13.5 SOLID STATE PHYSICS - ASHCROFT/MERMIN 5. Consider a metal in which thermal and electric currents flow simultancousl

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13 5 Solid State Physics Ashcroft Mermin 5 Consider A Metal In Which Thermal And Electric Currents Flow Simultancousl 1 (186.14 KiB) Viewed 20 times

13.5 SOLID STATE PHYSICS - ASHCROFT/MERMIN 5. Consider a metal in which thermal and electric currents flow simultancously. The rate at which heat is generated in a unit volume is related to the local energy and number densities by (cf. (13.39)): dq du du (13.80) dt dt dt where pt is the local chemical potential Using the equation of continuity, dn -Vj", (13.81) dt and the fact that the rate of change of the local energy density is determined by the rate at which electrons carry energy into the volume plus the rate at which the electric field does work, du dt V.j+ E.j. (13.82) show that (13.80) can be written in the form dK da Vj" + Ej, di (13.83) where j" is the thermal current (given by (13.38) and (13.40)), and & = E + (1/2)V 1. Assuming cubic symmetry, so that the tensors L' are diagonal, show that under conditions of uniform current flow (V.j = 0) and uniform temperature gradient (VPT = 0) that dq pj? + (VT) (13.8-4) dt dT dT where p is the resistivity, K is the thermal conductivity, and Q is the thermopower. By measuring the change in bulk heating as the current direction is reversed for fixed temperature gradient (known as the Thomson effect) one can therefore determine the temperature derivative of the thermopower, and thereby compute the value of Q at high temperatures, given its low-temperature value. Compare the numerical coefficient of VT-j with that of the crude estimate in Problem 3, Chapter 1 (VT)? - 7dQ Chapter 1 - Problem 3: 3. Thomson Effect Suppose that in addition to the applied electric field in Problem 2 there is also a uniform tem- perature gradient V Tin the metal. Since an electron emerges from a collision at an energy deter- mined by the local temperature, the energy lost in collisions will depend on how far down the temperature gradient the electron travels between collisions, as well as on how much energy it has gained from the electric field. Consequently the power lost will contain a term proportional to E. VT(which is easily isolated from other terms since it is the only term in the second-order energy loss that changes sign when the sign of E is reversed). Show that this contribution is given in the Drude model by a term of order (net/m) (dɛ/dT) (E-VT), where & is the mean thermal energy per electron. (Calculate the energy lost by a typical electron colliding at r, which made its last collision at r - d. Assuming a fixed (that is, energy-independent) relaxation time t, d can be found to linear order in the field and temperature gradient by simple kinematic arguments, which is enough to give the energy loss to second order.)