On May 10, 11:36 pm, Eric Gisse wrote:
On May 10, 9:29 pm, Dono wrote:



On May 10, 10:01 pm, Tom Roberts wrote:

Dono wrote:
How can I find the transformation that ties E,B,E',B'? Any reference
that you can suggest? Thank you.

Look up the Faraday two-form (its dual is called the Maxwell 2-form or
tensor). Its components, when projected onto an inertial frame, consist
of the 3-vector components of E and B:

[ 0 -Ex -Ey -Ez]
F = [ Ex 0 Bz -By]
[ Ey -Bz 0 Bx]
[ Ez By -Bx 0 ]

Thank you, I know all this, I even wrote the transform in its
vectorial form. I asked something different, how does all this FURTHER
transform when frames S and S' are ROTATING with a constant angular
speed?

Express x and y as a function of time then express Faraday's
components in that coordinate system.

Usehttp://mathworld.wolfram.com/RotationMatrix.htmlas a guide for
x(t) and y(t).



Thank you,
What I put down is the following :

a. In S

x=r*cos(theta)
y=r*sin(theta)


b. In S'

x'=r'*cos(theta')
y'=r'*sin(theta')


r'=gamma*(r-vt) (because S and S' move with relative speed v along
the common x-axis)
theta'=theta (because the axes of the two frames have aligned x
and y axis)
theta=w*t
theta'=w'*t'
t'=gamma(t-vx/c^2)

So:

x'=gamma (r-vt)*cos(theta')= gamma (r-vt)*cos(theta)=gamma (x-
vt*cos(theta))
y'=gamma (y-vt*sin(theta))

resulting into:

x'=gamma (x-vt*cos(wt))
y'=gamma (y-vt*sin(wt))
z'=z
t'=gamma(t-vx/c^2)

With the above, we can calculate:

dE/dt=dE/dx'*dx'/dt+dE/dy'*dy'/dt+dE/dz'*dz'/dt+dE/dt'*dt'/dt=

=dE/dx' gamma*(-v*cos(wt)+vwtsin(wt))+dE/dy'gamma*(-v*sin(wt)-
vwtcos(wt))+dE/dt' *gamma

and

dE/dx=dE/dx'*dx'/dx+dE/dy'*dy'/dx+dE/dz'*dz'/dx+dE/dt'*dt'/dx=dE/
dx'*gamma-dE/dt'*gamma*v/c^2

dE/dy=dE/dx'*dx'/dy+dE/dy'*dy'/dy+dE/dz'*dz'/dy+dE/dt'*dt'/dy=dE/
dy'*gamma

dE/dz=dE/dz


Does this look right?