From: pmb on
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

"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
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
...@..(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
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
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