On Feb 25, 10:35 pm, Koobee Wublee wrote:
On Feb 25, 12:17 pm, JanPB wrote:

Not only that, it's a trivial claim because it's a _definition_:
tensor is by definition a multilinear map from k copies of the given
linear space and l copies of its dual ("star") into real numbers:

T : V x ... x V x V* x ... x V* - R

("x" means Cartesian product). If there are k copies of V and l copies
of V*, we call T k times covariant and l times contravariant. So no
basis of V enters into it.

Yes, that is fine. However, it is the application of tensor that I
object. For instance, your T is a scalar which corresponds to ds^2,
the line element in space time.

T is not a scalar. There are two contexts:

(1) T as a tensor (multilinear map) on copies of a _single_ vector
space V and its dual V* - that's what I used above as a starting
point,

(2) T as a tensor _field_ (as in differential geometry and GR), i.e. a
family of tensors of type (1) indexed by manifold points. In other
words, at each manifold point p (e.g. spacetime event) there is a
tangent vector space V_p and its dual (V_p)* _and_ the multilinear map
T_p as described in (1). So at every point p of the manifold M we have
the multilinear map:

T_p : V_p x ... x V_p x (V_p)* x ... x (V_p)* - R

....subject to the usual smoothness constraints (so the maps T_p do not
change too wildly when the point p is changed).

A _scalar_ (short for _scalar field_) is a function F which assigns a
number to each manifold point p:

F : M - R

....i.e. its value is defined once the point p in M is given. Tensor
field of type (k,l) OTOH in order to produce a value requires a point
p _and_ k tangent vectors _and_ l cotangent vectors.

For example, spacetime metric is a tensor _field_ of type (2,0): it
requires a point and a pair of tangent vectors at that point to
produce a number. Which is of course obvious: (pseudo)-Riemannian
metric is a dot product by definition, so it _obviously_ requires a
pair of vectors at a point:

g_p : V_p x V_p - R

That's it. No basis and no matrix.

Of course one can specify the function g_p at every point p by fixing
a basis (e_1,...,e_n) of V_p and writing down the values of g_p at all
the basis vector pairs: g_p(e_i,e_j) - these numbers are denoted by
g_ij and there are n^2 of them at every point p.

The basis at each point p (e_1,...,e_n) of V_p can come from a
coordinate system in the neighbourhood of p (the d/dq^i basis, aka.
coordinate basis, aka. holonomic basis, etc.) or can be defined by
some other criterion (moving frame, aka. Cartan frame, aka. repere
mobile, aka. Vierbein (n=4), aka. tetrad (n=4), aka. non-holonomic
basis, etc.).

On the other hand, [g] with elements
g_ij is hardly what you have described above. shrug

g_ij(p) = g_p(e_i,e_j) (g_ij at p is the value of g at p of e_i
and e_j - IOW the dot product of the two basis vectors at p).

One can of course use various tools to
construct specific tensors, including linear space bases. This doesn't
make tensors any more coordinate-dependent than specifying a
particular map of the Earth would make airplane velocity map-
dependent. I suspect Koobee confuses basis choice with the choice of
units of measurement.

Your application of mathematics to physics is extraordinary faulty.
shrug Luckily, it is your problem, not mine. shrug

Anybody can _say_ "your whatever is faulty". Don't everybody's time
typing stuff like this in the middle of a technical argument which
alone can decide.

Dot product, for example, is a bilinear map from two copies of V:

T : V x V - R

T(u,v) = u . v

It's just a function so it doesn't depend on any basis, e.g. the usual
Euclidean dot product is:

T(u,v) = |u| * |v| * cos(alpha)

Of course if a basis is given, one may write down certain numbers
called components of T which can be used to reconstruct T out of those
numbers AND the basis. Let's say (e_1,...,e_n) is the basis of V. Then
the components of T are defined as:

T_ij = T(e_i, e_j) (real numbers)

Yes, agreed. Notice T_ij and T() depend on the choice of basis
vectors, e_i or e_j.

Of course.

...and T itself can then be reconstructed from the numbers T_ij:

T = T_ij * e_i# (x) e_j#

...where e_i# and e_j# are the covectors dual to vectors e_i and e_j,

This is wrong.

No, this is correct. Once again: e_i# is the covector dual to e_i.
Meaning: e_i# is a linear function from vectors to numbers, _defined_
on the basis vectors e_j as follows:

(e_i#)(e_j) = 0 for all j =/= i,

(e_i#)(e_i) = 1

This defines e_i# on all vectors by linearity: to any vector v it
assigns its i-th component with respect to the basis (e_1,...,e_n):

(e_i#)(v) = v^i whenever v = v^k e_k.

