Transformation of the Line-Element

[Anamitra]

Let us have a look at the metric:

{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}The left side is dot product between the identical vectors each being given by(dt,dx1,dx2,dx3)
Now the dot product is a scalar. The LHS ie ds^2 is a scalar since it is the result of a dot product. One should not apply covariant differentiation to it--ordinary differentiation should be applied to the LHS and hence to the RHS

 

[Anamitra]

Points to Observe:
1. ds^2 is a scalar . It is invariant wrt to coordinate transformations[ a class of transformations]
2. For parallel transport of the vector(dt,dr), ds^2 does not change.

The same points hold for the metric that I have introduced [in Relation 3 in Post #22]:

1. ds'^2 is a scalar
2. ds'^2 does not change for the parallel transport of(dt,dR) [see Posts #30 and #33]

 

[Anamitra]

Some Details in Relation to Post #30

Schwarzschild’s Metric for Radial Motion:
{ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}
Reduced metric:
{ds’}^{2}{=}{dt}^{2}{-}{dR}^{2}
Where,
{dR}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}^{2}
{ds’}^{2}{=}\frac{{ds}^{2}}{{1}{-}\frac{2m}{r}}
For the parallel transport of my vector (dt,dR)
1. dt=k1
=> {(}{1}{-}\frac{2m}{r}{)}^{-1/2}{(}{1}{-}\frac{2m}{r}{)}^{+1/2}{dt}{=}{k1}
=>{(}{1}{-}\frac{2m}{r}{)}^{+1/2}{dt}{=}{k1}{(}{1}{-}\frac{2m}{r}{)}^{+1/2}------------------------------(1)

2. dR=k2
=>{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}{=}{k2}
{(}{1}{-}\frac{2m}{r}{)}^{-1/2}{(}{1}{-}\frac{2m}{r}{)}^{-1/2}{dr}{=}{k2}
Therefore,
{(}{1}{-}\frac{2m}{r}{)}^{-1/2}{dr}{=}{k2}{(}{1}{-}\frac{2m}{r}{)}^{1/2}------------------------------(2)

From (1) and (2) ds^2 for the parallel transport of my vector is given by:
{ds}^{2}{=}{k1}^{2}{(}{1}{-}\frac{2m}{r}{)}{-}{k2}^{2}{(}{1}{-}\frac{2m}{r}{)}
Since k1 and k2 are constants for my parallel transport, ds^2 is not a constant for this transport

Rather,
\frac{{ds}^{2}}{{1}{-}\frac{2m}{r}}{=}{k1}^{2}{-}{k2}^{2}{=}{Const}
That is,

ds’^2 is constant for parallel transport of (dt,dR)

[It is to be noted that k1 and k2 are constants]

 

[Dale]

[For parallel transport the magnitude of the component vectors do not change-the individual magnitudes/norms are preserved at each point of the transport]

That is part of the definition of parallel transport. Although it is true, it is uninformative because it is true for all vectors.

What we are interested in is not the parallel transport of some vector or scalar, but the covariant derivative of the metric ds²=g and your ds'²=g', which are rank 2 tensors. If you compute the covariant derivatives you get:
\nabla_{\eta}g_{\mu \nu} = 0
and
\nabla_{\eta}g'_{\mu \nu} \ne 0
specifically, there are 4 non-zero components
\nabla_{r}g'_{tt} = -\frac{R}{r(r-R)}
\nabla_{r}g'_{rr} =\frac{rR}{(r-R)^3}
\nabla_{r}g'_{\theta \theta} =\frac{r^2 R}{(r-R)^2}
\nabla_{r}g'_{\phi \phi} =\frac{r^2 R \sin^2(\theta)}{(r-R)^2}

So the ds² is constant everywhere, and the dependence that you would expect by naively looking at the expression is simply the coordinates. But ds'² is not constant in a covariang sense; instead, ds'² changes wrt changes in r.

