Transformation of the Line-Element

[Anamitra]

Length of the line element does not change in transformations[subject to the conditions of continuity,differentiability,one-to-one correspondence etc]. "ds^2" is invariant with respect to transformations[subject to the conditions].

Now let us have a look at the following transformation--the projection transformation..

You are standing under a hemispherical roof. The floor is a flat surface. A curve is drawn on the ceiling and we take its projection on the floor[x-y plane].We get[by this transformation] another curve on the floor which is of unequal length.
Integral ds is different for the two curves.Obviously ds is also different.

Lets have a closer look:

Line element for the hemispherical ceiling:

{ds}^{2}{=}{R}^{2}{[}{d}{\theta}^{2}{+}{sin}^{2}{\theta}{d}{\phi}^{2}{]}

Transformation[for the projection]:

{x}{=}{R}{Sin}{\theta}{Cos}{\phi}

{y}{=}{R}{Sin}{\theta}{Sin}{\phi}

For an infinitesimal projection we get from geometrical conditions:

{ds'}^{2}{=}{dx}^{2}{+}{dy}^{2}{=}{R}^{2}{[}{cos}^{2}{\theta}{d}{\theta}^{2}{+}{Sin}^{2}{\theta}{d}{\phi}^{2}{]}
ds and ds' are not identical.

So in General Relativity we must consider only such transformations that leave ds unchanged

For an arbitrary well behaved transformation the the length of the line element may change

 

[PAllen]

Well, your construction is simply wrong. Given a coordinate transform, you have to derive the transformed metric, you don't just assume it. What you've done amounts to saying:

If I stretch and flatten a hemisphere to a plane, I've changed the geometry. Surprise!

Covariance, invariance etc. are based on the idea that: yes I can label points on the hemisphere using the floor, but since I want to preserve geometry, I derive a non-euclidean metric for the floor coordinates that accomplishes this. You've simply assumed a Euclidean metric and gotten nonsense.

To get the real metric, you can compute the Jacobian and do a matrix multiply, or you can manipulate differentials leading to a valid relationship between dx, dy and d theta, d phi (I'm assuming we consider R a constant in your construction; this is a 2-d coordinate transform on a 2-surface). To get the x,y metric you want to state the reverse transform (forgive my no latex):

phi = inverse tangent (y/x)
theta = inverse sin (sqrt(x^2+y^2)/R)

Take d of these, plug into the theta, phi line element. Then you get the correct x,y line element to preserve geometry.

 

[WannabeNewton]

I don't even get how you determined the metric components after the coordinate transformation. You don't have \theta and \phi explicitly in terms of the new coordinates so how did you evaluate g_{{\alpha }'{\beta }'} = \frac{\partial x^{\alpha }}{\partial {x^{\alpha}}'}\frac{\partial x^{\beta }}{\partial {x^{\beta}}'}g_{\alpha \beta }.

 

[Anamitra]

To get the x,y metric you want to state the reverse transform (forgive my no latex):

phi = inverse tangent (y/x)
theta = inverse sin (sqrt(x^2+y^2)/R)

Take d of these, plug into the theta, phi line element. Then you get the correct x,y line element to preserve geometry.

I this case you get:

{ds}^{2}{=}{f1}{(}{x}{,}{y}{)}{dx}^{2}{+}{f2}{(}{x}{,}{y}{)}{dy}^{2}

But dx and dy are orthogonal segments on the x-y plane[ a flat plane]. But ds, dx and dz are not following Pythagoras Theorem!

Actually what you get in a projection is,,

{ds'}^{2}{=}{dx}^{2}{+}{dy}^{2}

 

[PAllen]

I this case you get:

{ds}^{2}{=}{f1}{(}{x}{,}{y}{)}{dx}^{2}{+}{f2}{(}{x}{,}{y}{)}{dy}^{2}

But dx and dy are orthogonal segments on the x-y plane[ a flat plane]. But ds, dx and dz are not following Pythagoras Theorem!

Actually,

{ds'}^{2}{=}{dx}^{2}{+}{dy}^{2}

That's the whole point. A coordinate transform does not change geometry. Of course the x,y line element isn't Euclidean! Otherwise you've eliminated the curvature!

You can do any transform you want (essentially) but what preserves invariance / covariance is transforming the line element in the correct way - that preserves geometry.

