Ohm's Law derivation from Drude Model

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I'm having trouble understanding the derivation of Ohm's law from the drude model.

So you start with a simple sum of forces:

\Sigma F_x = - e \: E + F_{collision} = 0 (my understanding is that there are only two forces in Drude's model: those from electron-ion collisions and applied external electric field forces)

Now I'm confused about how F_{collision} is defined. The text I'm reading states

\bar F_{collision} = \frac{\Delta p_x}{\Delta t} \approx \frac{-mv}{\tau} (there is a bar over the F, which I'm assuming means averaged)

where tau is the relaxation time

It seems like they did something like this:

F=\frac{dp}{dt}=\frac{\Delta p}{\Delta t} =\frac{p}{t} =\frac{mv}{t}

This is the part i don't understand. how does \Delta p = p ?

Sure you can do that for say position:

\frac{\Delta x}{\Delta t} = \frac{x}{t}

but you must assume that the initial position and time were at x=0 and t=0 respectively. For momentum, how can you assume that the initial momentum is 0?

The derivation then continues with simple algebraic rearrangement.

- e \: E + \frac{-mv}{\tau} = 0 \: \: \Rightarrow \: \: eE = \frac{-mv}{\tau} \: \: \Rightarrow \: \: v=\frac{-eE \tau}{m} , where v is the drift velocity

Thanks guys.
 
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The Drude model is not at all "exact" in any way; the whole idea is to model what happens "on average" for a single particle in the electron gas and use that to model properties of the bulk. Hence, the details of every single collision is irrelevant.

\bar F_{collision} = \Delta p_x/\Delta t \approx \frac{-mv}{\tau} is just a common way of modelling collisions in classical mechanics. In this case it is not meant to be an exact description of what is happening, it is merely a good approximation where \tau is just essentially a free parameter with the dimension of time that has to be determined using experiments (i.e. \tau does not neccesarily have anything to do with the time it takes for a single, real, collsion, and the latter would obviously need to be modeled using QM and can't be described using classical mechanics).

\Delta p_x/\Delta t is just a first order approximation to dp/dt which would be exact if the momentum was changing in a linear fashion during the collision (this type of model is often introduced by considering a rubber ball hitting the floor). The reason why this works is because \Delta p is NOT the total momentum but the CHANGE in momentum that takes place during the time \Delta t=\tau.
 
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The equality \Delta p = p is, IIRC, made assuming that during a collision, the electron stops, or that any given electron might come out of a collision event in a random direction, so that the average p after a collision is 0.
 
In reality, the scattering time is just a phenomenological fitting parameter. You know that the collision force is going to be some function of the velocity, and so you expand it in a Taylor series, F = a_0 + a_1*v + higher order terms. However, you know that F=0 when v=0, and so the zeroth-order term must vanish. If you then assume that the velocity is small, then you can neglect the higher order terms and just get a term that is linear in v.

And, as others have noted, the Drude model is nowhere near exact. I can tell you from personal experience that more detailed models for carrier transport and scattering in solids are extremely important for a lot of solid state devices.
 
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Manchot said:
And, as others have noted, the Drude model is nowhere near exact. I can tell you from personal experience that more detailed models for carrier transport and scattering in solids are extremely important for a lot of solid state devices.

Exactly, see this post...

https://www.physicsforums.com/showthread.php?t=111335

My advisor, a couple of his collaborators and I at one time may have been the experts in transport theory in crystalline systems, I believe I did the only work in exact transport in anisotropic (tetragonal) lattices.

Drude is a reasonable 1st order estimate in cubic materials, you have to get into solving the Boltzmann Equation for really good estimates.
 
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