In deriving Euler's equations for fluid mechanics, in particular

$$f=\rho \partial_t v +\rho v\cdot \nabla v$$

for some body force $f$ (e.g Landau & Lifschitz 2.3) one assumes the continuity equation $\partial_t\rho=-\nabla\cdot\left(\rho v\right)$ and Newton's second law in the form $F=ma$ so $$f=\rho D_t v$$ where $D_t=\partial_t+v\cdot \nabla$ is the total time derivative (e.g Landau & Lifschitz 2.1).

If one instead uses $F=\dot p$ so that $$f = D_t\left(\rho v\right)$$ then in order to get to get our initial equation back one must assume $D_t \rho=0$ but of course $D_t \rho=\partial_t\rho+v\cdot \nabla\rho=\nabla\cdot\rho$ and so we have assumed the flow is incompressible.

Landau implies that he has not assumed incompressibility at this point so how is it that one can choose the first form for Newton's second law without loss of generality?

(If your answer would appeal to stress-energy tensors or Navier-Stokes etc., if possible please show how the assumption isn't implicitly made.)

To be more clear, as I have shown above, if you say that $f = D_t\left(\rho v\right)$ and also that $f=\rho \partial_t v +\rho v\cdot \nabla v$ then it follows that the flow is incompressible. I don't think that this does imply incompressibility so why is the assumption false?

  • $\begingroup$ I just realised it's all in the definition of the force density - it is just the force times the number density of particles $\vec{f}=\vec{F} n = \vec{F}\rho/m$, where $m$ is the single particle mass. You cancel $\rho$ and multiply both sides by $m$ and voila. I added this to my answer. Starting from the Boltzmann equation is maybe too lengthy, but well, for the sake of rigorous arguments... $\endgroup$
    – Void
    Jul 5, 2014 at 14:52

5 Answers 5


The main problem with continuum equations is that it is just a "lucky guess" of macroscopic dynamics otherwise governed by microscopic ones. You have encountered the moment where this "guessing" isn't entirely consistent.

Before delving into the technical derivation, the non-$D_t \rho$ in the Newton's equation is just a consequence of the fact that the actual force acts on microscopic particles whose mass doesn't change. Hence, only the macroscopic velocity of the fluid is affected by the force.

Let us define a one particle phase-space distribution function $f(\vec{x},\vec{p},t)$. We have the Boltzmann equation derived straightforward from classical mechanics (this is actually the only point where $\vec{F}=\dot{\vec{p}}$ enters) $$\frac{df}{dt} = \partial_t f + \{f,H\} = \partial_t f + \frac{\vec p \cdot \nabla_x f}{m} + \vec{F} \cdot \nabla_p f$$ Without colliding with other particles we would just have $df/dt = 0$ but with interaction we can describe the ensemble as N copies of a one-particle distribution with a collision term $$\partial_t f + \frac{\vec p \cdot \nabla_x f}{m} + \vec{F} \cdot \nabla_p f = \delta_t f|_c$$ We can now integrate the whole equation in momentum space $\int d^3p$ to get the continuity equation $$\partial_t \rho + \nabla \cdot (\rho \vec{v}) = \delta_t \rho|_c$$ With naturally $\rho \equiv m \int f d^3p$ and $\vec{v} = <\vec{p}>/\rho = \int \vec{p} f d^3 p/\rho$. The term $\delta_t \rho|_c$ is non-zero for example in chemical reactions. Integrating again the Boltzmann equation, this time like $\int \vec{p} ... d^3 p$, we get $$\partial_t(\rho v^j) + \sum \partial_{x_i} T_{ij} - \frac{\rho}{m} F^j = \delta_t p^j|_c$$ Where again $\delta_t p|_c$ is zero for momentum conserving collisions and no chemical reactions. The new symbol is $T_{ij} = <p_i p_j>/m = \int p_i p_j f d^3p/m$ and you can identify in it $ \rho v^i v^j$, isotropic pressure ($<\sum p_i^2>/3\rho$) and viscous stress ($(<p_i p_j>-<p_i><p_j>)/\rho$). For simplicity, let us now drop the viscous stress and using the continuum equation we can get $$\frac{\rho}{m} \vec{F} = \rho D_t \vec{v} + \nabla P ,$$ or a much more "Newton law" form $$\vec{F} = m D_t \vec{v} + \frac{m}{\rho} \nabla P.$$ So the Newton's law in continuum is derivable only as an intuition of how the sum of forces on individual particles will act on the average properties of the ensemble. It all make sense when you consider that $\vec{f}$ is not exactly the force but force time the number of particles per volume.

However, notice that this is the equation for mean velocity, but the force affects also higher momenta($\int \vec{p} \vec{p}...d^3p$) of the momentum equation and this is important for example in plasma physics.

In non-extreme fluids the higher momentum equations are usually neglected as the higher momenta of $\delta_t f|_c$ do not vanish so easily and are actually very difficult to derive in a non-perturbative regime (which is the case of a tightly-packed fluid rather than a loose plasma or gas). As a consequence, observed macroscopic properties of the fluid such as incompressibility are plugged in as extra conditions to get a practical model.

