On May 11, 4:09 pm, Dono wrote:
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.

Dono****o, your problem is bull****. If you don't know
how the source of the field is moving in S and S', your
problem has no solution. You must tie the source in a
fixed point, and see how that point moves in S while
the xy plane rotates, and see how it moves in S'.

On May 11, 8:35 am, Albert****o wrote:
On May 11, 4:09 pm, Dono wrote:



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.

Dono****o, your problem is bull****. If you don't know
how the source of the field is moving in S and S', your
problem has no solution. You must tie the source in a
fixed point, and see how that point moves in S while
the xy plane rotates, and see how it moves in S'.



The source is fixed in S, retardo.
S' moves along the common x-axis wit speed v
S and S' rotate together with fixed angular speed omega.
You can't even read the problem statement.

On May 11, 4:38 pm, Dono wrote:
On May 11, 8:35 am, Albert****o wrote:



On May 11, 4:09 pm, Dono wrote:

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.

Dono****o, your problem is bull****. If you don't know
how the source of the field is moving in S and S', your
problem has no solution. You must tie the source in a
fixed point, and see how that point moves in S while
the xy plane rotates, and see how it moves in S'.

The source is fixed in S, retardo.
S' moves along the common x-axis wit speed v
S and S' rotate together with fixed angular speed omega.
You can't even read the problem statement.

Ok, the source is fixed in S.
S' moves along the common x-axis wit speed v.
S and S' rotate together with fixed angular speed
omega. With respect to what are S and S' rotating
together with fixed angular speed, retardo? If
S rotates with angular speed omega with respect to
S', then S' rotates with minus omega wrt S. Where is
the axis of rotation located? And, how does the source
moves? Does the axis of rotation intersect the source
or not?

On May 11, 8:54 am, Albertito wrote:


You are a pathological cretin. Incurable.

Eric Gisse wrote:
On May 10, 9:01 pm, Tom Roberts wrote:
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?

No. Maxwell's equations are invariant under a 15-parameter group called
the conformal group, which is larger than the Poincare' group. The other
5 parameters are changes of scale, so if one keeps the same units only
the 10-parameter Poincare' group applies.


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

No. Maxwell's equations are invariant under time reversal and parity
inversion.


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

No. It does. Indeed, it transforms as a tensor under any coordinate
change whatsoever.


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?

I have no such hat.


Hint: remember that when one expresses a tensor F in terms of its
components {F_ij}, the tensor itself is:
F = F_ij e^i e^j
where the {e^i} are the basis vectors of the coordinate system. Ask
yourself how a translation affects the basis vectors.

Hint2: remember that the components of a tensor {F_ij} transform:
F_i'j' = dx^i/dx^i' dx^j/dx^j' F_ij
and compute dx^i/dx^i' for a translation.


Tom Roberts

On May 11, 4:20 pm, Dono wrote:
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.htmlasa 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=
Does this look right?

You must be quite a dumb writing such of that.

Please correct your mistakes and your misunderstandings immediately

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?

F is a tensor, so its components transform:
F_i'j' = dx^i/dx^i' dx^j/dx^j' F_ij
Compute dx^i/dx^i' for your rotation and apply. Note, however, that
interpreting those components in a rotating frame is non-trivial; in
particular I don't think the usual identification of them as components
of E and B applies. Indeed, I don't know what "E" and "B" mean in a
non-inertial coordinate system, and the only way I know to figure it out
is to relate them to E and B in an inertial system (so you're back where
you started).

IOW: this question is more complicated than how they transform, it is
also a question of what they mean.


Tom Roberts

On May 11, 9:10 am, Tom Roberts wrote:
I don't know what "E" and "B" mean in a
non-inertial coordinate system, and the only way I know to figure it out
is to relate them to E and B in an inertial system (so you're back where
you started).

IOW: this question is more complicated than how they transform, it is
also a question of what they mean.

Tom Roberts



Thank you, Tom

The problem has a physical meaning, I am trying to relate the (E,B)
measured at a certain latitude on the Earth surface to the values
predicted at a different latitude (same longitude) as seen by an
observer that follows the common meridian moving at tangential speed
v.

On May 11, 5:59 pm, Tom Roberts wrote:
Eric Gisse wrote:
On May 10, 9:01 pm, Tom Roberts wrote:
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?

No. Maxwell's equations are invariant under a 15-parameter group called
the conformal group, which is larger than the Poincare' group. The other
5 parameters are changes of scale, so if one keeps the same units only
the 10-parameter Poincare' group applies.

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

No. Maxwell's equations are invariant under time reversal and parity
inversion.

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

No. It does. Indeed, it transforms as a tensor under any coordinate
change whatsoever.

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?

I have no such hat.

Hint: remember that when one expresses a tensor F in terms of its
components {F_ij}, the tensor itself is:
F = F_ij e^i e^j
where the {e^i} are the basis vectors of the coordinate system. Ask
yourself how a translation affects the basis vectors.

Hint2: remember that the components of a tensor {F_ij} transform:
F_i'j' = dx^i/dx^i' dx^j/dx^j' F_ij
and compute dx^i/dx^i' for a translation.

Tom Roberts

Please dont ask him about tensors, he dont know his tensors.

On May 11, 9:18 am, Dono wrote:
On May 11, 9:10 am, Tom Roberts wrote:

I don't know what "E" and "B" mean in a
non-inertial coordinate system, and the only way I know to figure it out
is to relate them to E and B in an inertial system (so you're back where
you started).

IOW: this question is more complicated than how they transform, it is
also a question of what they mean.

Tom Roberts

Thank you, Tom

The problem has a physical meaning, I am trying to relate the (E,B)
measured at a certain latitude on the Earth surface to the values
predicted at a different latitude (same longitude) as seen by an
observer that follows the common meridian moving at tangential speed
v.

You should understand the canux own the North
Magnetic pole and are planning on putting a
man hole cover over it to shut it off, and
charge for its use.

Are you assuming the North Mag is located equally
to the North axial pole?
Ken
(We're not mind readers)