Vlasov equation, Maxwell distribution I have the Maxwellian distribution:
$$f(v)=n\left(\frac{m}{2\pi kT}\right)^{\frac{3}{2}}\exp\left(-\frac{mv^2}{2kT}\right)$$
I have to show that it is a solution to the Vlasov equation:
$$\frac{\partial f}{\partial t}+\vec{v}
\cdot \text{grad}(f)+\frac{q\vec{E}}{m}\cdot \text{grad}_v(f)=0$$
Since $f(v)$ depends on the velocity $v$ only, I assume that the first two terms are $0$. However, when I differentiate over $v$, I get something which is not $0$. So, am I on the right path? If not, any idea what can be done? 
 A: when you put the Maxwell equation in the vlasov equation, you calculate the averages and that is how the terms
$\left\langle \frac{\partial f}{\partial t}\right\rangle =0 $ since the distribution is not dependent on time and 
$\left\langle v.\nabla f\right\rangle =0$ because distribution is uniform on an average.
similarly if you differentiate the third term you will get the term 
$\left\langle E.v\right\rangle$ which will equate to zero since on the average velocity in the distribution do not change 
I think this will help
EDIT:
Regarding your comment that exponent also contain the electrostatic potential $\phi$. I would like to add that the exponential term containing the potential will look like
$n=n(0)\exp\left(\frac{e\phi}{kT}\right)$.
This term is independent of velocity hence the velocity derivative will vanish. Also if the system is in equilibrium the total number of charge particles will be constant which leads to 
$$\left\langle\frac{\partial n}{\partial t}+v.\frac{\partial n}{\partial x}\right\rangle=\left\langle\frac{\partial n}{\partial t}+\frac{\partial n.v}{\partial x}\right\rangle=0$$
which is just the conservation of charge i.e. number of particles changing within volume $dv$ will be equal to the current flowing through the enclosed surfaces.
A: A property of the Vlasov equation is that any distribution that is only a function of constants of motion is its solution. So if the velocity of the case you present is not a function of time, the distribution would trivially be a solution of Vlasov.
A: df/dt is 0 for stationary condition.
but that distribution function is for molecules. Without any charge or potential. However, in your third term you have dependence on intensity and charge, therefore, I suppose, your MB equation should have term for some potential in exponent. Electric or electrostatic. In case of electrostatic, then df/dt = 0.
maxwell-boltzmann is distribution function, v doesn't depend on the position. it just says what is the probability at given temperature that your particle will have certain speed.
