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 it, 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.
2.
${ 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 $p$.
Best Answer
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:
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."