Points to Observe:
1. ds^2 is a scalar . ...
1. ds'^2 is a scalar

Let us have a look at the metric:

{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}


The left side is dot product between the identical vectors each being given by(dt,dx1,dx2,dx3)
Now the dot product is a scalar. The LHS ie ds^2 is a scalar since it is the result of a dot product. One should not apply covariant differentiation to it--ordinary differentiation should be applied to the LHS and hence to the RHS

This is not correct, but the notation is indeed sloppy and can be confusing. For reference please see Sean Carrol's Lecture Notes on General Relativity (http://lanl.arxiv.org/abs/gr-qc/9712019v1) on page 48 in the first full paragraph following equation 2.32.

In the expression for the line element the dt and similar terms are not infinitesimal displacements, but rather they are basis dual vectors (one-forms) and the dt² refers to a tensor product, not a dot product. So the line element is a rank 2 tensor, not a scalar. Perhaps this is the source of some of your confusion.

 

[WannabeNewton]

But at each point p on a manifold g_{p}: T_{p}(M) \times T_{p}(M)\mapsto \mathbf{R} and ds^{2} = g_{\mu \nu }dx^{\mu }\otimes dx^{\nu } so
how could ds^{2} be a 2 - tensor when it is the result of the metric tensor mapping the members of the tangent space at p (or across the entire manifold by mapping all members of the tangent bundle if the distinction matters) to the reals?

 

[Dale]

Because the notation is sloppy and inconsistent. See the link I posted above. Sean Carroll explains it far better than I can. But in the form that Anamitra has been discussing ({ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}-{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}) it is clear that ds² is a rank 2 tensor, not a scalar.

 

[Anamitra]

But in the form that Anamitra has been discussing ({ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}-{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}) it is clear that ds² is a rank 2 tensor, not a scalar.

g(i,j) is a second rank tensor[g(i,j)=0 if j not eqaul to i:this is true for orthogonal systems].Again,(dt,dx1,dx2,dx3) is a tensor of rank one.
Therefore,

{g}_{ij}dx^{p}dx^{q}
is a fourth rank tensor.
Due to contraction between the upper and lower indices[putting p=i and q=j] we are getting a scalar=ds^2.

 

[Dale]

Read the section I linked to earlier:

"In fact our notation “ds²” does not refer to the exterior derivative of anything, or the square of anything; it’s just conventional shorthand for the metric tensor."

ds² is a rank 2 tensor, not a scalar.

Note that g_{\mu\nu}dx^{\mu}dx^{\nu} is very different from g_{00}dt^2+g_{11}dx1^2+g_{22}dx2^2+g_{33}dx3^2 which is clearly a rank 2 tensor. The notation is confusing and inconsistent, but we have been discussing the second case and not the first.

 

[Anamitra]

For Orthogonal Systems:

{g}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}{=}{g}_{00}{dt}^{2}{+}{g}_{11}{dx1}^{2}{+}{g}_{22}{dx2}^{2}{+}{g}_{33}{dx3}^{2}{=}{ds}^{2}

g(00) is positive;g(1,1),g(2,2),g(3,3) are negative [[each]

[We should remember that in flat spacetime:g(00)=1,g(11)=-1,g(22)=-1,g(33)=-1]

g(mu,nu) are the components of what we call the metric tensor. ds^2 is not a metric tensor itself. It is the norm of the tensor given by (dt,dx1,dx2,dx3).

If ds^2 itself were a tensor of rank 2[if I assume that] it would have 16 components in four dimensional spacetime. On transforming to some other system these components would change! The invariance of ds^2 would become a meaningless concept.

 

[Dale]

For Orthogonal Systems:

{g}_{\mu\nu}{dx}^{\mu}{dx}^{\nu}{=}{g}_{00}{dt}^{2}{+}{g}_{11}{dx1}^{2}{+}{g}_{22}{dx2}^{2}{+}{g}_{33}{dx3}^{2}{=}{ds}^{2}

No, this is incorrect. Did you read the reference?

