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-1

Even in Classical Mechanics the vector potential is defined in the whole space, i.e., it is a field, because it should be known wherever the charge in question goes. The charge may go everywhere we direct it with our initial conditions. In QM, where there is a wave function equation for a wave determined everywhere in space, the vector potential should also ...


-1

I agree with both your answers, but let me add one more answer, quite direct. One of the most fundamental operators needed in the quantum mechanics is the Hamiltonian, on which we base the Schrodinger equation. It is associated to the energy of the system, kinetic + potential. For the interaction between the e.m. field and an electron (neglecting the ...


0

No relation. The second law of thermodynamics can be proven on general statistical grounds. It is independent of what particles are in our universe or what equations describe their motion. That is why physicists are very confident about the second law of thermodynamics, even though nobody knows for sure what particles are in our universe or what equations ...


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Unquestionably the answer is yes. One would only have to look at the largest magnet in the solar system (the sun) to confirm this. If you have any doubt please go to the following site and learn http://wso.stanford.edu/Polar.html


3

The electric forces that we see in nature are due to separate charges (electric monopole charge), however, all matter ever isolated to date, including every atom on the periodic table and every particle in the standard model, has zero magnetic monopole charge. Therefore, the ordinary phenomena of magnetism and magnets have nothing to do with magnetic ...


4

You seek a 1-form $A$ on $\mathbb{R} - \{0\}$ such that $\mathrm{d}A = B$. On all of $\mathbb{R} - \{0\}$, $\mathrm{d}B = 0$, so this could exist. But, since you have magnetic flux, you require that the integral of $B$ over any 2-sphere around the origin should be $g$. Therefore, by Stokes' theorem, $$ g = \int_{S^2}B = \int_{S^2} \mathrm{d}A = ...


3

The magnetic flux through any closed surface enclosing the origin is just $g$ (the magnetic charge enclosed). If the magnetic field comes from a vector potential $\vec{B}=\nabla\times\vec{A}$, this surface integral by Stokes' theorem is an integral of $\vec{A}$ around the boundary of the surface. But the surface is closed, so has no boundary, so the answer ...


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user23873 answered my question in the comments. I quote: " Try reading the book 'Geometry, Topology and Gauge Fields: Foundations', the author (Naber) have this discussion right on the introductory chapter and points how the impossibility to define a proper vector potential on ℝ3−0 is linked with it's topology (the second homotopy group is non-trivial), and ...


0

If the curl of $\vec A = \vec B$, then $\vec A$ is never uniquely defined because you can add the gradient of a scalar function to it. As $\nabla \times \nabla \phi=0$, it does not change the B-field. However, I guess this is not what you mean. If we look at the curl of the vector potential $$ \nabla \times \vec A(\vec r) = \vec B$$ $$ \nabla \cdot \vec B ...



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