Re: Question on GR sources

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>
.