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https://meta-phys-thoughts.blogspot.com/2018/04/the-error-in-larmor-formula-for.html
In the blog post on April 5, 2018 we pointed out that Gauss's law for the electric field does not hold if the field is not static. That is, the electric field flux through a closed surface may differ from the total electric charge which is inside the surface.
We posted an example where a charge is held inside the bottle close to the neck, and quickly pulled inside the neck. The information of the movement of the charge spreads at the speed of light. If the main body of the bottle is large, its surface will not know too soon that the charge has moved. The electric flux through the main body will stay as it is. But the narrow neck will soon learn that the charge is now inside the neck, and the electric flux through the neck will increase quickly => the electric flux out of the bottle will increase for a moment.
Lines of Force of a Point Charge near a Schwarzschild Black Hole
Richard Squier Hanni and Remo Ruffini
Phys. Rev. D 8, 3259 – Published 15 November 1973
https://journals.aps.org/prd/abstract/10.1103/PhysRevD.8.3259
The authors seem to assume Gauss's law, even though the situation close to the horizon is not very "static". The charge, which is held at a fixed distance from the horizon, is in a rapid acceleration relative to inertial observers close to it.
Any ray of light which falls into the horizon crosses the horizon orthogonally. This means that the horizon is an equipotential surface for the electric field if the field lines follow paths of rays of light.
The authors assume that the total electric flux is zero through any sphere enclosing the horizon but not the point charge. The horizon in their analysis is like a conducting metal sphere, where the point charge induces an opposite charge at the area closest to the point charge. The rest of the sphere then has a charge of the same sign as the point charge. When the point charge is very close to the metal sphere, then the electric field is almost symmetric, emanating from the horizon.
But why should the total electric flux throught the horizon be zero?
Imagine that the geometry of spacetime is that the black hole is connected to a white hole. If an electric field line passes into the black hole, it will probably come out of the white hole. There is no reason why Gauss's law would hold for a surface which just encloses the black hole, but does not enclose the white hole.
If a black hole can devour electric field lines and not return them back out from the horizon, then we get an extra strong "no hair" theorem: a black hole loses all electric charge which falls into it. The field lines will decay as quickly as the radiation emitted by an object falling into the horizon. Soon the electric charge visible to outsiders is very close to 0.
Why the black hole does not devour the gravitational field, too? The gravitational field is really the spacetime manifold and its metric. We can extend the manifold to a future spacetime point x just by using the information in the incoming light cone of x. For the spherically symmetric case, Birkhoff's theorem guarantees that the metric is the static Schwarzschild metric outside the collapsing mass.
Cohen and Wald forgot to transform the potential to the local frame?
UPDATE May 23, 2019: it looks like Cohen and Wald have absorbed the potential transformation formula into the other coefficients of their equation. Their calculations are correct.
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https:/aip.scitation.org/doi/abs/10.1063/1.1665812
Cohen and Wald in equation (4) have denoted the electric potential A_0 by v.
The electric potential is the electric potential energy of a 1 coulomb test charge.
Is the potential defined relative to an observer at infinity? That would be logical. There is not much use for a "local" potential.
It is logical that equation (4) is the view of the local static observer. But then the chance in the global potential energy dv should be given in local energy. We should multiply dv by 1 / sqrt(1 - 2m/r) to get the view of the local observer. If we send 1 joule of energy from the infinity to the local observer, he sees the energy inflated by a blueshift.
If we have an observer on the surface of a neutron star, and he moves 1 coulomb over 1 volt potential in his own frame, he gets 1 joule of energy. But when he transmits the energy to the infinity, the energy is sapped by a redshift. The global potential difference is less than 1 volt.
Hanni and Ruffini in their paper omit the energy conversion just like Cohen and Wald.
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