Euclidean vs. Minkowski 4-space

**Ken S. Tucker** · #1 August 28th 03 posted to sci.physics.relativity

(Jose B. Almeida) wrote in message . com...
(Ken S. Tucker) wrote in message . com...
(Ken S. Tucker) wrote in message . com...

Now you may sub g_tt=1, g_xx=1 and recover Minkowski's

(ds)^2 = (c dt)^2 - (dx)^2.

The metric tensor matrix (absent of fields) for t and x is

g = |g_uv| = |1 -V(x) | = 1 - V(x)^2
| -V(x) 1 |

I like what I see so far; did you write something more formal?

Yes, I've used and studied it extensively, IMO it's simpler than
the conventional g_tt=1 g_xx=-1, it keeps ds perpendicular to
dx dy and dz, which is something you seem to want to do.

I would also welcome comments on these 2 eprints:
/physics/0201002; gr-qc/0107083.

gr-qc/0107083.

Well I think you'll have problems with
c^dt^2 = g_uv dx^u dx^v being accepted, the solution
I have suggested may help you there, (of course if you
want you can e-mail me on this), if I can help.

Eq. (47) is good looking, although I ended up with
antisymmetrical metrics such as g_01 = -g_10 in
the presence of electric fields.

You may want to check the spelling of Schwarzschild
throughout this posting, (I counted 3 different spellings,
and I figure the're all the same person!, see for instance
References #3)

Jose

Regards, Ken S. Tucker

**Jose B. Almeida** · #2 August 29th 03 posted to sci.physics.relativity

(Ken S. Tucker) wrote in message . com...
Well I think you'll have problems with
c^dt^2 = g_uv dx^u dx^v being accepted, the solution
I have suggested may help you there, (of course if you
want you can e-mail me on this), if I can help.

O.K. Let's clarify this point; the following was rejected in sci.physics.research.

(Jose B. Almeida) wrote in message
. com...

You can also redefine the proper time tau as

c^2dtau2 = dx2 + dy2 + dz2 + c^2dt2 (1)

Thanks David for posting this. In different words you are saying very
much the same as I said in thread "Euclidean vs. Minkowski 4-space"
and in several of my arxiv eprints.

I must clarify my position because with most people I also think
Eq. (1) is wrong but had it be written

c^2dt2 = dx2 + dy2 + dz2 + c^2dtau2 (2)

and it would have been right, in spite of the fact that I don't
know anyone who shares this opinion. Let me explain with an
example:

[Moderator's comment:
Are you saying that most physicists don't believe equation (2) is right?
That's clearly not true: equation (2) is a trivial algebraic
rearrangement of the equation below it, which every physicists
believes in.

Second, I don't understand the point of the argument that
makes up the rest of the post. If you're just trying to show
that equation (2) is correct, why not do the trivial one step
of algebra that derives it from the subsequent equation?]

The relativistic interval is given (even for me) by

c^2dtau2 = c^2dt2 - dx2 - dy2 - dz2

now consider a test particle moving with velocity v, such that

dt = dx/v (3)

by straightforward replacement we get, for this particle

c^2dtau2 = (c2/v2 -1)dx2 (4)

everyone agrees that this represents a straight line on the
(x,tau) plane, with slope sqrt(1/v2 - 1/c2); with need only the
ordinate at x=0 which we can choose as t0, the time at which the
test particle passed at the origin. We then have

tau = t0 + sqrt(1/v2 -1/c2) x

For this test particle we are allowed to write Eq. (2), as can be
verified by replacement with (3) and (4).

Now consider any other test particle, with different velocity in
any desired direction, and a similar verification can be made.
Hence, for all possible test particles Eq.(2) is verified. The
Universe can indeed be described in Euclidean 4-space and it looks
different from the relativistic description we are used to.

Jose

PS: If I hadn't read Kafka this would make a good substitute
Jose

**Ken S. Tucker** · #3 August 31st 03 posted to sci.physics.relativity

(Jose B. Almeida) wrote in message . com...
(Ken S. Tucker) wrote in message . com...

Sorry to be slow to respond, my mouse went flaky, and
can't get another one until Tuesday.

Yes, I've used and studied it extensively, IMO it's simpler than
the conventional g_tt=1 g_xx=-1, it keeps ds perpendicular to
dx dy and dz, which is something you seem to want to do.

Where can I get access to your studies?

I summarized the concept is this thread...

From: (Ken S. Tucker)
Newsgroups: sci.physics.relativity
Subject: CS group with all time axi parallel.
NNTP-Posting-Host: 209.91.148.56
Message-ID:

Then I did an acceleration application in this thread...

From: (Ken S. Tucker)
Newsgroups: sci.physics.relativity
Subject: More on Time Axi Parallel CS's
NNTP-Posting-Host: 209.91.148.88
Message-ID:

Naturally you can post questions or comments to these
threads. I'd be happy to elaborate.

Well I think you'll have problems with
c^dt^2 = g_uv dx^u dx^v being accepted, the solution
I have suggested may help you there, (of course if you
want you can e-mail me on this), if I can help.

You're right in guessing that nobody accepts this solution,
nevertheless I think it is a legitimate alternative.

The problem is that dt^2 is not invariant
but g_uv dx^u dx^v is. If you want 4 orthogonal
axes (ds dx dy dt) orthogonal then my suggestion
will do that.

You may want to
have a look at thread "4-dimensional optics".

Ok, mouse permitting.

Jose

Regards Ken S. Tucker

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