The Math is still Not Ready

On Jan 5, 8:54 am, Tom Roberts wrote:

Here is General Relativity:

On a 4-d Lorentzian manifold M,
G = T

where G is the Einstein curvature tensor and T is the energy-momentum tensor.

Please allow Koobee Wublee reminds Tom where that overly simplified
equation[s] above come from. Let’s follow Hilbert’s footsteps and
pull out the following so-called Lagrangian out of Hilbert’s ass.

** L = (R / K + rho) sqrt(-det[g])

Where

** L = Lagrangian
** R = Ricci scalar
** K = Constant
** rho = Mass density
** [g] = The metric (a 4x4 matrix)
** det[] = Determinant of a matrix

For the language of convention in this case, [A] means a matrix with
elements [A]_ijk... or [A]^ijk...

The field equations can be derived in just one step by taking the
partial derivative of the Lagrangian above with respect to [g^-1]^ij
where [g^-1], a matrix, is the inverse of [g], another matrix, and
after setting each of the partial derivative to null, the result is
the following relationships of matrices.

** [R] – R [g] / 2 = K rho [g] / 2

Where

** [R] = Ricci tensor (another 4x4 matrix)

You would call the following.

** [G] = [R] – R [g] / 2
** [T] = K rho [g] / 2

Thus,

** [G] = [T]

Koobee Wublee would also like to remind Tom that the above equation
has never been tested with any experimentations, and the best Tom can
hope for is the following where the energy momentum tensor is null.

** [G] = 0

Where

** [T] = 0

Using only diagonal [g], the equation above simplifies into the
following where the effect of the ever so celebrated trace term is
nullified. The null Ricci tensor was basically Nordstrom’s work where
Schwarzschild had been working on the solution for years. That is why
within a couple months after Hilbert presented the field equations,
Schwarzschild published a solution.

** [R] = 0, first proposed by Nordstrom as the field equations

Where

** R [g] / 2 = The trace term

shrug

To get SR from GR:

Riemann = 0
Top(M) ~ R^4

where Riemann is the Riemann curvature tensor on M, and Top(M) is the topology of M.

Nonsense, Tom. If the Riemann tensor is null, the Ricci tensor must
be null as well in which you end up with the null Ricci tensor above
where you can solve for the Schwarzschild metric and other equally
valid solutions that are able to degenerate into Newtonian law of
gravity at weak curvature in spacetime. shrug

The best way to get SR from GR is to set the gravitating mass, M, to
0, duh! shrug

** ds^2 = c^2 (1 – 2 U) dt^2 – dr^2 / (1 – 2 U) – r^2 dO^2

Where

** U = G M / c^2 / r

[Note that approximations are important in applying the theory,
as this is PHYSICS, not math. Based on your writings around
here, and your aversion to any intellectual effort, I estimate
you will never understand this.]

Tom, in GR, physics = math, and math = physics. So, start
understanding the mathematics involved instead of wishing for what you
believe in. shrug

Faith should not come into any equations of science, no? shrug

On Jan 6, 10:52 am, Jimmy Kesler wrote:
Koobee Wublee wrote:

For the language of convention in this case, [A] means a matrix with
elements [A]_ijk... or [A]^ijk...

three indices matrix?

Yes, Jimmy. Three indices indicate rank-three matrices. Riemann
curvature tensors are rank-four matrices while both the Ricci and the
metric tensors are rank-two matrices. shrug

where are the ten field equation then

R_{\mu \nu} - {1 \over 2}g_{\mu \nu}\,R + g_{\mu \nu} \Lambda = {8 \pi G
\over c^4} T_{\mu \nu}

The field equations actually involves with 4x4 matrices. Yes, they
are all the familiar rank-two matrices. These 4x4 matrices should
give you 16 elements, and each element is a partial differential
equations. When space and time are not intertwined like the
scientific communities have decided time and space should intertwine,
all correlations of elements between time and space become null. That
is what you are thinking of only 10 equations. However, for almost
all practical applications, only diagonal metrics are considered. In
doing so, there are only 4 differential equations you have to worry
about. shrug

and how come the expansion is metric tensor dependent, ie the universe
expand accordingly more at the regions with intense curvature (presence
of matter); why not the universe expands flat? what tells it must expand
like this?

Not sure what you are talking about, but the expansion of the universe
can be found with the de Sitter and the Friedman-LeMaitre-Robertson-
Walker metrics where both do not satisfy Newtonian law of gravity at
short distances. You can apply Koobee Wublee’s theorem of parallel
variations to find a modified de Sitter metric that satisfies the
Newtonian law of gravity. You can also modify the Schwarzschild
metric to allow an expanding universe at cosmic scales. shrug

this was the first question

and also, i dont understand a half of a metric tensor, lol, what on earth
is a half of a metric tensor!!

