• What are standard transformations between the fields inside and outside

    From john mcandrew@21:1/5 to All on Fri Mar 20 16:14:45 2020
    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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  • From Jos Bergervoet@21:1/5 to john mcandrew on Sat Mar 21 08:36:18 2020
    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

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  • From john mcandrew@21:1/5 to Jos Bergervoet on Sat Mar 21 15:47:28 2020
    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

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  • From Jos Bergervoet@21:1/5 to john mcandrew on Sun Mar 22 19:09:49 2020
    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

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  • From john mcandrew@21:1/5 to Jos Bergervoet on Mon Mar 23 20:14:00 2020
    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

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  • From Jos Bergervoet@21:1/5 to john mcandrew on Tue Mar 24 09:18:22 2020
    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

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  • From john mcandrew@21:1/5 to Jos Bergervoet on Tue Mar 24 18:24:53 2020
    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

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  • From john mcandrew@21:1/5 to Jos Bergervoet on Sat Apr 11 16:29:07 2020
    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

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  • From Jos Bergervoet@21:1/5 to john mcandrew on Tue Apr 14 09:00:40 2020
    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

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  • From Jos Bergervoet@21:1/5 to Jos Bergervoet on Tue Apr 14 21:35:38 2020
    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

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  • From john mcandrew@21:1/5 to Jos Bergervoet on Thu Apr 16 09:53:22 2020
    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

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  • From rubenfrutostalavera@gmail.com@21:1/5 to All on Sat Apr 18 16:30:05 2020
    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

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  • From rubenfrutostalavera@gmail.com@21:1/5 to All on Sat Apr 18 16:32:20 2020
    N factor dimensional puro b
    X log a_b =x
    A a

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