| Topic: |
Science > Physics |
| User: |
"mjb" |
| Date: |
09 Apr 2007 02:27:53 AM |
| Object: |
Inertia Tensor Fundamentals |
People seem quite pedantic about using the term 'inertia tensor', rather
than say 'inertia matrix', why is this?
Is it because its form changes depending on the number of dimensions that we
are working in? For instance:
When rotating in 2 dimensions, then the torque (scalar) is related to the
angular acceleration (scalar) by a scalar (grade 0 tensor) second moment
of mass.
When rotating in 3 dimensions, then the torque bivector is related to the
angular acceleration bivector by a 3x3 matrix (grade 2 tensor)
So what happens in 4 dimensions? (I'm not necessarily talking about
space-time here) I'm just trying to extend the above sequence to 4 space
dimensions. As I understand it, both torque and angular acceleration would
be represented by a 6 dimensional bivector? So how do we relate these two
6D bivectors? Do we use a 6x6 matrix? Or if we follow the above sequence
then it would be some sort of grade 4 hypermatrix? But I can't imagine how
this fits into the equation T = I a?
So my best guess so far is that the inertia tensor is always a matrix? Is
there a scientific way to work this out? I would do an experiment but I
don't know how to apply a torque in 4D?
Is the importance of the term 'inertia tensor' more to do with the way that
it changes with a change of coordinates? For instance if we rotate the
coordinates with the rotation matrix (or is it a tensor) [R]? then:
T' = [R] T
a' = [R] a
so:
T' = [R] I [R]^t a'
so
I' = [R] I [R]^t
I don't know much about tensors, do all tensors vary in this way with a
change of dimensions?
Does it make any difference that 'T' and 'a' are bivectors (as opposed to
vectors)?
Martin
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| User: "Eric Gisse" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 02:43:22 AM |
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On Apr 8, 11:27 pm, mjb <mjb4...@yahoo.com> wrote:
People seem quite pedantic about using the term 'inertia tensor', rather
than say 'inertia matrix', why is this?
A tensor is not a matrix.
A rank 2 tensor has a matrix representation, but that does not mean an
arbitrary tensor is representable as a matrix.
Is it because its form changes depending on the number of dimensions that we
are working in? For instance:
No.
[...]
So my best guess so far is that the inertia tensor is always a matrix? Is
there a scientific way to work this out? I would do an experiment but I
don't know how to apply a torque in 4D?
A rank 2 tensor is always representable as a nxn matrix where n is the
number of dimensions. The three dimensional inertia tensor has a 3x3
matrix representation, for example.
Is the importance of the term 'inertia tensor' more to do with the way that
it changes with a change of coordinates? For instance if we rotate the
coordinates with the rotation matrix (or is it a tensor) [R]? then:
T' = [R] T
a' = [R] a
so:
T' = [R] I [R]^t a'
so
I' = [R] I [R]^t
I don't know much about tensors, do all tensors vary in this way with a
change of dimensions?
Actually, that is exactly how they transform. It is just that it they
aren't representable as a single matrix in higher dimensions.
Does it make any difference that 'T' and 'a' are bivectors (as opposed to
vectors)?
Vector - a (0,1) tensor. Bivector - a (1,0) tensor.
The inertia tensor is a rank 2 symmetric tensor - and is not a
bivector.
Martin
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| User: "mjb" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 03:03:08 AM |
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A rank 2 tensor is always representable as a nxn matrix where n is the
number of dimensions. The three dimensional inertia tensor has a 3x3
matrix representation, for example.
So what form would an inertia tensor take in 4 dimensions? From what you say
its not a 6x6 matrix? is it a grade 4 tensor?
Vector - a (0,1) tensor. Bivector - a (1,0) tensor.
So is a bivector the same as a covector?
Martin
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| User: "Eric Gisse" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 10:26:22 AM |
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On Apr 9, 12:03 am, mjb <mjb4...@yahoo.com> wrote:
A rank 2 tensor is always representable as a nxn matrix where n is the
number of dimensions. The three dimensional inertia tensor has a 3x3
matrix representation, for example.
So what form would an inertia tensor take in 4 dimensions? From what you say
its not a 6x6 matrix? is it a grade 4 tensor?
4x4.
The rank of a tensor has to do with the number of indexes it carries.
However you don't have a four dimensional inertia tensor - there are
only 3 spatial dimensions.
Vector - a (0,1) tensor. Bivector - a (1,0) tensor.
So is a bivector the same as a covector?
Sounds like it.
Martin
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| User: "mjb" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 11:11:36 AM |
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Eric Gisse wrote:
So what form would an inertia tensor take in 4 dimensions? From what you
say its not a 6x6 matrix? is it a grade 4 tensor?
4x4.
So are you saying that in 4 dimensions the inertia tensor is represented by
a 4x4 matrix? (ie two indices) This would imply that an inertia tensor is
always a matrix? I thought (from reading Feynmans lectures) that rotation
in 'n' dimensions happens in a 'directed plane' which is represented by a
bivector. So in 4D there are 6 combinations of 2 from 4 so this bivector is
represented by 6 scalar values. So if it is a matrix it should be 6x6?
