The continuity equation in fluid mechanics states that $$ \frac{\partial\rho}{\partial t} + \nabla\cdot(ρ\mathbf u)=0 $$ Can you explain to me what is the physical meaning of each term of the equation and what are the direct outcomes?

Keep in mind that i do not have much knowledge about differential calculus.
Thank you!
After receiving some answers(for those who want to find out the answer,check the comments),i have only one question left.If the density gets bigger and bigger as time passes,then the first term of the equation is positive.What happens with the second term?What changes in that control volume?I mean,the density in the CV gets bigger,so what happens to velocity(intuitively)?I think that the velocity must get bigger in order to the fluid to get out of the CV.But that means that the second term is also positive and the equation is not right.So what is wrong with my thinking?

  • 2
    $\begingroup$ Have you read the Wikipedia entry on the continuity equation? $\endgroup$
    – Kyle Kanos
    Mar 28, 2015 at 13:18
  • $\begingroup$ Yes,but i think i need a more detailed explanation $\endgroup$ Mar 28, 2015 at 13:19
  • $\begingroup$ I think my answer to this question should give insight into each term. As might some of the answers to this question (be warned that the top rated answer there probably won't help if you are not mathematically inclined, the 2nd answer probably will do a better job). $\endgroup$
    – Kyle Kanos
    Mar 28, 2015 at 13:33
  • $\begingroup$ Your answer in the first link was really helpful.If you can just post here an answer which has the outcomes of the equation(like when ρ does not change in respect to time) i would be grateful. $\endgroup$ Mar 28, 2015 at 13:53
  • $\begingroup$ Hi Landos look at the equation you have in your question. Now move the second term over to the right hand side of the equation like in the first answer. If density does not change with respect to time, then the first term will equal 0. That means the other term on the right side also equals 0. Roughly speaking, if nothing comes in or out of the "box", then there is no velocity involved so the velocity vector u is also zero. Regards $\endgroup$
    – user74893
    Mar 28, 2015 at 14:47

1 Answer 1


Consider a control volume $\Sigma(t)$ of a fluid with density $\rho(\mathbf x,t)$. The mass inside $\Sigma(t)$ is clearly given by $$M(t):=\int_{\Sigma(t)}\rho(\mathbf x,t)\text d^3\mathbf x.$$ The way $\Sigma(t)$ is defined is that its mass content doesn't change with time, that is, a control volume is representing the time evolution of a certain amount of mass. Therefore the time derivative of the above integral should be zero. For a small time lapse $\epsilon$, the difference between $M(t)$ and $M(t+\epsilon)$ can be estimated by $$\int_{\Sigma(t)}\frac{\partial\rho}{\partial t}(\mathbf x,t)\epsilon + \int_{\partial\Sigma(t)}\epsilon\rho(\mathbf x,t)\mathbf v\cdot\hat{\mathbf n}\text dS,$$ where the first term is the integral of the difference of density at different times on the common domain, i.e. intersection, between $\Sigma(t+\epsilon)$ and $\Sigma(t)$, while the second term counts contributions from the bits that are in $\Sigma(t+\epsilon)$ and not in $\Sigma(t)$ and viceversa (the symmetric difference $\Sigma(t+\epsilon)\triangle\Sigma(t)$), whose volume is estimated by $\epsilon\mathbf v\cdot\hat{\mathbf n}\text dS$, $\hat{\mathbf n}$ being the normal to $\partial\Sigma(t)$.

Divide by $\epsilon$ and take the limit $\epsilon\to0$ together with Gauss theorem on the boundary term to conclude that $$\int_{\Sigma(t)}\left[\frac{\partial\rho}{\partial t}+\nabla\cdot(\rho\mathbf v)\right]\text dV=0$$ for any control volume $\Sigma(t)$, whence the continuity equation.

Physically, the integral form suggests that the continuity equation should be interpreted as relating the density at each point to the flow of matter through a fixed surface. In other terms, consider a fixed (not a control) volume $V$ of fluid. The motion of the fluid will bring some matter in and take some matter out of this volume, thus one can expect the density inside it, at a certain point $\mathbf x$ to change over time. The continuity equation is then saying that $$\frac{\partial\rho}{\partial t} = - \nabla\cdot(\rho\mathbf v)$$ since the RHS can be interpreted as the flow of the current $\rho\mathbf v$ across the boundary of $V$, we see that the variation of density in time inside $V$ is precisely related to the net flow of matter across the boundary of $V$ itself.

Example If the net flow is outward directed, the flow across the boundary will be positive (by definition of flow, since the velocity vector and the normal to the boundary form an acute angle) and as expected this lead to a drop in density, since the volume $V$ is losing matter.

  • 3
    $\begingroup$ Given that the OP asks for a physical picture and expresses that they're unfamiliar with differential calculus, how is this helpful? $\endgroup$
    – Kyle Kanos
    Mar 28, 2015 at 13:42
  • $\begingroup$ @KyleKanos how about now? $\endgroup$
    – Phoenix87
    Mar 28, 2015 at 13:49
  • $\begingroup$ i can understand a part of it,but the second term of the sum of the two integrals is a bit difficult to understand. $\endgroup$ Mar 28, 2015 at 13:50
  • 1
    $\begingroup$ @Phoenix87: I'd put your edit at the top, then add a statement like, From a mathematical perspective, it means this... to the original bit. $\endgroup$
    – Kyle Kanos
    Mar 28, 2015 at 13:52
  • 1
    $\begingroup$ I found this to be an excellent answer. $\endgroup$
    – Sean D
    Apr 13, 2016 at 12:38

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