Furthermore, tensor product "(x)" of two linear maps F and G is in
general a _bilinear map_ which assigns to a pair of vectors v and w
the following number:

(F (x) G)(v, w) = (by definition) = F(v) * G(w)

....where "*" is number multiplication. Thus T = T_ij * e_i# (x) e_j#
simply says that given two vectors v and w:

T(v, w) = T_ij * v^i * w^j - which is clearly true by
bilinearity of T. The right-hand side can be also written as:

[v^1 ... v^n] * [T_ij] * [w^1 ... w^n]^T

....i.e., a 1xn-matrix times an nxn-matrix times an nx1-matrix (which
is equal to a 1x1-matrix, or a number). You on the other hand write
the RHS as a "product":

[T_ij] "times" [v^i * w^j]

....which is not invariant hence useless (although correct in the fixed
coordinate system in which it was cooked up).

Given T_ij without a basis vectors, it is
meaningless. Have you not agreed with Professor Roberts?

The e_i#'s _are_ basis _covectors_. In finite dimensional spaces
vector bases determine covector bases and vice-versa. Exactly the same
information is given by specifying either.

[...]

In other words, we have simply collected the terms as if (x) were
commutative:

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

...where the new g_ij here are twice the previous values for i =/= j.

Finally, the symmetric product "(s)" is actually denoted by...
nothing, so the above is always written as:

g = g_ij dq^i dq^j

(NOTE: there is an analogous anti-symmetric tensor product defined
with the minus sign as 1/2 * [A (x) B - B (x) A] which is denoted by "/
\" and called "wedge product" - an extremely useful thing.)

All of the above is very basic. A bit tedious but quite
straightforward. It took no effort to write it from the top of my head
(I spent more time proofreading than composing), it's not where
difficulties of GR lie _at all_.

Basically, what you are doing is to take any [g] and reconstruct its
basis vectors.

Keep in mind that GR uses tensor _fields_ (context (2) above) and that
I used context (1) initially only to simplify the description.

This is where your matheMagic trick is exposed once
again.

Standard linear algebra, no tricks here.

Given the following geometry of a segment in space or
spacetime,

ds^2 = g_ij dq^i dq^j

We can easily write the above equation as the dot product of two
square matrices as defined below.

ds^2 = [g] * [dq dq]

Where

** g_ij = Elements of [g]
** dq^i dq^j = Elements of [dq dq]
** [g] * [dq dq] = g_ij dq^i dq^j

This is a correct but useless factoring because it immediately falls
apart when you change coordinates: the new matrix [dq dq] will no
longer have terms dq^i dq^j in the slot (i,j). The correct, invariant
factoring of the above product is into _three_ matrices as I wrote
above (again, it's all undergrad linear algebra):

ds^2 = ([dq] row vector) * [g] * ([dq] column vector).

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

You are telling me that you can reconstruct ds^2 using [g] only
without knowing what [dq dq] is.

No, I never said that. What I and everyone else has been telling you
for literally _months_ is that g - like any other tensor, vector, etc.
- can be reconstructed from [g] (the matrix of g_ij's) _AND_ the basis
vectors (or covectors).

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).

You are full of sh*t. Is it not
time for you to retire that matheMagical trick?

You are full of low-brow poetry.

--
Jan Bielawski

On Feb 29, 6:29 pm, JanPB wrote:
On Feb 25, 10:35 pm, Koobee Wublee wrote:

Yes, that is fine. However, it is the application of tensor that I
object. For instance, your T is a scalar which corresponds to ds^2,
the line element in space time.

T is not a scalar.

In the application of differential geometry, T is a scalar. However,
even if it is not, the concept is still the same. shrug

There are two contexts:

(1) T as a tensor (multilinear map) on copies of a _single_ vector
space V and its dual V* - that's what I used above as a starting
point,

(2) T as a tensor _field_ (as in differential geometry and GR), i.e. a
family of tensors of type (1) indexed by manifold points. In other
words, at each manifold point p (e.g. spacetime event) there is a
tangent vector space V_p and its dual (V_p)* _and_ the multilinear map
T_p as described in (1). So at every point p of the manifold M we have
the multilinear map:

T_p : V_p x ... x V_p x (V_p)* x ... x (V_p)* - R

...subject to the usual smoothness constraints (so the maps T_p do not
change too wildly when the point p is changed).

