Re: Simplier is more difficult

[David Bowman <dbowman@tiger.gtc.georgetown.ky.us>]

Ludwik Kowalski wrote:
> ....
> 1) Consider a point charge at rest. Its simple field E has no effect on
>   a magnetic probe (a small bar magnet or a current loop) situated nearby.
>   Move it with some speed v and the probe responds. We say B is a new 
>   field due to motion and we define it mathematically in terms of F, q 
>   and v.

So far so good.

> 2) But motion is relative. I can place myself in a frame of reference in
>   which the value of v is zero. In that frame B=0. This is true for any
>   v, even when v<<c. From my frame of reference the undeniable effect 
>   response of the probe must be described in terms of E only. I take it 
>   for granted (?) that this is always possible when a charge q moves 
>   along a short segment of its trajectory.

This is not necessarily true.  B=0 in a frame in which the particle is at
rest *only* if that particle is not undergoing any acceleration and has not
ever undergone any acceleration in the past.  Even if the particle is
currently not accelerating then there is still a region beyond a distance
equal to the light travel distance from the particle's current location
since the time when particle stopped accelerating.  Beyond this distance 
B /= 0.  If the particle is still currently accelerating then all magnetic
field effects are not zero even in close to the particle -- even in the
noninertial coordinate system in which the particle is always at rest.

> 3) An electric current in a wire (a collective motion of electrons) sets
>   a macroscopic magnetic field B around it. That B can be calculated at 
>   any point by integration of contributions made by each wire element.
>
>                         B=Integral(dB)
>
>   where dB is given by the Biot Savart formula.

This is only for filamentary currents which are in a time independent
*steady state* in a *complete circuit*.  Otherwise the BS law is incorrect.
Also the the dB integrand above is *not* rigorously correct for the
contribution of each individual current element -- even in the steady state
circuit case.  When considering the contributions of each tiny current
element there are transient and other time dependent effects as well as
relativistic corrections that depend on motional history of that current
element as well as its current state of acceleration (such as when it is
curving around a part of a solenoid).  In the special case of a global
steady state flow of a circuital current then the integral above is correct
for the *total* B field from the whole circuit, since in this case the
individual higher order (in 1/c^2) complications of the individual current
elements miraculously cancel out leaving a correct answer by just doing the
above naive BS law integral.

> 4) But dB is a magnetic field created by a moving charge in a short wire
>   segment dL. If each dB is a relativistic representation of an electric 
>   field then the same must be true for B. Thus a magnetic field of any
>   electric circuit can be viewed as a superposition of many relativistic
>   manifestations of electric fields.

Again you forgot about acceleration effects and other effects of velocity
of order of at least (v/c)^2.  Even if the extraneous complications did not
exist, it is still true that once a current configuration is set up which
does not correspond to a static charge configuration in at least *some*
Lorentz frame, then it *not* possible to transform the B field away by a
coordinate frame transformation, anyway.

> This kind of thinking applies to any set of electric circuits, including 
> a large solenoid imagined by David to produce a uniform magnetic field in 
> a finite region of space.
>                                             Ludwik Kowalski

Not quite.

I suggested the static uniform B field case originally because it was the
simplest case that I could think of.  For such a field there is no frame 
for which B --> 0.

If you insist on considering the sources of a given B field (i.e. the
currents in the solenoid) then I would rather have you consider a simpler
case for the sources.  Consider instead a steady state current flowing
through a uniform straight (thin) superconducting wire of infinite length.
In this case the current flow (at the macroscopic level) involves no
acceleration and the BS law for individual current elements has no
acceleration-dependent complications.  There are still order (v/c)^2
complications which cancel out when the full B field is calculated by
integrating over the global sum of the current elements.  In this case the
B field at any point is tangential to a circle surrounding the wire, whose
plane is perpendicular to the wire axis and whose center coincides with the
wire's central axis.  At each point on such a circle the B field has the
same magnitude.  The B field strength is inversely proportional to the
circle's radius (circumference) in the region outside the wire.  Consider
an attempt to make a Lorentz transformation boost along the wire axis to a
frame for which the electron current is zero.  In such a frame the wire's
positive metal ion cores move backward and a current is still present and
a B field is still present.  It is trivial (for this simple highly
symmetric problem) to transform the original B field configuration to any
such boosted frame for any boost velocity (along the wire) desired.  The
result is that in the boosted frame the B' field's directional pattern
remains unaffected.  Only the magnitude of the B field is affected giving:
B' = B/sqrt(1 - (v/c)^2)) where v is the boost velocity of the Lorentz
transformation.  The transformation results in a nonzero E' field, however,
which is radial outward (or inward depending on the relative signs of the
boost v and the current I) from wire (and locally perpendicular to the
local B' field) and has the value E' = v x B'.  For no such Lorentz frame
K' is B' ever zero.  In fact in all other such frames the B' field strength
actually *grows* with increasing v (although if one keeps v equal down
near typical electron drift current velocities the relativistic sqrt()
factors are so close to 1 that the B field is effectively invariant
under such a low speed Galilean-like boost).

BTW, if we went back to the original case of a pure static uniform B field
and transformed it to another inertial frame K' we would also observe B'
increase in magnitude with increasing v magnitude.  We would also see a
uniform E' field turn on and increase with the magnitude of v.  The only
exception to this case is if the direction of the boost velocity v happened
to be parallel (or antiparallel) to the original B direction.  In this
special case B' = B and E' = E = 0 for all speeds v.

David Bowman
dbowman@gtc.georgetown.ky.us