I doubt that you ever have understood this.
citation (probably by somebody else who also studied under the same supervisor) you can teach me physics! You are arrogant and boring at the same time. There is nothing that you have posted or can post on this thread that I do not already know about; and already have thought about intensely. So stop making a fool of yourself!
I do not stop them: Nature stops them since, even while they transport a current, they remain at the lowest energy that they can be at: They have
no energy to dissipate. Therefore, if they are accelerated, their energy will become larger than their lowest energy, and therefore they can dissipate this energy so that superconduction is not possible. Thus charge-carriers that can be accelerated can never be responsible for superconduction. This is spelled out very well on my website; but of course you do not read what another person’s writes: You just know that he must be wrong if what he is writing is not in the standard literature. Can you see why physics is becoming a fundamentalist religion? This is caused by people like you.
First asked 7th September 2011 (5 days and counting)
And for that matter - What are your charge carriers?
They cannot be electrons since they will always scatter (see below).
The charge carriers within
any conductor are
NOT electrons. Outside the conductor there is an energy level called the
vacuum level. Electrons are only present at energies above this vacuum level. The charge-carriers within a metal are at an energy W (the work function) below the vacuum level and therefore they are not electrons. They are wave packets that form by superposition of extended waves when you switch on the electric-field to cause a current to flow. If you meant these wave-packets by using the term “electrons”, then for a change you are correct. The wave packets are free to be accelerated and to, thus, gain kinetic energy which will dissipate: They can thus not be responsible for superconduction.
Any charge-carriers that move owing to Newton’s second law can thus not be responsible for superconduction. The motion of the charge-carriers must thus be purely a quantum-mechanical effect. On this all scientists, except the BZS scientists on this thread, agree. Where I disagree with BCS is on the mechanism for this quantum-mechanical motion. BCS advocates that it is sufficient to have a condensate of bosons. But forming a condensate does not prevent the bosons from being accelerated by an electric field, to thus gain kinetic energy above the lowest energy that they are in before the current started to flow, and to thus dissipate this kinetic energy.
In my model, the charge carriers within all superconductors are localised electron-states which have the lowest energy that they can have. Since they are anchored, they cannot be accelerated by an electric-field: i.e. the material is in essence an insulator. When the density of these localised states becomes high enough, they can exchange places by means of quantum fluctuations. Since the energy during a quantum fluctuation is used (to move on) and it then disappears, the charge-carriers have the same energy when they reach the target contact than the energy they have started off from the injection contact. This means that there cannot be an energy difference in energy for the charge carriers at these two contacts and thus not a voltage difference.
If there are no states for the charge-carriers to be scattered into then they cannot be scattered.
Exactly! You might even one day understand some physics! If the charge-carriers can gain kinetic energy by being accelerated, then the acceleration creates states into which the charge-carriers can be scattered.
The fermion or boson nature of the charge-carriers is crucial for this.
Nope! No matter whether the charge-carriers are fermions or bosons, if you can accelerate them, you create states into which the can scatter.
Fermions tend to occupy every state up to the Fermi surface. This means that they can scatter off the lattice into states above the Fermi surface. In addition thermal excitation empties states below the Fermi surface. This allows electrons to scatter into these states. Scattering = resistivity as you agree.
This is true for fermion wave-packets that can be accelerated. It will also be true for bosons which can be accelerated to be at energies that are higher than the energy at which the find themselves. They will then dissipate this energy.
Furthermore, you can have an ensemble of fermions which all have the same lowest energy: For example, donor electrons anchored at their donor atoms. They cannot be accelerated and can therefore not be scattered into higher energy states. They can, however, move by means of quantum fluctuations when their density becomes high enough: superconduction then occurs. This has been confirmed by experiment for donors in silicon as well as acceptors in diamond.
Bosons like to share states.
So what!?
They tend to leave an energy gap below the Fermi surface.
According to the BCS model this is so but it is unlikely that you will find bosons forming within any fermion-spectrum: In fact the fermion spectrum in a metal is in essence a boson spectrum since each wave with three quantum numbers n
1, n
2 and n
3 is occupied by having opposite spins.
Thus bosons such as Cooper pairs cannot scatter because there is not enough energy to allow them into allowable states. No scattering = no resistivity as you should agree.
No scattering within a material does not equal no resistance; only a constant drift speed caused by
no acceleration and no scattering can be equal zero resistance.
Adding thermal energy will allow the electrons to access states into which they can scatter. That is why the classical (BSC) superconductors have a critical temperature.
Wrong! The critical temperature is determined by the density of the localised anchored states at which they can start to move owing to quantum fluctuations. And since these localised states (similar to donor states) form at an energy-gap (delta)E below the conducting states, they increase in density with a decrease in temperature. The full gap is there right from the start (as proved by recent inelastic neutron scattering from lead): The change in energy you measure with decreasing temperature is the energy of the Fermi-level within the gap. As the temperature decreases the Fermi=level moves down to end up at the energy of the localised states at T=0. To expect that the Fermi-level will stay at the centre of an electronic energy gap while the temperature is lowered has been disproved over and over again in Solid State Statistical Thermodynamics.
Then you are out of luck. The actual BSC theory does explain superconductivity.
If you insist to believe this paranormal superstitious claptrap: Be my guest.