A _scalar_ (short for _scalar field_) is a function F which assigns a
number to each manifold point p:

F : M - R

...i.e. its value is defined once the point p in M is given. Tensor
field of type (k,l) OTOH in order to produce a value requires a point
p _and_ k tangent vectors _and_ l cotangent vectors.

For example, spacetime metric is a tensor _field_ of type (2,0): it
requires a point and a pair of tangent vectors at that point to
produce a number. Which is of course obvious: (pseudo)-Riemannian
metric is a dot product by definition, so it _obviously_ requires a
pair of vectors at a point:

g_p : V_p x V_p - R

That's it. No basis and no matrix.

Of course one can specify the function g_p at every point p by fixing
a basis (e_1,...,e_n) of V_p and writing down the values of g_p at all
the basis vector pairs: g_p(e_i,e_j) - these numbers are denoted by
g_ij and there are n^2 of them at every point p.

The basis at each point p (e_1,...,e_n) of V_p can come from a
coordinate system in the neighbourhood of p (the d/dq^i basis, aka.
coordinate basis, aka. holonomic basis, etc.) or can be defined by
some other criterion (moving frame, aka. Cartan frame, aka. repere
mobile, aka. Vierbein (n=4), aka. tetrad (n=4), aka. non-holonomic
basis, etc.).

Yes, this is all basic linear algebra stuff.

On the other hand, [g] with elements
g_ij is hardly what you have described above. shrug

g_ij(p) = g_p(e_i,e_j) (g_ij at p is the value of g at p of e_i
and e_j - IOW the dot product of the two basis vectors at p).

Your application of mathematics to physics is extraordinary faulty.
shrug Luckily, it is your problem, not mine. shrug

Anybody can _say_ "your whatever is faulty". Don't everybody's time
typing stuff like this in the middle of a technical argument which
alone can decide.

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

Yes, agreed. Notice T_ij and T() depend on the choice of basis
vectors, e_i or e_j.

Of course.

Excellent, we agree on this one, too.

...and T itself can then be reconstructed from the numbers T_ij:

T = T_ij * e_i# (x) e_j#

...where e_i# and e_j# are the covectors dual to vectors e_i and e_j,

This is wrong.

No, this is correct. [...]

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

Basically, what you are doing is to take any [g] and reconstruct its
basis vectors.

Keep in mind that GR uses tensor _fields_ (context (2) above) and that
I used context (1) initially only to simplify the description.

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

Given the following geometry of a segment in space or
spacetime,

ds^2 = g_ij dq^i dq^j

We can easily write the above equation as the dot product of two
square matrices as defined below.

ds^2 = [g] * [dq dq]

Where

** g_ij = Elements of [g]
** dq^i dq^j = Elements of [dq dq]
** [g] * [dq dq] = g_ij dq^i dq^j

This is a correct but useless factoring because it immediately falls
apart when you change coordinates: the new matrix [dq dq] will no
longer have terms dq^i dq^j in the slot (i,j).

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

The correct, invariant
factoring of the above product is into _three_ matrices as I wrote
above (again, it's all undergrad linear algebra):

ds^2 = ([dq] row vector) * [g] * ([dq] column vector).

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

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']).

As you said, this is very simple mathematics. It is the basics of
matrices. If you do not understand what matrices are, go back to
study them instead of giving them matheMagical properties. Math
cannot lie. shrug

You are telling me that you can reconstruct ds^2 using [g] only
without knowing what [dq dq] is.

No, I never said that. What I and everyone else has been telling you
for literally _months_ is that g - like any other tensor, vector, etc.
- can be reconstructed from [g] (the matrix of g_ij's) _AND_ the basis
vectors (or covectors).

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

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. What you wrote above says g is a scalar. We are back
to the beginning of the discussion. 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!

Given only 2 dimensions, the above equation becomes

g = g00 dq0 dq0 + g01 dq0 dq1 + g10 dq1 dq0 + g11 dq1 dq1

I can simply write the above equation into the following by inventing
an mathematical operator '$' as defined below.

g = [g] $ [dq dq]

Where

** g00, g01, g10, g11 = Elements of matrix [g]
** dq0 dq0, dq0 dq1, dq1 dq0, dq1 dq1 = Elements of matrix [dq dq]
** [g] $ [dq dq] = g00 dq0 dq0 + g01 dq0 dq1 + g10 dq1 dq0 + g11 dq1
dq1

You are full of sh*t. Is it not
time for you to retire that matheMagical trick?

You are full of low-brow poetry.

