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Re: [Phys-l] current vector



Bernard Cleyet wrote:

"I presume you mean discharge not by a wire connecting the concentric
spheres, but by a leaky dielectric. In which case you have "crystal
clarified" the discussion; thank you".


Yes, originally I meant discharge by a leaky dielectric. The irony
is that even though I think the example was most relevant and correct,
emphasizing the spherical symmetry may be misleading into thinking that
the absence of such symmetry would make the current somewhat vectorial.
That would mean the emergence of something that so horrified John Denker
in one of his previous messages, - something that may have any gradations
between a pure scalar and a pure vector. That would horrify me, too, so
now I want to strengthen my original example by stating that ANY CURRENT
as it is defined in the textbooks (I = dQ/dt), is not a vector. Therefore,
even when we consider a current in a wire, with a well defined direction
of the collective motion of charges along it, this is NOT a vector.
This apparently nonsensical statement can be easily proven. Current is
a surface integral of the current density j (which is a true vector) over
the corresponding surface. Each area element da of the surface is a vector
(in 3-D space) whose direction is represented by the corresponding local
normal. So current is the surface integral of the dot product j.da. A dot
product of two vectors is a scalar. Even when both vectors are parallel,
the moment we performed operation of the dot product, we loose all
information about their directions. This does appear to contradict the
experiment - after all, we see the charges flowing in a certain
direction. This is the case when relying on our senses mesleads us not
because the senses are wrong, but because the additional info they give
us is no longer relevant to the newly formed characteristic. We impose
onto current the additional info about current density.
This is typically human error, - a computer would have avoided it.

Moses Fayngold,
NJIT