Koobee Wublee wrote:
On Feb 22, 12:00 pm, Tom Roberts wrote:
There is no "matheMagic" here, just very basic tensor math.

Yes, you are still playing with matheMagic. What you refer to as the
tensor is T which is a scalar.

No! T is a rank-2 tensor, not a scalar. Just look at it:

Are the two following scalars the
same?
** T_ij e^i (x) e^j

Remember the definitions of those symbols: T_ij is a real number, e^i is
a basis VECTOR, as is e^j, and (x) is a TENSOR PRODUCT. The tensor
product of two vectors multiplied by a real number is NOT a scalar, it
is a rank-2 tensor; that expression is a sum of rank-2 tensors, and is
thus a rank-2 tensor.

** T'_ij e^i (x) e^j

You screwed up this one -- when using "T'_ij" (which I write with the
modern notation T_i'j') one must use the PRIMED basis vectors:
T_i'j' e^i' (x) e^j'
This RANK-2 TENSOR is the same tensor as your first ** equation above.

This is no different from the vector expressed in terms of
two different basis vectors and components:
v = v_i e^i = v_i' e^i'
In my example, {v_i}={1,0,0} with e^1 headed east;
{v_i'}={0,-1,0} with e^2' headed west. The vector v itself
is of course headed east.


It looks like you are still trapped in your matheMagic realm.

No, YOU are still trapped in your "lack of understanding" realm.

You seemed to understand that my vector example produces the same vector
using two different coordinate systems' basis vectors and vector
components. The rank-2 tensor is no different.


Tom Roberts

On Feb 21, 1:34*am, Koobee Wublee wrote:
On Feb 20, 7:17 pm, Edward Green wrote:

On Feb 20, 1:15 am, Koobee Wublee wrote:
Kinetic energy is observer dependent, and so is the momentum. *This is
also true for the observed mass. *Thus, the energy-momentum tensor
must be observer dependent as well. *Since it is the energy-momentum
tensor that equates with the Einstein tensor, the field equations
allow an observer dependent quantity to shape the curvature of
spacetime that is supposed to be observer independent or invariant.
Please give me a date when you personally will be in peace with this
particular self-inconsistency.

Even I can answer that like an establishment pro!

The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system.

Can you prove that mathematically? *Proof by faith does not count.

I'm hardly making an extraordinary claim. Your claim, on the other
hand, that the most basic machinery of GR is flawed is extraordinary.
Believe what you want -- I'm not playing.

On Feb 21, 1:34*am, Koobee Wublee wrote:
On Feb 20, 7:17 pm, Edward Green wrote:

concerning the pressure and energy density of a gas, and their
appearance in the stress-energy tensor

Either we can blindly plug in both terms, in or there is something
imperfect about the formulation.

Just a misunderstanding on your part.

A gas has an energy density, a gas has a pressure. Prima facie these
are different physical measures, even if they are connected behind the
scenes. If a theory has slots for pressure and for energy density,
but we are instructed by some ancillary set of rules to only use one
or the other in a given situation, then the original theory was
imperfectly specified.

Just what is my misunderstanding?

On Feb 24, 8:21 am, Tom Roberts wrote:

You seemed to understand that my vector example produces the same vector
using two different coordinate systems' basis vectors and vector
components. The rank-2 tensor is no different.

Adding a tiny bit of confusion here is the reversed index convention
Tom uses here. Koobee is now probably trying to see how all this fits
with the formula for g (the metric tensor) and he is about to discover
it doesn't fit because of the index positioning.

I recommended numerous times in the past to forget the so-called
"modern" (i.e., reversed up-down) convention for index placement as it
only serves to confuse everyone. It's practically never used. It is
also based on a misplaced pun of the meaning of "co-" and
"contravariant". There was a similar attempt in the '60s to rename
homology and cohomology based on identical arguments from category
theory but thankfully it failed and nobody even remembers it now.

--
Jan Bielawski

On Feb 24, 8:27 am, Edward Green wrote in many messages:
On Feb 21, 1:34 am, Koobee Wublee wrote:

The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system.

Can you prove that mathematically? Proof by faith does not count.

I'm hardly making an extraordinary claim.

Did you not have claimed the following?

"The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system."

I merely ask you to prove what you said.

Your claim, on the other
hand, that the most basic machinery of GR is flawed is extraordinary.

Why? I did back it up with rigorous mathematical analysis. shrug

Believe what you want -- I'm not playing.

There is no need to believe or play. You just have to do the
mathematics. shrug

A gas has an energy density, a gas has a pressure. Prima facie these
are different physical measures, even if they are connected behind the
scenes. If a theory has slots for pressure and for energy density,
but we are instructed by some ancillary set of rules to only use one
or the other in a given situation, then the original theory was
imperfectly specified.

Just what is my misunderstanding?