If ds^2 itself were a tensor of rank 2[if I assume that] it would have 16 components in four dimensional spacetime. On transforming to some other system these components would change! The invariance of ds^2 would become a meaningless concept.

This is correct. ds^2 has 16 components, and I misspoke when I said invariant. I meant constant, ie covariant derivative of 0.

 

[TrickyDicky]

If ds^2 itself were a tensor of rank 2[if I assume that] it would have 16 components in four dimensional spacetime.

Sure but in an orthonormal frame the second rank metric tensor only has 4 non-vanishing components, the diagonal ones.

 

[Anamitra]

What I have got out of the text[Sean M. Carroll,page 55,56] is that ds^2 is just a short hand notation of the metric tensor----that is ds^2 denotes the set [g(mu,nu)].This set has 16 members[tensor components] . The covariant derivative is of zero. We must take care not to include dt , dx1, dx2 and dx3 in this set in association with the metric coefficients.This is a relevant point,so far as the author's views are concerned [the metric coefficients contain all the information we need to describe the curvature of the manifold--the author remarks, apart from what I have asserted in the beginning of this post].I would request Dalespam to confirm or de-confirm me in case I have made a misinterpretation.

 

[Dale]

What I have got out of the text[Sean M. Carroll,page 55,56] is that ds^2 is just a short hand notation of the metric tensor----that is ds^2 denotes the set [g(mu,nu)].This set has 16 members[tensor components] . The covariant derivative is of zero.

Correct.

We must take care not to include dt , dx1, dx2 and dx3 in this set in association with the metric coefficients.

The tensor dx1 essentially just a coordinate basis (dual) vector for the metric, and dx1² is a rank-2 coordinate basis tensor. So their role in the expression for the line element/metric is similar to the use of i, j, and k as basis vectors in basic physics. You can think of dx1 as being the vector (0,1,0,0) and dx1² as being the tensor
\left(<br /> \begin{array}{cccc}<br /> 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 1 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0 \\<br /> 0 &amp; 0 &amp; 0 &amp; 0<br /> \end{array}<br /> \right)

 

[Anamitra]

So in your interpretation ds^2 is a 4*4 matrix with the main diagonal as [g(0,0),g(1,1),g(2,2),g(3,3)] where g(0,0) is positive and g(1,1),g(2,2),g(3,3) each being negative.[or you may take an alternative convention where the signs get reversed].

In such a situation how do you define the difference of proper time ,infinitesimal[in relation to curved spacetime] or of finite magnitude[curved spacetime]?

 

[Dale]

So in your interpretation ds^2 is a 4*4 matrix with the main diagonal as [g(0,0),g(1,1),g(2,2),g(3,3)] where g(0,0) is positive and g(1,1),g(2,2),g(3,3) each being negative.[or you may take an alternative convention where the signs get reversed].

Correct (in an orthogonal coordinate system).

However, more importantly, if I understand your goal correctly you want to figure out how to "flatten" a curved space. The curvature of a space is determined by the Riemann curvature tensor which consists of second derivatives of the rank 2 metric tensor. So if you were to change the expression for a mere scalar you would not accomplish that goal. In order to reach your goal you need to at least attempt manipulations of the rank 2 metric tensor.

In such a situation how do you define the difference of proper time ,infinitesimal[in relation to curved spacetime] or of finite magnitude[curved spacetime]?

\int \sqrt{g_{\mu\nu}dx^{\mu}dx^{\nu}}
Do you see the difference between the expression in the square root and this expression?
-dt^2+dx^2+dy^2+dz^2

 

[Anamitra]

What you have under the square root sign is just the dot product between two identical vectors[(dt,dx1,dx2,dx3) which is the square of proper-time for a timelike path. The values of g(mu,nu) are of a general nature that could represent the metric coefficients in curved spacetime or in flat spacetime.

In my formulation I can always use
{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2} --------------- (1)

ds is the proper time. As we have of a path we may use the "d" with "ds". [We should treat "dt","dx1",dx2" as infinitesimals and not as basis vectors. ]

By extremizing proper time I can arrive at the geodesic equation.

