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From: pmb on 27 May 2010 18:23 On May 25, 7:39 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > ..@..(Henry Wilson DSc) wrote: > > I would like a relativist to explain why the distance term ct is imaginary > > in 4D 'spacetime'. > > It isn't. > > Don't make stuff up, Ralph. Actually it "can" be imaginary. It just depends on the conventions an author chooses to use. In fact some modern textbooks still use that convention. For example, see "Basic Relativity" by Richard A. Mould, Springer Press. See section 3.7 "Four Vectors" on page 71. Usually the usage is reserved for special relativity. When this is the case the metric is g = diag(1, 1, 1, 1), i.e. the metric to a 4-D Euclidean space. Pete
From: Androcles on 27 May 2010 18:32 "pmb" <pmb61(a)hotmail.com> wrote in message news:eb141483-7202-4966-a50e-63ce6d16aebd(a)b21g2000vbh.googlegroups.com... On May 25, 7:39 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > ..@..(Henry Wilson DSc) wrote: > > I would like a relativist to explain why the distance term ct is > > imaginary > > in 4D 'spacetime'. > > It isn't. > > Don't make stuff up, Ralph. Actually it "can" be imaginary. It just depends on the conventions an author chooses to use. In fact some modern textbooks still use that convention. For example, see "Basic Relativity" by Richard A. Mould, Springer Press. See section 3.7 "Four Vectors" on page 71. Usually the usage is reserved for special relativity. When this is the case the metric is g = diag(1, 1, 1, 1), i.e. the metric to a 4-D Euclidean space. Pete ================================================ "Piggy Brown" <pmb61(a)hotmail.com> wrote in message news:7b6d6b55-b3b3-4f35-95bf-fc1fabba2cea(a)e38g2000yqa.googlegroups.com... "Don't just claim it. Nobody is interested in what you believe. Prove it." -- Piggy Mad Brown
From: Henry Wilson DSc on 27 May 2010 18:41 On Thu, 27 May 2010 12:47:34 -0700 (PDT), eon <ynes95v6(a)techemail.com> wrote: >On May 27, 9:22 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > >[mercifully ...] > >> >> There are no negative areas or volumes in relativity. You are making things >> up - again. > >why, are there negative lengths in nature? poor little eric reads too many books that he cannot understand. The Minkowski metric defines a perfectly Euclidean negative area.... QED Henry Wilson... ........Einstein's Relativity...The religion that worships negative areas.
From: eric gisse on 27 May 2010 20:59 ...@..(Henry Wilson DSc) wrote: [...] > > The metric defines a negative area. Have you seen any of those lately? ds^2 = -dt^2 + dx^2 + dy^2 + dz^2 Show us how to get from there to the negative area you claim exists. Show your work. [...]
From: pmb on 27 May 2010 22:05
On May 27, 6:38 pm, ..@..(Henry Wilson DSc) wrote: > On Thu, 27 May 2010 15:23:23 -0700 (PDT), pmb <pm...(a)hotmail.com> wrote: > >On May 25, 7:39 pm, eric gisse <jowr.pi.nos...(a)gmail.com> wrote: > >> ..@..(Henry Wilson DSc) wrote: > >> > I would like a relativist to explain why the distance term ct is imaginary > >> > in 4D 'spacetime'. > > >> It isn't. > > >> Don't make stuff up, Ralph. > > >Actually it "can" be imaginary. It just depends on the conventions an > >author chooses to use. In fact some modern textbooks still use that > >convention. For example, see "Basic Relativity" by Richard A. Mould, > >Springer Press. See section 3.7 "Four Vectors" on page 71. Usually the > >usage is reserved for special relativity. When this is the case the > >metric is g = diag(1, 1, 1, 1), i.e. the metric to a 4-D Euclidean > >space. > > Forget about the 4D jargon...just use x and t. It's a lot easier. > > Minkowski's s^2 is a negative area.....no such thing exists... > Multiply it by a length and you have a negative volume...Hahahhahha! It is illogical to claim that because a quantity is squared then it must represent an area. Pete |