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On Thu, 03 Jun 2004 16:18:36 -0500, Cecil Moore
wrote: Walter Maxwell wrote: Cecil, this is exactly what I've been trying to persuade you of, but always said no, there is no short developed. But you must also agree that under this condition the current doubles. Nope, for complete destructive interference, both the E-field and H-field associated with the two interfering waves collapse to zero. For the resulting complete constructive interference, the ratio of the E-field to the H-field equals the characteristic impedance of the medium. The two corresponding rearward-traveling superposing currents might be: (2a at 180 degrees) superposed with (2a at zero degrees) The superposed sum of the two rearward-traveling currents is zero. This acts like an open where currents go to zero. Of course, but the voltage doubles. Nope, again here are the two sets of reflected waves. #1 100v at zero degrees and 2a at 180 degrees = 200W #2 100v at 180 degrees and 2a at zero degrees = 200W Superposing those two reflected waves yields zero volts and zero amps. Well, Cecil, here's where we part company to a degree. Unlike voltage and current that can go to zero simultaneously only in the rearward direction, E and H fields can never go to zero simultaneously. For "complete destructive interference" as explained in _Optics_, by Hecht, the E-field and B-field (H-field) indeed do go to zero simultaneously. That is what causes a completely dark ring in a set of interference rings. Of course, a resulting corresponding complete constructive interference causes the brightest of rings. If Steve understands the action of the fields in the EM wave it's hard to understand why he finds it so erroneous to associate voltage and current with the their respective fields in impedance matching. Apparently he can't conceive that the voltages and currents in reflected waves can be considered to have been delivered by separate generators connected with opposing polarities. He pretty much ignored current. His power equations are exact copies of the light irradiance interference equations from optics, but he apparently didn't realize it until I pointed it out to him. Well, Cecil, perhaps I don't understand Melles-Griot. However, when an EM wave encounters a short circuit the E field goes to zero, but in its change from normal level to zero, that changing field develops a corresponding H field that adds to its original value, causing it to double--the H field does NOT go to zero. But as it returns to its normal value that change in turn develops a new E field propagating in the opposite direction. From there on the EM field reconstitutes itself with normal E and H fields of equal value, each supporting one half of the propagating energy. This sequence must also prevail in optics--are you sure you are interpreting your double zero at the correct point in the circles? Walt |