at tha time he invented and published his first
relativity he diden even knew his tensors at
that time

how strong bulivable a foundation such a theory can have
while you know his author dident knew his tensors?

On Sep 28, 5:57 pm, core duo wrote:
at tha time he invented and published his first
relativity he diden even knew his tensors at
that time

how strong bulivable a foundation such a theory can have
while you know his author dident knew his tensors?

xxein: There are no tensors that exist in the universe or beyond. We
make them up to suit our mathematical needs for representation of such
in a way that we wish to understand the universe as a math described
physic.

We can feel and understand without a mathematical measurement. It is
not so unlike a Relativity theory, but it is very local to our
feelings. It is also quite a different form of a relativity in that
it is only local.

We force a relativity beyond the local with an attempted physical
understanding of how it should work in the extended level. What we
miss is the interconnectiveness. We jump within, without, and in
between WITHOUT an interconnectivity. We fail to recognise that a
process that can happen on one scale, CAN happen on another. I am
referring to adiabacity - local processes that seem to go wildly
beyond the point of uniformly reversible reactions.

But adiabatics are not any wilder than a gravitatioinally bound-
produced universe as we now see it (from our local viewpoint of a
physic). It is just more recognizably produced as a time function for
extremely local phenonemae.

There is a lot of self-similarity that goes unrecognised as we
increase scale from Q to U.

On Sep 28, 5:57 pm, core duo wrote:
at tha time he invented and published his first
relativity he diden even knew his tensors at
that time

how strong bulivable a foundation such a theory can have
while you know his author dident knew his tensors?

See this smelly brown stuff. That's ****. See this brown stuff in
the can. That's shinola. Maybe someday you'll be able to tell the
difference.

On Sep 29, 7:45 pm, Igor wrote:
On Sep 28, 5:57 pm, core duo wrote:

at tha time he invented and published his first
relativity he diden even knew his tensors at
that time

how strong bulivable a foundation such a theory can have
while you know his author dident knew his tensors?

See this smelly brown stuff. That's ****.

what "this"?

keep you big smelly **** close on you?

isnt it unhealthy?


See this brown stuff in
the can. That's shinola. Maybe someday you'll be able to tell the
difference.

On Sep 28, 2:57 pm, core duo wrote:
at tha time he invented and published his first
relativity he diden even knew his tensors at
that time

At that time tensors were mostly a mathematician's thing.

how strong bulivable a foundation such a theory can have
while you know his author dident knew his tensors?

Forgive me but that's simply a dumb question. A theory's strength
depends on its content, not on random historical details surrounding
it.

--
Jan Bielawski

On Sep 29, 8:55 pm, JanPB wrote:
On Sep 28, 2:57 pm, core duo wrote:

at tha time he invented and published his first
relativity he diden even knew his tensors at
that time

At that time tensors were mostly a mathematician's thing.

how strong bulivable a foundation such a theory can have
while you know his author dident knew his tensors?

Forgive me but that's simply a dumb question. A theory's strength
depends on its content, not on random historical details surrounding
it.

--
Jan Bielawski

Einstein was really worried about his theories being accepted. This
worry made him grow tensor and tensor as time ( = change) went on.

On Sep 28, 2:57 pm, core duo wrote:

at tha time he invented and published his first
relativity he diden even knew his tensors at
that time

If your know GR at all, you will find that GR does not require tensor
calculus, and I do mean it.

how strong bulivable a foundation such a theory can have
while you know his author dident knew his tensors?

If you know history at all, you will find Einstein was no author of GR
or even SR, and once again I do mean this one. Here is a short course
on GR.

The development of GR first diverged from Newtonian physics around the
middle of the nineteenth century, when Riemann wrote down the
relationship of an actual displacement segment to how an observer
observes this same displacement segment.

ds^2 = g_ij dq^i dq^j

Where

** ds = Invariant geometry in displacement
** g_ij = Elements of the metric
** dg^i = Observer's choice of coordinate system
** i, j = 1, 2, 3 (3 spatial dimensions)

The shortest distance through the actual space (invariant geometry)
can now be computed using the calculus of variations. This was
exactly how Christoffel did it in the famous geodesic equations.

d^2q^n/ds^2 + g^nk (@g_ik/@q^j + @g_jk/@q^i - @g_ij/@q^k) @q^i/@s @q^j/
@s / 2 = 0

Where

** i, j, k, n = 1, 2, 3
** @ = Partial derivative operator

The quantities called the connection coefficients in the geodesic
equations become the Christoffel symbols of the second kind below.

Y^n_ij = g^nk (@g_ik/@q^j + @g_jk/@q^i - @g_ij/@q^k) / 2

Where

** d^2q^n/ds^2 + Y^n_ij @q^i/@s @q^j/@s = 0

However, due to the symmetry in the metric, there is at least another
way of presenting the geodesic equations.

d^2q^n/ds^2 + g^nk (@g_ik/@q^j - @g_ij/@q^k / 2) @q^i/@s @q^j/@s = 0

In doing so, the connection coefficients are very different from the
Christoffel symbols of the second kind.

