Let there be a closed boundary with a fixed charge density distributed upon the surface. Is there a standard way of transforming the fields between the inside and the outside? In particular, how is it best done when the boundary is a sphere?

Thanks in advance,

John McAndrew

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On 20/03/21 12:14 AM, john mcandrew wrote:

Let there be a closed boundary with a fixed charge density distributed upon the surface. Is there a standard way of transforming the fields between the inside and the outside? In particular, how is it best done when the boundary is a sphere?

No, there is absolutely no chance of doing that in general. Even
if the sphere is completely transparent, even if this charge is
zero. You simply cannot predict what happens elsewhere when you only
know what happens inside the sphere.

There could be a flash of light (coming in from a suddenly appearing
supernova, for instance) which is just about to enter your sphere,
even though inside the sphere nothing happens. You will then not
be able to use any knowledge about fields inside the sphere to
derive the presence of the light outside it.

What you ask for is clairvoyance!

--
Jos

--- SoupGate-Win32 v1.05
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On Saturday, March 21, 2020 at 7:36:20 AM UTC, Jos Bergervoet wrote:

On 20/03/21 12:14 AM, john mcandrew wrote:

Let there be a closed boundary with a fixed charge density distributed upon the surface. Is there a standard way of transforming the fields between the inside and the outside? In particular, how is it best done when the boundary is a sphere?

No, there is absolutely no chance of doing that in general. Even
if the sphere is completely transparent, even if this charge is
zero. You simply cannot predict what happens elsewhere when you only
know what happens inside the sphere.

There could be a flash of light (coming in from a suddenly appearing supernova, for instance) which is just about to enter your sphere,
even though inside the sphere nothing happens. You will then not
be able to use any knowledge about fields inside the sphere to
derive the presence of the light outside it.

What you ask for is clairvoyance!

--
Jos

I had in mind the fields generated by a fixed charge density on the surface of the boundary. For example, in the case of a spherical conducting boundary in free space, the field inside is zero, outside Coulombic with the charge uniformly distributed upon
the surface. With the sphere placed inside a uniform electric field E, the surface charge rearranges itself to create an opposing electric field -E, giving rise to a corresponding non-Coulombic electric field outside.

Of course, it's the surface charge on the boundary that directly defines the internal and external fields, but perhaps these fields can be related to one another directly via elegant transformations in special cases?

John McAndrew

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 20/03/21 11:47 PM, john mcandrew wrote:

On Saturday, March 21, 2020 at 7:36:20 AM UTC, Jos Bergervoet wrote:

On 20/03/21 12:14 AM, john mcandrew wrote:

Let there be a closed boundary with a fixed charge density distributed upon the surface. Is there a standard way of transforming the fields between the inside and the outside? In particular, how is it best done when the boundary is a sphere?

No, there is absolutely no chance of doing that in general. Even
if the sphere is completely transparent, even if this charge is
zero. You simply cannot predict what happens elsewhere when you only
know what happens inside the sphere.

There could be a flash of light (coming in from a suddenly appearing
supernova, for instance) which is just about to enter your sphere,
even though inside the sphere nothing happens. You will then not
be able to use any knowledge about fields inside the sphere to
derive the presence of the light outside it.

What you ask for is clairvoyance!

--
Jos

I had in mind the fields generated by a fixed charge density on the surface of the boundary. For example, in the case of a spherical conducting boundary in free space, the field inside is zero, outside Coulombic with the charge uniformly distributed

upon the surface. With the sphere placed inside a uniform electric field E, the surface charge rearranges itself to create an opposing electric field -E, giving rise to a corresponding non-Coulombic electric field outside.

So you restrict yourself to static fields? (Then no incoming
radiation can spoil the predictability.)

Of course, it's the surface charge on the boundary that directly defines the internal and external fields, but perhaps these fields can be related to one another directly via elegant transformations in special cases?

You could perhaps first split the field into the initial part from
external sources (before the sphere is present, like your example's
uniform field E) and the part created by the sphere's surface charge?

If you do allow that, then it's simple, the initial field is described
by a potential Vinit(r), the sphere's field by a potential Vsph(r) which
has the property:
Vsph(r) = Vsph(R^2/r) / |r|
where r is any position vector in 3D space. This relates outside and
inside positions. But if you know only the *field* inside the sphere
then the inside *potential* is still only known up to a constant, and
the external field only up to a Coulomb term, so the external field has
exactly one degree of freedom more than the internal field (this degree
of freedom is the total charge on the sphere, invisible from inside.)

