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Talk:Counterexamples to Relativity (view source)
Revision as of 01:33, October 15, 2011
, 01:33, October 15, 2011→Path Independence and Photons
As well, you state: "Relativity requires that anything traveling at the speed of light must have mass zero, so it must have momentum zero. But the laws of electrodynamics require that light have nonzero momentum." Yes, if you plug in the formula for non-relativistic, Newtonian momentum, you get zero, but SR gives you the equation E<sup>2</sup>=(pc)<sup>2</sup>+(mc<sup>2</sup>)<sup>2</sup>. Plug m=0 in. You get E=pc. So, the momentum is nonzero. That's a very well-known result... [[User:AndyFrankinson|AndyFrankinson]] 12:06, 1 October 2011 (EDT)
As well, you state: "Relativity requires that anything traveling at the speed of light must have mass zero, so it must have momentum zero. But the laws of electrodynamics require that light have nonzero momentum." Yes, if you plug in the formula for non-relativistic, Newtonian momentum, you get zero, but SR gives you the equation E<sup>2</sup>=(pc)<sup>2</sup>+(mc<sup>2</sup>)<sup>2</sup>. Plug m=0 in. You get E=pc. So, the momentum is nonzero. That's a very well-known result... [[User:AndyFrankinson|AndyFrankinson]] 12:06, 1 October 2011 (EDT)
:So should I take the "counter example" down...? Not to be rude but can someone at least respond to me? [[User:AndyFrankinson|AndyFrankinson]] 19:54, 14 October 2011 (EDT)
:So should I take the "counter example" down...? Not to be rude but can someone at least respond to me? [[User:AndyFrankinson|AndyFrankinson]] 19:54, 14 October 2011 (EDT)
::Take it down, then if someone reverts (which is likely) it'll provoke a short discussion that probably won't reach a consensus. Regardless, you're unlikely to succeed in keeping it changed. I'm not trying to discourage, go ahead and try changing it, I'm just letting you know what to expect. - [[User:JamesCA|JamesCA]] 21:33, 14 October 2011 (EDT)
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