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Why is the magnetic field of a superconductor normally excluded? 
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#1
Sep2312, 11:08 AM

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Why is the magnetic field of a superconductor normally excluded?



#2
Sep2312, 11:37 AM

P: 3,014

Because of the screening effect due to the dissipationless supercurrent flowing in the superconductor.



#3
Sep2312, 02:01 PM

P: 3,014




#4
Sep2312, 03:04 PM

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Why is the magnetic field of a superconductor normally excluded?
Naturally, the actual physics of how and why is a bit more complex and involves quantum mechanics. If you understand at least the Shroedinger's Equation, I can elaborate a bit more. 


#5
Sep2312, 03:18 PM

P: 3,014

It drops to zero in a very thin region on the surface of the SC, called London magnetic penetration depth. But, this expulsion of external magnetic field can only persist up to a point  critical magnetic field H_{c}, after which the SC goes into a normal metal state. This behavior is characteristic for so called type I SCs. For type II SCs, among which are all the high Tc SCs, the behavior is different. After the value of the external field exceeds a value, H_{c1}, there is the possibility of some of the magnetic field lines to penetrate the bulk of the superconductor in vortices which have a quantized magnitude of the magnetic flux passing through them. This is a mixed state, and persists up to a higher critical field H_{c2}, above which the material becomes a normal metal. 


#6
Sep2312, 03:46 PM

P: 52

Does the cooper pairs spins (up and down) cancelling out have anything to do with anything?



#7
Sep2312, 03:46 PM

P: 52

cheers for replying as well



#8
Sep2512, 12:48 AM

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#9
Sep2512, 04:01 AM

P: 673

There are typeI and typeII superconductors, depending on the ratio of the London penetration depth and the coherence length.
The magnetic field inside typeI superconductors is exactly zero. In typeII SC, the magnetic field can penetrate the SC, but it gets bundled into quantized "flux lines" http://en.wikipedia.org/wiki/TypeII_superconductor All superconducting magnets are built from typeII superconductors. 


#10
Sep2512, 04:31 AM

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P: 3,630

On a more abstract level, the expulsion of the magnetic field from a superconductor is due to the breaking of the U(1) gauge symmetry. Due to the AndersonHiggs mechanism, the electromagnetic field becomes massive and therefore cannot enter far into the superconductor.
The Anderson Higgs mechanism was just recently confirmed in elementary particle physics with the detection of the Higgs boson. In a superconductor, the Cooper pairs correspond to the Higgs boson. 


#11
Sep2512, 12:27 PM

P: 3,014

In the microscopic BCS theory, if one calculates the response of the current density J, to the vector potential A, one obtains: [tex] \mathbf{J} = Q \, \mathbf{A} [/tex] This constitutive relation, coupled with Ampere's Law: [tex] \nabla \times \mathbf{H} = \frac{4 \pi}{c} \, \mathbf{J} [/tex] and the expression for the magnetic field in terms of the vector potential [itex]\mathbf{H} = \nabla \times \mathbf{A}[/itex] leads to a Proca type equation. One can immediately see the breaking of gauge invariance, because a current is expressed in terms of a gaugedependent quantitiy, namely the vector potential. 


#12
Sep2512, 01:53 PM

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#13
Sep2512, 02:41 PM

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#14
Sep2612, 03:20 AM

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In post #2 and #3 you ascribe the Meissner effect to the screening effect of dissipationless supercurrent of Cooper pairs. However, the classical BCS explanation of the Meissner effect involves only single particle excitations, the reduced BCS hamiltonian does not even have collective excitations which could be described as a flow of Cooper pairs. 


#15
Sep2612, 08:26 AM

P: 3,014

It is true that the BCS Hamiltonian is a onebody Hamiltonian which may be diagonalized in terms of the so called Bogoliubov fermionic excitations (which have a gap Δ in their spectrum). However, it is precisely this gap that it is the effect of the Cooper pairs. That is why we have the anomalous vertex [itex]\Delta \, c^{\dagger}_{\mathbf{k}\uparrow} \, c^{\dagger}_{\mathbf{k}\downarrow} + h.c.[/itex]. Obviously, this term does not conserve the number of electrons. The explanation is that you may get two or destroy two electrons by breaking up a Cooper pair or binding them up into one from the "condensate". 


#16
Sep2612, 10:10 AM

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Ok, you are completely right. Where I got confused is here:
The Cooper pairs have zero momentum flow but [itex] 0=p=mveA[/itex] hence [itex] j=nev=ne^2/m A [/itex] so there is still a diamagnetic current. The point I wanted to make is the following: The appearance of the gap (which is responsible for the absence of the paramagnetic current) is not physical in the model of BCS, but is due to the artificial long range nature of the reduced interaction [itex]\sum_{kk'}V_{kk'}c_{k'\uparrow}^*c_{k'\downarrow}^*c_{k\downarrow}c_{k\uparrow}[/itex]. In a real superconductor there is a long range repulsive interaction between the electrons, namely the Coulomb interaction which gets collective excitations out of the gap. That's why many people didn't believe the explanation of the Meisner effect of BCS. 


#17
Sep2612, 11:13 AM

P: 3,014

[tex] V_{\mathbf{k},\mathbf{k}'} \neq \frac{4 \pi e^2}{\vert \mathbf{k}  \mathbf{k}' \vert^2} [/tex] which has a singularite for zero momentum transfer, but is something like: [tex] V_{\mathbf{k},\mathbf{k}'} \neq \frac{4 \pi e^2}{\vert \mathbf{k}  \mathbf{k}' \vert^2 + k^2_{TF}} [/tex] and saturates to a constant for small momentum transfers. Additionally, there an electronphonon interaction, which becomes attractive in the same region and overcompensates the screened Coulomb interaction. I think you mean that some people were puzzled what overcomes the Coulomb repulsion of the electrons to bind into a Cooper pair, but I find it pretty unbelievable that people suspected the BCS explanation of the Meissner effect. 


#18
Sep2612, 01:24 PM

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P: 3,630

No, the finite energy of the plasmons (irrespective of in the superconductor or normal metal) is due to the long range of the Coulomb forces. Screening does not work for this kind of collective modes.
The doubts on the validity of the BCS explanation of the Meissner effect spurred several important developments (I can give you references tomorrow). For example Nambu wrote in his Nobel lecture ( http://www.nobelprize.org/nobel_priz...ulecture.html ) "I will now recall the chain of events which led me to the idea of SSB and its application to particle physics. One day in 1956 R. Schrieffer gave us a seminar on what would come to be called the BCS theory [5] of supercon ductivity. I was impressed by the boldness of their ansatz for the state vector, but at the same time I became worried about the fact that it did not appear to respect gauge invariance. Soon thereafter Bogoliubov [6] and Valatin [7] independently introduced the concept of quasiparticles as fermionic excita tions in the BCS medium. The quasiparticles did not carry a definite charge as they were a superposition of electron and hole, with their proportion depending on the momentum. How can one then trust the BCS theory for discussing the electromagnetic properties like the Meissner effect? It actually took two years for me to resolve the problem to my satisfaction. There were a number of people who also addressed the same problem, but I wanted to understand it in my own way. Essentially it is the presence of a massless col lective mode, now known by the generic name of NambuGoldstone (NG) LPN 2008 ekvationer... boson, that saves charge conservation or gauge invariance." 


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