Joule Heating and the Thomson effect in Drude theory

[amjad-sh]

Homework Statement

$\mathbf A$.Joules heating
Consider a metal in a uniform temperature in a static uniform electric field E.An electron experiences a collision, and then after a time t, a second collision. In the drude model , Energy is not conserved for collisions, for the mean speed of an electron emerging from collision does not depend on the energy that the electron acquired from the field since the time of the preceding collision.(a)Show that the average energy lost to the ions in the second of two collisions separated by a time t is [(eEt)^2]/2m.(The average is over all directions in which the electron emerged from the first collision ).(b) show from the result of problem 1(b)[result is :show that the probability that the time interval between two successive collisions of an electron falls in a range between t and t+dt is $(dt/\tau) e^{-t/\tau}$] that the average energy loss to the ions per electron per collision is (eE$\tau$)^2/2m,and hence the average loss per cubic centimeter per second is (ne^2E$\tau$/m)E^2=$\sigma E^2$($\sigma$ is the conductivity). Deduce that the power loss in a wire of length L and cross section A is (I^2)R where R is the resistance of the wire and I is the current.$\mathbf B$.Thomson effect
Suppose in addition to the applied electric field in A there is a uniform temperature gradient ∇T in the metal. Since an electron emerges from a collision at an energy determined by the local temperature, the energy lost in collision s will depend on how far down the temperature gradient the electron travels between collisions, as well as how much energy it has gained from the electric field.Consequently the power lost will contain a term proportional to E.∇T( Which is easily isolated from the other terms which is the only term in the second order energy loss which changes sign when the sign of E is reversed).

Show that this contribution is given in the drude model by a term of order (ne$\tau$/m)dξ/dT(E.∇T) where ξ is the mean thermal energy per electron.( Calculate the energy lost by a typical electron colliding at $\mathbf r$, which made its last collision at $\mathbf r-\mathbf d$assuming a fixed (that is energy-independent) relaxation time $\tau$. $\mathbf d$ can be found to linear order in the field and temperature gradient by a simple kinematic arguments, which is enough to give the energy loss to second order .

2. Relevant equation
$\mathbf j=-ne\mathbf v$ where $\mathbf j$ is the current density
$\mathbf E=ρ\mathbf j$ where ρ is the resistivity.

The Attempt at a Solution

I actually solved whole of part A except of (b)( its last part)
if anybody can give me hints to solve it.
also if somebody can give me a hint to calculate the contribution in B.

 

[mark726]

For part (b) of question A, we can use the result from problem 1(b) to find the average energy loss per electron per collision. We know that the probability of the time interval between two successive collisions of an electron falling in a range between t and t+dt is given by (dt/tau)e^(-t/tau). So, the average energy loss per electron per collision can be calculated by multiplying this probability with the energy lost in a single collision, which we found in part (a). This gives us:

Average energy loss per electron per collision = [(eEt)^2/2m](dt/tau)e^(-t/tau)

To find the average loss per cubic centimeter per second, we need to multiply this by the number of collisions per second, which is given by n/t, where n is the number of electrons per unit volume and t is the average time between collisions. So, the average loss per cubic centimeter per second is:

Average loss per cubic centimeter per second = [(eEt)^2/2m](dt/tau)e^(-t/tau)(n/t)

We can also write this in terms of conductivity, where n is replaced by ne^2tau/m. This gives us:

Average loss per cubic centimeter per second = [(ne^2Etau/m)E^2](dt/tau)e^(-t/tau)(1/t)

Simplifying this, we get:

Average loss per cubic centimeter per second = (ne^2E^2tau/m)e^(-t/tau)

As we can see, this is the same as the conductivity (sigma) multiplied by the electric field squared. So, the average loss per cubic centimeter per second is:

Average loss per cubic centimeter per second = sigmaE^2

To deduce the power loss in a wire of length L and cross section A, we need to multiply this by the current density, which is given by j = nev. So, the power loss is:

Power loss = nev(sigmaE^2)

Since the current density is given by j = I/A, where I is the current and A is the cross section, we can write this as:

Power loss = (I/A)nev(sigmaE^2)

Using Ohm's law (E = rhoj), we can write this as:

Power loss = (I^2)(rho/A)

Finally, using