Phase transitions in a superconductor

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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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  • #2
nrqed
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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?
How did you get $$HdM=MdH?$$ Using an integration by parts and assuming vanishing surface term, we get $$HdM=-MdH$$
 
  • #3
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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)
 
  • #4
nrqed
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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?
 
  • #5
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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
 
  • #6
624
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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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