Re: Are There Unresolver Foundational Issues With GR

On Apr 28, 8:51 pm, Koobee Wublee <koobee.wub...@xxxxxxxxx> wrote:
On Apr 28, 12:14 am, JanPB <film...@xxxxxxxxx> wrote:

On Apr 27, 11:19 pm, Koobee Wublee <koobee.wub...@xxxxxxxxx> wrote:
OK. But you didn't answer the question that's really basic here:
how two functions assigning the same lengths and the same angles to
the same vectors can STILL be considered physically different. What
distinguishes them, physically?

This is a stupid question.

No, this is the root of your confusion. You claim that two different
expressions of the same geometry (what you call "two different
metrics") are to be treated as physically different.

This question remains the most stupid as I have seriously
encountered. It is not that I refuse to answer that question. I am
still appalled by your lack of understanding in the very fundamental
algebra. Sometimes, I am even debating with myself if I should
continue with this pointless discussion with you on the most basic of
the mathematics involved.

You are now gradually changing the topics of our discussion. Obviously
if one has two different "metrics" as you call them (meaning, two
different matrices of g_ij's) written wrt one and the same coordinate
system, then they represent different geometries.

This was never under the discussion.

What we are debating is your claim that there are infinitely many
solutions to EFE in the spherically symmetric vacuum case. You gave
several examples of different line elements claiming they were
physically different solutions. These line elements differed by a
coordinate change - so what you said above does not apply. Nobody here
ever claimed than in a _fixed coordinate system_ different line
elements yielded the same geometry.

Here is the problem. Since you keep insisting on existence of
different solutions besides Schwarzschild's I'm simply asking (for the
third time) HOW are they different? IOW, what can an observer
physically do to tell these "infinitely many" solutions of yours
apart?

You've agreed that geometry is determined by an assignment of lengths
and angles to tangent vectors. So obviously there must be something
ELSE in your mind besides lengths and angles that is physically
detectable and is implied by the EFE. WHAT IS IT? Is this some kind of
a secret?

So I'm asking a simple question: HOW can a person - some observer in
space or whatever - tell the difference?

Your question has no relevance to our discussion.

Just answer it, please. (Of course I think it's not only relevant,
it's the crux of the matter, that's why I'm asking it.)

I am not claiming
two different expressions for the same geometry. I am claiming two
different metrics using the same coordinate system must each
represents a different geometry. <shrug>

Of course. But that was never under the discussion. That is not what
Crothers claims. What you and Crothers have been saying was that there
was more than one solution to the EFE in the spherically symmetric
vacuum case.

Each such solution is a function assigning lengths and angles to
vectors. So you and Crothers claim that there is more than one such
function satisfying EFE. And yet ALL examples he and you produced so
far were of the following form: ONE AND THE SAME FUNCTION assigning
lengths and angles to vectors, ONLY WRITTEN IN DIFFERENT COORDINATES.
Same function assigning lengths and angles to vectors. Same solution.

I'm not at all surprised you refuse to answer it.

Before I answer this stupid question, you
have to answer the following question first.

Given a quadratic equation of the following,

x^2 - 3 x + 2 = 0

The solutions are (x = 2) or (x = 1).

What make all solutions equivalent? What makes (x = 1 = 2)?

This is of course irrelevant. Here you have an equation for points in
the parameter space R^1 - not a manifold. Einstein's equation OTOH is
an equation for an unknown function between two manifolds.

This is very relevant. Your question is like asking the above.

It's not. I've explained why.

Or
better yet let's throw in the units. Given the following,

x^2 - 3 dollars * x + 2 dollars^2 = 0

What are the solutions?

Your stupid question is like asking how (x = 1 dollar) and (x = 2
dollars) can be in co-existence at the same time. In your twisted
logic, you somehow conjured up another unit say 'gollar' where (1
gollar = 0.50 dollar). Therefore, continuing with your fouled
mathematics, you are claiming (x = 1 dollar) and (x = 2 gollars).
Since (x = 1 dollar = 2 gollars), therefore the above quadratic
equation really has only one solution. This is called matheMagic.

You really need to go back to the basics of algebra. I cannot help
you on this one.

Nonsense.

--
Jan Bielawski

.

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