We know that E and B transform as follows for S and S' moving with
relative speed V :

E'=E
B'=B for the components of E,B along the direction parallel with V

and

E'=gamma(V) (E+VxB)
B'=gamma(V) (B=VxE/c^2) for the components of E,B along the direction
perpendicular to V

Now, imagine that the axis x,y are rotating with the uniform angular
speed omega in the plane x-y while S' moves with uniform speed V along
the common axis x.
How can I find the transformation that ties E,B,E',B'? Any reference
that you can suggest? Thank you.

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 ]

NOTE: I'm using the sign conventions of MTW, section 3.
This section discusses SR, not GR, but in a general
geometrical framework.
MTW = Misner, Thorne, and Wheeler, _Gravitation_

The two form components, of course, transforms like the components of a
contravariant rank-2 tensor:

F_i'j' = L_i'^i L_j'^j F_ij

Where L_i'^i is the (homogeneous) Lorentz transform between two inertial
frames. NOTE: i' is distinct from i, because I use the modern notation
where the primes go on the indices, not the tensor symbol (here F),
because the tensor itself is unaffected by a coordinate change, only its
components change.

The major thing to notice is that in general E and B get intermixed by a
Lorentz transform; they are NOT independent. That's why we call it the
electromagnetic field.

Exercise for the reader: why is the HOMOGENEOUS Lorentz
transform used, and not the full Poincaré group (aka the
inhomogeneous Lorentz group, which includes translations)?
Can one use the full homogeneous Lorentz group, or only
the component of it which includes the identity?


Tom Roberts

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?

On May 10, 9:01*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 ]

* * * * NOTE: I'm using the sign conventions of MTW, section 3.
* * * * This section discusses SR, not GR, but in a general
* * * * geometrical framework.
* * * * MTW = Misner, Thorne, and Wheeler, _Gravitation_

The two form components, of course, transforms like the components of a
contravariant rank-2 tensor:

* * * * F_i'j' = L_i'^i L_j'^j F_ij

Where L_i'^i is the (homogeneous) Lorentz transform between two inertial
frames. NOTE: i' is distinct from i, because I use the modern notation
where the primes go on the indices, not the tensor symbol (here F),
because the tensor itself is unaffected by a coordinate change, only its
components change.

The major thing to notice is that in general E and B get intermixed by a
Lorentz transform; they are NOT independent. That's why we call it the
electromagnetic field.

* * * * Exercise for the reader: why is the HOMOGENEOUS Lorentz
* * * * transform used, and not the full Poincaré group (aka the
* * * * inhomogeneous Lorentz group, which includes translations)?

a) Electromagnetism is Lorentz, not Poincare, invariant?

b) The inhomogeneous Lorentz group contains parity inversions and only
continuous operations are allowed?

c) Because the Faraday tensor does not transform as a tensor under the
Poincare group?

F_i'j' = (L_i'^i + a^i) (L_j'^j + a^j) F_ij
F_i'j' = (L_i'^i L_j^j + a^i a^j + L_i'^i a^j + a^i L_j'^j) F_ij



* * * * Can one use the full homogeneous Lorentz group, or only
* * * * the component of it which includes the identity?

Are you wearing your continuous transformation only hat?


Tom Roberts

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.

Use http://mathworld.wolfram.com/RotationMatrix.html as a guide for
x(t) and y(t).

On May 11, 12:04 am, Dono wrote:
We know that E and B transform as follows for S and S' moving with
relative speed V :

E'=E
B'=B for the components of E,B along the direction parallel with V

and

E'=gamma(V) (E+VxB)
B'=gamma(V) (B=VxE/c^2) for the components of E,B along the direction
perpendicular to V

Now, imagine that the axis x,y are rotating with the uniform angular
speed omega in the plane x-y while S' moves with uniform speed V along
the common axis x.
How can I find the transformation that ties E,B,E',B'? Any reference
that you can suggest? Thank you.

Still at it, Dono****o?
http://img2.freeimagehosting.net/uploads/d5814688f0.jpg

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?

On May 11, 12:04 am, Dono wrote:
We know that E and B transform as follows for S and S' moving with
relative speed V :

E'=E
B'=B for the components of E,B along the direction parallel with V

and

E'=gamma(V) (E+VxB)
B'=gamma(V) (B=VxE/c^2) for the components of E,B along the direction
perpendicular to V

Now, imagine that the axis x,y are rotating with the uniform angular
speed omega in the plane x-y while S' moves with uniform speed V along
the common axis x.
How can I find the transformation that ties E,B,E',B'? Any reference
that you can suggest? Thank you.

Dono****o, the problem you've set up is
physically meaningless. The frames S and S'
have been left void. Where is the source of that
electric/magnetic field located? Is it at rest in frame
S' or in S. Which is the rest frame for that source,
Dono****o? Is that source's rest frame rotating?
Or is it that you have a third different frame S''
implicitly involved, where the source of the field
is at rest?

On May 11, Albert****o :
http://www.secularsobriety.org/mdavey_drunk2.jpg

On May 11, 8:03 am, Albertito wrote:
On May 11, 12:04 am, Dono wrote:



We know that E and B transform as follows for S and S' moving with
relative speed V :

E'=E
B'=B for the components of E,B along the direction parallel with V

and

E'=gamma(V) (E+VxB)
B'=gamma(V) (B=VxE/c^2) for the components of E,B along the direction
perpendicular to V

Now, imagine that the axis x,y are rotating with the uniform angular
speed omega in the plane x-y while S' moves with uniform speed V along
the common axis x.
How can I find the transformation that ties E,B,E',B'? Any reference
that you can suggest? Thank you.

Dono****o, the problem you've set up is
physically meaningless. The frames S and S'
have been left void. Where is the source of that
electric/magnetic field located? Is it at rest in frame
S' or in S. Which is the rest frame for that source,
Dono****o? Is that source's rest frame rotating?
Or is it that you have a third different frame S''
implicitly involved, where the source of the field
is at rest?



Read the problem statement, cretin.