• Is there a relationship between free radiation and the T-symmetry of Ma

    From john mcandrew@21:1/5 to All on Mon Jun 29 16:41:14 2020
    Imagine a closed electromagnetic system of charges interacting with one another at some time t. The field at every point can be written as the superposition of the Lienard-Wiechert field from every charge q_n at retarded time t_n = t - R_n/c.

    Now consider a time reversed system of the above via: dt -> -dt, j -> -j, B -> -B. At the same time t, the retarded time t'' = t - R*_n/c. In particular, R_n and R*_n are generally different for each charge q_n since their trajectories for the two cases
    are generally different. But this now creates a problem for my original assumption that the system is time reversible!

    A solution is to add another field that isn't coupled to the sources and hence is "free". This then alters the trajectories of the charges to ensure the system is time reversible at all t, but at the expense of no longer being coupled to the sources,
    becoming free radiation.

    Any comments correcting or adding to the above would be appreciated.

    John McAndrew

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  • From Jos Bergervoet@21:1/5 to john mcandrew on Sun Jul 5 21:35:35 2020
    On 20/06/30 1:41 AM, john mcandrew wrote:
    Imagine a closed electromagnetic system of charges interacting with one another at some time t. The field at every point can be written as the superposition of the Lienard-Wiechert field from every charge q_n at retarded time t_n = t - R_n/c.

    Your system is not "closed" unless you also give it a surrounding box,
    or some other hull, to keep any radiation from escaping. If you don't do
    that there will constantly be radiation escaping to infinity and also
    you have to add mechanical energy to your radiating charges all the
    time. I wouldn't call that a closed system.


    Now consider a time reversed system of the above via: dt -> -dt, j -> -j, B -> -B. At the same time t, the retarded time t'' = t - R*_n/c. In particular, R_n and R*_n are generally different for each charge q_n since their trajectories for the two
    cases are generally different. But this now creates a problem for my original assumption that the system is time reversible!

    Why? Logically the retarded time of the first system would become the *advanced* time for the reversed system. No reason why their reversed
    times need to be related.

    A solution is to add another field that isn't coupled to the sources and hence is "free".

    How would that solve a problem with the defined retarded time? It just
    is what it is: the particle's time where it crosses the retarded light
    cone seen from the observer.

    This then alters the trajectories of the charges to ensure the system is time reversible at all t,

    But it already was time-reversable, you actually did reverse it! (Or
    else your whole story about the "second" system that we get by time-
    reversal makes no sense.. Anyway we know that EM is time reversible,
    so that discussion seems irrelevant.)

    but at the expense of no longer being coupled to the sources, becoming free radiation.

    There *never was* a rule that EM radiation should be "coupled to
    sources" in general. If there is such a coupling in a certain system
    then that is a special case, but then don't be surprised if it is
    not the case in the time-reversed system.


    Any comments correcting or adding to the above would be appreciated.

    My correction would be that the system simply *is* time-reversible,
    that should not be the discussion to begin with. Apart from that, your conclusion seems basically valid: this system has sources for all its
    radiation when viewed in forward time, but does not have such a relation
    in backward time. (But in backward time it has a "sink" for all its
    occurring radiation!)

    --
    Jos

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  • From john mcandrew@21:1/5 to Jos Bergervoet on Sun Jul 5 16:16:04 2020
    On Sunday, July 5, 2020 at 8:35:37 PM UTC+1, Jos Bergervoet wrote:
    On 20/06/30 1:41 AM, john mcandrew wrote:
    Imagine a closed electromagnetic system of charges interacting with one another at some time t. The field at every point can be written as the superposition of the Lienard-Wiechert field from every charge q_n at retarded time t_n = t - R_n/c.

    Your system is not "closed" unless you also give it a surrounding box,
    or some other hull, to keep any radiation from escaping. If you don't do
    that there will constantly be radiation escaping to infinity and also
    you have to add mechanical energy to your radiating charges all the
    time. I wouldn't call that a closed system.


    Now consider a time reversed system of the above via: dt -> -dt, j -> -j, B -> -B. At the same time t, the retarded time t'' = t - R*_n/c. In particular, R_n and R*_n are generally different for each charge q_n since their trajectories for the two
    cases are generally different. But this now creates a problem for my original assumption that the system is time reversible!

    Why? Logically the retarded time of the first system would become the *advanced* time for the reversed system. No reason why their reversed
    times need to be related.

    A solution is to add another field that isn't coupled to the sources and hence is "free".

    How would that solve a problem with the defined retarded time? It just
    is what it is: the particle's time where it crosses the retarded light
    cone seen from the observer.

    This then alters the trajectories of the charges to ensure the system is time reversible at all t,

    But it already was time-reversable, you actually did reverse it! (Or
    else your whole story about the "second" system that we get by time-
    reversal makes no sense.. Anyway we know that EM is time reversible,
    so that discussion seems irrelevant.)

    but at the expense of no longer being coupled to the sources, becoming free radiation.

    There *never was* a rule that EM radiation should be "coupled to
    sources" in general. If there is such a coupling in a certain system
    then that is a special case, but then don't be surprised if it is
    not the case in the time-reversed system.


    Any comments correcting or adding to the above would be appreciated.

    My correction would be that the system simply *is* time-reversible,
    that should not be the discussion to begin with. Apart from that, your conclusion seems basically valid: this system has sources for all its radiation when viewed in forward time, but does not have such a relation
    in backward time. (But in backward time it has a "sink" for all its
    occurring radiation!)

    --
    Jos

    Yes, and I find this extraordinary, although too subtle for me previously to have seen its profound importance. Therefore this extra radiation in the time-reversed case is given by:

    Radiation field = Retarded field - Advanced field

    As defined by Dirac in his classic paper in equation (11): https://royalsocietypublishing.org/doi/pdf/10.1098/rspa.1938.0124

    But it's only now that I can see why it's obviously correct: as a consequence of the time reversibility of Maxwell's equations, and their solution in terms of the sources. Also, it shows that it's mathematically correct to treat the time reversed fields
    as composed of a source, and a sourceless field converging upon a different source; with their effects upon this source added together to give the time reversed effect.

    In the above I've quoted "extra radiation". In fact, it might be part of the external electromagnetic system required to time reverse the closed one, but has been sloppily unaccounted for. I'll have to think about this.

    John McAndrew.

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