You are still full of sh*t. shrug

On Mar 1, 8:55 pm, Koobee Wublee wrote:
On Feb 29, 6:29 pm, JanPB wrote:

On Feb 25, 10:35 pm, Koobee Wublee wrote:
Yes, that is fine. However, it is the application of tensor that I
object. For instance, your T is a scalar which corresponds to ds^2,
the line element in space time.

T is not a scalar.

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.

However,
even if it is not, the concept is still the same. shrug


There are two contexts:

(1) T as a tensor (multilinear map) on copies of a _single_ vector
space V and its dual V* - that's what I used above as a starting
point,

(2) T as a tensor _field_ (as in differential geometry and GR), i.e. a
family of tensors of type (1) indexed by manifold points. In other
words, at each manifold point p (e.g. spacetime event) there is a
tangent vector space V_p and its dual (V_p)* _and_ the multilinear map
T_p as described in (1). So at every point p of the manifold M we have
the multilinear map:

T_p : V_p x ... x V_p x (V_p)* x ... x (V_p)* - R

...subject to the usual smoothness constraints (so the maps T_p do not
change too wildly when the point p is changed).

A _scalar_ (short for _scalar field_) is a function F which assigns a
number to each manifold point p:

F : M - R

...i.e. its value is defined once the point p in M is given. Tensor
field of type (k,l) OTOH in order to produce a value requires a point
p _and_ k tangent vectors _and_ l cotangent vectors.

For example, spacetime metric is a tensor _field_ of type (2,0): it
requires a point and a pair of tangent vectors at that point to
produce a number. Which is of course obvious: (pseudo)-Riemannian
metric is a dot product by definition, so it _obviously_ requires a
pair of vectors at a point:

g_p : V_p x V_p - R

That's it. No basis and no matrix.

Of course one can specify the function g_p at every point p by fixing
a basis (e_1,...,e_n) of V_p and writing down the values of g_p at all
the basis vector pairs: g_p(e_i,e_j) - these numbers are denoted by
g_ij and there are n^2 of them at every point p.

The basis at each point p (e_1,...,e_n) of V_p can come from a
coordinate system in the neighbourhood of p (the d/dq^i basis, aka.
coordinate basis, aka. holonomic basis, etc.) or can be defined by
some other criterion (moving frame, aka. Cartan frame, aka. repere
mobile, aka. Vierbein (n=4), aka. tetrad (n=4), aka. non-holonomic
basis, etc.).

Yes, this is all basic linear algebra stuff.

On the other hand, [g] with elements
g_ij is hardly what you have described above. shrug

g_ij(p) = g_p(e_i,e_j) (g_ij at p is the value of g at p of e_i
and e_j - IOW the dot product of the two basis vectors at p).

Your application of mathematics to physics is extraordinary faulty.
shrug Luckily, it is your problem, not mine. shrug

Anybody can _say_ "your whatever is faulty". Don't everybody's time
typing stuff like this in the middle of a technical argument which
alone can decide.

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?

Yes, agreed. Notice T_ij and T() depend on the choice of basis
vectors, e_i or e_j.

Of course.

Excellent, we agree on this one, too.

...and T itself can then be reconstructed from the numbers T_ij:

T = T_ij * e_i# (x) e_j#

...where e_i# and e_j# are the covectors dual to vectors e_i and e_j,

This is wrong.

No, this is correct. [...]

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.

Basically, what you are doing is to take any [g] and reconstruct its
basis vectors.

Keep in mind that GR uses tensor _fields_ (context (2) above) and that
I used context (1) initially only to simplify the description.

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.

Given the following geometry of a segment in space or
spacetime,

ds^2 = g_ij dq^i dq^j

We can easily write the above equation as the dot product of two
square matrices as defined below.

ds^2 = [g] * [dq dq]

Where

** g_ij = Elements of [g]
** dq^i dq^j = Elements of [dq dq]
** [g] * [dq dq] = g_ij dq^i dq^j

This is a correct but useless factoring because it immediately falls
apart when you change coordinates: the new matrix [dq dq] will no
longer have terms dq^i dq^j in the slot (i,j).

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

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

Hence, this is wrong.

The correct, invariant
factoring of the above product is into _three_ matrices as I wrote
above (again, it's all undergrad linear algebra):

ds^2 = ([dq] row vector) * [g] * ([dq] column vector).

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

Correct.

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. 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.

As you said, this is very simple mathematics. It is the basics of
matrices. If you do not understand what matrices are, go back to
study them instead of giving them matheMagical properties. Math
cannot lie. shrug

You are telling me that you can reconstruct ds^2 using [g] only
without knowing what [dq dq] is.

