Sign up ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free.

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.

share|cite|improve this question
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 – Robert Mastragostino Dec 2 '12 at 17:27
Actually what you have to solve is this nonlinear differential equation $x''(t)+1/[2 x(t)]=0$ – Ana Dec 2 '12 at 17:59

1 Answer 1

up vote 3 down vote accepted

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.

share|cite|improve this answer
Actually you already applied the boundary conditions to get the integration limits and conclude that the last integral is in fact the Gamma function. – Ana Dec 2 '12 at 18:27
I didn't applied them in full, one still should find $c$. – TMS Dec 2 '12 at 18:30
Yes, you have used that $c=0$, but, as you said, incompletely. – Ana Dec 2 '12 at 18:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.