^n]t. }[/math]

The second boundary condition will be satisfied if [math]\displaystyle{ 144t^3(1+\beta t^\lambda)^n = 144 t^3 \beta^n t^{\lambda n}(1+\beta^{-1}t^{-\lambda})^n\sim 1 }[/math] as [math]\displaystyle{ t\rightarrow\infty }[/math]. This condition is satisfied if [math]\displaystyle{ \lambda n + 3 =0, \ 144 \beta^n =1 }[/math] and since [math]\displaystyle{ \lambda_1\lambda_2=-6 }[/math], Sommerfeld found the approximation as [math]\displaystyle{ \lambda = \lambda_1, \ n =-3/\lambda_1 = \lambda_2/2 }[/math]. Therefore, the approximate solution is

[math]\displaystyle{ y(x) = y_p(x) \{1+[y_p(x)]^{\lambda_1/3}\}^{\lambda_2/2}. }[/math]

This solution predicts the correct solution accurately for large [math]\displaystyle{ x }[/math], but still fails near the origin.

Solution near origin

Enrico Fermi provided the solution for [math]\displaystyle{ x\ll 1 }[/math] and later extended by Edward B. Baker. Hence for [math]\displaystyle{ x\ll 1 }[/math],

[math]\displaystyle{ \begin{align} y(x) = {} & 1 - Bx + \frac{1}{3} x^3 - \frac {2B} {15} x^4 + \cdots {} \\[6pt] & \cdots + x^{3/2} \left[\frac 4 3 - \frac{2B} 5 x + \frac{3B^2}{70} x^2 + \left(\frac{2}{27} + \frac{B^3}{252}\right) x^3 + \cdots\right] \end{align} }[/math]

where [math]\displaystyle{ B\approx 1.588071 }[/math].

It has been reported by Salvatore Esposito that the Italian physicist Ettore Majorana found in 1928 a semi-analytical series solution to the Thomas–Fermi equation for the neutral atom, which however remained unpublished until 2001. Using this approach it is possible to compute the constant B mentioned above to practically arbitrarily high accuracy; for example, its value to 100 digits is [math]\displaystyle{ B = 1.588 071 022 611 375 312 718 684 509 423 950 109 452 746 621 674825616765677418166551961154309262332033970138428665 }[/math].