Phase transitions in a superconductor

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SUMMARY

The discussion focuses on the phase transition of superconductors, specifically how the transition temperature decreases with increasing magnetic field, H. The magnetic moment, M, is defined as $$M=-\frac{HV}{4\pi}$$ for superconducting states (H < H_C(T)) and zero for normal states (H > H_C(T)). The participants derive the equality of heat capacities, C_H and C_M, and explore the Gibbs free energy relationship, ultimately seeking clarification on the derivation of $$dG=-SdT-MdH$$ and the implications of the negative sign in the context of thermodynamic identities.

PREREQUISITES
  • Understanding of superconductivity and phase transitions
  • Familiarity with thermodynamic concepts such as Gibbs free energy and heat capacity
  • Knowledge of magnetic properties in materials, specifically magnetic moment
  • Ability to perform calculus operations in thermodynamic equations
NEXT STEPS
  • Study the derivation of Gibbs free energy in phase transitions
  • Learn about the thermodynamic identities related to heat capacities in superconductors
  • Investigate the implications of magnetic fields on superconducting states
  • Explore advanced topics in superconductivity, including critical field theory
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Physicists, materials scientists, and students studying superconductivity and thermodynamics, particularly those interested in the mathematical modeling of phase transitions in superconductors.

Silviu
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Homework Statement


The phase transition to a superconducting state moves to lower temperatures as the applied magnetic field, H, increases. The magnetic moment, M, for a system of volume V is given by: $$M=-\frac{HV}{4\pi}$$
for ##H<H_C(T)## (superconducting) and $$M=0$$ ##H>H_C(T)## (normal). Changes in the internal energy, U, of such a system can be represented by the expression: $$dU=TdS+HdM$$ a) Show that for this system the constant field and constant magnetization heat capacities are equal: ##C_H=C_M## b) The phase transition between superconducting and normal phases takes place along a path ##H_C(T)##. Find an expression for the slope of the critical field, ##dH_C(T)/dT##, in terms of ##H_C(T)##, ##V## and the entropies of the two phases at the transitions.

Homework Equations

The Attempt at a Solution


a) So they give a solution, with a pretty long derivation and I think I am missing something. The way I was planning to do it was like this: $$C_H=\frac{dU}{dT}|_H=\frac{dU}{dT}|_{-4\pi M/V}$$ Assuming the volume is constant, which I assume is the case by "a system of volume V" this is equivalent to $$C_H=\frac{dU}{dT}|_M=C_M$$ I assume something is wrong, otherwise they wouldn't have a much longer derivation, but I am not sure what is wrong. b) I was trying to use the Gibbs free energy, making it equal for normal and superconducting at the boundary between the 2. So I have $$G=U+pV-TS$$ and going to the differential form I get $$dG=dU+pdV+Vdp-TdS-SdT = TdS+HdM-TdS-SdT=HdM-SdT$$ where I assumed that the pressure and volume are constant. However in their solution they get $$dG=-SdT-MdH$$ I assume that $$HdM=MdH$$ by the formula that connects them, but I am not sure how do they get a minus sign there. Can someone help me here?
 
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Silviu said:

Homework Statement


The phase transition to a superconducting state moves to lower temperatures as the applied magnetic field, H, increases. The magnetic moment, M, for a system of volume V is given by: $$M=-\frac{HV}{4\pi}$$
for ##H<H_C(T)## (superconducting) and $$M=0$$ ##H>H_C(T)## (normal). Changes in the internal energy, U, of such a system can be represented by the expression: $$dU=TdS+HdM$$ a) Show that for this system the constant field and constant magnetization heat capacities are equal: ##C_H=C_M## b) The phase transition between superconducting and normal phases takes place along a path ##H_C(T)##. Find an expression for the slope of the critical field, ##dH_C(T)/dT##, in terms of ##H_C(T)##, ##V## and the entropies of the two phases at the transitions.

Homework Equations

The Attempt at a Solution


a) So they give a solution, with a pretty long derivation and I think I am missing something. The way I was planning to do it was like this: $$C_H=\frac{dU}{dT}|_H=\frac{dU}{dT}|_{-4\pi M/V}$$ Assuming the volume is constant, which I assume is the case by "a system of volume V" this is equivalent to $$C_H=\frac{dU}{dT}|_M=C_M$$ I assume something is wrong, otherwise they wouldn't have a much longer derivation, but I am not sure what is wrong. b) I was trying to use the Gibbs free energy, making it equal for normal and superconducting at the boundary between the 2. So I have $$G=U+pV-TS$$ and going to the differential form I get $$dG=dU+pdV+Vdp-TdS-SdT = TdS+HdM-TdS-SdT=HdM-SdT$$ where I assumed that the pressure and volume are constant. However in their solution they get $$dG=-SdT-MdH$$ I assume that $$HdM=MdH$$ by the formula that connects them, but I am not sure how do they get a minus sign there. Can someone help me here?
How did you get $$HdM=MdH?$$ Using an integration by parts and assuming vanishing surface term, we get $$HdM=-MdH$$
 
nrqed said:
How did you get $$HdM=MdH?$$ Using an integration by parts and assuming vanishing surface term, we get $$HdM=-MdH$$
Oh, I just assumed $$HdM=(\frac{-4\pi M}{V})d(-\frac{HV}{4 \pi})=MdH$$ assuming constant ##V##. Is this wrong? (I mean I know it is wrong, but I am not sure why)
 
Silviu said:
Oh, I just assumed $$HdM=(\frac{-4\pi M}{V})d(-\frac{HV}{4 \pi})=MdH$$ assuming constant ##V##. Is this wrong? (I mean I know it is wrong, but I am not sure why)
Ah, I see your point. I had not looked carefully at the given info. I Have to think about it more then. Would it be long to show the solution the steps they follow to get their expression for dG?
 
nrqed said:
Ah, I see your point. I had not looked carefully at the given info. I Have to think about it more then. Would it be long to show the solution the steps they follow to get their expression for dG?
The statement is here (last problem) and the solution is here
 
nrqed said:
Ah, I see your point. I had not looked carefully at the given info. I Have to think about it more then. Would it be long to show the solution the steps they follow to get their expression for dG?
Any idea? At least for part a)?
 

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