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From the Orthogonal to the Non-orthogonal System
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Anamitra Palit
Jan 14, 2012, 4:22:25 PM1/14/12
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Let's consider a 2D orthogonal [x-y] system having origin at O in the
flat space context..'A' is a point on the x-axis and 'B' is another
on the y-axis.ABC is a right angled triangle with
AB^2=OA^2+OB^2
We now transform to a non-orthogonal system in the same manifold[flat
space] with the preservation of ds^2.We make the angle between the
axes x' and y' =theta not equal to a right angle.The axes are
maintained as straight lines.The value of the line element is
preserved .
A----->A'
B----->B'
O---->O'
Since ds^2 is preserved
OA=O'A'
OB=O'B'
AB=A'B'
AB is the shortest distance between A and B in the original space. If
you try to figure the situation on a piece of flat paper[after the
transformation], the distance from A' to B'' along a straight line
path in the new frame will not be equal to AB. If you imagine the
process of the axes getting gradually inclined the AB should get
converted to a curve A’B’. Then what about the straight line path
between A’ and B’ in the new frame. Is there really some path in the
old frame that corresponds to the “straight line” distance between A’
and B’?
[ You may consider the facts to decide the issue:(1)The line element
has been preserved (2)The straight line distance AB in the old frame
was the shortest one connecting A and B. is it possible to have a
shorter path connecting A' and B' in the new frame.?]
Indeed, we have the relation:
A'B'=O'A'^2+O'B'^2
in the transformed situation. We have a Pythagorean triplet for a
triangle without a right angle. This is not possible for flat space.
But we may curve the space to allow the Pythagorean relation in the
new situation. We have passed into a new manifold with the
preservation of ds^2 but with the non-preservation of angles. The
straight line distance between A’ and B’ as perceived on a flat piece
of paper simply does not exist in the present context. We may now do
some sort of re-labeling of coordinates to make the new system
orthogonal.
The significance of these Transformations:
Tensors are usually defined by relations like T'^[mu,nu]=[del
x'^[mu]]/]del x^[alpha]] * [del x'^[nu]]/]del x^[beta]] T^[alpha,beta]
Such definitions assume relations of bthe type:
x'^[mu]=f(x^[alpha],x^[beta])
ds^2 does not change in these relations. But angles between curves
might change.
The salient feature or the point of advantage is that we may pass from
one manifold to another, giving a mathematical foundation to the
universality of the physical laws--that they have the same tensor
form[covariant form] in all distinct manifolds.
Any failure to pass between different manifolds[by suitable
transformations] would restrict the tensor object to a particular type
of manifold.Such a situation would be a hindrance towards the claim
of the universality of the physical laws [in covariant form].
flat space context..'A' is a point on the x-axis and 'B' is another
on the y-axis.ABC is a right angled triangle with
AB^2=OA^2+OB^2
We now transform to a non-orthogonal system in the same manifold[flat
space] with the preservation of ds^2.We make the angle between the
axes x' and y' =theta not equal to a right angle.The axes are
maintained as straight lines.The value of the line element is
preserved .
A----->A'
B----->B'
O---->O'
Since ds^2 is preserved
OA=O'A'
OB=O'B'
AB=A'B'
AB is the shortest distance between A and B in the original space. If
you try to figure the situation on a piece of flat paper[after the
transformation], the distance from A' to B'' along a straight line
path in the new frame will not be equal to AB. If you imagine the
process of the axes getting gradually inclined the AB should get
converted to a curve A’B’. Then what about the straight line path
between A’ and B’ in the new frame. Is there really some path in the
old frame that corresponds to the “straight line” distance between A’
and B’?
[ You may consider the facts to decide the issue:(1)The line element
has been preserved (2)The straight line distance AB in the old frame
was the shortest one connecting A and B. is it possible to have a
shorter path connecting A' and B' in the new frame.?]
Indeed, we have the relation:
A'B'=O'A'^2+O'B'^2
in the transformed situation. We have a Pythagorean triplet for a
triangle without a right angle. This is not possible for flat space.
But we may curve the space to allow the Pythagorean relation in the
new situation. We have passed into a new manifold with the
preservation of ds^2 but with the non-preservation of angles. The
straight line distance between A’ and B’ as perceived on a flat piece
of paper simply does not exist in the present context. We may now do
some sort of re-labeling of coordinates to make the new system
orthogonal.
The significance of these Transformations:
Tensors are usually defined by relations like T'^[mu,nu]=[del
x'^[mu]]/]del x^[alpha]] * [del x'^[nu]]/]del x^[beta]] T^[alpha,beta]
Such definitions assume relations of bthe type:
x'^[mu]=f(x^[alpha],x^[beta])
ds^2 does not change in these relations. But angles between curves
might change.
The salient feature or the point of advantage is that we may pass from
one manifold to another, giving a mathematical foundation to the
universality of the physical laws--that they have the same tensor
form[covariant form] in all distinct manifolds.
Any failure to pass between different manifolds[by suitable
transformations] would restrict the tensor object to a particular type
of manifold.Such a situation would be a hindrance towards the claim
of the universality of the physical laws [in covariant form].
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