According to many sources (including Wikipedia, Stephani&Kluge, D.J. Acheson) a steady state ist:

In systems theory, a system in a steady state has numerous properties that are unchanging in time. This means that for those properties $p$ of the system, the partial derivative with respect to time is zero:

$$ \frac{\partial p}{\partial t} = 0 $$

But why is it defined like that? Why not ${d \over dt} p=0$ ? If only $ \frac{\partial p}{\partial t} = 0 $ then there will still be change in time if $p=p(\vec r(t))$ !

Since people seem to disagree that this is even a legitimate question here is a motivation for that, Wikipedia about total and partial derivative:

The total derivative of a function is different from its corresponding partial derivative ($\partial$). Calculation of the total derivative of f with respect to t does not assume that the other arguments are constant while t varies; instead, it allows the other arguments to depend on t.

So why can we assume the other variables are constant? If we are using $ \frac{\partial p}{\partial t} = 0 $ to determine if a steady state is present, why should we be able to assume no indirect dependence on time exists?

My try to explain this, if it is nonsense of course an explanation why would be great but just a good answer to my question would be too:

Reasons why ${\partial p \over \partial t} =0$ might make more sense than the total derivative to equal zero.

1. ${d p \over dt} =0 $ on it's own already indicates a conservation of the flux $j$ and density $\rho$ associated with $p$ by Reynolds theorem: $${d \over dt} p={d \over dt}\int_V \rho dV= \int_V \frac{\partial \rho}{\partial t} dV+ \int_{\partial V} \underbrace{\rho \vec v}_{= j} \cdot \vec n dA$$

since ${d p\over dt} =0$ normally holds independent of V if we use Gauss theorem $$\Rightarrow \frac{\partial \rho}{\partial t}+\nabla j=0$$

We see the total derivative being zero gives us a continuity equation.


${ d p \over dt} =0 $ implies if we write it out $${dp \over dt } = { \partial p \over \partial t} + ( v \nabla ) p=0$$ i.e. $${ \partial p \over \partial t} =- ( v \nabla ) p$$ So by the definition of the partial derivative: if we hold all other variables and only look at the change in $t$, we see, this derivative does not vanish and we even have a time dependend $p$.

  • $\begingroup$ Since the steady-state is a very general concept, I think the explanation should not rely on a specific theorem. Moreover, in the definition $p$ could be any physical property, not just the density. $\endgroup$ – AlQuemist Nov 30 '15 at 17:28
  • $\begingroup$ Well maybe there is a more general viewpoint but for most scalar and even vector object you could define a density and do this kind of calculation. Is it correct even for this case? That would be at least something. $\endgroup$ – pindakaas Nov 30 '15 at 21:19
  • $\begingroup$ All partial derivatives with respect to time are zero, including that of $\vec{r}$ (i.e. $\vec{r} \neq f(t)$), otherwise it cant be a steady-state by definition. So there isn't actually a problem here... $\endgroup$ – nluigi Nov 30 '15 at 21:55
  • $\begingroup$ I think the question is at a more fundamental level; at the level of the concept and definition, and the physical and mathematical picture which leads to such a definition.@pindakaas $\endgroup$ – AlQuemist Nov 30 '15 at 21:59
  • $\begingroup$ Did Liouville's theorem say that the absolute derivative of the density is always zero? Or I misunderstood your question?en.m.wikipedia.org/wiki/Liouville%27s_theorem_(Hamiltonian) $\endgroup$ – ophelia Nov 30 '15 at 22:23

Put simply, steady-state is a pointwise phenomenon, not a global system phenomenon. To answer your question, I will provide an example of a steady state system for which $\partial p / \partial t = 0$ but $d p / d t \neq = 0.$

Think of a horizontal stream of fluid on the $x$-axis flowing in the $+x$ direction. Suppose the velocity of the fluid may change at different points along the stream, as may the stream's width, but at any given point along the $x$ axis, the velocity and the width remain constant with time. This is a steady state system—if I take a picture of the stream at time $t = 0s$ and take another picture at time $t = 10 s$, the two pictures will look identical. In this time interval a great volume of fluid may have passed through each point in the stream, but the system as a whole looks the same as it did ten seconds in the past.