[Edit: note, your statement above is mathematically incorrect as well. You get dxdy term as well, not just dx^2, and dy^2. ]

 

[Anamitra]

[Edit: note, your statement above is mathematically incorrect as well. You get dxdy term as well, not just dx^2, and dy^2. ]

dxdy term exists only in non-orthogonal systems[link: https://www.physicsforums.com/showpost.php?p=3430803&postcount=5]

The transformations you are talking of do not correspond to an orthogonal projection.

 

[Anamitra]

Points to Observe

1.In a system which is not orthogonal Pythagoras Theorem is not supposed to hold between dx,dy and ds even in flat space.

2.Even for flat space the metric coefficients are not supposed to be the same for orthogonal and non orthogonal systems.

 

[Dale]

Anamitra you are correct, the metric on a flat projection is not the same as the metric on the sphere being projected. And yes, in GR we are explicitly not interested in projections, we are interested only in coordinate transformations as is mentioned in most GR texts and in PAllen's and WannabeNewton's responses. A projection is not merely a coordinate transformation, it changes vectors. Another way of saying it is that coordinate transformations are covariant, and a projection is not.

http://en.wikipedia.org/wiki/Covariant_transformation

 

[Anamitra]

The norm of the tensor will change in the projection type transformation[since ds is changing]. But what about the mutual relationship between the tensors--the laws?
The form of the law should be preserved.So we may work in the projected plane[a flat one] and finally go back to the curved space to get back the norm.

 

[Anamitra]

Let us define tensors by the transformation rule:

{{T'}^{\alpha}}_{\beta}{=}{k}\frac{{\partial}{x'}^{\alpha}}{{\partial}{x}^{\mu}}\frac{{\partial}{x}^{\nu}}{{\partial}{x}'^{\beta}}{{T}^{\mu}}_{\nu}
T' is in the theta,phi system while T is in the x,y system. K is a scale factor[due to change in ds].

Since ds is changing, norm of the tensor will change. But what about the laws--the mutual relationships?

Suppose in the x,y system we have the law:

{{A}^{\mu}}_{\nu}{=}{{B}^{\mu}}_{\nu}
In the transformed frame we have:
{{A'}^{\mu}}_{\nu}{=}{{B'}^{\mu}}_{\nu}

[We get this from the definition of tensors--their transformations]

If the form of the tensor equation does not change[that is if the law does not change,so far as mutual relationships are concerned],we may work in flat space and then go back to curved space to restore the norms.

[One may also consider the law:

{{A}^{\mu}}_{\nu}{=}{B}^{\mu} {C}_{\nu} ]

 

[Dale]

The norm of the tensor will change in the projection type transformation[since ds is changing]. But what about the mutual relationship between the tensors--the laws?
The form of the law should be preserved.

Not under a non-covariant transformation. I.e. the form of the laws are unchanged under a transformation to a different coordinate system on the same manifold, but there is no reason to expect that the form of the laws are unchanged under a transformation to a completely different manifold.

 

[PAllen]

Another observation is that for geometric theory like GR, physical observables (measurements) correspond to contractions to scalars. If these change, physics changes(e.g. the rest mass is the norm of 4-momentum; KE of x measured by y is dot product of x 4-momentum with y's 4 velocity). Another way of putting it: change the manifold geometry, and you've changed the physical universe (including rest mass, for example). Coordinate transforms are all about re-labeling points on the same manifold. Thus, once you've specified in any coordinates, it is uniquely determined for all other coordinates.

 

[Anamitra]

Not under a non-covariant transformation. I.e. the form of the laws are unchanged under a transformation to a different coordinate system on the same manifold, but there is no reason to expect that the form of the laws are unchanged under a transformation to a completely different manifold.

Let us assume that the law changes on passing from the old frame to the new frame[flat spacetime]. But the fact is that we have a law to be worked out in flat space considerations.Using the new law we can perform our calculations and then go back to the old frame to get the actual values of the variables in the original space

 

[Anamitra]

Another observation is that for geometric theory like GR, physical observables (measurements) correspond to contractions to scalars. If these change, physics changes(e.g. the rest mass is the norm of 4-momentum; KE of x measured by y is dot product of x 4-momentum with y's 4 velocity). Another way of putting it: change the manifold geometry, and you've changed the physical universe (including rest mass, for example). Coordinate transforms are all about re-labeling points on the same manifold. Thus, once you've specified in any coordinates, it is uniquely determined for all other coordinates.

I can always change the physical universe if I know how to go back to the old one.

 

[PAllen]

I can always change the physical universe if I know how to go back to the old one.