I don't know such a simple formula for arbitrary closed surfaces. Series expansion into multipoles for the potential would probably work but is
quite elaborate.. (and isn't closed-form, but an infinite series).

Conformal mapping to a sphere is not possible. In 3D, spheres only
map to spheres, the set of allowed conformal mappings is much smaller
in 3D than in 2D (and there is a nasty scaling factor involved..)

--
Jos

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Sunday, March 22, 2020 at 6:09:51 PM UTC, Jos Bergervoet wrote:

On 20/03/21 11:47 PM, john mcandrew wrote:

On Saturday, March 21, 2020 at 7:36:20 AM UTC, Jos Bergervoet wrote:

On 20/03/21 12:14 AM, john mcandrew wrote:

Let there be a closed boundary with a fixed charge density distributed upon the surface. Is there a standard way of transforming the fields between the inside and the outside? In particular, how is it best done when the boundary is a sphere?

No, there is absolutely no chance of doing that in general. Even
if the sphere is completely transparent, even if this charge is
zero. You simply cannot predict what happens elsewhere when you only
know what happens inside the sphere.

There could be a flash of light (coming in from a suddenly appearing
supernova, for instance) which is just about to enter your sphere,
even though inside the sphere nothing happens. You will then not
be able to use any knowledge about fields inside the sphere to
derive the presence of the light outside it.

What you ask for is clairvoyance!

--
Jos

I had in mind the fields generated by a fixed charge density on the surface of the boundary. For example, in the case of a spherical conducting boundary in free space, the field inside is zero, outside Coulombic with the charge uniformly distributed

upon the surface. With the sphere placed inside a uniform electric field E, the surface charge rearranges itself to create an opposing electric field -E, giving rise to a corresponding non-Coulombic electric field outside.

So you restrict yourself to static fields? (Then no incoming
radiation can spoil the predictability.)

Of course, it's the surface charge on the boundary that directly defines the internal and external fields, but perhaps these fields can be related to one another directly via elegant transformations in special cases?

You could perhaps first split the field into the initial part from
external sources (before the sphere is present, like your example's
uniform field E) and the part created by the sphere's surface charge?

If you do allow that, then it's simple, the initial field is described
by a potential Vinit(r), the sphere's field by a potential Vsph(r) which
has the property:
Vsph(r) = Vsph(R^2/r) / |r|
where r is any position vector in 3D space. This relates outside and
inside positions.

I'm assuming R is the radius of the sphere. I don't understand the Vsph(R^2/r) term where the vector r is in the denominator which isn't allowed in vector calculus as far as I know.

But if you know only the *field* inside the sphere
then the inside *potential* is still only known up to a constant, and
the external field only up to a Coulomb term, so the external field has exactly one degree of freedom more than the internal field (this degree
of freedom is the total charge on the sphere, invisible from inside.)

via Gauss's law?

I don't know such a simple formula for arbitrary closed surfaces. Series expansion into multipoles for the potential would probably work but is
quite elaborate.. (and isn't closed-form, but an infinite series).

Conformal mapping to a sphere is not possible. In 3D, spheres only
map to spheres, the set of allowed conformal mappings is much smaller
in 3D than in 2D (and there is a nasty scaling factor involved..)

Thanks for the info.

J McA

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 20/03/24 4:14 AM, john mcandrew wrote:

On Sunday, March 22, 2020 at 6:09:51 PM UTC, Jos Bergervoet wrote:

On 20/03/21 11:47 PM, john mcandrew wrote:

On Saturday, March 21, 2020 at 7:36:20 AM UTC, Jos Bergervoet wrote:

On 20/03/21 12:14 AM, john mcandrew wrote:

Let there be a closed boundary with a fixed charge density distributed upon the surface. Is there a standard way of transforming the fields between the inside and the outside? In particular, how is it best done when the boundary is a sphere?

No, there is absolutely no chance of doing that in general. Even
if the sphere is completely transparent, even if this charge is
zero. You simply cannot predict what happens elsewhere when you only
know what happens inside the sphere.

There could be a flash of light (coming in from a suddenly appearing
supernova, for instance) which is just about to enter your sphere,
even though inside the sphere nothing happens. You will then not
be able to use any knowledge about fields inside the sphere to
derive the presence of the light outside it.

