| Topic: |
Science > Physics |
| User: |
"Lee Herman" |
| Date: |
26 Jul 2003 05:49:51 PM |
| Object: |
orthogonal transformations |
Hi, I am a student who is doing independent studying on some classical
mechanic. I am having a question on the degree of freedom. In an
n-dimentional space, the rotation matrix will have n^2 elements.
However, why the orthogonality relations place (1/2)(n^2+n) conditions
upon the rotation matrix. If that is true, then the degree of freedom
in n-dimentional space will be, n^2-(1/2)(n^2+n)=(1/2)n(n-1). Thank
you for your help.
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| User: "Ian H" |
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| Title: Re: orthogonal transformations |
26 Jul 2003 08:40:27 PM |
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"Lee Herman" <ipensive@yahoo.com> wrote in message
news:dda267e8.0307261449.335ec429@posting.google.com...
: Hi, I am a student who is doing independent studying on some classical
: mechanic. I am having a question on the degree of freedom. In an
: n-dimentional space, the rotation matrix will have n^2 elements.
: However, why the orthogonality relations place (1/2)(n^2+n) conditions
: upon the rotation matrix. If that is true, then the degree of freedom
: in n-dimentional space will be, n^2-(1/2)(n^2+n)=(1/2)n(n-1). Thank
: you for your help.
Think of a rotation matrices for very small rotations (a neighbourhood
of
the identity matrix). Express such a matrix in the form I+D where D is a
matrix with very small coefficients.
Then orthogonality gives (I+D)^t (I+D) = I implying
D^t + D + DtD = 0 or D^t = -D (ignoring second order terms)
This argument shows that the rotation matrices very close to the
identity can be described in terms of antisymmetric matrices. (The
Lie Algebra of the rotation group is the algebra of antisymmetric
matrices). More exactly they are exponentials of antisymmetric matrices.
The antisymmetric matrices are determined by coefficients
above the diagonal, and there are (1/2)n(n-1) of these as you noted.
Alternatively it is possible to argue directly. The first column of a
rotation
matrix can be any vector in the space of length 1. There are n-1 choices
involved, (the last coefficient is determined by the length condition).
The
second vector can be any vector in the n-1-D subspace perpendicular to
the
first of length 1. There are n-2 degrees of freedom here. ETC. There are
(n-1) + (n-2) + ... + 1 degrees of freedom altogether, or (1/2) n(n-=1).
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| User: "Starblade Darksquall" |
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| Title: Re: orthogonal transformations |
26 Jul 2003 08:32:56 PM |
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(Lee Herman) wrote in message news:<dda267e8.0307261449.335ec429@posting.google.com>...
Hi, I am a student who is doing independent studying on some classical
mechanic. I am having a question on the degree of freedom. In an
n-dimentional space, the rotation matrix will have n^2 elements.
However, why the orthogonality relations place (1/2)(n^2+n) conditions
upon the rotation matrix. If that is true, then the degree of freedom
in n-dimentional space will be, n^2-(1/2)(n^2+n)=(1/2)n(n-1). Thank
you for your help.
Try to draw a rotation matrix. You find that, for a given angle and
direction, there is no way to do this for scalar space or vector
space, one way to do this for plane space, three ways to do this for
full space, ten ways to do this for fourth dimensional space, and so
forth. Timespace was excluded here because the lorentz transformation
does not rotate in the usual sense when including time as a dimension.
It transforms hyperbolically.
So the result you got was very good. Plug in numbers into it and see
what happens, just to check it out. That's the best way to test it.
(...Starblade Riven Darksquall...)
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| User: "" |
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| Title: Re: orthogonal transformations |
26 Jul 2003 08:36:05 PM |
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(Lee Herman) wrote in message news:<dda267e8.0307261449.335ec429@posting.google.com>...
Hi, I am a student who is doing independent studying on some classical
mechanic. I am having a question on the degree of freedom. In an
n-dimentional space, the rotation matrix will have n^2 elements.
However, why the orthogonality relations place (1/2)(n^2+n) conditions
upon the rotation matrix. If that is true, then the degree of freedom
in n-dimentional space will be, n^2-(1/2)(n^2+n)=(1/2)n(n-1). Thank
you for your help.
Well, this is pretty basic stuff. If your text does not answer
this question, and you don't have a teacher, you need a new
text.
But, as a hint: Why do you suppose the transform is called an
orthoganality transform in the first place? What is it that is
orothoganal, and to what?
As another approach, consider what the constraints are. It's
very easy to see in 2 dimensions. There, you have a generator
of rotations that looks like so.
0 1
-1 0
Basically, it's a function of this matrix that produce
rotations. There's some complication in getting from that to
sin(theta) cos(theta)
- cos(theta) sin(theta)
but if you think about it for a bit, and consult your text,
you should see that there is only one degree of freedom,
the angle. It's a fun old game to try to see how to get
from the first matrix to the second. You should play with
that for a bit.
