Highly collisional yet low resistivity plasmas

[TheCanadian]

In approximations for the applicability of ideal MHD to plasmas, it states that plasmas are considered highly collisional to permit the assumption that the plasma (i.e. electrons) follow a Maxwellian velocity distribution. Although is not resistivity based on collisions in the plasma? If there is high collisionality, how can this result in collisions that somehow have low resistivity?

 

[sophiecentaur]

I would expect the resistivity to depend on the number of free electrons as much as anything. Frequent collisions would ensure that there were plenty of parallel paths for the current can take. But would the collisions necessarily transfer much net energy to thermal once equilibrium was established?

 

[TheCanadian]

I would expect the resistivity to depend on the number of free electrons as much as anything. Frequent collisions would ensure that there were plenty of parallel paths for the current can take. But would the collisions necessarily transfer much net energy to thermal once equilibrium was established?

The number density of free electrons is certainly important in determining the rate of collisions. When considering momentum loss in plasmas, the rate is given in the first attachment. The actual magnitude for the energy loss would be calculated by multiplying the loss rate by the initial kinetic energy. Now if considering any collision in general (i.e. momentum loss rate), the ratio between the energy loss rate and momentum loss rate is given by the second attachment for thermal plasmas. If we're considering electron-electron collision to dominate (i.e. $m_1 = m_2 = m_e$), then the two rates are equivalent. Thus in ideal MHD, why is it permissible to consider these plasmas to be highly collisional yet of low resistively if energy exchange does occur between individual particles during collisions?

 

[sophiecentaur]

The actual magnitude for the energy loss would be calculated by multiplying the loss rate by the initial kinetic energy.

In a dense plasma, the KE, acquired from the overall field would be tiny? (I'm trying to look for reasons why the loss would be much less than you suggest.)

 

[TheCanadian]

In a dense plasma, the KE, acquired from the overall field would be tiny? (I'm trying to look for reasons why the loss would be much less than you suggest.)

Well ideal MHD is said to describe solar winds or neutron star magnetospheres reasonably well, yet to my knowledge these can have energies around 1-10 keV (on the order of a million Kelvin) and quite different densities.

 

[sophiecentaur]

I have no idea about such extreme conditions but what would be the Resistivity that you refer to above? Wouldn't it all be a pretty High Impedance phenomenon? Out perhaps the interactions wouldn't be losing much energy from the electrons - so not actual 'collisions'?

 

[TheCanadian]

I have no idea about such extreme conditions but what would be the Resistivity that you refer to above? Wouldn't it all be a pretty High Impedance phenomenon? Out perhaps the interactions wouldn't be losing much energy from the electrons - so not actual 'collisions'?

For the impedance to be negligible in MHD, we're considering the case: $\eta \bf\vec{J} \ll \bf\vec{E} + \bf\vec{v} \times \bf\vec{B}$, where $\eta$ is resistivity. Well that's what I'm trying to understand (and try to demonstrate in the above attachments): if collisions do take place, why would the collisions result in negligible energy loss?

 

[sophiecentaur]

For the impedance to be negligible in MHD, we're considering the case: $\eta \bf\vec{J} \ll \bf\vec{E} + \bf\vec{v} \times \bf\vec{B}$, where $\eta$ is resistivity. Well that's what I'm trying to understand (and try to demonstrate in the above attachments): if collisions do take place, why would the collisions result in negligible energy loss?

If the mean free path is low and the field is low then how much KE can be gained before a collision. If you can find those two values then you can find the velocity gained and hence the KE lost per interaction. Your equation is macroscopic whilst you need the microscopic model to give you the number you want.