On Mar 3, 7:20 pm, Koobee Wublee wrote:
On Mar 2, 11:32 pm, JanPB wrote:

On Mar 1, 8:55 pm, Koobee Wublee wrote:
In the application of differential geometry, T is a scalar.

T is not a scalar because specifying a manifold point (an event) does
not determine the value of T.

But the following describes how a point in space or spacetime is
curved relative to its neighbors, and it is a scalar. shrug

ds^2 = g_ij dq^i dq^j

I repeat, it is not a scalar because its numeric value is not
determined by the manifold point alone. It needs two vectors at that
point.

Not anybody. Only the ones who truly understand the mathematics and
the applications involved think your application of linear algebra is
indeed very faulty. shrug

Besides saying "it's faulty" are you saying anything else?

How about 'wrong'?

It's still just a word.

If (T = T_ij * e_i (x) e_j = T_ij * e_i# (x) e_j#),

It's not. Read what I wrote:

T = T_ij * e_i# (x) e_j#

I never said "T = T_ij * e_i (x) e_j".

then (e_i (x) e_j
= e_i# (x) e_j#).

False hypothesis, false conclusion.

If not, I still do know what your point is. shrug

Just writing a bilinear map of vectors (at a fixed manifold point) in
terms of a _covector_ basis {e_i#}. These _covectors_ are defined as
the duals to some given fixed basis _vectors_ {e_i}.

The word "dual" here means that each e_i# is a linear map defined on
as follows:

(e_i#)(e_j) = (by def.) = delta^i_j (the Kronecker delta)

(defining a linear map by specifying its values on a basis).

Note that this operation (of assigning a covector to a vector) is NOT
what's called "lowering the index". What we have here is the standard
linear space dual defined by a basis.

Now writing a symmetric bilinear map T in terms of T_ij and e_i# and
the symmetric tensor product (this product is traditionally denoted by
a _blank space_ between the e_i#'s):

T = T_ij * e_i# e_j# [ sum over i=j only ]

....is exactly the same as this (except T is written at a single manifold
point):

ds^2 = g_ij * dq^i dq^j [ sum over i=j only ]

....i.e., at each p (manifold point) dq^i is by definition the i-th
covector dual to the standard coordinate basis {d/dq^k} at that point:

(dq^i)(d/dq^j) = (by def.) = delta^i_j

Unravelling the notation, the above implies then, by the way:

for f - a function (scalar) defined in a neighbourhood of the
point p - we have:

(d/dq^i)(f) = df/dq^i

and for any tangent vector v at p:

(dq^j)(v) = dq^j/dv (directional derivative of the j-th
coordinate function in the v direction)

What GR uses is not a tensor field. That is the mistake you are
making over and over again. Grade school mathematical logic proves
you wrong over and over again. shrug

I'm not going to answer silly assertions pulled out of blue. Anyone
interested can check what differential geometry uses.

That is fine. You have the right to remain silent. Any nonsense you
come up will be used against you.

No, it is not useless, and it does not fall apart. What falls apart
is the set of field equations themselves. At the stage of the field
equations already laid out to be solved for their solutions, the basis
vectors are already well established. In doing so, any solution must
be applied to the basis vector. Thus, different metric as solutions
to the field equations with the same basis vector yields a completely
different and independent geometry.

Obviously given a _fixed_ basis, different tensor coefficients imply
different tensors.

That is exactly my point. Finally, you are starting to understand
that.

I have never claimed otherwise. This is an obvious point.

But one can change the basis before the "laying
out" as you called it and then continue solving.

Yes, you can. However, in doing so, you end up with a different set
of field equations that is only valid for the new basis vectors.

No. Field equations don't know anything about bases just like standard
scalar PDEs. The latter are equations for _scalar functions_ (smoothly
varying numbers), the former are equations for _tensors_ (smoothly
varying multilinear maps). In either case bases don't enter into it as
they are at the end of the day merely assignments of numbers to the
input data (points in the scalar case, points and vectors/covectors in
the second).

Einstein's equation is a constraint on the multilinear maps (actually,
bilinear and symmetric) we seek. But you are free to use linear
algebra and simply decompose the multilinear maps on both sides in
terms of any basis you want (any moving frame will do, it doesn't even
have to be derived from a coordinate system - this is what Cartan does
with orthonormal frames or Newman-Penrose with null frames).