Or, for another approach, think about what the independent parts
of a member of SO(n) are. SO(n) means matrices with determinate
one, that are orthoganal. If you think about it you will see
that what you have are that only the off-diagonal elements on
one side of the diagonal are free. So, you've fot n^2 to start.
You take off the diagonal, that's -n, and then divide by 2,
to get n(n-1)/2.
Socks
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| User: "John E. Prussing" |
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| Title: Re: orthogonal transformations |
26 Jul 2003 07:40:57 PM |
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In <dda267e8.0307261449.335ec429@posting.google.com> (Lee Herman) writes:
Hi, I am a student who is doing independent studying on some classical
mechanic. I am having a question on the degree of freedom. In an
n-dimentional space, the rotation matrix will have n^2 elements.
However, why the orthogonality relations place (1/2)(n^2+n) conditions
upon the rotation matrix. If that is true, then the degree of freedom
in n-dimentional space will be, n^2-(1/2)(n^2+n)=(1/2)n(n-1). Thank
you for your help.
I'm not exactly sure what your question is, but you seem to be confusing
an orthogonal matrix with a symmetric one. For a symmetric matrix (the
transpose equals the matrix) (1/2)(n^2+n) elements are independent.
For an orthogonal matrix, the inverse equals the matrix, the columns are
mutual orthonormal. In general a rotation matrix is orthogonal but not
symmetric. Look at a single axis rotation matrix in three dimensions: a11
= a22 = cos(x), a33 = 1, a12 = -a21 = sin(x), a13 = a23 = a31 = a32 = 0.
Because a12 = -a21, it is not symmetric.
--
John E. Prussing
University of Illinois at Urbana-Champaign
Department of Aerospace Engineering
http://www.uiuc.edu/~prussing
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| User: "Igor" |
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| Title: Re: orthogonal transformations |
27 Jul 2003 02:16:00 PM |
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On 26 Jul 2003 15:49:51 -0700, (Lee Herman) wrote:
Hi, I am a student who is doing independent studying on some classical
mechanic. I am having a question on the degree of freedom. In an
n-dimentional space, the rotation matrix will have n^2 elements.
However, why the orthogonality relations place (1/2)(n^2+n) conditions
upon the rotation matrix. If that is true, then the degree of freedom
in n-dimentional space will be, n^2-(1/2)(n^2+n)=(1/2)n(n-1). Thank
you for your help.
Just remember that the orthogonal transformation must preserve the
scalar product of two vectors. If I have two n -dim vectors v and w
and transform them as v' = M v and w' = M w, where M is a nxn matrix,
their scalar products wTv and w'Tv' must be the same (where T denotes
transpose). This leads to the necessary condition that MTM = MMT =
I, which automatically defines M to be a orthogonal matrix. Now, any
nxn orthogonal matrix starts out with n^2 independent components, but
the equation MTM = I provides n(n+1)/2 independent equations (since I
is a symmetric matrix) that must be satisfied in the components of M.
Thus, there will be n^2 - n(n+1)/2 = n(n-1)/2 independent
components of M left.
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| User: "Timo Nieminen" |
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| Title: Re: orthogonal transformations |
27 Jul 2003 09:39:42 PM |
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On Sun, 27 Jul 2003, Dirk Van de moortel wrote:
Check up on Euler angles:
http://mathworld.wolfram.com/EulerAngles.html
Ack! Euler angles! An evil best avoided ...
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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| User: "Sam Wormley" |
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| Title: Re: orthogonal transformations |
27 Jul 2003 11:03:18 PM |
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Timo Nieminen wrote:
On Sun, 27 Jul 2003, Dirk Van de moortel wrote:
Check up on Euler angles:
http://mathworld.wolfram.com/EulerAngles.html
Ack! Euler angles! An evil best avoided ...
Nothing particularly difficult about Euler Angles
Pay attention to who's convention you are using.
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| User: "Timo Nieminen" |
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| Title: Re: orthogonal transformations |
27 Jul 2003 11:36:06 PM |
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On Mon, 28 Jul 2003, Sam Wormley wrote:
Timo Nieminen wrote:
On Sun, 27 Jul 2003, Dirk Van de moortel wrote:
Check up on Euler angles:
http://mathworld.wolfram.com/EulerAngles.html
Ack! Euler angles! An evil best avoided ...
Nothing particularly difficult about Euler Angles
Pay attention to who's convention you are using.
'Tis true, but I never found anything particular difficult about using
rotation vectors, or rotation axis + angle, instead, which always made
more sense to me.
--
Timo Nieminen - Home page: http://www.physics.uq.edu.au/people/nieminen/
Shrine to Spirits: http://www.users.bigpond.com/timo_nieminen/spirits.html
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