So far as your interpretation is concerned dt,dx1,dx2 and dx3 in equation (1) are basis vectors[ like i,j,k in the three dimensional orthogonal system.] and ds^2 is the short-hand notation for the metric tensor itself. Then how are you getting proper time? The problem remains.

ds^2=dt^2-dx^2-dy^2-dz^2 is a particular case of equation (1) so far as my formulation is concerned.

This is again the dot product between a pair of identical vectors--(dt,dx,dy,dz)

The space is flat spacetime--the Christoffel tensors are zero and and the Curvature tensor evaluates to zero value at each point.

 

[Dale]

Anamitra, do you want to examine if the space is flat or not? If so, then you cannot use a scalar, you must examine the metric tensor. If not, then I don't understand your point of this whole exercise.

Btw, stop trying to blame me for the sloppy and inconsistent notation. It isn't my fault that it is inconsistent and it is well explained by Carroll. Now all you have to decide is what you are interested in looking at: the metric tensor or the spacetime interval scalar. Please make a clear choice and then we can proceed from there.

 

[Mentz114]

The inverse metric is the sum of the tensor products of the tangent space cobasis \sigma

<br /> g_{\mu\nu}=-\vec{\sigma}_0 \otimes \vec{\sigma}_0 + \vec{\sigma}_1 \otimes \vec{\sigma}_1+ \vec{\sigma}_2 \otimes \vec{\sigma}_2+ \vec{\sigma}_3 \otimes \vec{\sigma}_3<br />
and the metric is got the same way from the basis vectors. I just thought I'd mention that.

[edited ]

 

[Anamitra]

Anamitra, do you want to examine if the space is flat or not? If so, then you cannot use a scalar, you must examine the metric tensor. If not, then I don't understand your point of this whole exercise.

The curvedness or the flatness of space has to be understood/analyzed with the help of the metric tensor

In the original metric the the values of g(mu,nu) corresponded to that of curved space while in the transformed metric[Post #22,Relation 3] it corresponds to that of flat space .

 

[Anamitra]

Some Queries:
1. Equation 2.45 page 53 in the text,page 60 in the pdf[Sean M. Carroll]

{d}^{n}{x}{=}{dx}^{0}\wedge{dx}^{1}\wedge{dx}^{2}{\wedge}{dx}^{3}{...}{\wedge}{dx}^{n-1} ---------- (1)

The quantities dx0,dx1,dx2 etc in the wedge product are not differentials as we know them. They are simply vectors in the dual space.

We may take a set of actual differentials as we know them:(dx0,dx1,dx2...dx(n-1))----(2)
They transform like contravariant tensors:
{dx&#039;}^{i}{=}\frac{{\partial}{x&#039;}^{i}}{{\partial}{x}^{k}}{{dx}^{k}}

We may consider a set of linear operators,which [each of them] on operating on an arbitrary member of the set comprising contravariant tensors produces a scalar belonging to some field.
These linear operators themselves form a vector space ,which is the dual space wrt the original set [which contains contravariant tensors].[The vectors in the dual space are supposed to be covariant while the original space contains contravariant tensors]

When we consider a wedge product of type:

{dx}^{1}\wedge{dx}^{2}----(3)

dx1 and dx2 in the wedge product are not differentials as we know them--they are simply vectors/linear operators[covariant vectors] from the dual space.They are not supposed to be from the original space. Expression (3) is actually an antisymmetric combination of such vectors from the dual space.

Equation (1) is actually is a relative tensor or tensor density.
The author remarks:
"It is clear that the naive volume element dnx transforms as a density, not a tensor, but
it is straightforward to construct an invariant volume element by multiplying by..."
The word naive has been used to mean that the volume element is a tensor density and not a tensor.But what about the volume element term.Its simply causing me a lot of confusion even if it is used as a short-hand notation.

