Phase transitions in a superconductor

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Homework Help Overview

The discussion revolves around the phase transition to a superconducting state and its relationship with applied magnetic fields. The original poster presents a problem involving the magnetic moment of a system and seeks to demonstrate the equality of heat capacities under constant field and magnetization conditions, as well as to find an expression for the slope of the critical field in relation to temperature and entropy.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants explore the derivation of heat capacities and question the assumptions made regarding constant volume and the relationships between magnetic moment and field. They also discuss the application of Gibbs free energy in the context of the phase transition.

Discussion Status

There is active engagement with the problem, as participants share their reasoning and seek clarification on specific steps in the derivation. Some participants express uncertainty about their assumptions and the implications of the relationships they are exploring, indicating a productive dialogue without a clear consensus yet.

Contextual Notes

Participants are navigating complex relationships in thermodynamics and magnetism, with some expressing confusion over the derivation steps and the signs in equations. The original problem context includes specific constraints related to the system's volume and the nature of the phase transition.

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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