Say I have a system of charges interacting with one another, but constrained by additional local forces: what are the necessary conditions for finding a valid solution to Maxwell's equations by only reversing the velocity of the charges? I can think of
two cases:

1. Each charge is constrained to travel at an arbitrary constant velocity generally different to one another. The system doesn't radiate from Lamor's formula.

2. Each charge is constrained to accelerate at a constant acceleration such that each charge maintains a constant proper distance from one another. The system radiates via Lamor's formula, but this radiation is reversible and given the name "acceleration
energy/momentum", "Schott energy/momentum" etc.

Are there any other examples?

Thanks in advance,

JMcA

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On 20/08/06 2:51 AM, john mcandrew wrote:

Say I have a system of charges interacting with one another, but constrained by additional local forces: what are the necessary conditions for finding a valid solution to Maxwell's equations by only reversing the velocity of the charges?

If you just use time reversal your velocities will be reversed.
In classical physics time reversal is a symmetry so your solution
with the reversed velocities will be a "valid solution" as required,
without the need to impose any "necessary conditions".

In the non-classical (QFT standard model) case, you would not
have time reversal symmetry for all possible forces (there you
only have CPT-invariance). So there you would have to impose
extra conditions, for instance that the forces are only from
the strong interaction and the electromagnetic interaction, but
not coming from the weak interaction. (This of course would make
it a bit unphysical, because the existing particles we know of
actually do feel the weak interaction..)

--
Jos

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On Thursday, August 6, 2020 at 12:41:41 PM UTC+1, Jos Bergervoet wrote:

On 20/08/06 2:51 AM, john mcandrew wrote:

Say I have a system of charges interacting with one another, but constrained by additional local forces: what are the necessary conditions for finding a valid solution to Maxwell's equations by only reversing the velocity of the charges?

If you just use time reversal your velocities will be reversed.
In classical physics time reversal is a symmetry so your solution
with the reversed velocities will be a "valid solution" as required,
without the need to impose any "necessary conditions".

In the non-classical (QFT standard model) case, you would not
have time reversal symmetry for all possible forces (there you
only have CPT-invariance). So there you would have to impose
extra conditions, for instance that the forces are only from
the strong interaction and the electromagnetic interaction, but
not coming from the weak interaction. (This of course would make
it a bit unphysical, because the existing particles we know of
actually do feel the weak interaction..)

--
Jos

Yes, here I'm interested in the special case where the fields at any point in the time-reversed case can still be traced back to a time-reversed source that generated it. More generally, the forward case creates "irreversible radiation" that can still be
traced back to the sources, but not so when time reversed where this sourceless field has to be added in. I think what I'm after is the non-radiation condition:
https://en.wikipedia.org/wiki/Nonradiation_condition#:~:text=Classical%20nonradiation%20conditions%20define%20the,will%20not%20emit%20electromagnetic%20radiation.&text=In%20some%20classical%20electron%20models,that%20no%20radiation%20is%20emitted.

This gives a few other examples of accelerated systems of charges that don't radiate and, I'm guessing, is therefore time-reversible via the sources alone. Or put another way: retarded-field - advanced-field = 0 everywhere.

JMcA

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On 20/08/07 1:47 AM, john mcandrew wrote:

On Thursday, August 6, 2020 at 12:41:41 PM UTC+1, Jos Bergervoet wrote:

On 20/08/06 2:51 AM, john mcandrew wrote:

Say I have a system of charges interacting with one another, but constrained by additional local forces: what are the necessary conditions for finding a valid solution to Maxwell's equations by only reversing the velocity of the charges?

If you just use time reversal your velocities will be reversed.
In classical physics time reversal is a symmetry so your solution
with the reversed velocities will be a "valid solution" as required,
without the need to impose any "necessary conditions".

In the non-classical (QFT standard model) case, you would not
have time reversal symmetry for all possible forces (there you
only have CPT-invariance). So there you would have to impose
extra conditions, for instance that the forces are only from
the strong interaction and the electromagnetic interaction, but
not coming from the weak interaction. (This of course would make
it a bit unphysical, because the existing particles we know of
actually do feel the weak interaction..)

Yes, here I'm interested in the special case where the fields at any point in the time-reversed case can still be traced back to a time-reversed source that generated it. More generally, the forward case creates "irreversible radiation" that can still

be traced back to the sources, but not so when time reversed where this sourceless field has to be added in. I think what I'm after is the non-radiation condition:

https://en.wikipedia.org/wiki/Nonradiation_condition#:~:text=Classical%20nonradiation%20conditions%20define%20the,will%20not%20emit%20electromagnetic%20radiation.&text=In%20some%20classical%20electron%20models,that%20no%20radiation%20is%20emitted.

You will then still not get time reversal "by only reversing the
velocity of the charges" as you write in the title. You also have
to change the sign of the magnetic field if you do that (and other
things as well perhaps..)

