- #1
Plant_Boy
- 15
- 1
I'm reading a number of papers, journals, reports and what not trying to grasp at what the actual definition of [itex]\tau[/itex].
[itex] \tau = \frac {m_{e}}{\rho e^{2} n} [/itex]
Am I correct in thinking:
[itex] m_{e}[/itex] - mass of an electron
[itex] \rho [/itex] - resistivity
[itex] e [/itex] - charge of electron
[itex] n [/itex] - number of electrons per unit mass
In one I read it is Mean Free Time between Collisions of Electrons. Another states it is Relaxation Time. Are these two definitions dependent on whether you use a DC or AC energy source?
I understand from the equation as the mass in kilogram of an electron is divided by an ohm, kilogram meter squared per second[itex]{}^3[/itex] per ampere[itex]{}^2[/itex], times an ampere second times a number. The values cancel out to leave an ampere second[itex]{}^2[/itex] per meter[itex]{}^2[/itex].
Or,
[itex]\tau = (A s^2 m^{-2})[/itex]
So, one amp takes s[itex]{}^2[/itex] seconds to decay into an area?
[itex] \tau = \frac {m_{e}}{\rho e^{2} n} [/itex]
Am I correct in thinking:
[itex] m_{e}[/itex] - mass of an electron
[itex] \rho [/itex] - resistivity
[itex] e [/itex] - charge of electron
[itex] n [/itex] - number of electrons per unit mass
In one I read it is Mean Free Time between Collisions of Electrons. Another states it is Relaxation Time. Are these two definitions dependent on whether you use a DC or AC energy source?
I understand from the equation as the mass in kilogram of an electron is divided by an ohm, kilogram meter squared per second[itex]{}^3[/itex] per ampere[itex]{}^2[/itex], times an ampere second times a number. The values cancel out to leave an ampere second[itex]{}^2[/itex] per meter[itex]{}^2[/itex].
Or,
[itex]\tau = (A s^2 m^{-2})[/itex]
So, one amp takes s[itex]{}^2[/itex] seconds to decay into an area?