No, I never said that. What I and everyone else has been telling you
for literally _months_ is that g - like any other tensor, vector, etc.
- can be reconstructed from [g] (the matrix of g_ij's) _AND_ the basis
vectors (or covectors).

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?

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.

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.

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!

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

--
Jan Bielawski

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

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'?

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

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.

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.

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

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

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

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

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!

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?

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

On Mar 7, 1:49 pm, JanPB wrote:

A quick typo correction (important). I wrote:

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).

In the last sentence the word "they" refers to "functions" and
"tensors" in the previous sentence, NOT to "bases".

(This is what happenes when you edit quickly.)

--
Jan Bielawski

On Mar 7, 1:49 pm, JanPB wrote:
On Mar 3, 7:20 pm, Koobee Wublee wrote:

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.

How can you argument with the mathematics? ds^2 is a scalar according
to the mathematics above. shrug

How about 'wrong'?

It's still just a word.

It also carries a meaning. You have the right to bury your head in
the ground and avoid the meaning of the word 'wrong'. shrug

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

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).

Then, this is rather irrelevant to your discussions. shrug

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 ]

Whatever you do, ds^2 still ends up to be a scalar. shrug

...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)

It is elementary. shrug

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.

Well, you have been speaking with a forked tongue for quite
sometimes. shrug

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.

This is wrong. When you actually write down the field equations that
prepares you to solve for the Schwarzschild metric, the basis vectors
are already set and "cast in concrete". In doing so, you are confused
with the general way of writing the field equations that is because
you choose to *believe in* tensors. In GR, what you called tensors
are merely matrices. shrug

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).

shrug

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).

shrug

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.

Whatever you do at this stage, you cannot change the basis vectors.
shrug

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.

Yes, this different set of field equations is only valid for this new
basis vectors. shrug

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.

After the solutions are solved from the field equations, the part
where you change to a different set of basis vector for each solution
remains matheMagical. shrug

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.

Did you not have agreed the basis vectors are already set and
unchangeable at this stage?

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.

If not, back to the beginning and play your matheMagic trick again.
If so, you are deeply confused, and luckily that is your problem
only. shrug

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.

This g is still a scalar. It describes the invariance in the
geometry. So, using a different set of coordinate system, dq^i, you
must also use another set of g_ij. Thus, the metric, g_ij, is merely
a matrix. shrug

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".

This is where you are wrong. shrug

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.

Nonsense.

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. [...]

Only in the matheMagic realm. shrug

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"

But I do.

...your instinctive reaction is:

"Therefore, it's wrong".

I don't believe in matheMagic like yourself. shrug

Normal people first instinctively assume instead:

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

Yes, that would be the approach I will take if this is the case.
shrug

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

Yes, you are still confused. shrug

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.)

I repeat.

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

If ([g] != [g']), then ([dq] != [dq']) to preserve the equality.

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".

Are you forgetting something. The product (dq^i dq^j) is a scalar.
It looks like you are confusing a scalar, a 1-by-1 matrix, with an n-
by-n matrix where n 1. shrug

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

See above for clarification.

You are still confused. shrug

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.

That is correct.

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.

An asylum is probably the best for you. shrug

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

Yes. shrug

This follows immediately from the assumed g_ij=g_ji

Yes. shrug

and the definition
of the symmetric tensor product of A and B as
1/2*(A (x) B + B (x) A).

You are losing it now. Isn't it rather too early for you to do so?
Are you not belonging in the flower-children generation?

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

Well, it is not a solution to the field equations. shrug

Besides, the above equation allows the observed to move beyond the
speed of light. This is wrong. Of course, some crackpot came up with
the nonsense of space is able to move beyond the speed of light. Do
you get it yet? The whole religion of SR and GR is based on
interpretations and conjectures. If things don't fit, make it fit.

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) ]

Another to say this is that the metric is symmetrical. It is a
symmetrical matrix. shrug

(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".

It is elementary. shrug

I have enough nonsense for one day. Good night, your majesty.

On Mar 7, 11:10 pm, Koobee Wublee wrote:
On Mar 7, 1:49 pm, JanPB wrote:

On Mar 3, 7:20 pm, Koobee Wublee wrote:
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.

How can you argument with the mathematics? ds^2 is a scalar according
to the mathematics above. shrug

Take a 2D sphere of radius 1 (say). Pick a point p on it. Tell us,
what real number is ds^2 equal to at p?

--
Jan Bielawski