I Some help with deriving svedberg equilibrium equatio please

[HS-experiment]

Hello Physics Forums,

I am studying the equations related to analytical ultracentrifugation. The equation I am interested in describes centrifugation equilibrium, ie at the point in time when sedimentation is balanced with diffusion in the ultracentrifuge. Fortunately, unlike the Lamm equation, this one can be solved analytically.

I found the following in a textbook. Can someone please explain to me the steps taken going from the first equation to the next?

$\frac {RT}{Nf } \frac{dC}{dr} = \frac{M(1-\bar vp)w^2rC}{Nf}$

We now write the derivative as dC/dr, because at equilibrium, C is a function only of r, not also of t. We can factor out the Nf on both sides and again rearrange to

$\frac{dln(C)}{d(r^2)}=\frac{M(1-\bar vp)w^2}{2RT}$

where
R = gas constant
T = temperature
N = avogadro's number
f = frictional coefficient
C = concentration of solute
r = distance from center of rotor
M = molar mass
$\bar v$ = partial specific volume
p = density of solvent
$w^2$ = rotational velocity
In particular, I don't know what to do after I remove Nf and move RT to the right hand side of the equation.

$\frac{dC}{dr}=\frac{M(1-\bar vp)w^2rC}{RT}$

I think that it is something simple, but my math ability is really bad.

Many thanks.

BTW This is taken fom the textbook 'Principles of Physical Biochemistry', second edition.

 

[Charles Link]

This is a little first year calculus: $\frac{d \, \ln{C}}{dC}=\frac{1}{C}$ so that $\frac{dC}{C}=d ( \ln{C})$. Meanwhile $\frac{d (r^2)}{dr}=2r$ so that $r \, dr=\frac{1}{2} d( r^2)$. $\\$ Recommendation is to use the homework template next time and post in the homework section. I think this one is ok where it is, but the Mentors might move it to the homework section.

 

[HS-experiment]

Hi Charles. Thanks a lot for your explanation and sorry if I placed this question in the wrong directory. My calculus course is coming back to me through a haze .. I forgot that when you are taking a derivative you need to add the fraction in front $\frac{1}{2}d(r^2)$ to end up with the correct answer. I should really take a refresher course some time