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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?

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  • $\begingroup$ I'm not sure if this covers your excercise, but it will depend on your v(t): if v is e.g. a constant of motion, dv/dt = 0, your equation is (trivially) an equilibrium solution (but in contrast to Boltzmann equation, where the Maxwell distribution is the only solution, this is not the only one). In general (for random v(t)), the Maxwellian distribution is not a solution. $\endgroup$ – Solarflare Jun 14 '16 at 11:15
  • $\begingroup$ No, $\partial_{t} f$ is not generally equal to zero for a collisionless gas governed by the Vlasov equation. In steady-state, then yes this is zero but it is not a general solution. $\endgroup$ – honeste_vivere Jun 14 '16 at 14:53
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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.

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    $\begingroup$ This is not generally true. You can solve the Vlasov equation numerically without performing an ensemble average in the manner you show. $\endgroup$ – honeste_vivere Jun 14 '16 at 18:06
  • $\begingroup$ @honeste_vivere That is an addition to my little knowledge. Thanks for your comment, I will try to read more on the topic. As far as I know, the distribution is generally defined for a large number of particles and shows the probability of finding any particle in a particular state. Also the equilibrium means that on average the distribution do not change but velocity of individual particles can change. $\endgroup$ – hsinghal Jun 14 '16 at 18:12
  • $\begingroup$ Thermodynamics is an approximation to kinetic theory and/or statistical mechanics, where non-equilibrium gases can exist. It is often the case that we do perform ensemble averages for various reasons, but that is not something that is general or fundamental. $\endgroup$ – honeste_vivere Jun 14 '16 at 18:41
  • $\begingroup$ I think I get it now, also I can now imagine that for any realistic numerical solution of Vlasov equations how huge computing power would be required. $\endgroup$ – hsinghal Jun 14 '16 at 18:44
  • $\begingroup$ Thank you! Yes, the exponent also contains electrotatic potential $\phi$. $\endgroup$ – Martin Nikovski Jun 14 '16 at 20:58
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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.

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  • $\begingroup$ Alright, thank you. However, I struggle with the proof of that property. $\endgroup$ – Martin Nikovski Jun 14 '16 at 12:54
  • $\begingroup$ I don't think v is constant, but the energy mv^2/2 is... $\endgroup$ – Martin Nikovski Jun 14 '16 at 13:29
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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.

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