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jcap
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The following is an improved version of my previous post https://www.physicsforums.com/threa...contradicts-the-equivalence-principle.964594/
Consider the following system comprising a particle on the left with charge ##+q## that is a large distance ##d## away from two oppositely charged particles on the right, with charges ##+q##,##-q##, held apart by a spring of length ##L## and spring constant ##k##.
Let us assume that the left-hand particle is sufficiently far from the right-hand particles so that the "static" horizontal component of the electric field that it produces near those particles is negligible. (Actually this assumption is not necessary. We could have another charge ##+q## on the other side of the vertical dipole connected to the first by a horizontal rod. They could then be as close as we like as the horizontal electrostatic forces that they produce on the dipole will cancel out.)
To start with consider the system at rest in empty space. The only electric forces acting on the right-hand particles are "static" attractive forces that are balanced by the compressed spring.
Now let us assume that the whole system is falling in a gravitational field with acceleration ##g##. According to the equivalence principle this situation should be indistinguishable from the system in empty space.
But now as the left-hand charge ##+q## has an acceleration ##g## it should produce a "radiative" vertical component to the electric field in the vicinity of the right-hand particles.
Each right-hand particle is subjected to an extra vertical electromagnetic force whose magnitude is given by
$$F_{EM}=\frac{q^2}{4\pi\epsilon_0 d c^2}g.$$
As the forces on the differently-charged right-hand particles point in opposite directions, the spring is stretched by an amount given by
$$\Delta L=\frac{2F_{EM}}{k}.$$
Thus a local observer can tell that he is falling in a gravitational field which contradicts the equivalence principle.
What's gone wrong?
Consider the following system comprising a particle on the left with charge ##+q## that is a large distance ##d## away from two oppositely charged particles on the right, with charges ##+q##,##-q##, held apart by a spring of length ##L## and spring constant ##k##.
Let us assume that the left-hand particle is sufficiently far from the right-hand particles so that the "static" horizontal component of the electric field that it produces near those particles is negligible. (Actually this assumption is not necessary. We could have another charge ##+q## on the other side of the vertical dipole connected to the first by a horizontal rod. They could then be as close as we like as the horizontal electrostatic forces that they produce on the dipole will cancel out.)
To start with consider the system at rest in empty space. The only electric forces acting on the right-hand particles are "static" attractive forces that are balanced by the compressed spring.
Now let us assume that the whole system is falling in a gravitational field with acceleration ##g##. According to the equivalence principle this situation should be indistinguishable from the system in empty space.
But now as the left-hand charge ##+q## has an acceleration ##g## it should produce a "radiative" vertical component to the electric field in the vicinity of the right-hand particles.
Each right-hand particle is subjected to an extra vertical electromagnetic force whose magnitude is given by
$$F_{EM}=\frac{q^2}{4\pi\epsilon_0 d c^2}g.$$
As the forces on the differently-charged right-hand particles point in opposite directions, the spring is stretched by an amount given by
$$\Delta L=\frac{2F_{EM}}{k}.$$
Thus a local observer can tell that he is falling in a gravitational field which contradicts the equivalence principle.
What's gone wrong?
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