# Derivation of drift velocity treating electrons as doing work on the surroundings as a gas enclosed in a container [closed]

So,

the original derivation of drift velocity can be found in this link; Method of averaging velocities of electron when deriving drift velocity

Now I used(tried to use, to be more precise) an alternate method because this derivation expressed drift velocity as a function of mean relaxation time and not as a function of time

Now using thermodynamics and Kinetic theory of gases

$$P = 1/3nmv^2$$

Now, $$P_(new) = 1/3nmv_{rmsnew}^2$$

Since, it is rms, Mean of $$(u_1vector \pm{atvector})^2 + (u_2vector \pm{atvector})^2....$$

simplifies as $$v_{rms}^2 + a^2t^2 + 2v_{rms}.a.t.cos\theta$$

(I took $$v_{rms}.at$$ as zero because they aren't correlated even though I'm not sure about a lot about these steps

therefore $$P_{int}(t) = 1/3nm(v_{rms}^2 + a^2t^2)$$

Now, as the gas was initially at equilibrium therefore $$P_{gasintial} = P_{ext}$$ Therefore if we take the gas as enclosed in any container with a piston At time t, $$V = V_{0} + 1/36.A.a^2t^4$$

Therefore, drift Velocity as a function of Time = $$1/A(dV/dt) = (1/9)a^2t^3$$

I kind of imagined it as a column of a hypothetical gas where a piston is compressing the gas with a constant force and as it keeps compressing the gas simultaneously some of the particles are accelerating but also some particles also come to rest at end of the column so the particle number density and root mean square velocity remains the same and hence the pressure remains the same as the initial conditions $$(1/3nmv^2)$$

this piston is being pushed by another gas column whose initial pressure was $$1/3nmv^2$$ and so on till infinity.

All the other columns can be considered as one column being pushed by a piston from the negative terminal with a pressure that is changing with time unlike the first column in contact with the positive terminal which has constant pressure due to constant root mean square velocity. This column also should maintain the same number density because there is a influx of new particles as it is expanding

Now I'm highly skeptical about this since pressure should be equal in all hypothetical columns which it's not

Obviously, the biggest problem is that the units are coming out as m/s which is good but drift velocity is coming proportional to the square of magnitude of electric field which doesn't match with the more straightforward formula

Also, I'm not sure that we can apply this in a condition where electrons are coming to rest(at the ends of the wire) and accelerating at the same time and whether external Pressure can be taken as $$1/3nmv^2$$

• Welcome to Physics! Please use MathJaX to enter equations; as written, your question is hard to read. You can find a tutorial on how to use MathJaX here. Commented Aug 26, 2022 at 14:35
• I'll see and edit as much as is possible. Thnx:) Commented Aug 26, 2022 at 14:52
• @MichaelSeifert forgot to tag you, just in case you're willing to help Commented Aug 27, 2022 at 6:25