The point of covariantly transforming the metric is so that calculations in the new coordinates directly lead to the same physical observables as the old coordinates. The aim is to keep the manifold geometry the same while relabeling or shifting points around.

What's the point of changing the manifold geomertry? Everything you conclude is false for the original manifold. In your 2-d example:

- lengths different
- angles different
- sum of angles of triangle different
- geodesics diffferent

So you then throw away the new manifold and go back to the old for any computations. So why introduce the new at all?

Your original post said only certain transforms are valid in GR. This is false, as normally understood. The normal understanding is that you transform the metric using the same transform applied to points. The result is that any transform then preserves the line element and all metric features.

 

[Anamitra]

So you then throw away the new manifold and go back to the old for any computations. So why introduce the new at all?

Calculations are much simpler in flat spacetime context--and we can always revert back to the old system at will. In the new system[transformed system--flat spacetime] we have specific laws[corresponding to the old manifold]

Your original post said only certain transforms are valid in GR. This is false, as normally understood. The normal understanding is that you transform the metric using the same transform applied to points. The result is that any transform then preserves the line element and all metric features.

In GR we are used to taking only those transformations where ds^2 does not change.
We can of course consider arbitrary well behaved transformations where ds^2 changes provided we are sure of the transformation rules /procedures so that we may switch between the two manifolds[and enjoy the comfort of working in flat spacetime]

 

[Ben Niehoff]

Here is a quantity you can consider on your "projected" manifold:

g^{\mu\nu} R^\lambda{}_{\mu\lambda\nu}

where R^\lambda{}_{\mu\nu\rho} is the Riemann tensor. The quantity defined above is a scalar (notice all indices are contracted), so it should be invariant under coordinate transformations. Calculate this quantity in your flat "projected" space, and then use your calculation to tell me what this quantity is in the curved space.

 

[PAllen]

Calculations are much simpler in flat spacetime context--and we can always revert back to the old system at will. In the new system[transformed system--flat spacetime] we have specific laws[corresponding to the old manifold]

But no calculation you make in the new manifold has any value at all. You have to recompute everything from scratch in the old manifold.

In GR we are used to taking only those transformations where ds^2 does not change.
We can of course consider arbitrary well behaved transformations where ds^2 changes provided we are sure of the transformation rules /procedures so that we may switch between the two manifolds[and enjoy the comfort of working in flat spacetime]

GR covers arbitary tranformation. It requires only the simple consistency that you transform vectors, tensors, etc, consistent with the way you transform coordinates.

 

[Dale]

Let us assume that the law changes on passing from the old frame to the new frame[flat spacetime].

You are not changing from an old frame to a new frame, that is just a coordinate transformation. You are changing to a new Riemannian manifold. You cannot do what you are talking about simply by changing frames.

 

[Dale]

Calculations are much simpler in flat spacetime context

That is highly unlikely. I would have to see a clear example to believe that.

 

[Anamitra]

Lets come to a basic issue:

Consider two infinitesimally close spacetime points A ans B[on the spacetime surface].

The separation between them is given by:

{ds}^{2}{=}{g}_{00}{dt}^{2}{-}{g}_{11}{dx1}^{2}-{g}_{22}{dx2}^{2}{-}{g}_{33}{dx3}^{2} -------------------- (1)

Now I connect A and B by several paths[long ones or infinitesimally small ones] lying on the spacetime surface and calculate integral ds for different paths. In all likelihood I would be getting different values.

In fact if you connect A and B by one million infinitesimal paths [on the spacetime surface] we may get different values of pathlength.

Is there any guarantee that a physical path should exist between A and B whose value is given by (1)
To what extent ds itself is a perfect differential is an issue to rekcon with!

The General Relativity metric at most is a valid concept only in a path dependent situation.

 

[Anamitra]

Radial Motion in Schwarzschild's Geometry

Metric for Radial Motion

{ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2} --- (1)
\frac{{ds}^{2}}{{1}{-}\frac{2m}{r}}{=}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}^{2} --- (2)
{ds’}^{2}{=}{dt}^{2}{-}{dR}^{2} -------------- (3)
LHS of the above corresponds to the fact that ds itself is a path dependent quantity.
dR is defined by:
{dR}{=}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}
{R-R0}{=}{\int}{(}{1}{-}\frac{2m}{r}{)}^{-2}{dr}
The integral on RHS is a definite integral and so the constant which may contain time gets eliminated.
[It should be noted that \frac{{\partial}{R}}{{\partial}{t}}{=}{0} ]
Equation (3) is a flat spacetime metric which allows non local velocities, relative velocities etc
[One may take the definition of radial speed as V=dR/dt]
ds and ds’ are not exact differentials[they are path dependent quantities] and one must not use the the formula for exact differentials to expand them.

