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?
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
On Thursday, August 6, 2020 at 12:41:41 PM UTC+1, Jos Bergervoet wrote: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:
On 20/08/06 2:51 AM, john mcandrew wrote:
If you just use time reversal your velocities will be reversed.
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?
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
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.
On 20/08/07 1:47 AM, john mcandrew wrote: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:
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:
If you just use time reversal your velocities will be reversed.
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?
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
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..)
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
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