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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.

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  • $\begingroup$ You mention in your question that the second equation is in steady-state. So why are you confused about not having a time derivative in the second equation? In steady state $\partial/\partial t=0$. Also, can you make the fonts of the second equation consistent with the first? $\endgroup$
    – Newbie
    Jan 13, 2022 at 0:31
  • $\begingroup$ Fixed the fonts. Yes I understand now that the time derivative term is missing in the logarithmic form because it's steady state -- I think it's the algebra and log-derivatives I need to work through to go from the steady-state version of the first equation to the logarithmic equation. $\endgroup$ Jan 13, 2022 at 0:46
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    $\begingroup$ I'm providing the answer. $\endgroup$
    – Newbie
    Jan 13, 2022 at 0:46
  • $\begingroup$ The fonts in the second equation still don't match the fonts in the first equation. Compare for instance $r$ and $v_{r}$. $\endgroup$
    – Newbie
    Jan 13, 2022 at 1:10

2 Answers 2

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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 $$\frac{1}{r}[2+\frac{\partial \ln v_{r}}{\partial(\ln r)}+\frac{\partial \ln Q}{\partial(\ln r)}]=0$$

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  • $\begingroup$ Oh wow, thank you! So this is why I couldn't figure it out -- multiple clever uses of the chain rule and log-derivatives... Is this standard in fluid dynamics textbooks -- how did you get it so quickly? $\endgroup$ Jan 13, 2022 at 1:36
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    $\begingroup$ @quantumflash I haven't read any fluid dynamics textbooks. As you said it's really a bunch of math operations. Thanks for accepting the answer. $\endgroup$
    – Newbie
    Jan 13, 2022 at 1:42
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    $\begingroup$ You can use $\int_{V} \nabla \cdot (Q\vec v)dV'=\int_{\partial V}Q\vec v\cdot d\vec S\,'$ which is the divergence theorem. Now assume the shell in between $r_{1}$ and $r_{2}$ with $r_{2}>r_{1}$. You know that there is no source in this region and you are in steady state. Also assume that $\vec v=\vec v_{r}\hat r$. Then you know fluxes are only in the radial direction and should cancel out, otherwise you will have build up of mass in the shell and that's not steady state. This is another way to interpret the identity in the solution. $\endgroup$
    – Newbie
    Jan 13, 2022 at 13:41
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    $\begingroup$ For interpretation write the result in the form of $\frac{\partial \ln v_{r}}{\partial(\ln r)}+\frac{\partial \ln Q}{\partial(\ln r)}=-2\rightarrow \frac{\partial \ln (Qv_{r})}{\partial(\ln r)}=-2$. Continuity equation is a conservation equation. You are thinking about the mass or density that passes through a spherical area (although the $4\pi$) are missing from the continuity equation its just a constant factor. You're in steady state and there are no sources such that $\nabla\cdot (Q\vec v)=0$. Thus, As you go to larger $r$ this area increase should be cancelled out by decrease in flux. $\endgroup$
    – Newbie
    Jan 13, 2022 at 13:44
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    $\begingroup$ Okay thank you. I think I get it. I was getting held up by thinking the flux $\rho v$ is constant but it is NOT -- it's actually the product $r^2 \rho v$ that is constant (as you say, at large $r$, the spherical area is bigger hence in a steady state flow, the flux $\rho v$ must be smaller at large $r$ so that you get the same $\dot{M} = 4 \pi r^2 \rho v$ at all $r$). And in fact the logarithmic form makes clear how the flux profile should fall off with radius: as $r^{-2}$ since $\frac{\partial \rm{ln}(\rho v)}{\partial \rm{ln}r}=-2$. This allows $r^2 \rho v$ to be constant at all $r$. $\endgroup$ Jan 13, 2022 at 18:54
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If it is a steady flow (so that only the 2nd term is present), and we use the product rule for differentiation, we get $$2r\rho v+r^2\frac{d\rho}{dr}v+r^2\rho \frac{dv}{dr}=0$$Dividing this by $r\rho v$ gives $$2+\frac{d\ln{\rho}}{d\ln{r}}+\frac{d\ln{v}}{d\ln{r}}=0$$

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  • $\begingroup$ Thank you! And yes the trick to go from your first equation to the second is to remember (as the previous answer also said) that $d \rm{ln} r = (1/r) dr$ for substitution. Can you please check my interpretation of this logarithmic continuity equation in words in the my new comment under the previous answer? Is it just saying that the slope of the (logarithmic) mass flux profile is just -2? How to reconcile this with the non-log equation saying the linear slope is exactly 0 (flat $\dot{M}(r)=4\pi r^2\rho(r) v(r)$ profile)? $\endgroup$ Jan 13, 2022 at 13:10
  • $\begingroup$ $\dot{M}$ is not a function of r. $\endgroup$ Jan 13, 2022 at 17:22

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