Hello
I have a question about electromagnetic field transformation between
different inertial frames.
Consider the following case:
A charge q1 is kept stationary with respect to another charge q2 which
is moving with velocity v. When the moving charge is in a position such

that the line connecting it with the other charge is normal to the
direction of motion, Coulomb's law says that the field in the 2 frames
=q1 q2/4pi.epsilon.r^2. This expression has the same value in both
frames since all the quantities in it are invariant. (r is normal to
the velocity) While Lorentz transformation of the electric field says
that the field will differ by a factor of gamma (Here the electric
field has one component : in the direction normal to velocity)


We can imagine further that there is a shield between the 2 charges
having a hole just in the line connecting them in this position to
eliminate the field at other positions of the moving charge and hence
time retardation effects.
Thanks.


www.sultan.org

"Tareq" wrote in message
ups.com...
Hello
I have a question about electromagnetic field transformation between
different inertial frames.
Consider the following case:
A charge q1 is kept stationary with respect to another charge q2 which
is moving with velocity v. When the moving charge is in a position such

that the line connecting it with the other charge is normal to the
direction of motion, Coulomb's law says that the field in the 2 frames
=q1 q2/4pi.epsilon.r^2. This expression has the same value in both
frames since all the quantities in it are invariant. (r is normal to
the velocity)

That is not correct. In fact it is only components normal to the motion
that are changed - see for example page 497 - Griffith - Introduction to
Electrodynamics.

Thanks
Bill

While Lorentz transformation of the electric field says
that the field will differ by a factor of gamma (Here the electric
field has one component : in the direction normal to velocity)


We can imagine further that there is a shield between the 2 charges
having a hole just in the line connecting them in this position to
eliminate the field at other positions of the moving charge and hence
time retardation effects.
Thanks.


www.sultan.org

I don't think you can do this. Coulomb's law is valid in
electrostatics, but you're not dealing with electrostatics here.

I checked Jackson (3rd ed, eq.11.152) but it says that the field is
radial in both frames ( field lines emanate from the present position
of charge ) but this is in contrast to retardation. If we calculated
the field from a retarded position, it would be in a different
direction !!

You either need to compute the retarded A field explicitly (and then
find E and B from that), which you didn't do,

There will be indeed a magnetic field in both frames but its direction
is in the direction of the current of the other charge (the direction
of motion) and so will be the electric field component due to it, but I
am interested in the normal component only.

Yes, but remember that if you want to work out the forces acting on
each charge in a frame in which they're in motion, there's also a
magnetic field present which can't be ignored. And, in a frame in
which they're not at rest, the magnetic field will be exerting a force
on them.

No. In a frame of one charge, this charge is at rest and will not be
affected by any magnetic field.

Please specify exactly what the "shield" is made of and how it works.

Just a very large conductive plane of large thickness.

---------------------------------------------------------------
Give your soul a chance to breathe
www.sultan.org

Tareq wrote:

I don't think you can do this. Coulomb's law is valid in
electrostatics, but you're not dealing with electrostatics here.

I checked Jackson (3rd ed, eq.11.152) but it says that the field is
radial in both frames ( field lines emanate from the present position
of charge ) but this is in contrast to retardation. If we calculated
the field from a retarded position, it would be in a different
direction !!

You either need to compute the retarded A field explicitly (and then
find E and B from that), which you didn't do,

There will be indeed a magnetic field in both frames but its direction
is in the direction of the current of the other charge (the direction
of motion) and so will be the electric field component due to it, but I
am interested in the normal component only.

Yes, but remember that if you want to work out the forces acting on
each charge in a frame in which they're in motion, there's also a
magnetic field present which can't be ignored. And, in a frame in
which they're not at rest, the magnetic field will be exerting a force
on them.

No. In a frame of one charge, this charge is at rest and will not be
affected by any magnetic field.

Please specify exactly what the "shield" is made of and how it works.

Just a very large conductive plane of large thickness.

---------------------------------------------------------------
Give your soul a chance to breathe
www.sultan.org

The motion of the charge is indeed coupled through the
shield but you need to consider the field propagating
on the shields surface at c and how quickly it will equalise
a gradient. A small hole would appear the same as a
solid shield but a slot perpedicular to the charges
motion would nearly cause the shield to vanish.

Remember... the charges are coupled through volumes of
space that include an area of the shield not by lines of
force that can penetrate a hole.

"Coulombs Law"
http://farside.ph.utexas.edu/teachin...es/node28.html

Sue...

On Wed, 29 Mar 2006 05:36:28 -0800, Tareq wrote:


I don't think you can do this. Coulomb's law is valid in
electrostatics, but you're not dealing with electrostatics here.

