Gamma Function Integral in Lagrange Mechanics?

[kcirick]

Question:
A particle of mass m starting from rest at x=1 moves along the x-axis toward the origin. Its potential energy is V=\frac{1}{2}mlnx. Write the Lagrange equation and integrate it to find the time required for the particle to reach the origin.

Lagrange Equation in 1-D:
\frac{d}{dt}\frac{\partial L}{\partial\dot{x}}-\frac{\partial L}{\partial x}=0
L = T - V = \frac{1}{2}mv^{2}-\frac{1}{2}mlnx =\frac{1}{2}m\dot{x}^{2}-\frac{1}{2}mlnx

Substitute L in Lagrange Equation:
\frac{d}{dt}\frac{\partial}{\partial\dot{x}}\left(\frac{1}{2}m\dot{x}^{2}\right)-\frac{\partial}{\partial x}\left(-\frac{1}{2}mlnx\right)=0
\frac{d}{dt}\frac{\partial}{\partial\dot{x}}\left(\frac{1}{2}m\dot{x}^{2}\right)=\frac{\partial}{\partial x}\left(-\frac{1}{2}mlnx\right)
\frac{d}{dt}m\dot{x}=-\frac{m}{2x}

... And I don't really know what to do from here. The answer is given and it is supposed to be \Gamma\left(\frac{1}{2}\right). Can someone tell me where to go from where I left off? Thank you!
-Rick

 

[tim_lou]

i don't know if this will help... but you can do this problem without using the Lagrangian. since the motion is in one dimension, energy is always conserved.

then integrate through dt=dx/v, express v as a function of x using energy. then integrate both side from x=1 to x=0

 

[kcirick]

Well, the question says specifically to use the Lagrange Equation. My problem is I don't know what to integrate and what I'm integrating over. Also, I'm assuming I have to use either one of these relations:

\Gamma\left(p\right)=\int^{\infty}_{0}x^{p-1}e^{-x}dx

or

\Gamma\left(p+1\right) = p\Gamma\left(p\right)

 

[OlderDan]

Well, the question says specifically to use the Lagrange Equation. My problem is I don't know what to integrate and what I'm integrating over. Also, I'm assuming I have to use either one of these relations:

\Gamma\left(p\right)=\int^{\infty}_{0}x^{p-1}e^{-x}dx

or

\Gamma\left(p+1\right) = p\Gamma\left(p\right)

I think there is another form that will help you. See

http://mathworld.wolfram.com/GammaFunction.html

equation (5)

or

http://numbers.computation.free.fr/Constants/Miscellaneous/gammaFunction.html

equation (1)

Your equation

\frac{d}{dt}m\dot{x}=-\frac{m}{2x}

can be integrated by separating variables

\frac{dv}{dt}=-\frac{1}{2x}

\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=-\frac{1}{2x}

2vdv =-\frac{dx}{x}

Integrate this and set v = 0 at x = 1 and then separate variables and integrate again. That will give you that Euler form for \Gamma\left(\frac{1}{2}\right)