The rank of a tensor has to do with the number of indexes it carries.
However you don't have a four dimensional inertia tensor - there are
only 3 spatial dimensions.
Well I must admit I haven't seen 4 spacial dimensions but I thought one of
the big advantages of tensors is that the concepts can be extended to 'n'
dimensions, as a mathematical exercise at least, even if it does not
represent a proven physical theory. Are you saying that tensors can't be
used in this way?
Vector - a (0,1) tensor. Bivector - a (1,0) tensor.
So is a bivector the same as a covector?
This might be an issue of terminology? I have been using the term bivector
to mean the result of cross product of two vectors, which represents a
directed plane, which can represent (infinitesimal or continuous) rotation?
And I've been using covector to represent covariant index or 'linear
operator'.
Is this correct terminology? and is there a way to show if these two are the
same thing?
Thanks,
Martin
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| User: "Eric Gisse" |
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| Title: Re: Inertia Tensor Fundamentals |
11 Apr 2007 04:40:51 PM |
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On Apr 9, 8:11 am, mjb <mjb4...@yahoo.com> wrote:
Eric Gisse wrote:
So what form would an inertia tensor take in 4 dimensions? From what you
say its not a 6x6 matrix? is it a grade 4 tensor?
4x4.
So are you saying that in 4 dimensions the inertia tensor is represented by
a 4x4 matrix? (ie two indices) This would imply that an inertia tensor is
always a matrix? I thought (from reading Feynmans lectures) that rotation
in 'n' dimensions happens in a 'directed plane' which is represented by a
bivector. So in 4D there are 6 combinations of 2 from 4 so this bivector is
represented by 6 scalar values. So if it is a matrix it should be 6x6?
Rotations in general can be described by a tensor.
There isn't a 4 dimensional inertia tensor anyway - there are only 3
spatial dimensions. This does not change in relativity.
Look - each index grants you n entries. A rank 1 tensor - n. Rank 2 -
n^2. Rank 3 - n^3. You can describe a rank 3 tensor with n nxn
matricies, but not with one single matrix.
The rank of a tensor has to do with the number of indexes it carries.
However you don't have a four dimensional inertia tensor - there are
only 3 spatial dimensions.
Well I must admit I haven't seen 4 spacial dimensions but I thought one of
the big advantages of tensors is that the concepts can be extended to 'n'
dimensions, as a mathematical exercise at least, even if it does not
represent a proven physical theory. Are you saying that tensors can't be
used in this way?
It can, it just doesn't have any physical meaning.
Vector - a (0,1) tensor. Bivector - a (1,0) tensor.
So is a bivector the same as a covector?
This might be an issue of terminology? I have been using the term bivector
to mean the result of cross product of two vectors, which represents a
directed plane, which can represent (infinitesimal or continuous) rotation?
Ah. That makes more sense now.
A bivector in 3 dimensions is a 3-vector, but that is not true in
general. In general, it is a tensor.
It does _NOT_ represent rotation. It represents something orthogonal
to the two things you are wedging.
And I've been using covector to represent covariant index or 'linear
operator'.
Is this correct terminology? and is there a way to show if these two are the
same thing?
No, because they are not.
Thanks,
Martin
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| User: "Jim Black" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 07:55:11 PM |
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On Apr 9, 2:43 am, "Eric Gisse" <jowr...@gmail.com> wrote:
On Apr 8, 11:27 pm, mjb <mjb4...@yahoo.com> wrote:
Does it make any difference that 'T' and 'a' are bivectors (as opposed to
vectors)?
Vector - a (0,1) tensor. Bivector - a (1,0) tensor.
You're thinking of covectors, and I think your order is opposite to
the standard one, in which a vector is of type (1,0) and a covector of
type (0,1). A bivector is an antisymmetric tensor of type (2,0).
--
Jim E. Black
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| User: "mjb" |
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| Title: Re: Inertia Tensor Fundamentals |
10 Apr 2007 02:23:07 AM |
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Jim Black wrote:
You're thinking of covectors, and I think your order is opposite to
the standard one, in which a vector is of type (1,0) and a covector of
type (0,1). A bivector is an antisymmetric tensor of type (2,0).
Jim,
I came across bivectors through Geometric Algebra/Clifford Algebra where
they represent directed areas and rotations in 'n' dimensions. In Geometric
Algebra they appear as if they have one index, but then I don't think it
has the concept of multiple indices? How would I define the bivector in
tensor algebra? Is it the outer product (or do I mean the wedge product?)
of two basis vectors in the plane required?
I guess I need to find out about the various types of multiplication in
tensor algebra. Do you know of any sources that don't assume too much maths
background?