Z^n_ij = g^nk @g_ik/@q^j - @g_ij/@q^k / 2

Where

** d^2q^n/ds^2 + Z^n_ij @q^i/@s @q^j/@s = 0

About a decade before the transition of the nineteenth and the
twentieth centuries, Ricci defined the covariant derivative based on
the geodesic equations and the connection coefficients. However,
Ricci did not know there is another set of connection coefficients
that are equally valid to describe the geodesic equations as the
Christoffel symbols of the second kind.

DX^n/Ds = dX^n/ds + Y^n_ij dq^i/ds X^j

Where

** DX^n/DS = Covariant derivative on X, a vector

The idea is to allow the covariant derivative of (X = dq^n/ds) to be
null in accordance with the geodesic equations. However,
mathematically there exists another operator that can achieve the
exact same thing.

EX^n/Es = dX^n/ds + Z^n_ij dq^i/ds X^j

Where

** EX^n/ES = Another operator on X, a vector

Ricci went on to derive (invent) the Riemann tensor which just like
the metric is merely a matrix. The derivation takes us through the
null geodesic variations.

R^n_ikj = @Y^n_ij/@q^k - @Y^n_ik/@q^j + Y^n_kl Y^l_jk - Y^n_jl Y^l_ik

Or

R^n_ikj = @Y^n_ij/@q^k - @Y^n_ik/@q^j + Y^n_jl Y^l_ik - Y^n_jl Y^l_ik

Ricci, however, only discovered the first tensor above while the
second one is also very mathematically valid in accordance with the
method of null geodesic variations. Ricci's student Levi-Civita then
invented the Ricci tensor based on the Riemann tensor derived by Ricci
(the first equation above).

R_ij = @Y^k_ij/@q^k - @Y^k_ik/@q^j + Y^k_kl Y^l_ij - Y^k_jl Y^l_ik

Where

** R_ij = R^k_ikj

The Ricci scalar follows as described below.

R = g^ij R_ij

Where

** g^ij = inverse of the matrix g_ij the metric

After the introduction of the Lorentz transformation, the Goettingen
group of physicists including Minkowski, Hilbert, Schwarzschild, and
Klein extended Riemann's description of curved space into a four-
dimensional spacetime.

ds^2 = g_ij dq^i dq^j

Where

** ds = Invariant geometry in spacetime
** g_ij = Elements of the metric
** dg^i = Observer's choice of coordinate system
** i, j = 0, 1, 2, 3 (1 temporal and 3 spatial dimensions)

In 1915, Hilbert finally invented the following Lagrangian which does
not even satisfy as a Lagrangian according to the variations of
calculus.

L = (H R + p c^2) sqrt(-det(g^ij))

Where

** L = Hilbert's Lagragian
** R = Ricci scalar
** p = density of matter
** det() = determinant of the matrix as operand
** H = a constant

Hilbert then went on to take the partial derivative of this Lagrangian
with respect to each element of the metric represented by g^ij and
setting it to zero.

@L/@g^ij = H sqrt(-det(g^ij)) @R/@g^ij - H R @det(g^ij)/@g^ij / sqrt(-
det(g^ij)) / 2 - p c^2 @det(g^ij)/@g^ij / sqrt(-det(g^ij)) / 2 = 0

Where (mathematical identity)

** @R/@g^ij = R_ij
** @det(g^ij)/@g^ij = g_ij det(g^ij)

The result is the set of Einstein field equations.

R_ij - R g_ij / 2 = c^2 p g_ij / H / 2

Or

G_ij = T_ij

Where

** G_ij = R_ij - R g_ij / 2
** T_ij = c^2 p g_ij / H / 2

Einstein played no role. His rediscovery of the equivalence principle
also finds no role in the derivation. The derivation of GR is totally
based on mathematical nonsense.

Very soon after the introduction of the field equations, Schwarzschild
discovered the following static and spherically symmetric solution
(metric).

ds^2 = c^2 (1 - R / (r^3 + R^3)^(1/3)) dt^2 - r^4 dr^2 / (r^3 + R^3) /
((r^3 + R^3)^(1/3) - R) - (r^3 + R^3)^(2/3) dO^2

Where

** R = G M / c^2
** dO^2 = cos^2Phi dTheta^2 + dPhi^2

There are actually an infinite number of solutions (metric) to the
field equations using the same set of coordinate system. The most
popular one was derived by Hilbert in 1916 now called the
Schwarzschild metric.

ds^2 = c^2 (1 - 2 R / r) dt^2 - dr^2 / (1 - 2 R / r) - r^2 dO^2

Notice Schwarzschild's original solution does not manifest black holes
but Schwarzschild metric does. The following solution also as simple
as the Schwarzschild metric does not manifest black holes as well.