To have only reversed velocities and nothing else changed in the
time reversed solution you would need a case where the magnetic
field is zero everywhere, it seems..

One possibility is of course a pure electrostatic case, but then
all velocities are zero and you don't have to change anything to
get the time-reversed solution!

Another case would be a current distribution (ignoring individual
particles, only looking at bulk currents) where the current is
such that it has no magnetic field. For instance a purely radial
current. To avoid the need for an infinite charge reservoir at
the center, you could take a "breathing mode" time-dependence
for the radial current. An AC current flowing radially in some
medium between two concentric spheres. It will create an ac E-
field, also radial, and 90 degrees out of phase, but no B-field.
For time reversal you now *only* have to reverse the current, i.e.
reverse the velocities of the carriers.

But if you actually do look at the microscopic fields between the
individual charge carriers, you will see that the B-field is not
really zero and the E-field also not really invariant under time-
reversal, I'm afraid..

But in any case, non-radiation is insufficient. A coil with ac
current in a perfect Faraday cage is a non-radiating system, but
for time-reversal it needs more than velocity-reversal, also the
B-fields inside the cage will need a sign flip!

--
Jos

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On Wednesday, August 12, 2020 at 6:26:23 PM UTC+1, Jos Bergervoet wrote:

On 20/08/07 1:47 AM, john mcandrew wrote:

On Thursday, August 6, 2020 at 12:41:41 PM UTC+1, Jos Bergervoet wrote:

On 20/08/06 2:51 AM, john mcandrew wrote:

Say I have a system of charges interacting with one another, but constrained by additional local forces: what are the necessary conditions for finding a valid solution to Maxwell's equations by only reversing the velocity of the charges?

If you just use time reversal your velocities will be reversed.
In classical physics time reversal is a symmetry so your solution
with the reversed velocities will be a "valid solution" as required,
without the need to impose any "necessary conditions".

In the non-classical (QFT standard model) case, you would not
have time reversal symmetry for all possible forces (there you
only have CPT-invariance). So there you would have to impose
extra conditions, for instance that the forces are only from
the strong interaction and the electromagnetic interaction, but
not coming from the weak interaction. (This of course would make
it a bit unphysical, because the existing particles we know of
actually do feel the weak interaction..)

Yes, here I'm interested in the special case where the fields at any point in the time-reversed case can still be traced back to a time-reversed source that generated it. More generally, the forward case creates "irreversible radiation" that can

still be traced back to the sources, but not so when time reversed where this sourceless field has to be added in. I think what I'm after is the non-radiation condition:

https://en.wikipedia.org/wiki/Nonradiation_condition#:~:text=Classical%20nonradiation%20conditions%20define%20the,will%20not%20emit%20electromagnetic%20radiation.&text=In%20some%20classical%20electron%20models,that%20no%20radiation%20is%20emitted.

You will then still not get time reversal "by only reversing the
velocity of the charges" as you write in the title. You also have
to change the sign of the magnetic field if you do that (and other
things as well perhaps..)

I'm under the impression that for a charge travelling at a constant velocity, say: when its velocity is reversed, the E field it creates is the same as before but the sign of the B field flips compared to the forwards direction the charge was travelling
at. Feynman has a section on this for anyone else interested: https://www.feynmanlectures.caltech.edu/II_26.html

B = v x E (26.9) in the above link.

Likewise for two charges travelling at constrained constant velocities wrt one another, the B fields they create will flip sign, hence cancelling the sign change in v in the Lorentz force equation, giving the same EM force on the charges as in the
forward case.

[snipped]

JMcA

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On 20/08/13 9:02 AM, john mcandrew wrote:

On Wednesday, August 12, 2020 at 6:26:23 PM UTC+1, Jos Bergervoet wrote:

On 20/08/07 1:47 AM, john mcandrew wrote:

...

You will then still not get time reversal "by only reversing the
velocity of the charges" as you write in the title. You also have
to change the sign of the magnetic field if you do that (and other
things as well perhaps..)

I'm under the impression that for a charge travelling at a constant velocity, say: when its velocity is reversed, the E field it creates is the same as before but the sign of the B field flips

Exactly, so "only reversing the velocity" in your title is
incorrect. Both the B-field and the velocities flip sign.

You can argue that change of B is somehow "implicit" in what
you write, but it is the opposite: what you write simply is
not possible. You cannot "only" change the velocity without
changing something else.

Or else you could also argue that the change is correctly
phrased as: "by only reversing the sign of the B-field" and
that the necessary change in velocity is then implicit. It
always follows from the fields which source there must be
present, like it follows from the sources which fields there
must be (if incoming fields are excluded).

What you mean is not "by only reversing the velocities" but
"by only specifying that the velocities are reversed" (which
leaves room for other changes.)

--
Jos

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