Re: Axis of rotation

On Sat, 3 Nov 2006, Timothy Golden BandTechnology.com wrote:


Timo A. Nieminen wrote:
On Fri, 3 Nov 2006, Timothy Golden BandTechnology.com wrote:
Aren't these really two different concepts? They are both projections
via a n->n matrix product. However rotation inherently implies not
stretching things about(conserving distances)
and the Lorentz implies no rotation but plenty of distance scaling.

Recall that the metric tensor tells you the distance between 2 points.
Given X1, X2 as column vectors of the Cartesian coordinates [aside: note
that we can do this because Cartesian coordinates are special - the
coordinates stuck into an Nx1 matrix will in general not be a "vector"],
the distance between them is sqrt((X1-X2)^T G (X1-X2)) where G is the
NxN metric tensor and ^T indicates transpose. More general we can write
the scalar product of 2 vectors a and b as a.b = a^T G b.

While it might not be the only definition, the most useful definition of
"rotation" here is "homogeneous linear transformations" under which the
scalar product a.b is invariant.
[cut]

I have to disagree but perhaps the disagreement is a matter of context.
The context which I am approaching the comparison in is that of a rigid
object passing through a rotational transform versus a Lorentz
transform.
So We have an object O, A transform R(O), and a tranform L(O).
L(O) will generally contain points on the object whose distance have
changed.
This is also called length contraction.
R(O) will conserve all distances for all points on the object.

As I study I am seeing a rotational interpretation of The Lorentz
transform on wikipedia:
http://en.wikipedia.org/wiki/Length_contraction
I fail to see this as rotation. Consider instead the issue of
orthogonality. In effect the Lorentz transform can destroy
orthogonality. This is a scaling operation; it adjusts the dimension of
the object in a specific way. In their cuboid sketch they are seeing
angles change but these angles are not obeying a rotational algorithm.
They change in parallel. The axes are folding, not rotating.

Consider geometric objects under change of coordinate axis. The geometric objects themselves cannot change - only the description of them in the new coordinate system when compared with the original coordinate system. That is, we're really talking about "passive transformations", where the object stays the same and the coordinate system changes.

Now, if we want to describe the object or some properties of the object using geometric language, some of the terms that are the most useful are those that don't depend on the choice of coordinate system. Compare this, for example, with writing everything in terms of the coordinates in some given coordinate system - it's useful to write del^2 a + k^2 a = 0 rather than writing this in terms of derivatives of the coordinates (and vector components of a if a is a vector and not a scalar).

Now, a lot of these coordinate-free forms of writing equations etc use operators like the Laplacian, div, curl, grad, the d'Alembertian. Note that these operators look quite different when written in non-Cartesian coordinate systems (eg cylindrical and spherical, as can be seen eg on the inside back cover of Jackson). Why are they different? Ans: the metric tensor is not an identity matrix. See eg S. Visnovsky, Matrix representations for vector differential operators in general orthogonal coordinates, Czechoslovak Journal of Physics, 54, 793-888, 2004.

If you're doing rotations in 3D space in, for example, parabolic coordinates, the rotation transformation isn't going to look like a Cartesian rotation. The effect, however, will be the same.

So, what about Lorentz transformations? Lorentz transformations are rotations under the Minkowski metric. What happens in the 3D Euclidean space part will not in general be a rotation. You can see the same thing happen in 3D Euclidean space. Consider a general 3D rotation, but only look a 2D projection onto the xy-plane. If you choose a 3D rotation about the z-axis, it will still look like a rotation in the 2D projection, otherwise not. Likewise, a Lorentz transformation when the boost parameters are zero (ie the relative velocity between the original and new coordinate systems is zero) looks like a 3D Euclidean rotation, but otherwise will not. The Lorentz transformation is a 4D rotation under the Minkowski metric - don't expect it to look like a 3D Euclidean rotation except in special cases. Do expect, nay demand, that the 4D scalar product remain invariant under Lorentz transformations. Remember that the geometric object left unchanged by Lorentz transformations are 4-scalars, 4-vectors, 4-tensors etc.

--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
E-prints: http://eprint.uq.edu.au/view/person/Nieminen,_Timo_A..html
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html

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