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Talk:Counterexamples to Relativity (view source)
Revision as of 12:58, August 16, 2011
, 12:58, August 16, 2011→A point of view on this article
: On point (7), I think there is circularity in your argument, assuming what it is at issue by insisting that E remain constant.--[[User:Aschlafly|Andy Schlafly]] 20:05, 15 August 2011 (EDT)
: On point (7), I think there is circularity in your argument, assuming what it is at issue by insisting that E remain constant.--[[User:Aschlafly|Andy Schlafly]] 20:05, 15 August 2011 (EDT)
::It's not so easy as you think. To invalidate a theory with an experiment you have to carefully consider every other aspect that could influence the measure. If a theory works good, but a new experiment finds that in some conditions the theory doesn't work anymore, the experiments has not falsified the theory, but only proven that this theory has a finite range of applications. Examples are Newtonian mechanics, classical electromagnetism, even nonrelativistic quantum mechanics. Today general relativity has not been proved correct on big scales, while it works correctly within the solar system. Special relativity is thought not to have similar (distance) limits, but may have corrections due to quantum gravity, which today are only conjectured and not seen experimentally. I'll tell you a secret: "true" theories do not exist, and all the theories have limits of applicability. Today
we don't know the limits of both relativity theories, because they agree with ''all'' current experiments.
::(16) The truth is that who doesn't know relativity is ignoring 100 years of history of physics.
::Andy, let me explain to you some math stuff. Suppose you have a function of two variables, e.g. ''f(x,y)''. Suppose you want to change the value of the first variabile: what does the second? Well, you have to decide it: you can keep ''y'' to a fixed value, or you can change ''y'' as ''x'' varies: in this case ''y'' become a function of ''x'': ''y=g(x)''. The original expression becomes ''f(x,g(x))'' and is a 1-variable function, that you can call ''h(x)''. In other words, you have to specify a trajectory in the ''(x,y)'' plane. This is also the way limits work: you have to specify the trajectory.
::In our case we want to send ''m'' to zero. And the other variabiles (energy, momentum, speed)? Since the relation ''E²=(pc)²+(mc²)²'' holds, energy and momentum are not indipendent. If you vary ''m'', ''E'' and ''p'' cannot be both costant. A possible choice is keeping fixed ''E'' and vary ''p'', but you can also vary ''E'' keeping ''p'' fixed. Or you can vary both, specifying a function ''E(m)'' or ''p(m)'', as before. You might think that the final result can depend on the limit choice, and this is indeed the case for ''E'' and ''p''. BUT, when ''m=0'' you have ''E²=(pc)²'' for every path choice. A a consequence, the speed of the particle equals the speed of light for any energy or momentum value, since the relation ''v=c(pc/E)'' holds for every mass. If you take the limit keeping ''p'' fixed you'll obtain exactly the same speed for the particle. No circularities.
::Andy, what will you do with these points? --[[User:MM87|<font color="#E56717">'''MM'''</font>]]''[[User_talk:MM87|<font color="#FBB117">'''''87'''''</font>]] 08:58, 16 August 2011 (EDT)
== Attempt at Helping the Consnesus Along ==
== Attempt at Helping the Consnesus Along ==