My derivations are very brief, for details see e.g. http://www.astronomy.ohio-state.edu/~dhw/A825/notes2.pdf (with a different notation).

EDIT: Combining my memory using the phase-space notation with the notation of the linked text, I have done some possibly quite confusing typos. Now the equations should be all correct. I also added a commentary to the much clearer form of the final "Newton's law".


If one instead uses $F=\dot p$ so that $$f = D_t\left(\rho v\right)$$

The formulation of the second law by the equation

$$ \mathbf F=\frac{d\mathbf p}{dt} $$ where $\mathbf p$ is total momentum of the system is valid only provided the system considered does not lose parts and thus has constant mass. This is usually the case in hydrodynamics, because we think of a fluid element which keeps all of its parts inside a boundary surface that is moving and transforming with the fluid.

But it is incorrect to think this implies the above quoted equation. On the contrary, the equation implied is the one where $\rho$ is not subject to differentiation with respect to time. This is always so in fluid mechanics whether the fluid is compressible or not.

We can rewrite the above equation into familiar form $$ \mathbf F = m \mathbf a, $$

i.e. force equals mass times acceleration of the center of mass of the system.

To find the expression for force per unit volume, we assume that total force acting on the element with volume $\Delta V$ is accurately enough given by $$ \mathbf F = \mathbf f\Delta V $$ and that its mass is accurately enough given by $$ m = \rho\Delta V. $$ It follows that $$ \mathbf f = \rho \mathbf a. $$ The quantity $\mathbf a$ is acceleration of the fluid element - a rate of change of its velocity $\mathbf u(t)$. Since this velocity can be expressed via the velocity field function $\mathbf v(x,y,z,t)$ as $$ \mathbf u(t) = \mathbf v(x(t),y(t),z(t),t), $$ where $x(t),...$ are coordinates of the fluid element at time $t$, we can express $\mathbf a = d\mathbf u/dt$ as

$$ d\mathbf u/dt = \partial_t \mathbf v + \partial_x (\mathbf v )\dot{x} + \partial_y (\mathbf v )\dot{y} + \partial_z (\mathbf v )\dot{z} $$

This is more commonly written in the form

$$ \mathbf a = \partial_t \mathbf v + \mathbf v \cdot \nabla\mathbf v. $$

  • $\begingroup$ Isn't "we think of a fluid element which keeps all of its parts inside some (variable) boundary surface S. Then m is constant" equivalent to saying $D_t\rho=0$ though, the density of the fluid element doesn't change as it flows along? $\endgroup$
    – user21433
    Jul 1, 2014 at 20:19
  • 2
    $\begingroup$ No, that is a very different thing. Think of the water element as of a marble made of polyuretane (foam). When you squeeze it, its mass stays the same - no parts leave it. The are only packed more closely now so the volume decreases and density increases. $\endgroup$ Jul 1, 2014 at 21:01

It is easier if you start from a modern continuum mechanics perspective.

Recall that Newton's 2nd Law applies to a system of fixed mass. To follow fluid volumes of fixed material around, we'll have to introduce a reference configuration $B_0$ at t=0.

So the fluid's deformation is an invertible map $\chi$ from $B_0(X,Y,Z)$ to $B(x,y,z)$ and depends on X,Y,Z and time $t$. $B$ is a deformed configuration of the fluid.

The Cauchy momentum balance in the deformed configuration is

$$ \sigma_{ij,j} + \rho b_i = \rho \dot{v_i} $$

where the comma denotes partial differentiation with respect to the deformed variable $x_j$ and $b_i$ is body-force per unit mass and $\rho$ is the density in the deformed configuration.

Note that no assumption of incompressibility is made so far.

For an inviscid fluid, the stress tensor does not contain shear terms, so we can replace $\sigma_{ij}$ by $-p \delta_{ij}$ where $p$ is the pressure. The momentum balance now reads

$$ -\frac{\partial p}{\partial x_i} + \rho b_i = \rho \dot{v_i} $$

The right hand side contains a material derivative, which expands as $$ \frac{Dv_i}{Dt} = \frac{\partial v_i}{\partial t} + v_k \frac{\partial v_i}{\partial x_k} $$

If you introduce a body force per unit deformed volume, $f_i = \rho b_i$ and re-arrange, you have

$$ -\frac{\partial p}{\partial x_i} + f_i = \rho \frac{\partial v_i}{\partial t} + \rho v_k \frac{\partial v_i}{\partial x_k} $$

Finally, in vector form this is

$$ -\nabla p + \mathbf{f} = \rho \frac{\partial \mathbf{v}}{\partial t} + \rho \mathbf{v} \cdot \nabla \mathbf{v} $$

which is the momentum balance in Euler's equation. (Your equation is missing the pressure gradient term.)

Note that incompressibility is not involved at any stage of the proceedings. If you want to impose incompressibility, the accompanying continuity equation is just $$ div \mathbf{v}=0 $$

and $\rho$ is treated as a constant in the momentum balance.


