Reynolds transport theorem gives the time variation of the amount of a magnitude $\mathbf f = \mathbf f(\mathbf x,t)$ contained within a time-dependent control volume $\Omega(t)$ by

$$ \frac{d}{d t} \int_{\Omega(t)} \mathbf{f} d V =\underbrace{\int_{\Omega(t)} \frac{\partial \mathbf{f}}{\partial t} d V}_{1}+\underbrace{\int_{\partial \Omega(t)}\left(\mathbf{v}^{b} \cdot \mathbf{n}\right) \mathbf{f} d A}_{2} $$

where $\mathbf n(\mathbf x,t)$ is the outward-pointing unit normal vector, $\mathbf x$ is a point in the region and is the variable of integration, $dV$ and $dA$ are volume and surface elements at $\mathbf x$, and $\mathbf v^b(\mathbf x,t)$ is the velocity of the area element.

I see that the term $2$ is the amount of $\mathbf f$ that flows across the surface of the control volume $\partial \Omega(t)$. However, what would the meaning of the term 1 be, and how does it differ from the LHS term?

I have seen this related question, but it does not fully answer my doubt.

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    $\begingroup$ The LHS is a rate of change which includes the changing boundary. Term 1 is a rate of change which does not account for the changing boundary. These are my observations as a novice, hopefully they helps you down the right path. $\endgroup$ Commented Dec 6, 2021 at 11:07

3 Answers 3


It's actually pretty simple.

LHS: Rate of change in a quantity $\bf f$ in a volume $\Omega$ is equal to

(1): the rate of change in quantity $\bf f$ in the volume itself, plus

(2): the net influx of the quantity across the boundaries of $\Omega$

The easiest example is mass. If you have a fluid flowing, and an imaginary box (the control volume), the total change in mass inside the box equals the change in density of the fluid inside (usually zero for liquids or low-speed flows of gases) plus the total of the inflow and outflow of fluid across the boundaries.

In the case of mass which cannot have a source or sink in a 3 dimensional fluid, this means that e.g. if you have density decreasing in a control volume, like due to an expanding gas, any decrease in the amount of mass inside the volume must correspond to a net outflow across the boundaries. The mass has to go somewhere. In an incompressiblec flow of a liquid, say, the density doesn't change, so your term (1) is zero, and the equation will require that any inflow must be balanced by an equal outflow. "What goes in must come out elsewhere."

Other things that can be tracked on this way are: momentum, which gives us the velocity equations; energy, which gives us the energy/heat transfer equation; entropy, which gives us yet another equation, usually used more in compressible or turbulent flows.

Unlike mass, momentum and energy can have sources and sinks in the flow, which is why we need to add terms accounting for forces (viscosity, pressure, gravity) that add or remove momentum, and heat sources (viscous dissipation, external heating, internal heat generation due to chemical reaction) that add or remove energy.

But the full set of Navier Stokes equations are taking 3 conversation laws:

Mass is conserved

Momentum is conserved

Energy is conserved

And then saying "each one of these three in any control volume is the sum of the change in that quantity in the CV, plus the net influx/outflux to the CV."


Perhaps the simplest way to understand the difference between the two terms is to place oneself in a situation for which the integral noted 1 is zero, ie in steady state.

Imagine for example a nozzle with a gas which enters on the left with a low speed and this same gas which leaves on the right with a high speed. If we are in steady state, the quantity of gas in the nozzle does not change, nor does its momentum. If we apply Reynolds' theorem to the momentum of the nozzle, the integral (1) represents the derivative with respect to time of the momentum in the nozzle. But there is always the same momentum inside this nozzle: it does not change and therefore the integral (1) is zero.

Now isolate in thought the matter that is in the nozzle at any given time and follow it for a short while. It is a closed system. He moved a little left and right. The small part on the right has a high speed while on the left, we "lost" a part which had a low speed. In total, the momentum of the closed system that we follow over time has changed and its derivative is non-zero. It is the LHS of the Reynolds equation.

It should be understood that even in steady state, while the fields do not depend on time, a closed system moves within these fields and the quantities associated with it can vary. We find the same idea for the Eulerian acceleration which can be non-zero even if the velocity field is permanent. A fluid particle moves within this permanent field and its speed changes over time.

Hope it can help and sorry for my poor english.


I have the answer to your question. First, I'll note that while the wiki proof is no doubt correct, there are much better, i.e. comprehensible, illustrated proofs available at https://www.youtube.com/watch?v=hpCYVluvC_o and https://www.youtube.com/watch?v=hpCYVluvC_o The conceptual proof is easy but the math symbolism obscures it big time in the absence of illustrative diagrams, IMO.

The lhs of the RTT in the wiki proof represents the time derivative of integral of f over the domain while first integral on the rhs integrates the partial derivative of f over the domain. When the theorem is proved the integral on the right also has the derivative outside the integral and the integrals look the same, but they are not.

To apply the theorem to determine the lhs of the Navier-Stokes equations, for example, the derivative is moved inside the integral on the right, justified by the Leibniz integration rule, and evaluated for in infinitesimal control volume.

If you could move the derivative under the integral of the lhs of the RTT then the integrals for the control volume would be equal. However, you cannot do that because boundaries of the lhs integral change with time and the simple form of the Leibniz integration rule used on the rhs integral does not apply to the lhs integral.


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