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A particle moving towards the origin has initial conditions $x(t=0) = 1$ and $\dot{x}(t=0)=0$.

If the Lagrangian is $$L:=\frac{m}{2}\dot{x}^2 -\frac{m}{2}\ln|x|$$

This should satisfy Euler Lagrange Equation $$\frac{d}{dt} (\frac{\partial L}{\partial \dot{x}}) = \frac{\partial L}{\partial x}$$

Prove the particle reaches the origin at $\Gamma(1/2) = \sqrt{\pi}$.

1) okay to begin I simply plug in and expand the D.E.

$$\frac{d}{dt}[(\frac{\partial \frac{m}{2} \dot{x}^2}{\partial \dot{x}}) - \frac{\partial\frac{m}{2}ln|x|}{\partial \dot{x}}] = \frac{\partial \frac{m}{2}\dot{x}^2}{\partial x} - \frac{\partial \frac{m}{2}ln|x|}{\partial x}$$

2) since $x(t)$ and $\dot{x}(t)$ are functions of time, the cross partials dissapear and I am left with:

$$\frac{d}{dt}(\frac{\partial \frac{m}{2} \dot{x}^2}{\partial \dot{x}}) = - \frac{\partial \frac{m}{2}ln|x|}{\partial x}$$

Which reduces to:

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

This is equivalent to: $$\ddot{x} = - \frac{1}{2x}$$

3) Now I will separate and Integrate (Keeping in mind the particle starts from Rest):

$$\dot{x} = \frac{dx}{dt} = -\frac{1}{2}ln|x|$$ $$t = -2 \int^{x}_{x_0} \frac{dx}{ln|x|}$$

All I really want to know is that Up until this point, Have I done everything correct? Because I feel like I haven't. I don't think I can even Integrate this because I put it into wolfram and I got a mess.

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  • $\begingroup$ Try writing out your "separate and integrate step" in more detail: actually write out what variable you're integrating with respect to on each side. What I would recommend is multiplying each side by $2 $\endgroup$ Dec 2, 2012 at 17:27
  • $\begingroup$ Actually what you have to solve is this nonlinear differential equation $x''(t)+1/[2 x(t)]=0$ $\endgroup$
    – Ana S. H.
    Dec 2, 2012 at 17:59

1 Answer 1

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Something not right with your third step, you have:

$\ddot{x}=-\frac{1}{2x}\Rightarrow\dot{x}\ddot{x}=-\frac{\dot{x}}{2x}\Rightarrow\frac{1}{2}\left(\dot{x}^{2}\right)^{.}=-\frac{1}{2}\left(\ln\left(x\right)\right)^{.}\Rightarrow\dot{x}^{2}=-\ln\left(x\right)+c\Rightarrow t=\int\frac{dx}{\sqrt{c-\ln\left(x\right)}}=\left|\begin{array}{c} c-\ln\left(x\right)=z\\ dx=-e^{c-z}dz \end{array}\right|=-e^{c}\int e^{-z}z^{\frac{1}{2}-1}dz=-e^{c}\Gamma(\frac{1}{2})$

now just apply your boundary conditions.

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  • $\begingroup$ Actually you already applied the boundary conditions to get the integration limits and conclude that the last integral is in fact the Gamma function. $\endgroup$
    – Ana S. H.
    Dec 2, 2012 at 18:27
  • $\begingroup$ I didn't applied them in full, one still should find $c$. $\endgroup$
    – TMS
    Dec 2, 2012 at 18:30
  • $\begingroup$ Yes, you have used that $c=0$, but, as you said, incompletely. $\endgroup$
    – Ana S. H.
    Dec 2, 2012 at 18:41

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