Another look at Thomson's argument. (i) Starting from...

Another look at Thomson's argument. (i) Starting from equations (28.15)-(28.18), obtain $$ \frac{\mathrm{d} S}{\mathrm{~d} t}=f_U \frac{\mathrm{d} U_1}{\mathrm{~d} t}+f_N \frac{\mathrm{d} N_1}{\mathrm{~d} t} $$ and hence $$ \frac{\mathrm{d} S}{\mathrm{~d} t}=L_{U U} f_U^2+L_{\mathrm{NN}} f_N^2+\left(L_{U N}+L_{\mathrm{NU}}\right) f_U f_N $$ (making no assumption about the symmetry of $L_{i j}$ ). (ii) From (28.18), obtain $f_N=\left(\dot{N}_1-L_{N U} f_U\right) / L_{N N}$ and hence $$ \begin{aligned} \frac{\mathrm{d} S}{\mathrm{~d} t}= & \left(L_{U U}-\frac{L_{U N} L_{N U}}{L_{N N}}\right) f_U^2 \\ & +\left(\frac{L_{U N}-L_{N U}}{L_{N N}}\right) f_U \dot{N}_1+\frac{\dot{N}_1^2}{L_{N N}} \end{aligned} $$ Applying this to thermoelectricity, we may note that when no current flows, $\dot{N}_1=0$ and in this case only the first term on the right-hand side survives. It follows that this term must describe entropy production associated with ordinary thermal conduction owing to a temperature gradient (the temperature gradient is expressed by $\left.f_U\right)$. Similarly, when $f_U=0$ only the third term survives, so this must be the term describing Ohmic or resistive heating. To derive the second Thomson relation using equation (28.9), Thomson assumed that these two processes account for all the entropy production-that is, the reversible processes do not themselves generate entropy, and this is still true even in the presence of the irreversible processes. We can now see that Thomson's assumption is valid if and only if $L_{U N}=L_{N U}$.

Another look at Thomson's argument.
(i) Starting from equations (28.15)-(28.18), obtain
$$
\frac{\mathrm{d} S}{\mathrm{~d} t}=f_U \frac{\mathrm{d} U_1}{\mathrm{~d} t}+f_N \frac{\mathrm{d} N_1}{\mathrm{~d} t}
$$ and hence
$$
\frac{\mathrm{d} S}{\mathrm{~d} t}=L_{U U} f_U^2+L_{\mathrm{NN}} f_N^2+\left(L_{U N}+L_{\mathrm{NU}}\right) f_U f_N
$$
(making no assumption about the symmetry of $L_{i j}$ ).
(ii) From (28.18), obtain $f_N=\left(\dot{N}_1-L_{N U} f_U\right) / L_{N N}$ and hence
$$
\begin{aligned}
\frac{\mathrm{d} S}{\mathrm{~d} t}= & \left(L_{U U}-\frac{L_{U N} L_{N U}}{L_{N N}}\right) f_U^2 \\
& +\left(\frac{L_{U N}-L_{N U}}{L_{N N}}\right) f_U \dot{N}_1+\frac{\dot{N}_1^2}{L_{N N}}
\end{aligned}
$$
Applying this to thermoelectricity, we may note that when no current flows, $\dot{N}_1=0$ and in this case only the first term on the right-hand side survives. It follows that this term must describe entropy production associated with ordinary thermal conduction owing to a temperature gradient (the temperature gradient is expressed by $\left.f_U\right)$. Similarly, when $f_U=0$ only the third term survives, so this must be the term describing Ohmic or resistive heating. To derive the second Thomson relation using equation (28.9), Thomson assumed that these two processes account for all the entropy production-that is, the reversible processes do not themselves generate entropy, and this is still true even in the presence of the irreversible processes. We can now see that Thomson's assumption is valid if and only if $L_{U N}=L_{N U}$.

Key Concepts

Entropy Production

Entropy production in nonequilibrium thermodynamics quantifies the irreversible generation of entropy due to processes such as thermal conduction and electrical conduction. It is analyzed by expressing the rate of change of entropy in terms of generalized fluxes and forces, highlighting how different transport processes contribute to the overall increase in entropy in a system undergoing irreversible processes.

Thermodynamic Forces and Fluxes

Thermodynamic forces, often gradients of thermodynamic potentials like temperature or chemical potential, drive the corresponding fluxes such as heat or particle currents. In linear irreversible thermodynamics, the rates of these fluxes are assumed to be linearly related to their conjugate forces through phenomenological coefficients, creating a framework for analyzing diverse nonequilibrium phenomena.

Phenomenological Coefficients

These coefficients describe the linear relationship between thermodynamic fluxes and forces and capture how processes such as heat conduction and electrical conduction interrelate within a system. They are typically denoted by L_ij, and their values determine the extent to which one type of force can influence a different flux, such as a temperature gradient affecting particle current in thermoelectric systems.

Onsager Reciprocal Relations

The Onsager reciprocal relations assert that the cross-coefficients in the linear response relations (L_ij) are equal to their transposed counterparts (L_ji) under conditions of microscopic reversibility. This symmetry is fundamental in ensuring consistency between the different irreversible processes, and it is critical for validating theoretical assumptions such as those made by Thomson in simplifying the entropy production analysis.

Thomson Relations in Thermoelectricity

Thomson's argument in the context of thermoelectricity links the contributions to entropy production from thermal conduction and electrical conduction. By imposing the condition L_UN = L_NU, Thomson ensures that reversible processes do not contribute to net entropy production, thereby cleanly separating the irreversible sources. This relation is central to understanding how thermal and electrical phenomena are coupled in a thermoelectric system.

Transcript

00:01 Hi everyone this is the problem based on change in entropy of the object.

00:05 Here it is given the antrophy increase in the first object where temperature increase from t1 to t2 is given by the change in antrophy in the first object if temperature increases from t1 to t, m1 c1 lawn of t upon t one change in antrophic in second object if temperature decreases from t2 to t -dus delta s2 will be m2c2 lon of t -dus upon t two so total entropy change delta skel two delta skel two plus delta s 2 m1 c1 lown of t upon t 1 plus m 2 c2 lon of t d d d d 2 if the heat given by the second object is equal to heat received by the first object according to conservation of energy that is q1 is called to minus q2 m1 c1 t minus t 1 that is m1 c1 t minus 2 m2 c2 t 2 minus tess now b part we have m1 c1 t minus t d d d d d s t2 minus t d -d -s.

04:16 So t2 minus t -dus we will get m1c1 t -1 -t -1 upon m2 -z2 therefore total efficiency change, delta as equal to m1c1 of t upon t1 plus m2c2 loan of t -dus upon t2 m1c1lon of t -t -2 m1c1 lon of t2 t upon t -1 sorry plus m2c2 lon of t 2 minus m1 c1 upon m2 c2 upon t 2 so delta s equal to m1 c1 m1 c1 plus m2 c2 non of 1 minus m1 c1 m2 t 2 t 2 minus t 1 for maximum for maximum delta s d upon d t of delta s must be 0 differentiating the above equation with respect to t you will get m1 c1 upon t plus m2c2 into m1 c1 divided by m2 c2 1 1 minus 1 upon t 2 divided by 1 minus m1 c1 into t upon t must be 0 so on solving it you will get t is called to t 1 c1 c 4 if t is less than t -des, then it is possible to approach each other.

07:36 In this process, the total entropy of the system will keep on increasing.

08:13 The final stage of the system will be that for which no further entropy changes possible...

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