Let us consider the wedge product in expression (3) when it is expanded in terms of the basis vectors in a two dimensional vector space vector space[dual space]

{dx}^{1}\wedge{dx}^{2}{=}{C1}{(}{e1}\wedge{e1}{)}{+}{C}_{12}<br /> {(}{e1}\wedge{e2}{)}{+}{C}_{21}{(}{(}{e2}\wedge{e1}{)}{+}{C}_{2}{(}{e2}\wedge{e2}{)} ------------------------ (4)

Now,

{(}{e1}\wedge{e1}{)}{=}{0}
and
{(}{e2}\wedge{e2}{)}{=}{0}

due to the antisymmetric nature of the wedge product.
Again,
{e1}\wedge{e2}{=}{-}{e2}\wedge{e1}
The expressions denoted by (3) or (4) are not commutative wrt to the interchange of the quantities dx(mu) and dx(nu).[Because of antisymmetric combination in the wedge product]
So that was my first query is in relation to the volume element being denoted by the wedge product. We may have similar issues regarding equations 2.47 and 2.48. in the text
The problem is not in relation to the transformations shown but regarding what the volume element really means in this situation--[what is the physical nature of the volume element that is being transformed if we are considering vectors from the dual space instead of actual infinitesimals as we know them]. Incidentally, the dx(i) s given in a wedge product are simply dual vectors and not infinitesimals --dx(i) in the original space as we know them.

My next query[which is an allied one] is in relation to equation to 3.48[page 69 in the text,page 76 in the pdf file]

Propertime has been defined by the author as:
{\tau}{=}{\int}{(}{(-1)}{g}_{\mu\nu}\frac{{dx}^{\mu}}{d\lambda}\frac{{dx}^{\nu}}{d\lambda}{)}^{1/2}{d}{\lambda}
"We therefore consider the proper time functional where the integral is over the pathTo search for shortest-distance paths, we will do the
usual calculus of variations treatment to seek extrema of this functional. (In fact they will
turn out to be curves of maximum proper time.)"--the author has remarked.

It looks like a statement we find in the traditional texts[A clear mention of the term path is there--this makes dtau meaningful].

Does the author consider dual vectors here for the dx(mu) and dx(nu)? I am not sure.Rather I am feeling confused.
I would request DaleSpam to explain the whole issue.

 

[Anamitra]

Points to Observe:

1. 1-form ,2-form etc[p-forms in general] are associated with covariant tensors which are antisymmetric wrt the interchange of any pair of indices.

2. One may try to make the following interpretation: that the infinitesimals on the RHS of equation (1) in post #50 are components of a contravariant tensor while the whole thing [the wedge product]is a p-form then the question that naturally arises is---what specifically is the operation wedge product in this case since we are considering the components of the same tensor and we have to get a covariant tensor as the result of the wedge product?

[In the dual space we don’t have any difficulty in dealing with components of the vectors
In fact each component vector will have a linear operator[covariant tensor] in the dual space.]

 

[Dale]

Does the author consider dual vectors here for the dx(mu) and dx(nu)? I am not sure.Rather I am feeling confused.
I would request DaleSpam to explain the whole issue.

I don't think that I can explain the "whole issue", but I can at least point out Dr. Carroll's very notation. If you look carefully at equation 3.48 you will notice that the font for the "d" is different than the font for the "d" on the rhs of equation 2.45. One is in italics and the other is non-italicised. The non-italicised "d" is introduced on page 12 in equation 1.40, and he uses it consistently from then on. The italicised "d" is never formally introduced, and is inconsistently used. Sometimes it represents an ordinary total derivative and other times it doesn't seem to mean anything and other times it is used to represent a small vector.

In any case, pay close attention to the font of the "d"s in Sean Carroll's notes.

 

[Dale]

The curvedness or the flatness of space has to be understood/analyzed with the help of the metric tensor

OK, so going back to the previous discussion. If you are interested in curvedness or flatness then as you say we need to look at the metric tensor. If we do the operation you suggest on the metric tensor, specifically if we multiply it by the tensor (1-2m/r)^(-1), then the result is another rank 2 tensor, which you have called ds²'. Agreed?

