X-Git-Url: https://hackdaworld.org/gitweb/?p=lectures%2Flatex.git;a=blobdiff_plain;f=solid_state_physics%2Ftutorial%2F2_04s.tex;fp=solid_state_physics%2Ftutorial%2F2_04s.tex;h=7391ad03ebcc7da137b4e0309d53929564a3d67b;hp=11e16f1fd48fe7af94d000aa7d570ded2e8cd23e;hb=75000006ad7eda5e22f1ca8fb47b619eb92c3cac;hpb=d4433169938510c3c6e8898649c0586cb6c48013 diff --git a/solid_state_physics/tutorial/2_04s.tex b/solid_state_physics/tutorial/2_04s.tex index 11e16f1..7391ad0 100644 --- a/solid_state_physics/tutorial/2_04s.tex +++ b/solid_state_physics/tutorial/2_04s.tex @@ -152,26 +152,81 @@ \frac{1}{L_3}\frac{\partial L_3}{\partial T}=3\alpha_L\nonumber \end{eqnarray} \item \[ - dF=-pdV-SdT \Rightarrow p=-\left.\frac{\partial}{\partial V}\right|T - \] - \[ - dE=TdS-pdV \Rightarrow - \] - Find an expression for the pressure as a function of the free energy - $F=E-TS$. - Rewrite this equation to express the pressure entirely in terms of - the internal energy $E$. - Evaluate the pressure by using the harmonic form of the internal energy. - {\bf Hint:} - Step 2 introduced an integral over the temperature $T'$. - Change the integration variable $T'$ to $x=\hbar\omega_s({\bf k})/T'$. - Use integration by parts with respect to $x$. - \item The normal mode frequencies of a rigorously harmonic crystal - are unaffected by a change in volume. - What does this imply for the pressure - (Which variables does the pressure depend on)? - Draw conclusions for the coefficient of thermal expansion. - \item Find an expression for $C_p-C_V$ in terms of temperature $T$, + dF=-pdV-SdT \Rightarrow p=-\left.\frac{\partial F}{\partial V}\right|T + \] + \[ + \left.\frac{\partial E}{\partial T}\right|_V= + \left.\frac{\partial E}{\partial S}\right|_V + \left.\frac{\partial S}{\partial T}\right|_V= + T\left.\frac{\partial S}{\partial T}\right|_V + \Rightarrow + \left.\frac{\partial S}{\partial T}\right|_V= + \frac{1}{T}\left.\frac{\partial E}{\partial T}\right|_V + \] + \[ + \textrm{Using } F=E-TS \textrm{ and } + TS=T\int_0^T\frac{\partial S}{\partial T'}dT' + \textrm{ (Entropy density vanishes at $T=0$)} + \] + \[ + \Rightarrow + p=-\frac{\partial}{\partial V}\left( + E-T\int_0^T\frac{dT'}{T'}\frac{\partial E}{\partial T'} + \right) + \] + Harmonic approximation of the internal energy: + \[ + E=E^{\text{eq}}+\frac{1}{2}\sum_{{\bf k}s}\hbar\omega_s({\bf k})+ + \sum_{{\bf k}s} + \frac{\hbar\omega_s({\bf k})}{e^{\beta\hbar\omega_s({\bf k})}-1} + \] + \[ + \ldots + \] + \[ + x=\hbar\omega_s({\bf k})/T' + \] + \[ + \ldots + \] + \[ + \Rightarrow + p=-\frac{\partial}{\partial V}\left( + E^{\text{eq}}+\frac{1}{2}\sum_{{\bf k}s}\hbar\omega_s({\bf k}) + \right)+ + \sum_{{\bf k}s}\left(-\frac{\partial}{\partial V}\hbar\omega_s({\bf k}) + \right)\frac{1}{e^{\beta\hbar\omega_s({\bf k})}-1} + \] + \item The pressure depends on temperature + only if the normal mode frequencies depend on the volume. + However, the normal mode frequencies of a rigorously harmonic crystal + are unaffected by a change in volume.\\ + $\Rightarrow$ + The pressure solely depends on the volume.\\ + $\Rightarrow$ + The pressure required to maintain a given volume + does not vary with temperature. + \[ + \left.\frac{\partial p}{\partial T}\right|_V=0 + \] + \[ + \left.\frac{\partial V}{\partial T}\right|_p= + -\frac{\left.\frac{\partial p}{\partial T}\right|_V} + {\left.\frac{\partial p}{\partial V}\right|_T}=0 + \] + \item $\frac{1}{B}=-\frac{1}{V}\left.\frac{\partial V}{\partial p}\right|_T$ + and $\alpha_V=\frac{1}{V}\left.\frac{\partial V}{\partial T}\right|_p$ + \[ + C_p-C_V=\left.\frac{\partial E}{\partial T}\right|_p- + \left.\frac{\partial E}{\partial T}\right|_V= + \frac{\partial E}{\partial S} + \left.\frac{\partial S}{\partial T}\right|_p- + \frac{\partial E}{\partial S} + \left.\frac{\partial S}{\partial T}\right|_V= + T\left.\frac{\partial S}{\partial T}\right|_p- + T\left.\frac{\partial S}{\partial T}\right|_V + \] + Find an expression for $C_p-C_V$ in terms of temperature $T$, volume $V$, the coefficient of thermal expansion $\alpha_V$ and the inverse bulk modulus (isothermal compressibility) $\frac{1}{B}=-\frac{1}{V}\left.\frac{\partial V}{\partial p}\right|_T$.\\