That's kind of how it looks.....particles that have bosons, like protons and electrons have mass, and particles that do not, like photons, don't.
Did I get that correct? I may not be older than quantum physics, but I am older than the first high school text books that covered it.
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Cheers and Stay Safe,
Marky the Sparky
OSHA 1910.304(g)(2)(iv) "One phase conductor of a multiphase system where one phase is grounded shall be grounded"
Kind of sorta depending on how picky you want to be. Send a photon into a volume sphere and that sphere will experience an increase in equivalent mass because of the mass-energy equivalence. That is not technically perfect, but close enough. I have a physics paper somewhere that gets the terminology technically and pickily correct if it is important.
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I don't know. I've subscribed to Scientific American for maybe 40 years. After throwing out all of the left wing kitty-cat articles, I can read and sort of understand maybe half of the rest. Particle physics, quantum mechanics, is way beyond my ken.The first I recall of hearing about "up, down, charm, strange, top, bottom" was maybe while I was in college, in the early 70s.
Luckily for me, string theory and gravitrons have never affected my work. The math models that cover power and control fit the real world very well. And I understand those models fairly well.
ice
Harmless flakes working togther can unleash an avalanche of destruction
You are the one bringing up mass and the change in mass via the change in energy. I'm willing to discuss a topic if you want and we can discuss invariant mass, relativistic mass, energy mass equivalants, etc. but don't put words in my mouth just so you can have something to argue against. If you want to discuss something I have said in particular, that will be fine.
You posted E=mc2 but if you want to get down to the particulars, we need the full equation for the mass conversion that also has momentum: m = sqrt(E2/c4 - p2/c2)
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Where did I promise that? IIRC, you were looking for the equation for the axial electric field in a non-ideal conductor and I promised to help you if you could not do it on your own. Since you could not, I provided that equation: E=V/L.
There is no need for me to provide the equation for the energy flow outside the ideal conductor because you already have it: the equation for the Poynting Vector, which was the basis for that whole line of discussion. In an ideal conductor, the axially-directed electric field, which is external to the conductor surface, crossed with the magnetic field encircling the conductor gives the energy vector flowing towards the load.
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