Gamma function application

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

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

Substitute L in Lagrange Equation:
[tex]\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[/tex]
[tex]\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)[/tex]
[tex]\frac{d}{dt}m\dot{x}=-\frac{m}{2x}[/tex]

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

Answers and Replies

  • #2
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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
 
  • #3
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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:

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

or

[tex] \Gamma\left(p+1\right) = p\Gamma\left(p\right) [/tex]
 
  • #4
OlderDan
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kcirick said:
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:

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

or

[tex] \Gamma\left(p+1\right) = p\Gamma\left(p\right) [/tex]
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

[tex]\frac{d}{dt}m\dot{x}=-\frac{m}{2x}[/tex]

can be integrated by separating variables

[tex]\frac{dv}{dt}=-\frac{1}{2x}[/tex]

[tex]\frac{dv}{dx}\frac{dx}{dt}=v\frac{dv}{dx}=-\frac{1}{2x}[/tex]

[tex]2vdv =-\frac{dx}{x}[/tex]

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