Impact parameter dependence of classical scattering angle

[otg]

Homework Statement

The problem is to get the classical turning point as a function of the impact parameter b for the Lennard-Jones potential.

Homework Equations

The Lennard-Jones potential is given as V(r)=4\epsilon[(\frac{\sigma}{r})^{12}-(\frac{\sigma}{r})^6][\itex].<br /> The effective potential is V_{eff}(r)=V(r)+E\frac{b^2}{r^2}[\itex].&lt;br /&gt; Also, \frac{m}{2}\dot{r}+V_{eff}(r)=E[\itex].&amp;lt;br /&amp;gt; &amp;lt;br /&amp;gt; &amp;lt;h2&amp;gt;The Attempt at a Solution&amp;lt;/h2&amp;gt;&amp;lt;br /&amp;gt; Since the turning point occurs when \dot{r}=0[\itex], we have that 4\epsilon[(\frac{\sigma}{r})^{12}-(\frac{\sigma}{r})^6]+E\frac{b^2}{r^2}=E[\itex]. Even though \sigma,\,\epsilon\,\text{and }\,E[\itex] are given, I find it quite hard if not impossible to solve the equation to get r(b)[\itex]. There has to be some other way even if this was the indicated strategy.&amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;gt; The exercise says that the equation V_{eff}(r)=E[\itex] can be rewritten as a polynomial and solved with a computer program, but since it becomes (at least that&amp;amp;amp;amp;amp;amp;amp;#039;s what I get) a twelve degree polynomial in r[\itex] it&amp;amp;amp;amp;amp;amp;amp;amp;#039;s not possible to get a decent answer.&amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;gt; &amp;amp;amp;amp;amp;amp;amp;lt;br /&amp;amp;amp;amp;amp;amp;amp;gt; Grateful for any kind of help how to proceed. And also how to get the latex working in the posts...

 

[andrien]

Homework Statement

The problem is to get the classical turning point as a function of the impact parameter b for the Lennard-Jones potential.

Homework Equations

The Lennard-Jones potential is given as V(r)=4\epsilon[(\frac{\sigma}{r})^{12}-(\frac{\sigma}{r})^6].
The effective potential is V_{eff}(r)=V(r)+E\frac{b^2}{r^2}.
Also, \frac{m}{2}\dot{r}+V_{eff}(r)=E.

The Attempt at a Solution

Since the turning point occurs when \dot{r}=0, we have that 4\epsilon[(\frac{\sigma}{r})^{12}-(\frac{\sigma}{r})^6]+E\frac{b^2}{r^2}=E. Even though \sigma,\,\epsilon\,\text{and }\,E are given, I find it quite hard if not impossible to solve the equation to get r(b). There has to be some other way even if this was the indicated strategy.

The exercise says that the equation V_{eff}(r)=E can be rewritten as a polynomial and solved with a computer program, but since it becomes (at least that's what I get) a twelve degree polynomial in r it's not possible to get a decent answer.

Grateful for any kind of help how to proceed. And also how to get the latex working in the posts...

use / sign not \ after ending( i don't think there is a simple way)

 

[otg]

use / sign not \ after ending( i don't think there is a simple way)

ha ha ok my fingers must've gone into LaTeX mode :)

To reply on my own post then, no there is no simple way, I think I interpreted the question "solve for r as function of b" as "find an expression" but the task was as simple as plot the roots to find the functional behavior. Thanx for pointing out the /-thing though (probably would have worked next time without me knowing what went wrong) :)