Why you thing that the electric field lines (electric force) will be
compressed?
If electric field lines are homogeneous space (seam) bonded to the
heterogenous seam of the 'particle',
{Or each the 'boundary condition of the other}
then homogeneous space, unlike
heterogeneous space, permits "two, or more, points to be in the same
place at the same time
(xref: Einstein - Botzmann vs. Fermi - Dirac statistics).
Therefore when [exterior sound] the
heterogeneous seam (boundary condition, fermion, 1/2 spin,
only one point in anyone place at one time) is
'displaced', it compresses the homogeneous space against itself. Since
homogenous space permits two points to be in the same place at the same
time, to what degree (xref: elasticity), or since homogeneous space
can be actively homogenous by definition seeking for all points in that
direction, or seam, or space, to actually be in the same place at the
same time, or how close to that ideal symmetry, THEN you would expect a
compression of the 'field' upon motion of the 'charge'.
I hadn't thought to be able to express it that way until your
question. Thanks.
But E-M charge, unlike gravity, is 'reverse' homogeneous, or heterogeneous, to itself, except 'opposite' charges are actively homogeneous to one another while remaining passively homogeneous to all other E-M neutral points . So besides the 'local' infinity of Euclid's zero dimensional point, is there a 'distant' infinity where [^] all the horizons meet? {But they are not parallel}. Is that the 'other side' of the local, zero dimensional, infinity? Is that what
E-M charge 'converges upon' regarding 'like' charge?
Rigid vs. elastic
Elastic in which direction on which side of the 'curve'? Deviation
from perfect symmetry expressed as a 'curve'.