Well, pressure is force. I never said force is completely observer
dependent, did I?

On Feb 24, 8:21 am, Tom Roberts wrote:
Koobee Wublee wrote:

Yes, you are still playing with matheMagic. What you refer to as the
tensor is T which is a scalar.

No! T is a rank-2 tensor, not a scalar. Just look at it:

Are the two following scalars the
same?
** T_ij e^i (x) e^j

Remember the definitions of those symbols: T_ij is a real number, e^i is
a basis VECTOR, as is e^j, and (x) is a TENSOR PRODUCT. The tensor
product of two vectors multiplied by a real number is NOT a scalar, it
is a rank-2 tensor; that expression is a sum of rank-2 tensors, and is
thus a rank-2 tensor.

In the content of discussing GR, it makes sense to describe the
following.

** T = ds^2
** T_ij = g_ij
** e^i (x) e^j = dq^i dq^j

Thus, we end up with the following describing a segment of spacetime.

ds^2 = g_ij dq^i dq^j

Now, I know you are referring to something else. We can also talk
about the irrelevant subject you have brought up. Since it is
irrelevant, it should not apply to GR. So, let's continue with our
discussion.

** T'_ij e^i (x) e^j

You screwed up this one -- when using "T'_ij" (which I write with the
modern notation T_i'j') one must use the PRIMED basis vectors:

You need to learn to read what I wrote. I have already said in order
to describe the same matrix, the following must be true.

T = T_ij e^i (x) e^j = T'_ij e'^i (x) e'^j

However, if using the same basis vectors, the following two matrices
cannot be the same.

** T_ij e^i (x) e^j
** T'_ij e^i (x) e^j

T_i'j' e^i' (x) e^j'

The notation above is no different from the following. Who are you
trying to kid? Do you not understand the rules of the lazy-man's
summation notation?

T_ij e^i (x) e^j

This RANK-2 TENSOR is the same tensor as your first ** equation above.

This is no different from the vector expressed in terms of
two different basis vectors and components:
v = v_i e^i = v_i' e^i'
In my example, {v_i}={1,0,0} with e^1 headed east;
{v_i'}={0,-1,0} with e^2' headed west. The vector v itself
is of course headed east.

Your rank-2 tensor T does not pertain to anything the metric is
involved in. The metric, [g] in my notation of a matrix with square
brackets, has no meaning if the basis vectors are not specified. So,
when we write down a segment of space or spacetime, we establish a
relationship between the metric and the basis vectors as described
below.

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

Where

** ds^2 = Segment displacement squared
** g_ij = Elements to the matrix [g], the metric
** dq^i dq^j = Elements to the matrix [dq^2]
** [] * [] = Dot product of two matrices as defined above

We know that ds^2 is invariant by the reason of it being the very
geometry itself. Since the observer can choose any coordinate system
where each with its own basis vectors, [dq^2] is different from the
choice of coordinate system to another. Now, this is a very basic
mathematics. In order for ds^2 to remain invariant, [g] must be
different for every [dq^2].

It looks like you are still trapped in your matheMagic realm.

No, YOU are still trapped in your "lack of understanding" realm.

sigh

You seemed to understand that my vector example produces the same vector
using two different coordinate systems' basis vectors and vector
components. The rank-2 tensor is no different.

Yes, I do understand what you are doing. The problem is that what you
are doing does not apply to the metric. If you think it does, you are
still trapped in the very matheMagic realm for quite sometime.
shrug

On Feb 24, 8:03 pm, Koobee Wublee wrote:
On Feb 24, 8:27 am, Edward Green wrote in many messages:

On Feb 21, 1:34 am, Koobee Wublee wrote:
The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system.

Can you prove that mathematically? Proof by faith does not count.

I'm hardly making an extraordinary claim.

Did you not have claimed the following?

"The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system."

I merely ask you to prove what you said.

Why should we have to prove anything to you? You routinely make stupid
assertions then run away when asked for proof - or even a REFERENCE.


Your claim, on the other
hand, that the most basic machinery of GR is flawed is extraordinary.

Why? I did back it up with rigorous mathematical analysis. shrug

Believe what you want -- I'm not playing.

There is no need to believe or play. You just have to do the
mathematics. shrug

A gas has an energy density, a gas has a pressure. Prima facie these
are different physical measures, even if they are connected behind the
scenes. If a theory has slots for pressure and for energy density,
but we are instructed by some ancillary set of rules to only use one
or the other in a given situation, then the original theory was
imperfectly specified.

Just what is my misunderstanding?

Well, pressure is force. I never said force is completely observer
dependent, did I?

Pressure....is force? Might want to check the units on pressure then
force.

On Feb 25, 12:03*am, Koobee Wublee wrote:
On Feb 24, 8:27 am, Edward Green wrote in many messages:

On Feb 21, 1:34 am, Koobee Wublee wrote:
The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system.

