neutralseer
Jul16-08, 05:21 PM
A year ago we had a HW problem about galactic rotation curves:
If the dark matter density is
\rho= \frac{\rho_0}{1+\left(\frac{r}{r_0}\right)^2},
Then how does velocity depend radius at large r ( r>>r_o)?
You want to use the virial theorem here, so you calculate M(r) and then finally calculate the potential:
\phi(r)=\int_{-\infty}^{r} \frac{k}{s}ds=\infty
with k = constant.
Oops, phi(r) diverges logarithmically. Oh well, you just have to set the zero point somewhere else besides infinity, say at a > r_0 , so we get,
\phi(r)=k\ln{\frac{r}{a}}
Now, since the force between any two particles is an inverse square law, the virial theorem says:
2KE+PE=0 . Thus we can write:
v=\sqrt{-\phi(r)} =\sqrt{ -k\ln{\frac{r}{a}}} .
Obviously, something is wrong with this answer. An observable quantity like the velocity, ‘v’, cannot depend on an arbitrarily chosen constant, ‘a’. Where did my logic go south?
If the dark matter density is
\rho= \frac{\rho_0}{1+\left(\frac{r}{r_0}\right)^2},
Then how does velocity depend radius at large r ( r>>r_o)?
You want to use the virial theorem here, so you calculate M(r) and then finally calculate the potential:
\phi(r)=\int_{-\infty}^{r} \frac{k}{s}ds=\infty
with k = constant.
Oops, phi(r) diverges logarithmically. Oh well, you just have to set the zero point somewhere else besides infinity, say at a > r_0 , so we get,
\phi(r)=k\ln{\frac{r}{a}}
Now, since the force between any two particles is an inverse square law, the virial theorem says:
2KE+PE=0 . Thus we can write:
v=\sqrt{-\phi(r)} =\sqrt{ -k\ln{\frac{r}{a}}} .
Obviously, something is wrong with this answer. An observable quantity like the velocity, ‘v’, cannot depend on an arbitrarily chosen constant, ‘a’. Where did my logic go south?