What you ask for is clairvoyance!

--
Jos

I had in mind the fields generated by a fixed charge density on the surface of the boundary. For example, in the case of a spherical conducting boundary in free space, the field inside is zero, outside Coulombic with the charge uniformly distributed

upon the surface. With the sphere placed inside a uniform electric field E, the surface charge rearranges itself to create an opposing electric field -E, giving rise to a corresponding non-Coulombic electric field outside.

So you restrict yourself to static fields? (Then no incoming
radiation can spoil the predictability.)

Of course, it's the surface charge on the boundary that directly defines the internal and external fields, but perhaps these fields can be related to one another directly via elegant transformations in special cases?

You could perhaps first split the field into the initial part from
external sources (before the sphere is present, like your example's
uniform field E) and the part created by the sphere's surface charge?

If you do allow that, then it's simple, the initial field is described
by a potential Vinit(r), the sphere's field by a potential Vsph(r) which
has the property:
Vsph(r) = Vsph(R^2/r) / |r|
where r is any position vector in 3D space. This relates outside and
inside positions.

I'm assuming R is the radius of the sphere. I don't understand the Vsph(R^2/r) term where the vector r is in the denominator which isn't allowed in vector calculus as far as I know.

Sorry that was sloppy (and the way I wrote it is correct only for the
unit sphere). I only meant to say that in that case the well-known
radius inversion is a simple solution. For better notation let's use
3D position vector x:

x = r * n

with r the magnitude of x and a unit vector, n, giving the direction.
Then:

Vsph(r * n) = Vsph(R^2/r * n) * R/r

or:

Vsph(x) = Vsph(R^2/r^2 * x) * R/r

Correct notation (I hope) and valid for any R.

But if you know only the *field* inside the sphere
then the inside *potential* is still only known up to a constant, and
the external field only up to a Coulomb term, so the external field has
exactly one degree of freedom more than the internal field (this degree
of freedom is the total charge on the sphere, invisible from inside.)

via Gauss's law?

Gauss law can only be applied *outside* the sphere.

I don't know such a simple formula for arbitrary closed surfaces. Series
expansion into multipoles for the potential would probably work but is
quite elaborate.. (and isn't closed-form, but an infinite series).

Conformal mapping to a sphere is not possible. In 3D, spheres only
map to spheres, the set of allowed conformal mappings is much smaller
in 3D than in 2D (and there is a nasty scaling factor involved..)

Thanks for the info.

J McA

--
Jos

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Tuesday, March 24, 2020 at 8:18:24 AM UTC, Jos Bergervoet wrote:

On 20/03/24 4:14 AM, john mcandrew wrote:

On Sunday, March 22, 2020 at 6:09:51 PM UTC, Jos Bergervoet wrote:

On 20/03/21 11:47 PM, john mcandrew wrote:

On Saturday, March 21, 2020 at 7:36:20 AM UTC, Jos Bergervoet wrote:

On 20/03/21 12:14 AM, john mcandrew wrote:

Let there be a closed boundary with a fixed charge density distributed upon the surface. Is there a standard way of transforming the fields between the inside and the outside? In particular, how is it best done when the boundary is a sphere?

No, there is absolutely no chance of doing that in general. Even
if the sphere is completely transparent, even if this charge is
zero. You simply cannot predict what happens elsewhere when you only >>>> know what happens inside the sphere.

There could be a flash of light (coming in from a suddenly appearing >>>> supernova, for instance) which is just about to enter your sphere,
even though inside the sphere nothing happens. You will then not
be able to use any knowledge about fields inside the sphere to
derive the presence of the light outside it.

What you ask for is clairvoyance!

--
Jos

I had in mind the fields generated by a fixed charge density on the surface of the boundary. For example, in the case of a spherical conducting boundary in free space, the field inside is zero, outside Coulombic with the charge uniformly

distributed upon the surface. With the sphere placed inside a uniform electric field E, the surface charge rearranges itself to create an opposing electric field -E, giving rise to a corresponding non-Coulombic electric field outside.

So you restrict yourself to static fields? (Then no incoming
radiation can spoil the predictability.)

Of course, it's the surface charge on the boundary that directly defines the internal and external fields, but perhaps these fields can be related to one another directly via elegant transformations in special cases?