Can you prove that mathematically? *Proof by faith does not count.

I'm hardly making an extraordinary claim.

Did you not have claimed the following?

"The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system."

Yes I did. But in context, that hardly seems like an extraordinary
claim; it's a quite ordinary claim.

You're the one who seems to be making the extraordinary claim: that
the standard mathematical machinery of GR is horribly broken -- even
before we use it for anything.

I merely ask you to prove what you said.

Your claim, on the other
hand, that the most basic machinery of GR is flawed is extraordinary.

Why? *I did back it up with rigorous mathematical analysis. *shrug

That emoticon... where have I seen it before...

In terms of "proof", what I meant by I'm not playing, is that I'm not
inclined to try to prove anything to you, nor argue whether I have to
prove something to you, or you have to prove something to me, for it
is a matter of no import to me what you believe, or affect to believe,
for I half believe IHBT.

I mention to the rafters that there are two ways of identifying a
tensor -- by its transformation properties under change of
coordinates, or else by some more abstract pre-coordinate properties
(like some general linear map jazz), which then determine the
transformation properties which -- surprise! -- are the same as
before. Physicists are reputed to use the first approach,
mathematicians the second, though I think the distinction tends to
blur.

Believe what you want -- I'm not playing.

There is no need to believe or play. *You just have to do the
mathematics. *shrug

Sure there is... I have to agree to play with you. I suppose I have,
by replying, but I'm not going to play the "burden of proof" game.

Just what is my misunderstanding?

Well, pressure is force. *I never said force is completely observer
dependent, did I?

That shows just about null reading comprehension, given my write-up.

Have fun.

On Feb 25, 9:39 am, Edward Green wrote:
On Feb 25, 12:03 am, Koobee Wublee wrote:

On Feb 24, 8:27 am, Edward Green wrote in many messages:

On Feb 21, 1:34 am, Koobee Wublee wrote:
The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system.

Can you prove that mathematically? Proof by faith does not count.

I'm hardly making an extraordinary claim.

Did you not have claimed the following?

"The _components_ of the tensor may be observer dependent, but the
abstract tensor itself is independent of any coordinate system."

Yes I did. But in context, that hardly seems like an extraordinary
claim; it's a quite ordinary claim.

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. 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.

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)

....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,
i.e. they are linear functions assigning to any vector v its i-th
(resp. j-th) component with respect to the basis (e_1,...,e_n), like
so:

if v = v^k e_k then (e_i#)(v) = v^i

.... and the tensor product (x) is defined as a bilinear map acting of
vector pairs like so:

(e_i# (x) e_j#)(v,w) = (e_i#)(v) * (e_j#)(w) =(in other words)=
v^i * w^j

In the manifold case we have the above situation repeated at every
manifold point (like at every spacetime event). So for every event we
have a separate vector space V (the tangent space at that event) and
separate bilinear function T. The requirement is that T depend
smoothly on the manifold point. So still no basis dependence.

One can use bases in the tangent spaces to define dot products there
and one standard way to do so is to pick a coordinate system on the
manifold. It can be an arbitrary smooth function (e.g. curvilinear).
This fixes a linear basis in each tangent space - if (q^1,...,q^n) is
the coordinate system then the induced linear basis vectors at a point
q on the manifold consist of vectors denoted by d/dq^1,...,d/dq^n
(partial derivative operators at the point q). The dual covector basis
is then (d/dq^i)#, and it's denoted by dq^i:

dq^i =(by definition)= the covector dual of d/dq^i = (d/dq^i)#

So at each point the smoothly varying dot product g on a manifold can
be written with respect to the bases induced by a coordinate system
as:

g = g_ij dq^i (x) dq^j (this equation holds at each manifold
point separately)

Because the dot product g is presumed symmetric, g(u,v) = g(v,u), the
above is almost always written using the symmetric tensor product
"(s)" rather than the tensor product "(x)", where "(s)" is defined as:

A (s) B = 1/2 * [ A (x) B + B (x) A ]

We can rewrite "(x)" in terms of "(s)" because of the symmetry of g:

g(d/dq^i, d/dq^j) = g(d/dq^j, d/dq^i)

i.e.

g_ij = g_ji

....so we can factor out g_ij from each expression of the type:

g_ij dq^i (x) dq^j + g_ji dq^j (x) dq^i (for i =/= j)

....so we get:

g_ij * (dq^i (x) dq^j + dq^j (x) dq^i) = 2g_ij dq^i (s) dq^j

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_.

--
Jan Bielawski

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

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

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.

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

[...]

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. This is where your matheMagic trick is exposed once
again. 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

You are telling me that you can reconstruct ds^2 using [g] only
without knowing what [dq dq] is. You are full of sh*t. Is it not
time for you to retire that matheMagical trick?