# How do You Define “Coordinate Transformation”?

## What is the exact definition of "Coordinate Transformation"?

Issues to Consider in Relation to the Question:

1. Let us first consider an orthogonal system. The dot product of the vectors along the coordinate curves at some point. We consider two such vectors $e_1=(0,1,0,0)$ and $e_2=(0,0,1,0)$. $e_1.e_2=0$ implies $g_{12}=0$. For orthogonal systems cross terms of the form $dx^i dx^j$ should not be present if $i\ne j$. Their presence can only indicate a non-orthogonal system. An arbitrary transformation[consistent with continuity, differentiability etc] on an orthogonal system can lead to a non-orthogonal one[generally speaking].

2. In a transformation converting an orthogonal system to a non-orthogonal system the angles will not be preserved. Lets take a practical view of the situation. We start with a rectangular Cartesian system in two-dimensions in flat space[the x and the y-axes being perpendicular to each other]. If we pass to a non-orthoganal system with the preservation of the metric, the inclination of the old x and y axes will change .But they will remain straight lines sine we are still on flat space[the metric being preserved].We choose these inclined lines as our new axes. The old unit-vectors are no more orthogonal. It would be interesting to consider a pair of points A(1,0) and B(0,1) in the old rectangular frame. The straight-line distance between them will change in the new frame—indicating a change in the value of $ds^2$. If the value of $ds^2$ does not change we are simply passing into curved space. The metric has to change.

Points to Observe:

A. Flat space is characterized by the successful application of the Pythagoras theorem.In the original[orthogonal system] we have:$$OA^2+OB^2=AB^2$$ If the line element remains unchanged on a transformation to a non-orthogonal system we have the relation$$O'A'^2+O'B'^2=A'B'^2$$ in the new [non-orthogonal]system.But the inclination between the axes will not be 90 degrees in the transformed frame if it corresponds to a non-orthogonal-system.The consistent option would be to consider the transformed region to be a curved surface.
[The relation $O'A'^2+O'B'^2=A'B'^2$ might hold on a curved surface despite none of the angles being a right angle. Also,the nature of a geodesic in the physical sense, should not change if the nature of the manifold does not change. If the nature of the geodesic changes, the manifold itself is changing]

B. If the manifold changes with a non-orthogonal transformation,allowing transformations between arbitrary manifolds where the value of $ds^2$ is preserved, all physical laws should have the same form[covariant or tensor form] in different/distinct manifolds[having different metric composition]

It is always possible to create a transformation that converts a straight line to a curved line. If the straight line in the first frame represents the path of a light ray in flat space-time, the same path in the second one, ie, the transformed one is a curved line. No harm, if we treat the second frame as a mathematical work-space. But is might be possible to locate a true physical manifold of a distinct nature where the light ray follows the exactly same curved path.

The following Link is related to the issue in consideration:

http://physics.stackexchange.com/questions/19429/the-light-ray-bends-round

[Post: The Light Ray Bends Round]

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More generally, en.wikipedia.org/wiki/Smooth_manifold#Atlases --- coordinate transformations are called "transition functions" in this context. –  Neal Dec 6 '11 at 5:01
$ds^2=g_{\mu\nu}dx^\mu dx^\nu$--(1). For an orthogonal system the dot product between the unit vectors at the intersection of the coordinate curves is zero.Let $e_1=(0,1,0,0)$ and $e_2=(0,0,1,0)$. N0w, $e_1.e_2=0 => g_{12}=0$.That is, the $dx_1dx_2$ should be missing in (1). If the coordinates(initially orthogonal) are transformed using relations like$x^\alpha=x^\alpha(x’^\mu)$ we end up with terms like $dx_idx_j,i\nej$. Angles are not being preserved though ds^2 is being preserved. –  Anamitra Palit Dec 27 '11 at 6:18
Are non-orthogonal transformations included in coordinate transformations? Is the metric preserved for non-orthogonal transformations?Incidentally the definition/transformation of tensors like:$T'^{\mu\nu}=\frac{\partial x'^\mu}{\partial x^\alpha}\frac{\partial x'^\nu}{\partial x^\beta}T^{\alpha\beta}$ seems to entertain the preservation of ds^2 but the definiton does not ensure the preservation of angles. –  Anamitra Palit Dec 27 '11 at 6:48
@Anamitra: The metric is not part of the data of a differentiable manifold, so the definition of coordinate transformation certainly cannot have anything to do with the metric. Can you please quote a specific definition you have in mind and ask a question about that definition? –  Zhen Lin Dec 30 '11 at 8:52
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