# Transformation of the Line-Element

1. Aug 3, 2011

### 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

Last edited: Aug 3, 2011
2. Aug 4, 2011

### 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.

3. Aug 4, 2011

### 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 $g_{{\alpha }'{\beta }'} = \frac{\partial x^{\alpha }}{\partial {x^{\alpha}}'}\frac{\partial x^{\beta }}{\partial {x^{\beta}}'}g_{\alpha \beta }$.

4. Aug 4, 2011

### Anamitra

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,,

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

5. Aug 4, 2011

### PAllen

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. ]

Last edited: Aug 4, 2011
6. Aug 4, 2011

### Anamitra

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.

7. Aug 4, 2011

### 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.

8. Aug 4, 2011

### Staff: Mentor

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

Last edited: Aug 4, 2011
9. Aug 4, 2011

### 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.

10. Aug 4, 2011

### 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:

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

Last edited: Aug 4, 2011
11. Aug 4, 2011

### Staff: Mentor

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.

12. Aug 4, 2011

### 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.

13. Aug 4, 2011

### Anamitra

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

14. Aug 4, 2011

### Anamitra

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

15. Aug 4, 2011

### PAllen

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.

16. Aug 4, 2011

### Anamitra

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]

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]

17. Aug 4, 2011

### 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.

18. Aug 4, 2011

### PAllen

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.

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

19. Aug 4, 2011

### Staff: Mentor

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.

20. Aug 4, 2011

### Staff: Mentor

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