This is with regards to problem 3.19 from Goldstein's Classical Mechanics,

A particle moves in a force field described by the Yukowa potential $$ V(r) = -\frac{k}{r} e^{-\frac{r}{a}}, $$ where $k$ and $a$ are positive.

where I bolded the assumptions as this is the only information I can imagine helps me resolve this.

A solution due to Professor Laura Reina at Florida State Uni, as well as a solution due to Slader.com

both use the following expression for the force felt by a particle in the given Yukawa potential:

$$ F(r) = -\frac{k}{r^2} e^{-\frac{r}{a}} $$

I am struggling to wrap my head around this. This is clearly not the result of

$$ -\frac{\partial V(r)}{\partial r} $$

which evaluates to

$$ -\frac{k}{r^2} e^{-\frac{r}{a}} - \frac{k}{ar} e^{-\frac{r}{a}} $$

Can anyone help me understand why the second term $-\frac{k}{ar} e^{-\frac{r}{a}}$ can be excluded here? I tried plotting some various example of this, varying k and a which are allowed to be any positive numbers, but I've no insight.

There was a question regarding this same topic which was not answered Deriving potential from central force

  • 1
    $\begingroup$ I looked at Reina’s solution. Her $F(r)$ is wrong but she isn’t using it. The force in the 3rd equation on page 5 is correct. $\endgroup$ – G. Smith Jul 25 '20 at 22:35
  • 2
    $\begingroup$ Your $V(r)$ is a repulsive potential and doesn’t have orbits. Doesn’t Goldstein have a negative sign? Reina is using $V(r)=-\frac{k}{r}e^{-r/a}$. $\endgroup$ – G. Smith Jul 25 '20 at 22:40
  • 2
    $\begingroup$ And be careful: her $V’(r)$ doesn’t mean $dV/dr$! Very confusing! $\endgroup$ – G. Smith Jul 25 '20 at 22:46
  • 1
    $\begingroup$ Don’t forget that $F_r=-dV/dr$ has a negative sign. $\endgroup$ – G. Smith Jul 25 '20 at 22:50
  • $\begingroup$ Thank you very much for the clarification! The pointer to the correction expression on page 5 helps! :D and indeed I'm missing a minus sign! $\endgroup$ – Lopey Tall Jul 25 '20 at 23:29

I think the solutions you have posted are just incorrect. The exact expression should include the $-\frac{k}{ar}e^{-ar}$ term. You could approximately ignore this term in the limit $r\ll a$, but in this limit one should also Taylor expand the exponential to be consistent.

In fact, looking at Prof. Reina's solutions, the third equation in the solution to 4.a does contain both terms you wrote down.

  • $\begingroup$ AH! You're absolutely right! Thanks for the pointer to the correct expression! $\endgroup$ – Lopey Tall Jul 25 '20 at 23:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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