You could perhaps first split the field into the initial part from
external sources (before the sphere is present, like your example's
uniform field E) and the part created by the sphere's surface charge?

If you do allow that, then it's simple, the initial field is described
by a potential Vinit(r), the sphere's field by a potential Vsph(r) which >> has the property:
Vsph(r) = Vsph(R^2/r) / |r|
where r is any position vector in 3D space. This relates outside and
inside positions.

I'm assuming R is the radius of the sphere. I don't understand the Vsph(R^2/r) term where the vector r is in the denominator which isn't allowed in vector calculus as far as I know.

Sorry that was sloppy (and the way I wrote it is correct only for the
unit sphere). I only meant to say that in that case the well-known
radius inversion is a simple solution. For better notation let's use
3D position vector x:

x = r * n

with r the magnitude of x and a unit vector, n, giving the direction.
Then:

Vsph(r * n) = Vsph(R^2/r * n) * R/r

or:

Vsph(x) = Vsph(R^2/r^2 * x) * R/r

Correct notation (I hope) and valid for any R.

For anyone else reading, this is the Kelvin transform, used in potential theory from my initial search on the topic:
https://en.wikipedia.org/wiki/Kelvin_transform https://en.wikipedia.org/wiki/Potential_theory

Kelvin transformation and inverse multipoles in electrostatics https://arxiv.org/pdf/1611.05942.pdf

But if you know only the *field* inside the sphere
then the inside *potential* is still only known up to a constant, and
the external field only up to a Coulomb term, so the external field has
exactly one degree of freedom more than the internal field (this degree
of freedom is the total charge on the sphere, invisible from inside.)

via Gauss's law?

Gauss law can only be applied *outside* the sphere.

To get a feel for your basic points, I found a constant charge Q arbitrarily distributed on a spherical surface helped. Just inside the boundary Gauss's law gives integrate(E.ds) = 0, just outside integrate(E.ds) = Q. So the external field has one more
degree of freedom, parameterized by Q, over the internal field.

[snipped]

JMcA

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On Tuesday, March 24, 2020 at 8:18:24 AM UTC, Jos Bergervoet wrote:

[snipped]

Sorry that was sloppy (and the way I wrote it is correct only for the
unit sphere). I only meant to say that in that case the well-known
radius inversion is a simple solution. For better notation let's use
3D position vector x:

x = r * n

with r the magnitude of x and a unit vector, n, giving the direction.
Then:

Vsph(r * n) = Vsph(R^2/r * n) * R/r

or:

Vsph(x) = Vsph(R^2/r^2 * x) * R/r

Correct notation (I hope) and valid for any R.

Do you have a reference for this?

I've come across two references, the first from L.D. Landau & E.M. Lifshitz, page 11:
https://archive.org/details/ElectrodynamicsOfContinuousMedia/page/n21/mode/2up

The second from the first edition of Jackson, page 35: https://archive.org/details/ClassicalElectrodynamics/page/n53/mode/2up

To me, they don't look exactly the same and I have more confidence with Jackson's version. In particular, Landau expresses phi' in terms of r', whereas Jackson does this in terms of r, as you've done, although you've left V the same on both sides, rather
than as V' on one side:

V'sph(r * n) = Vsph(R^2/r * n) * R/r

or:

V'sph(x) = Vsph(R^2/r^2 * x) * R/r

That is, according to Jackson, the inversion theorem enables us to find another valid potential function V'(x) from V(x) and the corresponding different sources for it. I can't see how this theorem can be used to relate potentials inside and outside a
sphere for the same potential function V(x).

[snipped]

John McAndrew

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

On 20/04/12 1:29 AM, john mcandrew wrote:

On Tuesday, March 24, 2020 at 8:18:24 AM UTC, Jos Bergervoet wrote:

[snipped]

Sorry that was sloppy (and the way I wrote it is correct only for the
unit sphere). I only meant to say that in that case the well-known
radius inversion is a simple solution. For better notation let's use
3D position vector x:

x = r * n

with r the magnitude of x and a unit vector, n, giving the direction.
Then:

Vsph(r * n) = Vsph(R^2/r * n) * R/r

or:

Vsph(x) = Vsph(R^2/r^2 * x) * R/r

Correct notation (I hope) and valid for any R.

Do you have a reference for this?