As I showed in 34, this tensor has non-zero covariant derivatives.

 

[Anamitra]

For the first case[Radial motion in Schwarzschild's Geometry] The metric tensor[Eqn (1) in the next post ,#55] has the components:

{g}_{tt}{=}{(}{1}{-}\frac{2m}{r}{)}

{g}_{rr}{=}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}

If you divide both sides of the first metric representing Radial motion in Schwarzschild's Geometry by 1-2m/r,the resulting metric tensor[Eqn (3) in post #55 has the components:

{g}_{tt}{=}{1}
{g}_{rr}{=}{-}{(}{1}{-}\frac{2m}{r}{)}^{-2}
This metric corresponds to a curved spacetime different from the first one.
Finally the metric tensor[Eqn (4) in Post #55] has the components:

{g}_{tt}{=}{1}
{g}_{rr}{=}{-1}
This represents the metric for flat space-time.

ds^2 ,ds'^2 are simply dot products and are not to be treated as tensors themselves in my formulation. I am not treating dt,dr as basic vectors. They are simply infinitesimal changes in t or r.
[DaleSpam in Post #43 has suggested considering dt,dx etc as basis-vectors in order to treat ds^2 as a second rank tensor]

With, the path specified we may use "d" with ds or ds'

 

[Anamitra]

First we write the metric[for radial motion is Schwarzscild’s Geometry] :

{ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{+}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}-------------------------------- (1)
Or,
\frac{{ds}^{2}}{{1}{-}\frac{2m}{r}}{=}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}^{2}---------------------------- (2)
Or,
{ds’}^{2}{=}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}^{2} ---------------------------- (3)


Transformed metric:
{ds’}^{2}{=}{dt}^{2}{-}{dR}^{2} -------------------- (4)

Where,
{ds’}^{2}{=}\frac{{ds}^{2}}{{1}{-}\frac{2m}{r}}

{dR}^{2}{=}\frac{1}{{(}{1}{-}\frac{2m}{r}{)}^{-2}}{dr}^{2}
The original metric is independent of t .
Therefore,
(1,0) is a killing vector.
Now dot product between a Killing vector and v is preserved for extremal paths
Therefore,
{(}{1}{-}\frac{2m}{r}{)}\frac{dt}{ds}{=}{constant}
This is the conservation of energy[for unit rest mass] for geodetic motion[space-time geodesics being considered] in the original equation.

If we look at equation(4) it is independent of t and R[if we change t and R by constant amounts the metric does not change] We have two killing vectors (1,0) and (0,1)
For geodetic motion:
1.dt/ds’ = const [conservation of energy for unit rest mass]
2. dR/dt = const [this makes the motion uniform: gamma= const]

Energy is conserved for geodetic paths[space-time geodesics]. The law remains unaltered for both the manifolds.

But if geodetic motion in metric represented by (1) is transformed to the flat spacetime situation, we do not get geodetic motion wrt the flat spacetime metric.Energy changes in this case

In the flat spacetime situation the particle undergoes motion with a variable speed. Energy is not conserved for nongeodetic path[spacetime path]. For the purpose of acceleration it must be receiving energy from some inertial source/agent[supplying inertial interaction] whose energy must decrease. So the total energy of particle+inertial agent must remain constant in the flat spacetime context for non-geodetic motion.

The effect of spacetime curvature is being replaced by inertial interaction in the flat spacetime context.

 

[WannabeNewton]

ds^2 ,ds'^2 are simply dot products and are not to be treated as tensors themselves in my formulation. I am not treating dt,dr as basic vectors. They are simply infinitesimal changes in t or r.
[DaleSpam in Post #43 has suggested considering dt,dx etc as basis-vectors in order to treat ds^2 as a second rank tensor]

No, if you treat the quantities as the dual basis vectors then the metric tensor can map the tensor product of the dual basis vectors so that you get a scalar. If you treat them as infinitesimals then the metric tensor doesn't map them and you still have a 2 - tensor.