The expression $$ f = \rho D_t(v) $$ holds for both compressible and incompressible fluids, while the expression $$ f = D_t(\rho v) $$ only holds for an incompressible fluid.

To see why this is so, first note that Newton's force law relates the mass of an object to its acceleration, not its mass density. So you will need to convert mass densities to masses. To do this, you will multiply densities by volumes.

For an incompressible fluid, the volume of a given packet of mass ("fluid element") may change as the fluid flows. In general, you need to multiply densities in fluid elements by the co-moving volume, which I'll denote by $\cal{V}$. Then the total mass $M$ equals $\cal{V} \rho$ and the total force $F$ equals $f \cal{V}$.

The above statements imply $D_t(\cal{V}) \ne 0$ in general for a compressible fluid, but $D_t(\cal{V}) = 0$ generally holds for an incompressible fluid. Also, by definition, $D_t{\rho} \ne 0$ in general for an incompressible fluid, while $D_t{\rho} = 0$ generally holds for an compressible fluid.

Now lets examine the two equations again, and convert densities to masses by multiplying through by $\cal{V}$. The first equation becomes

$$ F = M D_t(v) \qquad {\rm or\,\,equivalently \,\, } \qquad F = M a \,\, , $$

which is the general statement of Newton's force law.

The second equation becomes

$$ F = \cal{V} D_t(\rho v) \, \, . $$ If the fluid is incompresible, then $\rho$ can be brought outside the time derivative and we again arrive at Newton's force law. But otherwise, we are stuck because we can't move either $\cal V$ or $\rho$ to the other side of the derivative.

Added: As another answer notes here, you can't forget that pressure gradients also exert forces on the fluid. The above discussion is correct as long as the quantities $f$ and $F$ are assumed to denote the combined effects of pressure forces and body forces such as gravity. Although, I will admit that it is more standard to use $f$ to denote the body force only, and you should explicitly include the pressure force term.


With this answer I'll try to note the mistake committed in the original formulation

"Landau implies that he has not assumed incompressibility at this point so how is it that one can choose the first form for Newton's second law without loss of generality?"

Well the problem is originated when you assert:

"If one instead uses $F=\dot p$ so that $$f = D_t\left(\rho v\right)"$$

It so happens that $\mathbf{F}=\mathbf{\dot{p}}$ is not translated correctly to a continuum as $\mathbf{f}=D_t(\rho\mathbf{v})$, to get the correct expression you must apply the the second law to a finite portion of continumm and treat it as a system of particles and to do that you must take the material derivative of and integral expression: $$\int_V f_i\,d^3r=\frac{D}{Dt}\int_V\rho v_i\,d^3r=\int_V\frac{\partial(\rho v_i)}{\partial t}\,d^3r+\oint_{\partial V}\rho v_iv_kn_k\,ds$$ $$\int_V f_i\,d^3r=\frac{D}{Dt}\int_V\rho v_i\,d^3r=\int_V\left(\frac{\partial(\rho v_i)}{\partial t}+(\rho v_iv_k)_{,k}\right)d^3r$$ so that the second law correctly stated in differential form looks like: $$f_i=\frac{\partial(\rho v_i)}{\partial t}+(\rho v_iv_k)_{,k} $$ this is certainly different from $f_i=D_t(\rho v_i)$ which treacherously may pass as the correct formulation of $F=\dot{p}$ for a continuum. Finally we observe that after some trivial algebra and application of continuity equation it does reduce to $f_i=\rho D_t v_i$ as Landau stated without any assumption of incompressibility

Here's the algebra: $$f_i=\frac{\partial(\rho v_i)}{\partial t}+(\rho v_iv_k)_{,k}=\partial_t\rho v_i+\rho\partial_tv_i+v_i(\rho v_k)_{,k}+v_{i,k}(\rho v_k) $$

$$f_i=v_i[\partial_t\rho)+(\rho v_k)_{,k}] +\rho\partial_tv_i+v_{i,k}(\rho v_k) $$ $$f_i=\rho(\partial_tv_i+v_{i,k} v_k)=\rho D_tv_i $$

the expression in brakets is zero because it's the continuity equation

  • $\begingroup$ You've assumed $v^i{}_{,i}=0$ in your algebra. $\endgroup$
    – user21433
    May 19, 2015 at 16:07
  • $\begingroup$ No, I did not assume that $\endgroup$
    – facenian
    May 19, 2015 at 16:14
  • $\begingroup$ $D_t\left(\rho v^i\right)=\partial_t\left(\rho v^i\right)+v^i\partial_i\left(\rho v^j\right)=\partial_t\left(\rho v^i\right)+\partial_i\left(\rho v^iv^j\right)-\left(\rho v^j\right)\partial_iv^i$ $\endgroup$
    – user21433
    May 19, 2015 at 17:02
  • $\begingroup$ I added the algebra, maybe I exaggerated when I said it's trivial $\endgroup$
    – facenian
    May 19, 2015 at 17:56

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