Half a metric tensor is simply ([g] / 2) where [g] is a 4x4 matrix
describing spacetime. Not sure what is not to understand? shrug

On Sat, 5 Jan 2013 21:36:26 -0800 (PST), Koobee Wublee
wrote:

On Jan 5, 8:54 am, Tom Roberts wrote:

Here is General Relativity:

On a 4-d Lorentzian manifold M,
G = T

where G is the Einstein curvature tensor and T is the energy-momentum tensor.

Please allow Koobee Wublee reminds Tom where that overly simplified
equation[s] above come from. Let’s follow Hilbert’s footsteps and
pull out the following so-called Lagrangian out of Hilbert’s ass.

** L = (R / K + rho)sqrt(-det[g])

sqrt(-det[g])?
Why should it be necessary to first make the determinant negative? (we
can all see the algebraic requirement of course).

Don't you have any suspicions about such a fictitious looking term?

I have pointed this out befo the metric tensor g is invalid. The
term g00 = -1 is purely fraudulent, an arrangement calculated to avoid
the product ict x ict and make it look like other real dimensions:
e.g. ct x ct.
This is gloatingly described in Gravitation by MTWheeler, "Farewell
to ict".
I think we can agree that it is invalid to make major changes in the
coefficients of a matrix like g, just to make up for the defects in
the vector field.
g is Diagonal and is meant strictly for stretching, but at the same
time With a negative determinant it is thereby inadvertently
converting positive volumes into negative ones, which is clearly
impermissible.
It is regrettable that this duplicity has not been challenged
anywhere, but it should be up for discussion.
The time coordinate has to be retained as ict and it can never legally
be promoted as an additional dimension that can be matched up with the
real XYZ.

Snip

Faith should not come into any equations of science, no? shrug
No.

John Polasek

On Mar 17, 2:15 pm, wrote:
On Sat, 5 Jan 2013, Koobee Wublee wrote:

Please allow Koobee Wublee reminds Tom where that overly simplified
equation[s] above come from. Let’s follow Hilbert’s footsteps and
pull out the following so-called Lagrangian out of Hilbert’s ass.

** L = (R / K + rho)sqrt(-det[g])

sqrt(-det[g])?

The determinant of the metric is negative. So, sqrt(-det[g]) is a
real number. shrug

Why should it be necessary to first make the determinant negative? (we
can all see the algebraic requirement of course).

That is because nothing can travel beyond the speed of light. shrug

Don't you have any suspicions about such a fictitious looking term?

Yes, of course. shrug

I have pointed this out befo the metric tensor g is invalid. The
term g00 = -1 is purely fraudulent, an arrangement calculated to avoid
the product ict x ict and make it look like other real dimensions:
e.g. ct x ct.

[g]_00 (your g00) is not -1. It is +1 --- (1 – 2 U) thing. shrug

This is gloatingly described in Gravitation by MTWheeler, "Farewell
to ict".

If nothing can travel beyond the speed of light, the signature of the
metric ought to be (+1, -1, -1, -1). shrug

I think we can agree that it is invalid to make major changes in the
coefficients of a matrix like g, just to make up for the defects in
the vector field.

The fault of GR starts way before the construction of spacetime.
shrug

g is Diagonal and is meant strictly for stretching, but at the same
time With a negative determinant it is thereby inadvertently
converting positive volumes into negative ones, which is clearly
impermissible.

[g] (your g) does not have to be diagonal. It is made diagonal to
simplify the already complex math. If [g] is not diagonal, it would
be relatively impossible to solve for the field equations. shrug

It is regrettable that this duplicity has not been challenged
anywhere, but it should be up for discussion.
The time coordinate has to be retained as ict and it can never legally
be promoted as an additional dimension that can be matched up with the
real XYZ.

There is no ict thing if the signature is (+1, -1, -1, -1). shrug

Faith should not come into any equations of science, no? shrug

No.

amen

On Sun, 17 Mar 2013 21:10:16 -0700 (PDT), Koobee Wublee
wrote:

On Mar 17, 2:15 pm, wrote:
On Sat, 5 Jan 2013, Koobee Wublee wrote:


If nothing can travel beyond the speed of light, the signature of the
metric ought to be (+1, -1, -1, -1). shrug
Look at what you have gone and done, proved negative volume. This
format is just as flaky.
John Polasek