I've come across two references, the first from L.D. Landau & E.M. Lifshitz, page 11:
https://archive.org/details/ElectrodynamicsOfContinuousMedia/page/n21/mode/2up

The second from the first edition of Jackson, page 35: https://archive.org/details/ClassicalElectrodynamics/page/n53/mode/2up

To me, they don't look exactly the same and I have more confidence with Jackson's version.

I would expect Jackson to be correct, his book is the gold standard and
has seen many reprints so elementary errors would have been corrected.

In particular, Landau expresses phi' in terms of r', whereas Jackson does this in terms of r, as you've done,

Perhaps Landau then is dealing with a different transform than the straightforward inversion? With inversion the value of phi will not
change, only r is mapped to 1/r (with some scaling factors).

although you've left V the same on both sides, rather than as V' on one side:

V'sph(r * n) = Vsph(R^2/r * n) * R/r

I use the same symbol Vsph because the soultion *is* the actual
potential that we are talking about. (See below..)

or:

V'sph(x) = Vsph(R^2/r^2 * x) * R/r

That is, according to Jackson, the inversion theorem enables us to find another valid potential function V'(x) from V(x) and the corresponding different sources for it. I can't see how this theorem can be used to relate potentials inside and outside a

sphere for the same potential function V(x).

It is of no help to find just another *valid* solution. Because then you
still don't know whether that is the actually occurring solution for
your sphere. And using two different names (like V and V') would then
be appropriate.

What you need is to proof that this valid solution is the only one that
can occur. In our reasoning in this thread (IIRC) we first excluded the
field of external sources. So the remaining problem has only charge on
the sphere's surface, none outside or inside it.

In that case, taking the solution inside the sphere and using the
inversion procedure, does give you not just a 'valid' solution outside
the sphere but it gives you the only possible solution.

You look for a solution in the r>R region that must match the solution
in the r<R region at the boundary, so at r=R they both must be the same.
The inversion-based solution does this. So you know that it is a correct solution of the Laplace operator in your source-free region r>R and that
it fits the boundary condition. That is enough for uniqueness.

Of course the uniqueness proof will assume *another* solution exists,
then form the difference between the two purported solutions, and then
this difference must be 0 on the r=R surface and also will have no
sources for r>R, so it must be identically zero.

--
Jos

--- SoupGate-Win32 v1.05
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On 20/04/14 9:00 AM, Jos Bergervoet wrote:

On 20/04/12 1:29 AM, john mcandrew wrote:

On Tuesday, March 24, 2020 at 8:18:24 AM UTC, Jos Bergervoet wrote:

[snipped]

Sorry that was sloppy (and the way I wrote it is correct only for the
unit sphere). I only meant to say that in that case the well-known
radius inversion is a simple solution. For better notation let's use
3D position vector x:

    x = r * n

with r the magnitude of x and a unit vector, n, giving the direction.
Then:

    Vsph(r * n) = Vsph(R^2/r * n) * R/r

or:

    Vsph(x) = Vsph(R^2/r^2 * x) * R/r

Correct notation (I hope) and valid for any R.

Do you have a reference for this?

I've come across two references, the first from L.D. Landau & E.M.
Lifshitz, page 11:
https://archive.org/details/ElectrodynamicsOfContinuousMedia/page/n21/mode/2up


The second from the first edition of Jackson, page 35:
https://archive.org/details/ClassicalElectrodynamics/page/n53/mode/2up

To me, they don't look exactly the same and I have more confidence
with Jackson's version.

I would expect Jackson to be correct, his book is the gold standard and
has seen many reprints so elementary errors would have been corrected.

In particular, Landau expresses phi' in terms of r', whereas Jackson
does this in terms of r, as you've done,

Perhaps Landau then is dealing with a different transform than the straightforward inversion? With inversion the value of phi will not
change, only r is mapped to 1/r (with some scaling factors).

although you've left V the same on both sides, rather than as V' on
one side:

V'sph(r * n) = Vsph(R^2/r * n) * R/r

I use the same symbol Vsph because the soultion *is* the actual
potential that we are talking about. (See below..)

or:
V'sph(x) = Vsph(R^2/r^2 * x) * R/r

That is, according to Jackson, the inversion theorem enables us to
find another valid potential function V'(x) from V(x) and the
corresponding different sources for it. I can't see how this theorem
can be used to relate potentials inside and outside a sphere for the
same potential function V(x).

It is of no help to find just another *valid* solution. Because then you still don't know whether that is the actually occurring solution for
your sphere. And using two different names (like V and V') would then
be appropriate.