Let $v(x, t)$ be the velocity of the stream at a given $x$ position, and $w(x, t)$ be its width. The loose notion of "steady state" we gave above is put more rigorously as follows:

This stream is in a steady state if at any given point $x$ in the stream, the quantities of interest $w(x, t)$ and $v(x, t)$ are unchanging with time.

Since we are fixing a point $x$ in the stream, the above is equivalent to demanding $\frac{\partial v}{\partial t} = \frac{\partial w}{\partial t} = 0.$

The total derivatives may not be zero anywhere. For example, we have

$$ \frac{d v}{d t} = \frac{\partial v}{\partial x} \frac{d x}{d t} + \frac{\partial v}{\partial t}. $$

If the velocity gradient $\partial v / \partial x$ is nonzero, and the velocity $d x / d t$ at a given point is nonzero, then the total derivative $dv / dt$ is nonzero. That is, if I look at a single particle of the fluid, of course its velocity changes with time. It moves along the stream, and the velocity changes at different points in the stream.

But at any given point, the velocity of all particles passing through that point is constant for all time. This is what is meant by "steady state."

  • $\begingroup$ If you add the reason that fixing the point is equivalent to the partial derivatives vanishing I think I'll accept this. Probably for any quantity ${d p\over dt} = {\partial p \over \partial t} + (v \nabla) p$ this just always means that $(v \nabla) p=0$ as wee look at $p(x=const)$ meaning a specific point? $\endgroup$ – pindakaas Dec 14 '15 at 16:32

This discussion feels familiar; I figure this question is a follow up of some comments to an answer i provided to one of your previous questions. Specifically:

thanks for the great answer. Although I have to say i find steady state to be a bit confusing here since this in fluis mechanics normally means $∂_t v=0$ but I venture you mean steady mass state? So $\frac{dM}{dt}=0$ right ? – pindakaas Nov 25 at 10:33

My response:

@pindakaas - at steady-state any time derivatives vanish, i.e. $\partial_t\rho=0$ and $\partial_t \boldsymbol{v}=0$. The implication is that density is constant and by the continuity equation $\boldsymbol{\nabla}\cdot\boldsymbol{v}=0$. The term $$\boldsymbol{\nabla}\cdot[\rho\boldsymbol{v}\otimes\boldsymbol{v}]=\rho\boldsymbol{\nabla}\cdot[\boldsymbol{v}\otimes\boldsymbol{v}]=\rho\boldsymbol{v}(\boldsymbol{\nabla}\cdot\boldsymbol{v})+\rho(\boldsymbol{v}\cdot\boldsymbol{\nabla})\boldsymbol{v}$$ using the identity shown in my answer. – nluigi Nov 25 at 10:55

and with you concluding:

I don't think I am making myself clear. Assuming mass conservation is enaugh. Assuming $\partial_t \rho=0$ and assuming incompressibility is simply not necessary. – pindakaas Nov 25 at 18:11

At the time i didn't really understand what you were talking about (even after our talk in chat) but i let it go because i figured we were talking about the same thing just from different perspectives. But this question here confirms my suspicion that we in fact weren't talking about the same thing. Hopefully i can now clear it up.

Consider some conserved quantity $\theta=f\left(t,\boldsymbol{r}\right)$ (where $\theta$ could be $\rho$ or $\boldsymbol{v}$), its total derivative (often known as the material derivative in continuum mechanics) is by definition: $${\mathrm{D}\theta \over \mathrm{D}t} = {\partial\theta \over \partial t}{dt \over dt} + {\partial \theta \over \partial \boldsymbol{r}}\cdot{d\boldsymbol{r} \over dt} = {\partial\theta \over \partial t} + \boldsymbol{v} \cdot {\partial \theta \over \partial \boldsymbol{r}}$$ Here any terms may be time-dependent but if we were to consider just $\theta=f\left(\boldsymbol{r}\right)$ than the partial derivative wrt time would not exist in the above equation; this means if $\theta\neq f\left(t\right)$ then by definition $\partial_t\theta =0$. The case $\theta=f\left(\boldsymbol{r}(t)\right)$ is equivalent to $\theta=f\left(t,\boldsymbol{r}\right)$, i.e. $\theta$ cannot be considered time-independent.

