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## Main Question or Discussion Point

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

So you start with a simple sum of forces:

[tex] \Sigma F_x = - e \: E + F_{collision} = 0 [/tex] (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 [tex] F_{collision} [/tex] is defined. The text i'm reading states

[tex] \bar F_{collision} = \frac{\Delta p_x}{\Delta t} \approx \frac{-mv}{\tau} [/tex] (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:

[tex]F=\frac{dp}{dt}=\frac{\Delta p}{\Delta t} =\frac{p}{t} =\frac{mv}{t} [/tex]

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

Sure you can do that for say position:

[tex] \frac{\Delta x}{\Delta t} = \frac{x}{t} [/tex]

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.

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

Thanks guys.

So you start with a simple sum of forces:

[tex] \Sigma F_x = - e \: E + F_{collision} = 0 [/tex] (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 [tex] F_{collision} [/tex] is defined. The text i'm reading states

[tex] \bar F_{collision} = \frac{\Delta p_x}{\Delta t} \approx \frac{-mv}{\tau} [/tex] (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:

[tex]F=\frac{dp}{dt}=\frac{\Delta p}{\Delta t} =\frac{p}{t} =\frac{mv}{t} [/tex]

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

Sure you can do that for say position:

[tex] \frac{\Delta x}{\Delta t} = \frac{x}{t} [/tex]

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.

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

Thanks guys.

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