What you need is to proof that this valid solution is the only one that
can occur. In our reasoning in this thread (IIRC) we first excluded the
field of external sources. So the remaining problem has only charge on
the sphere's surface, none outside or inside it.

In that case, taking the solution inside the sphere and using the
inversion procedure, does give you not just a 'valid' solution outside
the sphere but it gives you the only possible solution.

You look for a solution in the r>R region that must match the solution
in the r<R region at the boundary, so at r=R they both must be the same.
The inversion-based solution does this. So you know that it is a correct solution of the Laplace operator in your source-free region r>R and that
it fits the boundary condition. That is enough for uniqueness.

Of course the uniqueness proof will assume *another* solution exists,
then form the difference between the two purported solutions, and then
this difference must be 0 on the r=R surface and also will have no
sources for r>R, so it must be identically zero.

Actually I should again mention the extra degree of freedom for the
total charge on the sphere! (We are in fact revisiting the complete
earlier discussion..) The addition that can be made to the external
potential is:

V_add(r) = C - C R/r

which is zero at r=R so the addition gives another solution which
shows that the result from the inversion procedure is actually *not*
a unique exterior solution, but as we have seen the parameter C is
the only freedom left in it.

--
Jos

--- SoupGate-Win32 v1.05
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On Tuesday, April 14, 2020 at 8:00:46 AM UTC+1, Jos Bergervoet wrote:

On 20/04/12 1:29 AM, john mcandrew wrote:

[snipped]

or:

V'sph(x) = Vsph(R^2/r^2 * x) * R/r

That is, according to Jackson, the inversion theorem enables us to find another valid potential function V'(x) from V(x) and the corresponding different sources for it. I can't see how this theorem can be used to relate potentials inside and outside

a sphere for the same potential function V(x).

It is of no help to find just another *valid* solution. Because then you still don't know whether that is the actually occurring solution for
your sphere. And using two different names (like V and V') would then
be appropriate.

What you need is to proof that this valid solution is the only one that
can occur. In our reasoning in this thread (IIRC) we first excluded the
field of external sources. So the remaining problem has only charge on
the sphere's surface, none outside or inside it.

In that case, taking the solution inside the sphere and using the
inversion procedure, does give you not just a 'valid' solution outside
the sphere but it gives you the only possible solution.

You look for a solution in the r>R region that must match the solution
in the r<R region at the boundary, so at r=R they both must be the same.
The inversion-based solution does this. So you know that it is a correct solution of the Laplace operator in your source-free region r>R and that
it fits the boundary condition. That is enough for uniqueness.

Of course the uniqueness proof will assume *another* solution exists,
then form the difference between the two purported solutions, and then
this difference must be 0 on the r=R surface and also will have no
sources for r>R, so it must be identically zero.

The solution is staring me in the face. Both Jackson and Landau state that the transformation generally moves and changes the value of the source charges:

q'_i = q_i R/r_i, r'_i = R^2/r_i using spherical coords (r_i, theta, phi)

where q_i and q'_i is the ith charge at positions ~r = (r_i, theta, phi) and ~r' = (r'_i, theta, phi) respectively.

In the case where all the sources lie on r_i = R, these becomes the same sources for V'(~r) so that V'(~r) = V(~r) confirming the correctness of your expression. I'm posting the following link as a reminder to look further into this remarkable
transformation when I have time, and for anyone else vaguely interested:

On the Advancements of
Conformal Transformations and their Associated Symmetries
in Geometry and Theoretical Physics1
https://arxiv.org/pdf/0808.2730.pdf

JMcA

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

I am FX FRT, according to all those people the greater radius has a widening of unit proportions in cyq that the intensity transfers in cyq from another different r, that is linear differential calculation of fourth revolution but try practical physics
in exits from space, gravity is the best impulse under survival conditions but a log x -0 / log ax matrix is ​​equivalent to saying that logarithm is equal to its division by its subtraction and that the base is conjugated with the exponential, that
is equivalent to redesigning the equation and making the parametric of two solutions cyq and that the q is innerrente to the concept of prestige, that is to say N ^ ax / N ^ x being the material conjugate of dimensions and something similar

--- SoupGate-Win32 v1.05
* Origin: fsxNet Usenet Gateway (21:1/5)

N factor dimensional puro b
X log a_b =x
A a

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