If a steady state is defined by ${\mathrm{D}\theta \over \mathrm{D}t} =0$ then besides the time-dependence also the convective flux of the quantity $\theta$ by the velocity $\boldsymbol{v}$ is lost. Obviously in a field like fluid dynamics this is counter-productive. Instead by steady state we imply that all relevant quantities do not change with time, i.e. ${\partial\theta \over \partial t}=0$. Now it may seem that some terms (e.g. $\boldsymbol{v}$) may still be time-dependent but with an example i will show this isn't the case.

Consider the continuity and Navier-Stokes equations which need to be simultaneously solved in fluid dynamics: $$\partial_t \rho + \boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\right)=0$$ $$\partial_t \left(\rho\boldsymbol{v}\right) + \boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\otimes\boldsymbol{v}\right)=-\boldsymbol{\nabla}p + \mu\Delta \boldsymbol{v}$$

In steady-state we say no quantities change with time, i.e. $\partial_t \rho=0$ and $\partial_t \boldsymbol{v}=0$ such that $\rho\neq f\left(t\right)$ and $\boldsymbol{v} \neq f\left(t\right)$: $$\boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\right)=0$$ $$\boldsymbol{\nabla}\cdot\left(\rho\boldsymbol{v}\otimes\boldsymbol{v}\right)=-\boldsymbol{\nabla}p + \mu\Delta \boldsymbol{v}$$ No terms in these equations are now dependent on time but some are spatially dependent, $\mu$ is a material constant and $p$ is related to the velocity field (in incompressible flows) which is by definition steady.

A steady flow doesn't mean time is non-existent; if your were to do a simulation of a steady flow and released tracer particles into the flow then at subsequent time intervals the particles will move along the streamlines of the flow. The fact the flow is steady just means that the streamlines are not changing in time not that the $\boldsymbol{v}=0$.

  • $\begingroup$ This seems like a reasonable line of thought to me. But i wish there was a more rigorous way of expressing it. I will have a think on it. $\endgroup$ – pindakaas Dec 1 '15 at 13:02
  • $\begingroup$ maybe you would like to add this to your answer mybe I'll wright my own in the end. One thing that makes the connection steadystate <=> ${d \over dt} P=0$ questionable aswell is that for any quanity $P$ that has a density $\rho$ associated to it and a continuity equation. One can easily see that always ${d \over dt} \rho = - \rho \nabla v$ follows. Meaning if ${d \over dt} \rho =0$ we have incompressibility which should not follow from steady state and continuity. So defining the steady state with the total derivative would be problematic at least.. $\endgroup$ – pindakaas Dec 5 '15 at 10:04
  • $\begingroup$ further such a quantity $P$ is conserved as a concequence of ${d \over dt} P =0$ and will yield a continuity equation for it's density so the total derivative just means conservation not steady state $\endgroup$ – pindakaas Dec 5 '15 at 10:19

Wikipedia is totally unreliable for maths and science. You are right that it is the total differential that has to be zero, not the partial derivative. This is simply common sense, and has nothing to do with transport or anything specific, as Khwarazmi points out.

  • 1
    $\begingroup$ I am a Wikipedia editor, so I should know. $\endgroup$ – joseph f. johnson Nov 30 '15 at 17:34
  • $\begingroup$ Definitely, it is not fair to say that Wikipedia is “totally unreliable” for mathematics or science. It is reliable for many cases, although it is not something one can use as “the proof”. See e.g., “What's Wrong with Wikipedia?”. @joseph f. johnson $\endgroup$ – AlQuemist Nov 30 '15 at 19:43
  • $\begingroup$ Well, "reliable" means you can trust it even when you cannot tell the difference. Being unreliable doesn't mean you are never right. It is a question of trust. I appreciate the feedback, though. $\endgroup$ – joseph f. johnson Nov 30 '15 at 20:22
  • $\begingroup$ well wikipedia is not the only source that states this. Edited the quesion to reflect that. $\endgroup$ – pindakaas Nov 30 '15 at 21:15

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