Can every global conservation law be written as following? Consider a physical quantity $\phi$ that is globally conserved.
From Feynman's argument (in his volume 2 I think), which states that local conservation follows from global conservation due to special relativity, we can say that $\phi$ is locally conserved also.
Mathematically this can be written as,
$$\frac{\partial\rho}{\partial t}+\vec\nabla\cdot\vec J=0$$
where $\rho=$ density of $\phi$
and, $\vec J=\rho\vec v$
The above law, with $\vec\nabla=(\frac{\partial}{\partial q_i},\frac{\partial}{\partial p_i})$ in phase can be written as: (this is how Louiville's theorem in statistical mechanics is derived from microstate conservation)
$$\frac{d\rho}{dt}=0$$
Firstly, Is this equation true for the density for any conserved quantity?
Secondly, I want to know what exactly this equation means?
If the above was $\frac{d(\text{total charge})}{dt}=0$ then that would have been obvious.
But how come the total time-derivative of the density being zero implies that total quantity is conserved?
P.S: I know how the above analysis can come from the Hamiltonian mechanics where the time derivative of any function can be written as the sum of a partial time derivative plus the Poisson bracket with the Hamiltonian. I am more interested in knowing how the equation can be understood in clear physical/visual/intuitive terms.
 A: I'm not sure if this is what you're after but to me the most intuitive explanation comes from Gauss' Law: integrating the continuity equation over some arbitrary volume $V$
\begin{align*}
0 &= \int_V \partial_t \rho + \nabla\cdot\mathbf{j} \,\mathrm{d}V \\
0 &= \int_V \partial_t \rho \,\mathrm{d}V + \int_{\partial V} \mathbf{j}\,\mathrm{d}\mathbf{S} \\
\int_{V}\frac{\partial \rho}{\partial t}\,\mathrm{d}V &= - \int_{\partial V} \mathbf{j}\,\mathrm{d}\mathbf{S}
\end{align*}
Or in other words: if the density $\rho$ inside the volume $V$ changes, then it must be through some flux $\mathbf{j}$ leaving the volume through the surface $\partial V$.
A: Maybe it is of interest to think in terms of the "formula" for the total derivative:
$$\frac{d}{dt}=\vec{v}\cdot\nabla + \frac{\partial}{\partial_t} $$
The second term takes into account the explicit dependence on time. You could think of a radioactive decay, in which a sample is losing mass over time. The first term encodes the implicit dependence, meaning that the temporal dependence could be hidden in other variables. Think of a mass of water moving in a closed container, there is no gain or loss of water, but the local density changes because the position of the particles does depend on time ($x(t),y(t),z(t)$).
Now your doubts:

Firstly, Is this equation true for the density for any conserved quantity?

Yes, if the total derivative with respect to time is different from zero, then the quantity will change over time meaning that is not constant. The only way that the quantity can remain constant over time is that the implicit and explicit terms are balanced (your first equation).

Secondly, I want to know what exactly this equation means?

This is rather conceptual. I would think on the "wet observer": say that you want to check if $\rho(x,y,z,t)$ is constant over time and that you can measure it at any point and at any moment. In this setting, $\frac{d\rho}{dt}=0$ would mean that if you measure $\rho$ along a trajectory ($x(t),y(t),z(t)$), then you will always measure the same value.

If the above was d(total charge)dt=0 then that would have been obvious. But how come the total time-derivative of the density being zero implies that total quantity is conserved?

By definition, the density is always the intensive version of an extensive quantity. In other words, it is always true that $\rho=\frac{M}{V}$ being $\rho$ the density (of particles, charges, probability, ...), $M$ a mass (your total quantity) and $V$ a volume (an extensive parameter). Since the extensive parameter usually does not depend on time, you can conclude that if the density is conserved then the total quantity is conserved.
