A Useful Variation of the Lorentz Field Transformations G.R.Dixon, 11/17/03
In general, the Lorentz transformations answer the following question: Given the results of a mutual measurement, made using instruments at rest in inertial frames K and K’, how do the results of these measurements relate? The measurements might be something as simple as determining the "where" and "when" of a physical particle. Or they might be more involved, e.g. a determination of the electric and magnetic fields at two momentarily coincident test charges’ positions. It is usually assumed (ideally of course) that inertial frames K and K’ are laid out in rectangular coordinates, with synchronized clocks at rest at every point in each frame. The corresponding axes are parallel, and the origins coincide when the clocks there read t=t’=0. K’ moves in the positive x-direction of K at speed v. (And K moves in the negative x’-direction of K’ at speed v.) If K’ observes a particle to be at (x’,y’,z’) when the K’-clock there reads t’, then the transformation states that the K space-time coordinates are: , (1a) , (1b) , (1c) . (1d) Similarly, if K measures the electric and magnetic field
components to be (E , (2a) (2b) , (2c) , (2d) , (2e) . (2f) One of the limitations of the standard transformations is that measurements, made at different points in space and all at a given instant in one frame, map to measurements made at different times in another frame. For example, K might note which K’-clocks coincide with x Perhaps this limitation of the basic transformations is never more bothersome than in field theory. Typically some law (e.g. Coulomb’s law) provides knowledge of the electric and magnetic fields at all space-time points in K. The standard field transformations
(Eqs. 2a-f) then give the field components in K’ at many different moments. But K’ (like K) might want to know what the fields are in K’ all at a given moment t’. The question becomes: Given a knowledge of , (3a) (3b) (3c) , (3d) (3e)
. (3f) Eqs. 3a-f are simpler at the instant t’=0. And quite often matters are further simplified if the fields are static (independent of t) in frame K.
Example. A point charge q is permanently at rest at the origin of K. Thus at t’=0 it is at the origin of K’ and is moving at constant speed v in the negative x’-direction. At t’=0 let the field evaluation point be , (4a) , (4b) . (4c) What are the K’ electric field components? In K there is only an electrostatic field. Thus substituting in Eqs. 3a-f: , (5a) . (5b) For the special case q=0, . (6) Thus "behind" and "in front of" the moving charge the electric field is less than the resting case by a factor of
g For the special case q=p/2, . (7) Thus "to the side of" the moving charge the electric field is greater than the resting case by a factor of g. Since . (8) At a point "to the side" and in the x’y’-plane, . (9)
Although the transformations in Eqs. 3a-f are more cumbersome than the standard field transformations of
Eqs. 2a-f, they are readily programmed and evaluated by a computer program. And (at the risk of belaboring a point), given a knowledge of |