When a given arbitrary Fadeev-Popov-Trace eigenstate is to be isotropically stable, its directly corresponding first-order light-cone-gauge eigenstate, will consequently tend to have a relatively enhanced probability, of being isotropically stable. Whereas; When a given arbitrary Fadeev-Popov-Trace eigenstate is to lack isotropic stability, its directly corresponding first-order light-cone-gauge eigenstate, will consequently tend to have a relatively lowered probability, of being isotropically stable. Sincerely, SAM.
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