When you do that, you can write down each component separately - you
get a _set_ of standard PDEs for unknown _functions_ (components wrt
to the basis you picked). You can solve this set of PDEs and the
solution functions together with your basis give you the multilinear
map you sought.

You can repeat this process beginning with the original Einstein
tensor equation for some other basis, get a different set of scalar
PDEs and a different set of solution functions which together with
your different basis sum up algebraically to _the same_ multilinear
map as in the first basis. It's not any "magic", it's a simple
_algebraic_ sum which is easily verified to always yield the same
multilinear map no matter what basis you originally started with.

In particular, no matter what coordinate system {q^i} (hence covector
basis {dq^i}) you choose, you end up with a set of "solution
functions" g_ij which together with your basis {dq^i} sum up to _the
same_ bilinear map g_ij*dq^i dq^j as your solution.

Given any coordinate system {q^i}, you get:

(1) vector {d/dq^i} and covector {dq^i} basis fields,
(2) the component version with respect to (1) of Einstein's equation
in the form of a _system_ of standard scalar PDEs,
(3) solutions (functions) g_ij to this system of PDEs.

The quantities in (1), (2), and (3) are all dependent on your choice
of the coordinate system {q^i} and yet they always sum up to a
_fixed_, _unchanging_ (independent of your choice of {q^i}) field of
bilinear symmetric maps:

g = g_ij dq^i dq^j [ sum over i=j ]

This g (usually denoted by "ds^2") is the solution of the original
(single, tensor) equation.

A different set of
functions results which nevertheless together with the different basis
describes the same tensor. The field equations are equations for
unknown tensor(s).

Staying with the same field equations after changing the basis vectors
is just mathematically wrong. shrug

You are talking about two different things. The tensor EFE does not
change because it doesn't even _have_ any concept of "basis". OTOH if
you select a basis _and_ write the original tensor equation (which is
a single equation) in terms of that basis, you obtain a _set_ of
standard functional PDEs which _do_ change with the basis.

One more time: an infinite family of _sets of standard PDEs_
corresponds to the single tensor equation (the EFE): one _set of
standard scalar PDEs_ per basis.

And every pair:

(a basis, set of standard PDEs corresponding to this basis)

....yields as solution:

(a basis, solution functions) [ for example: ({dq^i}, g_ij) ]

....which always add up algebraically to one and the same (smoothly
varying) multilinear map.

Since the field equations yield
an infinite numbers of solutions, they are indeed faulty right from
the ground up.

Hence, this is wrong.

That is correct. The field equations are just nonsense. That is
because the mathematical foundation of the field equations is a total
nonsense --- all man-made stuff. shrug

What I find puzzling is that when faced with:

"I don't understand why this is"

....your instinctive reaction is:

"Therefore, it's wrong".

Normal people first instinctively assume instead:

"I won't say it's wrong until my feeling of non-understanding is
gone".

See, one may not understand something but most people at that instant
_understand very well that they don't understand_.

It's only natural. Can you imagine, e.g., a court system based on your
approach to judging hypotheses? It would be pure nightmare.

Yes, keep in mind that ds^2 is the invariant quantity --- not [g] or
[dq] and [dq]^T.

Correct.

That is good. You only have to work on one small area to understand
the whole picture.



In this, standard, factoring, the slots g_ij and the slots of both n-
tuples retain their identity upon coordinate changes.

No, it does not. Given the following,

ds^2 = [dq]^T * [g] * [dq] = [dq']^T * [g'] * [dq']

Where

** [dq], [dq'] = Column vectors
** [dq]^T, [dq']^T = Row vectors
** [g], [g'] = Square matrices

For (ds^2 != 0), if ([g] = [g']), then ([dq] = [dq']).

What's [g] = [g'] doing here.

[g] and [g'] are defined according to the equation above.

What I said was if coordinates are
changed then the slots maintain their identity. And if coordinate are
changed, then typically [g] =/= [g'], so your statement above is off-
topic.

No, it is not off topic. It explains where your thought process goes
hay-wire. shrug

Explain why ds^2 = [dq]^T * [g] * [dq] = [dq']^T * [g'] * [dq'] is
wrong.
(Hint: you can't do that.)

g, the tensor, is not applicable in GR. What is applicable is [g] the
matrix. shrug

So you are now saying g is a tensor after all?

