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Problem Statement:
when a superconductor of the first order is under the influence of an external magnetic field, the field is repelled from the the supercondutin material until the field reaches a critical value called the critical field H_c(T). for H>H_c the material turns into regualr metal (normal state) and for H<H_C the material turns into a superconductor.
The coexsitence line separating between the two phases is a parabola (as a function of the temprature) given by:
H_c(T)=H_0+aT+bT^2
1.Get an analogous equation to clausius -clpareon equation for the coexistence line and show that 'a' must be 0, in the two phases the ground state is non degenrate, and gibbs energy is G=U-TS-HM where H is the magnetic field, U is the thermal energy, T temp, S entropy, M magentization, and dU=TdS+Hdm+\mudN.
2.from the codnition H_c(T_c)=0 express b with H_0 and T_c.
3. use the fact that B=0 in the superconductin state, and the connection B=H+4\pi m (this is a vectorial equation), to compute the latent heat per molecule in the transition from superconductor to normal state as function of T, you can assume that the magnetization at normal state is zero.
4. use the connection: c_H=T\frac{dS}{dT}_H to find the jump in heat capacity per molecule when crossing the coexistence line in a constant field.
My attempt at solution
for (1) I think the equation is something like this:
\frac{dH}{dT}=\frac{(\frac{d\mu}{dT}_H_{super}-\frac{d\mu}{dT}_H_{Normal})}{(\frac{d\mu}{dH}_T_{Normal}-\frac{d\mu}{dH}_T_{Super})}
which form what is given I think that dG=\mu dN-SdT-mdH
I guess the equation in question shoule look something like this:
\frac{dH}{dT}=\frac{S_{super}-S_{normal}}{m_{normal}-m_{super}}.
And now a=0 cause dH/dT=a+2bT and at T=0 the entropies are equal while the magnetizations aren't so we get that a=0.
for (2), ofcourse it should be: b=-H0/T_c^2.
for (3), here I'm using the fact L=T_c*(S_sup-S_nor)/N (N the number of particles) is our latent heat and H=4pi*m in the superconductor, so we get that:
dH/dT=-L/(T_c*m_{super})=2bT_c
and we know what is m_super, it equals H_c/4pi=H0+bT_c^2
after plugging the equations I get: L=-2bT^2_c(H_0+bT^2_c)/4\pi, is this even correct?
for (4), for c_H I'm just taking the derivative of L wrt T ( I'm repalcing T_c with some T in the coexistnece line), and after some rearrangements I get to what is \delta c_H.
So what do you think of this essay of mine, any good?
My solution is valid?
thanks in advance.
when a superconductor of the first order is under the influence of an external magnetic field, the field is repelled from the the supercondutin material until the field reaches a critical value called the critical field H_c(T). for H>H_c the material turns into regualr metal (normal state) and for H<H_C the material turns into a superconductor.
The coexsitence line separating between the two phases is a parabola (as a function of the temprature) given by:
H_c(T)=H_0+aT+bT^2
1.Get an analogous equation to clausius -clpareon equation for the coexistence line and show that 'a' must be 0, in the two phases the ground state is non degenrate, and gibbs energy is G=U-TS-HM where H is the magnetic field, U is the thermal energy, T temp, S entropy, M magentization, and dU=TdS+Hdm+\mudN.
2.from the codnition H_c(T_c)=0 express b with H_0 and T_c.
3. use the fact that B=0 in the superconductin state, and the connection B=H+4\pi m (this is a vectorial equation), to compute the latent heat per molecule in the transition from superconductor to normal state as function of T, you can assume that the magnetization at normal state is zero.
4. use the connection: c_H=T\frac{dS}{dT}_H to find the jump in heat capacity per molecule when crossing the coexistence line in a constant field.
My attempt at solution
for (1) I think the equation is something like this:
\frac{dH}{dT}=\frac{(\frac{d\mu}{dT}_H_{super}-\frac{d\mu}{dT}_H_{Normal})}{(\frac{d\mu}{dH}_T_{Normal}-\frac{d\mu}{dH}_T_{Super})}
which form what is given I think that dG=\mu dN-SdT-mdH
I guess the equation in question shoule look something like this:
\frac{dH}{dT}=\frac{S_{super}-S_{normal}}{m_{normal}-m_{super}}.
And now a=0 cause dH/dT=a+2bT and at T=0 the entropies are equal while the magnetizations aren't so we get that a=0.
for (2), ofcourse it should be: b=-H0/T_c^2.
for (3), here I'm using the fact L=T_c*(S_sup-S_nor)/N (N the number of particles) is our latent heat and H=4pi*m in the superconductor, so we get that:
dH/dT=-L/(T_c*m_{super})=2bT_c
and we know what is m_super, it equals H_c/4pi=H0+bT_c^2
after plugging the equations I get: L=-2bT^2_c(H_0+bT^2_c)/4\pi, is this even correct?
for (4), for c_H I'm just taking the derivative of L wrt T ( I'm repalcing T_c with some T in the coexistnece line), and after some rearrangements I get to what is \delta c_H.
So what do you think of this essay of mine, any good?
My solution is valid?
thanks in advance.