For the Boltzman equation, why is df/dt=0 when collisionless?

In summary, the general equation for the evolution of a system's density function includes terms for external force, diffusion, and collisions. However, if the external force and diffusion are the same for all particles, the equation can be simplified to only include the first two terms.
  • #1
TimeRip496
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From wikipedia,
The general equation is $$\frac{df}{dt} = (\frac{∂f}{∂t})_{force}+(\frac{∂f}{∂t})_{diff}+(\frac{∂f}{∂t})_{coll}$$
where the "force" term external force, the "diff" term represents the diffusion of particles, and "coll" is the collision term.

So shouldn't be df/dt=0 when it is not just collisionless but also force and diffusion are zero too?

Wikipedia also mention that since collisions do occur, the particle density in the phase-space volume d3rd3p changes. So will df/dt be non-zero when d3rd3p changes?
In that case for df/dt=0 without having a zero external force and diffusion, the external force and diffusion must be the same for all the particles in order to mantain the same density, am I right to say this?
 
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  • #2
Yes, df/dt can be zero without having a zero external force and diffusion if the external force and diffusion are the same for all particles in order to maintain the same density. In this case, the collisions will not change the density of the system, and the equation of motion can be written as:df/dt = (∂f/∂t)_force + (∂f/∂t)_diff + 0where the "force" term represents the external force, the "diff" term represents the diffusion of particles, and the 0 term represents the absence of collisions.
 

Related to For the Boltzman equation, why is df/dt=0 when collisionless?

1. Why does df/dt equal 0 in the Boltzman equation when there are no collisions?

When there are no collisions, it means that the particles in the system are not interacting with each other. This is because in the Boltzman equation, f represents the distribution function of particles, which describes the probability of finding a particle with a certain energy at a given time. When there are no collisions, the particles do not exchange energy or momentum, and therefore the distribution function does not change over time. This results in df/dt, which represents the rate of change of the distribution function, equaling 0.

2. How does the Boltzman equation account for collisions?

The Boltzman equation takes into account collisions through the collision term, which is represented by the symbol C[f]. This term describes the rate at which particles collide and exchange energy and momentum with each other. This allows the distribution function to change over time, accounting for the effects of collisions in the system.

3. What is the significance of df/dt=0 in the Boltzman equation?

The fact that df/dt=0 in the Boltzman equation when there are no collisions is significant because it means that the distribution function remains constant over time. This allows for the calculation of various physical quantities, such as the average energy or number of particles in the system, which would not be possible if the distribution function was constantly changing.

4. Can the Boltzman equation be applied to both classical and quantum systems?

Yes, the Boltzman equation can be applied to both classical and quantum systems. However, in classical systems, the distribution function is described by the positions and momenta of the particles, while in quantum systems, it is described by the wave function of the particles.

5. How does the Boltzman equation relate to the second law of thermodynamics?

The Boltzman equation is closely related to the second law of thermodynamics, which states that the entropy of a closed system will tend to increase over time. In the Boltzman equation, the distribution function is directly related to the entropy of the system, and the collision term causes the distribution function to evolve towards a state of maximum entropy. This shows that the Boltzman equation is a fundamental equation for understanding the behavior of systems in thermodynamic equilibrium.

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