I checked Jackson (3rd ed, eq.11.152) but it says that the field is radial
in both frames ( field lines emanate from the present position of charge )
but this is in contrast to retardation. If we calculated the field from a
retarded position, it would be in a different direction !!

Sez who?

The retarded E field from a uniformly moving charge is radial, and in fact
matches what you get by transforming the E field in the charge's own rest
frame. See the result for a moving charge in any electrodynamics text.

Calculating the retarded field is a lot messier than just figuring out
where it was and saying, "Oh, the E field must be radial with respect to
that position".

If you get a different result, all you've proved is that you did the
calculation wrong.



You either need to compute the retarded A field explicitly (and then
find E and B from that), which you didn't do,

There will be indeed a magnetic field in both frames but its direction
is in the direction of the current of the other charge (the direction of
motion) and so will be the electric field component due to it, but I am
interested in the normal component only.

Yes, but remember that if you want to work out the forces acting on
each charge in a frame in which they're in motion, there's also a
magnetic field present which can't be ignored. And, in a frame in
which they're not at rest, the magnetic field will be exerting a force
on them.

No. In a frame of one charge, this charge is at rest and will not be
affected by any magnetic field.

Please specify exactly what the "shield" is made of and how it works.

Just a very large conductive plane of large thickness.

--------------------------------------------------------------- Give
your soul a chance to breathe
www.sultan.org

--
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I can be also contacted through http://www.physicsinsights.org

On Mon, 03 Apr 2006 02:39:08 -0700, Tareq wrote:

Sal, see:
http://groups.google.com/group/sci.p...77441a5001e817

Right. The field of a uniformly moving point charge is radial with
regard to the current position, but it's also pancaked, as a proper
calculation of the retarded integrals will show.

But I really don't understand what point you're trying to make
here, because...

You said:

I checked Jackson (3rd ed, eq.11.152) but it says that the field is
radial in both frames ( field lines emanate from the present
position of charge ) but this is in contrast to retardation. If we
calculated the field from a retarded position, it would be in a
different direction !!

That is not correct as regards the retarded A field. When the field
of a uniformly moving charge is determined by evaluating the retarded
integral of the A field over all space, the resulting E field is
radial with respect to the current position of the charge.

As I said, you can't just look at the retarded position of the charges
and use Coulomb's law to find the "retarded E field" directly; you get
the wrong answer (as I believe you have shown). The E field is not
"complete" by itself, it's not described by a 4-vector field, and you
can't, in general, just ignore that fact and consequently ignore the
"rest" of the field. The E field is _part_ _of_ a rank 2 tensor. The
A field, in contrast, _is_ a proper 4-vector field, and _is_ complete
in and of itself. But in either case you can't just treat the moving
point charge as though it's stationary when you try to evaluate the
field from the retarded position!

I haven't got a copy of Jackson here. However, I have got a couple of
other references which cover this. See, for instance, Griffiths,
Intro to Electrodynamcs, 3rd edition, pp 429-440. Here's a nice
quote (emphasis as in the original text -- Griffiths likes italics):

[Griffiths, pp 430-431, excerpt:]
Now, a naive reading of the formula

V(r,t) = 1/(4 pi eps_0) Int(rho(r',t_r)/|r| dtau') (10.35)

might suggest to you that the retarded potential of a point charge
is simply

(1/(4 pi eps_0)) (q/|r|)

(the same as in the static case, only with the understanding that
|r| is the distance to the _retarded_ position of the charge). But
this is wrong, for a very subtle reason: It is true that for a point
source the denominator |r| comes outside the integral, but what
remains,

Int(rho(r',t_r)dtau')

is _not_ equal to the charge of the particle. To calculate the
total charge of a configuration you must integrate rho over the
entire distrubution at _one_ _instant_ _of_ _time_, but here the
retardation, t_r = t-|r|/c, obliges us to evaluate rho at
_different_ _times_ for different parts of the configuration. If
the source is moving, this will give a distorted picture of the
total charge. You might think that this problem would disappear for
_point_ charges, but it doesn't. In Maxwell's electrodynamics,
formulated as it is in terms of charge and current _densities_, a
point charge must be regarded as the limit of an extended charge,
when the size goes to zero.

And that's also the limit of what I'm willing to type in from the
text. The derivation of the retarded potential for a point charge
goes on for pages.

Another reference which covers this is Rindler, "Introduction to
Special Relativity", pp 115-116, section 41, "Field of a uniformly
moving charge". It's far more concise, which is nice if you don't
have the patience to wade through Griffiths' mass of verbiage, but the
downside is that it's kind of dense.


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