Minkowski Metric.
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From: PD on 28 May 2010 11:44 On May 28, 10:37 am, kenseto <kens...(a)erinet.com> wrote: > On May 25, 6:09 pm, ..@..(Henry Wilson DSc) wrote: > > > I would like a relativist to explain why the distance term ct is imaginary in > > 4D 'spacetime'. > > > Does this imply that the whole theory is just a figment of Einstein's > > imagination? > > SR uses a rubber meter stick (1 meter=1/299,792,458 light-second) to > measure length and rubber second a rubber second to measure > time.....as such it cannot not be disproved. First question is, why do you feel the need to disprove relativity? Second comment is, it's simple to disprove a theory. A theory makes predictions of measurable quantities -- that is, what will be measured under certain circumstances -- and then you go make the measurement. If they measurement disagrees with the prediction, then the theory is disproven. > Definition for a rubber second: Every SR observer assumes that his > clock second is a standard unit of time and yet at the same time the > passage of a clock second in A's frame does not correspond to the > passage of a clock second in B's frame. What this mean is that a clock > second in different frame have different duration (different time > content). What you THINK a term is defined as, and what it REALLY is defined as, are two completely different things. You haven't caught onto that yet.
From: eric gisse on 28 May 2010 07:28 ...@..(Henry Wilson DSc) wrote: > On Wed, 26 May 2010 19:10:32 -0700, eric gisse <jowr.pi.nospam(a)gmail.com> > wrote: > >>..@..(Henry Wilson DSc) wrote: >> >>> On Wed, 26 May 2010 17:51:50 -0700, eric gisse >>> <jowr.pi.nospam(a)gmail.com> wrote: >>> > >>>>> >>>>> Its 'pathlength' in spacetime should be given by s^2 = (vt)^2 + >>>>> (ct)^2. >>>> >>>>Except it isn't, nor should it be. >>> >>> Why not? >> >>Because the geometry that SR requires is not Euclid. >> >>Minkowski figured this out over a century ago. > > He figured how to make a quick dollar by backing the idiot Einstein. Really, a 'quick dollar'? > > > Henry Wilson... > > .......A relativist's IQ = his snipping ability.
From: eric gisse on 28 May 2010 07:34 PD wrote: [...] >> Minkowski's s^2 is a negative area.....no such thing exists... >> Multiply it by a length and you have a negative volume...Hahahhahha! > > Hmmm, Pythagoras's rule is that c^2 = a^2 + b^2. So according to you, > c^2 must be related to the area of the triangle. And this in turn > implies that any two triangles that have the same value of c^2 must > have common area. Hmmmm.... I have to admit, Henri is being wrong & stupid in a new and interesting way. The stupidity of folks like Androcles is static, but Henri is always innovating. [...]
From: PD on 28 May 2010 16:29 On May 28, 6:34 am, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > PD wrote: > > [...] > > >> Minkowski's s^2 is a negative area.....no such thing exists... > >> Multiply it by a length and you have a negative volume...Hahahhahha! > > > Hmmm, Pythagoras's rule is that c^2 = a^2 + b^2. So according to you, > > c^2 must be related to the area of the triangle. And this in turn > > implies that any two triangles that have the same value of c^2 must > > have common area. Hmmmm.... > > I have to admit, Henri is being wrong & stupid in a new and interesting way. > > The stupidity of folks like Androcles is static, but Henri is always > innovating. It's like he is easily distracted by shiny, sharp objects, and then he picks them up and runs with them until he runs into a door or a piece of furniture and hurts himself. PD
From: Koobee Wublee on 29 May 2010 01:13
On May 28, 8:44 am, PD <thedraperfam...(a)gmail.com> wrote: > First question is, why do you feel the need to disprove relativity? It is not a need to but what experiment shows. <shrug> > Second comment is, it's simple to disprove a theory. Yes, it is. Just go back to the null results of the MMX. <shrug> > A theory makes > predictions of measurable quantities -- that is, what will be measured > under certain circumstances -- and then you go make the measurement. > If they measurement disagrees with the prediction, then the theory is > disproven. Let's not confuse with completely disproving a theory versus a necessary modification to that theory. <shrug> In particular, http://groups.google.com/group/sci.physics.relativity/msg/c5a0a3c587fd4df4?hl=en |