Question

A particle starting from rest at $\mathit{x}\mathbf{=}\mathbf{1}$ moves along the xaxis toward the origin.

Its potential energy is $\mathit{V}\mathbf{=}\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}\mathit{l}\mathit{n}\mathit{x}$. Write the Lagrange equation and integrate it

to find the time required for the particle to reach the origin.

Short answer

The solution to this problem is $t_{\mathrm{x}=0}=\Gamma \left(\frac{1}{2}\right).$

Step-by-step solution

Given Information

The given information is the gamma function to be used in the solution.

The position of the particle is and the potential energy is .

.

Definition of Gamma function

To extend the factorial to any real number (whether or not the real number is a whole number), the gamma function is defined as for .

Find the solution

Write initial position and velocity.

Write Lagrangian of the system.

, where T is kinetic, and V is potential energy.

Write Euler-Lagrange equations.

The acceleration of this system is equal to .

Reduce the differential equation to lower degree.

Integrate both sides.

Since x is decreasing in time (particle going towards origin), pick minus sign, because it is the physical one even though mathematically, both are correct.

Use supposition method to solve the integral.

The integral is now in the form of .

Express the integral in gamma function, and then evaluate the gamma function and integral numerically.

The Gamma function is for .

Hence, the solution to this problem is

Given Information

The given information is the gamma function to be used in the solution.

The position of the particle is and the potential energy is .

Definition of Gamma function

To extend the factorial to any real number $\mathit{x}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$(whether or not the real number is a whole number), the gamma function is defined as$\mathit{\Gamma }\mathbf{\left(}\mathbf{p}\mathbf{\right)}\mathbf{=}{\mathbf{\int }}_{\mathbf{0}}^{\mathbf{\infty }}{\mathit{x}}^{\mathbf{p}\mathbf{-}\mathbf{1}}{\mathit{e}}^{\mathbf{-}\mathbf{x}}\mathit{d}\mathit{x}$ for $\mathit{p}\mathbf{}\mathbf{>}\mathbf{}\mathbf{0}$.

Find the solution

Write initial position and velocity.

$\begin{array}{l}{x}_0=1\\ {\dot{x}}_0=0\end{array}$

Write Lagrangian of the system.

$L=T-V$, where T is kinetic, and V is potential energy.

$\begin{array}{l}T=\frac{1}{2}m{x}^2\\ v=\frac{1}{2}m\ln \left(x\right)\\ L=\frac{1}{2}m{x}^2-\frac{1}{2}m\ln \left(x\right)\end{array}$

Write Euler-Lagrange equations.

$\begin{array}{c}\frac{d}{dt}\left(\frac{\partial L}{\partial q}\right)=\frac{\partial L}{\partial q}\\ \frac{d}{dt}\left(m\dot{x}\right)=-\frac{m}{2x}\\ \frac{d}{dt}\dot{x}=-\frac{1}{2x}\end{array}$

The acceleration of this system is equal to $-\frac{1}{2x}$.

Reduce the differential equation to lower degree.

$\begin{array}{c}\dot{v}=-\frac{1}{2x}\\ \dot{v}=\frac{dv}{dx}·\frac{dx}{dt}\\ \frac{dv}{dx}v=-\frac{1}{2x}\end{array}$

integrate both sides.

$\begin{array}{c}{\int }_0^{v}v\text{'}dv\text{'}=-\frac{1}{2}{\int }_1^x\frac{dx\text{'}}{x\text{'}}\\ \frac{v^2}{2}=-\frac{1}{2}\ln \left(x\right)\\ v^2=\ln \left(\frac{1}{x}\right)\\ v=\pm {\left(\ln \left(\frac{1}{x}\right)\right)}^{\frac{1}{2}}\end{array}$

Since x is decreasing in time (particle going towards origin), pick minus sign, because it is the physical one even though mathematically, both are correct.

$\begin{array}{l} \frac{dx}{dt}=-{\left(\ln \left(\frac{1}{x}\right)\right)}^{\frac{1}{2}}\\ {\int }_0^{t_{x=0}}dt=-{\int }_1^0{\left(\ln \left(\frac{1}{x}\right)\right)}^{\frac{-1}{2}}dx\\ t_{x=0}={\int }_0^1{\left(\ln \left(\frac{1}{x}\right)\right)}^{\frac{-1}{2}}dx\end{array}$

Use supposition method to solve the integral.

$\begin{array}{c}x=e^{-u}\\ u=\ln \left(\frac{1}{x}\right)\\ dx=-e^{-u}du\end{array}$

The integral is now in the form of $t_{x=0}={\int }_0^{\infty }{u}^{\frac{-1}{2}}{e}^{-u}du$.

Express the integral in gamma function, and then evaluate the gamma function and integral numerically.

The Gamma function is$\Gamma \left(p\right)={\int }_0^{\infty }{x}^{p-1}{e}^{-x}dx$ for $p>0$.

$\begin{array}{c}{t}_{x=0}={\int }_0^{\infty }{u}^{\frac{-1}{2}}{e}^{-u}du\\ =\Gamma \left(\frac{1}{2}\right)\end{array}$

Hence, the solution to this problem is$t_{x=0}=\Gamma \left(\frac{1}{2}\right).$