Whenever we apply a General Relativity metric we have a path in our mind. The metric itself expresses a relationship between quantities which are not perfect differentials[though we use "d" for ds]

 

[Dale]

ds' is no longer invariant, i.e. it is not a metric

 

[Anamitra]

We are supposed to choose transformations wrt which ds' is invariant.
Suppose:
t=F(k)
R=F1(m)

{ds'}^{2}{=}{d}{(}{F}{(}{k}{)}{)}^{2}{-}{d}{(}{F1}{(}{m}{)}{)}^{2} ---- (1)
[(k,m) are the new coordinates.]
{ds'}^{2} is not changing.

Arbitrary transformations like orthogonal projections must be excluded!

[This has already been suggested by you]

 

[Dale]

Yes, it is changing. ds' is now a function of r. In other words, according to ds' two identically constructed clocks at different r will measure different amounts of proper time per tick.

ds' is not the line element, it is not the metric. The expression on the rhs is not a flat spacetime, and you cannot do parallel transport or covariant differentiation with it. Furthermore, your laws of physics now change from place to place.

 

[Anamitra]

s was a function of r ant t
ds was a function of r and t

ds'=ds/F(r)

s' is a function of r and t.
Alternatively ds' is a function of t and R
From equation (3) of post #22 ds' is a line element in flat spacetime.

 

[Dale]

No, the line element is not a function of r and t because the connection is metric-compatible. This is, in fact, the whole point of having a metric-compatible connection.

 

[Anamitra]

Yes, it is changing. ds' is now a function of r. In other words, according to ds' two identically constructed clocks at different r will measure different amounts of proper time per tick.

Lets take Schwarzschild's Equation for radial motion:

{ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}
We take two points along the radial line for which
1.The values of r are different[say r1 and r2]
2. Same values dt and dr are considered.

ds will be different for the two points.

The only thing is that ds has to be invariant--that too wrt to a certain class of transformation-----for which ds is invariant

[For any pair of events dt and dr should be the same for all observers]

 

[Dale]

Lets take Schwarzschild's Equation for radial motion:

{ds}^{2}{=}{(}{1}{-}\frac{2m}{r}{)}{dt}^{2}{-}{(}{1}{-}\frac{2m}{r}{)}^{-1}{dr}^{2}
We take two points along the radial line for which
1.The values of r are different[say r1 and r2]
2. Same values dt and dr are considered.

ds will be different for the two points.

That shows that dt and dr functions of r, not that ds is.

If you want to show whether or not something is constant wrt some variable you take a derivative, in this case a covariant derivative. The Levi-Civta connection is metric compatible, meaning that the covariant derivative of the metric is 0, but not your modified metric.

 

[Anamitra]

The parallel Transport issue:

Suoppse you are parallel transporting (dt,dr) in your system and I am parallel transporting (dt,dR) in my rectangular system. In my case the components dt and dR are not supposed to change in magnitude individually at any point of the transport[parallel-transport]

For my parallel transport:
{dR}{=}{[}{{1}{-}\frac{2m}{r}}{]}^{-1}{dr}
Should not change at any point of the journey. “dt” should not change also.[Norms of the component vectors are preserved in a parallel transport for each and every point of the movement]
For your case:
{[}{1}{-}\frac{2m}{r}{]}^{-1/2}{dr} should not change
And {(}{1}{-}\frac{2m}{r}{)}^{1/2}dt should not change. The norm of each vector dt and dr should not change

Now let me go back to my parallel transport for a moment again:
{dR}{=}{[}{{1}{-}\frac{2m}{r}}{]}^{-1}{dr}
{dR}{=}{[}{{1}{-}\frac{2m}{r}}{]}^{-1/2}{dr}{*}{[}{{1}{-}\frac{2m}{r}}{]}^{-1/2}{=}{Const}

Which means:
{[}{1}{-}\frac{2m}{r}{]}^{-1/2}{dr} is not constant

One may apply similar logic to the time component
So far as my motion[parallel transport] is concerned ds is not constant . So far as your parallel transport is concerned ds is constant!

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

 

[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.