A Question related to energy losses in superconductors

When we look at a superconductor, the conductivity is infinite. And when we apply a potential difference across the ends of the superconductor, the electrons have to start accelerating because there is no resistance. But once electrons start accelerating, they must produce electromagnetic waves which can lead to power losses. Then why is it that books mention that there are no power losses in a superconductor?

• I'm no expert, but I suspect that at least part of the explanation is the same as that for why electrons don't radiate away their orbital energy as they "orbit" in an atom. Quantum electrodynamics is a strange beast and on the scale of an electron, classical "electromagnetic waves" is not always a useful or correct way to think about what is going on. With any luck, someone who took more than an undergrad quantum mechanics course will come along and give a real answer. Feb 20, 2018 at 0:43
• Also, I believe that there are some tiny but noticeable energy losses in a superconductor. Superconducting magnet coils in particle accelerators have to be aggressively cooled while in use to prevent them warming themselves above their superconducting temperature. Again, I don't remember the details, but someone surely does. Feb 20, 2018 at 0:48

There are losses in the situation you describe, as electromagnetic radiation is radiated away. These are not resistive losses, however. That is to say, no energy is lost to heat.

Once you remove the potential difference, however, the current persists. It is in this situation that there are truly no losses. The current remains without any applied potential, and does not decay over time.

• So is it a current without potential difference? Looks like Maxwell's equations would not work there, right? Sep 30, 2021 at 3:23
• @Xfce4 Maxwell's equations don't require there to be a potential difference for there to be a current. Thee very simplest problem in magnetostatics is an isolated current, which has a magnetic field but no electric field and thus no potential difference.
– Chris
Sep 30, 2021 at 6:59
• Doesn't a contradiction occur between Faraday's Law and Ampere's Law in the case of $0V$ ($0\vec{E}$) and $I \neq{0A}$ ? Sep 30, 2021 at 7:31
• @Xfce4 No. Why would you think it does? In the steady state, Faraday's law gives $\nabla\times\vec{E}=0$ (since the magnetic field is time-invariant) and Ampere's Law gives $\nabla\times\vec{B}=\mu_0\vec{J}$. Meanwhile the other two give $\nabla\cdot\vec{E}=0$ and $\nabla\cdot\vec{B}=0$.
– Chris
Sep 30, 2021 at 7:37
• Ok. I rushed into it. I can delete the comments if you wish. Sep 30, 2021 at 16:27

Superconductivity comes from a condensation of Cooper pairs into a super-fluid state. Cooper pairs are composed of two electrons which are bonded by some attractive forces (the origin of the force depends of the materials, but in classical superconductors, the force is electron-phonon interactions). Cooper pairs are bosons (at large scale) even though they are composed of two fermions. That is why they can "condense" in a single quantum state. You can see this like a Bose-Einstein condensation that happens at low temperature and where all particles behave the same and are strongly correlated.

This single quantum state is the ground state of a many-body system with interactions between particles, the particles here are at their minimal energy value (they cannot go lower). Thus, the particles cannot radiate energy even though they are accelerated or decelerated. The analogy with a super-fluid is that the electrons acts like they are a liquid with absolute zero viscosity (this is why there is no resistivity at that point). Technically, once the electrons in a superconductor starts flowing, they can flow like this for eternity even without any applied voltage (experiments were done to test this idea and could only result in a lower boundary for the current lifetime).