Well, that depends on how you apply this to GR. shrug

Since the metric is merely a matrix not a tensor because it obeys the
equation of ds^2 above, what builds on top of it such as the Riemann
and the Ricci curvature tensors are merely matrices as well. Tensors
are not applicable in differential geometry. shrug

Your delirium tremens is getting boring.

Likewise. shrug



If a coordinate system (q^1,...,q^n) is given on a (portion of)
manifold, then the standard _vector_ basis for all tangent spaces in
that manifold portion is given by the vector fields (d/dq^1,...,d/
dq^n), and the standard _covector_ basis for all cotangent spaces
there is given by (dq^1,...,dq^n) (dq's are the "sharps" (#) of d/
dq's). Hence:

g = g_ij dq^i (x) dq^j

in exact analogy to my tensor T at a single point (context (1)):

T = T_ij e_i# (x) e_j#.

Finally, since g is symmetric, one can always rearrange the terms of
the RHS of g so that only _symmetric tensor products_ of dq^i's
appear:

g = g_ij dq^i dq^j

(in other words, one can commute like so: dq^i dq^j = dq^j dq^i, which
is false for tensor product: dq^i (x) dq^j =/= dq^j (x) dq^i).

Ah, matheMagic.

It's called "linear algebra".

What you wrote above says g is a scalar.

Again, you are not reading before posting. What I wrote above says g
is a symmetric tensor.

You first wrote

g = g_ij dq^i (x) dq^j

Then, you wrote 'if g is symmetric,' then

g = g_ij dq^i dq^j

Which is a scalar.

No. You simply don't read what I write. I said earlier that "dq^i
dq^j" (a _blank space_ between dq^i and dq^j) _was_ a different type
of tensor product defined like so:

dq^i dq^j = 1/2 * ( dq^i (x) dq^j + dq^j (x) dq^i )

Traditionally this product (called "the symmetric tensor product") is
denoted by _nothing_, hence "dq^i dq^j".

So, you are once again speaking with a forked
tongue contradictory of yourself from the start.

See above for clarification.

We are back
to the beginning of the discussion.

Therefore, no.

The following equation does not
have to dictate the symmetric in [g]. If (g_ij != g_ji), the above
equation is still very valid. Your argument is total BS!

Now I noticed what you were referring to - you mean the equation:

g = g_ij dq^i dq^j

....is still valid with g_ij != g_ji.

The answer is no. You should have noticed it yourself and I should
have made it clear that the equation with the symmetric tensor product
(the one above) has the summation over all i,j such that i=j. So the
assumption g_ij=g_ji is implicit.

Let me state it again in detail:

g = g_ij dq^i (x) dq^j [ sum over all i, j ]

and:

g = h_ij dq^i dq^j [ sum over all i, j such that i=j ]

....describe the same (symmetric) tensor provided that in the first
equation g_ij = g_ji _and_:

h_ij = 2 * g_ij for ij,
h_ii = g_ii

This follows immediately from the assumed g_ij=g_ji and the definition
of the symmetric tensor product of A and B as
1/2*(A (x) B + B (x) A).

One frequently sees this done in the GR context. People write metrics
like:

g = -(1 - 2M/r) dt^2 + 2 dt dr + r^2 dtheta^2 + r^2 sin^2(theta)
dphi^2

This is written using symmetric tensor product ("i=j" - notice the
presence of "dt dr" and the absence of "dr dt") and that's why when
people write down the matrix of tensor components, they divide the
"ij" terms by 2 - this is necessary when passing back from h_ij to
g_ij:

[ -(1 - 2M/r) 1 0 0 ]
[ ]
[ 1 0 0 0 ]
[ ]
[ 0 0 r^2 0 ]
[ ]
[ 0 0 0 r^2 sin^2(theta) ]

(I'm using the coordinate ordering of (t,r,theta,phi). NOTE that the
"dr dt" and the "dt dr" slots in the matrix are equal to 1, not 2.
This is usually done without explanation in most texts as a "sort of
an obvious thing to do".

Again, if g_ij != g_ji then one cannot rewrite the original tensor
product "(x)" equation in terms of the symmetric ("blank") tensor
product.

Of course the equation would still be valid but again I never claimed
otherwise.

So, when you said 'if g is symmetric', do you have a different meaning
in mind?

See above.

--
Jan Bielawski