 

[Anamitra]

Precisely speaking I am treating (dt,dr) as a vector.

{ds}^{2}{=}{g}_{tt}{dt}^{2}{-}{g}_{rr}{dr}^{2}

is a dot product [between the identical vectors each being given by (dt,dr)]if you consider the definition of dot product .

In the four-dimensional picture:

{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}{-}{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2}
is a dot product[between the identical vectors (dt,dx1,dx2,dx3)] if one considers the definition of dot product.

In my formulation I am using this dot product interpretation of ds^2.

If you choose some other type of formulation,it is of course your choice

[The infinitesimals dt,dx1,dx2 and dx3 do follow the transformation rules required by tensors. They behave like the components of a contravariant tensor.]

 

[Dale]

If you divide both sides of the first metric representing Radial motion in Schwarzschild's Geometry by 1-2m/r,the resulting metric tensor[Eqn (3) in post #55 has the components:

{g}_{tt}{=}{1}
{g}_{rr}{=}{-}{(}{1}{-}\frac{2m}{r}{)}^{-2}
This metric corresponds to a curved spacetime different from the first one.
Finally the metric tensor[Eqn (4) in Post #55] has the components:

{g}_{tt}{=}{1}
{g}_{rr}{=}{-1}
This represents the metric for flat space-time.

No, the resulting tensor is not a metric tensor at all. (1-2m/r)^-1 is a scalar. The operation of multiplying a tensor of some rank by a scalar is another tensor of the same rank. That resulting tensor may have components that look like the flat spacetime metric, but it is not.

First, tensor multiplication is not a coordinate transformation. You have not changed coordinates, you have merely found a certain tensor with a specific form in the original coordinates.

Second, because you have not done a coordinate transform the Christoffel symbols are unchanged, so the curvature may still be non-zero even though the form of this tensor is superficially the same as the metric tensor in flat spacetime.

Third, the resulting tensor does not, in general, have a 0 covariant derivative as I showed above.

In order to find a different expression for the metric tensor (rather than finding a completely different tensor as you have done here or a completely different manifold as you did earlier) you must do a coordinate transform. However, you can prove in general that a non-zero tensor will remain non-zero under all coordinate transforms, so the curvature tensor cannot be made zero through a coordinate transform. Your task is fundamentally impossible.

 

[Anamitra]

First let us consider the Right –hand sides of the relations (1) and (3) of Post #55.
The metric coefficients are different—they represent different manifolds.[the coefficients of dt and dr are different[individually] in the two cases
Now the path described by the two metrics are identical—so we are considering the intersection of two spacetime surfaces[two manifolds ] at the line of their intersection—and the same path is being described by the two metrics but in separate manifolds.
Needless to say that the two manifolds have been described in the same system of coordinates so far as the metrics in (1) and (3) are concerned.

Between (3) and (4) of Post #55
The transformation (dt,dr)-->(dt,dR) follows the transformation rule of the contravariant tensors:

 

[Anamitra]

A familiar example of two different/distinct manifolds that may exist on the same coordinate grid:

1. {ds}^{2}{=}{dt}^{2}{-}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{)}

2. {ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}{-}{r}^{2}{(}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{)}

The first metric represents flat spacetime in spherical system of coordinates.[Diagonal matrix [1,-1,-1,-1] for flat spacetime is true only in the rectangular system of coordinates.]
The second one represents Schwarzschild's Geometry.

The t,r,theta phi coordinate grid is the same for both the systems but the physical separations are different.
For example the same differences of dr[ie coordinate separation]in the two systems would imply different physical separations for the two systems.
In flat spacetime the physical separation would be simply dr.
But in Schwarzschild's Geometry the physical separation corresponding corresponding to the same dr would be Sqrt[(1-2m/r)^(-1)](dr)

We may think of considering the two manifolds in the same t,r,theta,phi system where the coordinate separations would be identical but the physical separations would be different.