ds^2 = c^2 dt^2 / ( 1 + 2 R / r) - (1 + 2 R / r) dr^2 - (r + R)^2 dO^2

Although not all the static and spherically symmetric solutions
degenerate to Newtonian law of gravity, all these three metrics above
do. This means the universe must be expanding and finally collapsing
back to itself. After observing the universe to be static, Einstein
correctly identified the field equations and Newtonian law of gravity
as not fit this observation. He cleverly introduced (pull out of his
*ss) a negative mass density to counter the attraction of gravity.
The reason is very simple. Positive mass manifests attraction in
gravity; negative mass manifests repulsion in gravity. In order to
hide the embarrassment of introducing negative mass in vacuum, he
again cleverly called this quantity as the Cosmological constant. The
development of GR at this stage is a total joke, but the nonsense did
not end here. Friedman, Lemaitre, Robertson, and Walker discovered a
non-static but spherically symmetric solution to the field equations
called the Friedman-Lemaitre-Robertson-Walker (FLRW) metric.

ds^2 = c^2 dt^2 - a^2 (dr^2 / (1 - r^2 / R^2) + r^2 dO^2)

Where

** a = Function of t only
** R = Constant

This means two of the field equations are

** (da/dt)^2 / a^2 + c^2 / R^2 / a^2 = 8 pi G p / 3
** 2 d^2a/dt^2 / a + (da/dt)^2 / a^2 + c^2 / R^2 / a^2 = 8 pi G p

We can very easily solve these differential equations.

If R^2 = 0,

** a^2 = c^2 cosh^2(w(t+T)) / (w^2 R^2)
** p = 3 w^2 / (4 pi G)

If R^2 0,

** a^2 = - c^2 cos^2(w(t+T)) / (w^2 R^2)
** p = - 3 w^2 / (4 pi G)

Where

** w, T = Integration constants

The density of the universe, p, must always remain constant. This
means the universe must be static as observed back then. Even with
the introduction of the Cosmological constant, the basic form of the
solution above remains the same. The Cosmological becomes totally
useless. The introduction of the Cosmological constant is the only
blunder in Einstein's contribution in GR.

There are two problems with this FLRW metric.

** There is no solution combining the Schwarzschild metric and the
FLRW metric. This means the FLRW metric does not satisfy the
Newtonian law of gravity. Gravity is not caused by a curvature in
spacetime in general but only the gravitational time dilation.

** When Lemaitre first then Hubble discovered the red shift of
distant galaxies, there is no remedy for the FLRW metric to satisfy
this observation even with the Cosmological constant.

There are so many problems with GR right from the start. The most
basic is even more embarrassing. Any diligent grade school children
can identify the mathematical relationship below.

Given that

A = B C

If (A = constant and B != 0), then (C = A / B).

This blunder came as early as during Ricci's time when the Riemann
tensor which is merely a matrix was incorrectly identified as a tensor
which means it becomes invariant to any coordinate transformation.
Similarly, the metric is merely a matrix. Ricci deified the metric
into a tensor. Mathematically, this can easily be proven wrong.

d[s] = [Q] d[q] = [Q'] d[q']

Where

** d[s] = Invariant geometry in displacement vector
** [Q], [Q'] = Matrices
** d[q], dq[q'] = Coordinate systems

The above equation squared is

ds^2 = [g] * d[q^2] = [g'] * d[q'^2]

Where

** [g] = [Q]^Transpose [Q]
** [g'] = [Q']^Transpose [Q']
** d[q^2] = d[q] d[q]^Transpose
** d[q'^2] = d[q'] d[q']^Transpose
** [A] * [b] = SUM(SUM(A_ij B_ij)), dot product
** ds^2 = Invariant, still

The metric [g] and the metric [g'] cannot be the same if the choice of
coordinate system [q] is different from [q'].

ds^2 = [g] * d[q^2] = g_ij dq^i dq^j = Invariant

The geometry, ds^2, must be invariant due to obvious reason. The
choice of coordinate system, d[q^2], is observer dependent. This can
only mean the metric, [g], must also be observer dependent. The
metric, the Riemann, and the Ricci tensors cannot be tensors after
all. All solutions to the field equations must be unique and
independent of each one where all solutions must reference to the same
choice of coordinate system in describing vastly different invariant
geometries. What good is the set of field equations that can generate
an infinite numbers of solutions to describe infinitely different and
independent universes? What good is the set of field equations can
either generate a solution that manifests black holes and also ones
that don't?

On Sep 29, 9:32 pm, Koobee Wublee wrote:

[mostly nonsense]

You don't help your case at all by posting a big long idiocy like
this.

--
Jan Bielawski

"JanPB" wrote in message
: [entirely nonsense]

On Sep 30, 6:32 am, Koobee Wublee wrote:
On Sep 28, 2:57 pm, core duo wrote:

at tha time he invented and published his first
relativity he diden even knew his tensors at
that time

If your know GR at all, you will find that GR does not require tensor
calculus, and I do mean it.

yes, you are right

i snip tha rest of your post, it looks like
tensors to me anyways, thanks