I am having trouble understanding how to convert the standard spherically symmetric fluid equations into logarithmic form (which can be useful, e.g., when you're working with many orders of magnitude in density as discussed in this post).
For example, the mass continuity equation for a spherically symmetric fluid is
$ \frac{\partial \rho}{\partial t} + \frac{1}{r^2}\frac{\partial(r^2 \rho v_r)}{\partial r} = 0$
which is sometimes also simply written as $\dot{M}=4\pi r^2 \rho v$ if you further assume a steady state.
But then I have also seen the logarithmic form of the continuity equation for a spherically symmetric fluid (in a steady state) given as
$ \frac{\rm \partial\,ln\,\rho}{\rm \partial\,ln\,r} + \frac{\rm \partial\,ln\,v}{\rm \partial\,ln\,r} = -2$
How do you get to this logarithmic form from the original non-steady-state, non-logarithmic continuity equation above? Apologies if this is super simple but I think the omission of the time derivative term in the logarithmic form is confusing me. Any clarification would be greatly appreciated.
Best Answer
As already mentioned in my comment the second equation holds in steady-state since in this case $\partial/\partial t=0$. In the following I have used $Q$ instead of $\rho$ for notation clarity. We start with $$\frac{\partial}{\partial r}(r^{2}Qv_{r})=0\rightarrow \frac{\partial}{\partial r}[(rv_{r})(rQ)]=0\rightarrow \frac{\partial(rv_{r})}{\partial r}(rQ)+(rv_{r})\frac{\partial(rQ)}{\partial r}=0$$ Before calculating the derivatives divide the terms by $r^{2}Qv_{r}$ to get $$\frac{1}{rv_{r}}\frac{\partial(rv_{r})}{\partial r}+\frac{1}{rQ}\frac{\partial(rQ)}{\partial r}=0$$ Since $\frac{d}{dx}\ln[f(x)]=\frac{1}{f(x)}f'(x)$ we get $$\frac{\partial}{\partial r}\ln[rv_{r}]+\frac{\partial}{\partial r}\ln[rQ]=\frac{\partial}{\partial r}\ln r+\frac{\partial}{\partial r}\ln v_{r}+\frac{\partial}{\partial r}\ln r+\frac{\partial}{\partial r}\ln Q=0$$ We also know that $$\frac{\partial}{\partial r}=\frac{\partial}{\partial(\ln r)}\frac{\partial(\ln r)}{\partial r}=\frac{1}{r}\frac{\partial}{\partial(\ln r)}$$
Thus $$\frac{1}{r}+\frac{1}{r}\frac{\partial \ln v_{r}}{\partial(\ln r)}+\frac{1}{r}+\frac{1}{r}\frac{\partial \ln Q}{\partial(\ln r)}=0$$ In other words $$\frac{1}{r}[2+\frac{\partial \ln v_{r}}{\partial(\ln r)}+\frac{\partial \ln Q}{\partial(\ln r)}]=0$$