Thanks,
Martin
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| User: "Jim Black" |
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| Title: Re: Inertia Tensor Fundamentals |
10 Apr 2007 11:51:44 PM |
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On Apr 10, 2:23 am, mjb <mjb4...@yahoo.com> wrote:
Jim Blackwrote:
You're thinking of covectors, and I think your order is opposite to
the standard one, in which a vector is of type (1,0) and a covector of
type (0,1). A bivector is an antisymmetric tensor of type (2,0).
Jim,
I came across bivectors through Geometric Algebra/Clifford Algebra where
they represent directed areas and rotations in 'n' dimensions. In Geometric
Algebra they appear as if they have one index, but then I don't think it
has the concept of multiple indices? How would I define the bivector in
tensor algebra? Is it the outer product (or do I mean the wedge product?)
of two basis vectors in the plane required?
Unfortunately I'm not familiar with Clifford Algebra. In tensor
notation, you can do a wedge product by taking the tensor product and
antisymmetrizing:
A_uv = (1/2) (B_u C_v - B_v C_u)
I guess I need to find out about the various types of multiplication in
tensor algebra. Do you know of any sources that don't assume too much maths
background?
I don't know how much I can help you with picking a good textbook,
although you may find this online text helpful:
http://www.math.odu.edu/~jhh/counter2.html
Fortunately there's not too much to learn. The tensor product is done
just by multiplying components, and everything else you can get by
permuting of indices and contractions.
--
Jim E. Black
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| User: "Jim Black" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 08:29:36 PM |
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On Apr 9, 2:27 am, mjb <mjb4...@yahoo.com> wrote:
People seem quite pedantic about using the term 'inertia tensor', rather
than say 'inertia matrix', why is this?
Is it because its form changes depending on the number of dimensions that we
are working in? For instance:
When rotating in 2 dimensions, then the torque (scalar) is related to the
angular acceleration (scalar) by a scalar (grade 0 tensor) second moment
of mass.
When rotating in 3 dimensions, then the torque bivector is related to the
angular acceleration bivector by a 3x3 matrix (grade 2 tensor)
So what happens in 4 dimensions? (I'm not necessarily talking about
space-time here) I'm just trying to extend the above sequence to 4 space
dimensions. As I understand it, both torque and angular acceleration would
be represented by a 6 dimensional bivector? So how do we relate these two
6D bivectors? Do we use a 6x6 matrix? Or if we follow the above sequence
then it would be some sort of grade 4 hypermatrix? But I can't imagine how
this fits into the equation T = I a?
You want a linear map from rank-2 tensors to rank-2 tensors. You need
a rank-4 tensor. The equation T = I a would now look like:
T_mn = (1/2) I_mnpr a_pr
Repeated indices indicate summation, so what I've written above
actually means:
T_mn = (1/2) sum_{p=1}^{4} sum_{r=1}^{4} I_mnpr a_pr
I've thrown in the factor of (1/2) to compensate for duplicate terms
such as I_1212 a_12 and I_1221 a_21.
--
Jim E. Black
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| User: "mjb" |
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| Title: Re: Inertia Tensor Fundamentals |
10 Apr 2007 02:50:58 AM |
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Jim Black wrote:
You want a linear map from rank-2 tensors to rank-2 tensors. You need
a rank-4 tensor. The equation T = I a would now look like:
T_mn = (1/2) I_mnpr a_pr
Repeated indices indicate summation, so what I've written above
actually means:
T_mn = (1/2) sum_{p=1}^{4} sum_{r=1}^{4} I_mnpr a_pr
I've thrown in the factor of (1/2) to compensate for duplicate terms
such as I_1212 a_12 and I_1221 a_21.
Does this mean that, in general, Inertia Tensors are always rank-4?
Is it just in 3 dimensions we can take advantage of the duality between
vectors and bivectors and therefore reduce the torque and ang acceleration
to rank 1 and inertia tensor to rank 2?
Thanks,
Martin
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| User: "Jim Black" |
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| Title: Re: Inertia Tensor Fundamentals |
10 Apr 2007 11:55:39 PM |
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On Apr 10, 2:50 am, mjb <mjb4...@yahoo.com> wrote:
Jim Black wrote:
You want a linear map from rank-2 tensors to rank-2 tensors. You need
a rank-4 tensor. The equation T = I a would now look like:
T_mn = (1/2) I_mnpr a_pr
Repeated indices indicate summation, so what I've written above
actually means:
T_mn = (1/2) sum_{p=1}^{4} sum_{r=1}^{4} I_mnpr a_pr
I've thrown in the factor of (1/2) to compensate for duplicate terms
such as I_1212 a_12 and I_1221 a_21.
Does this mean that, in general, Inertia Tensors are always rank-4?
Is it just in 3 dimensions we can take advantage of the duality between
vectors and bivectors and therefore reduce the torque and ang acceleration
to rank 1 and inertia tensor to rank 2?
Yes.
--
Jim E. Black
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| User: "John C. Polasek" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 08:24:54 PM |
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On Mon, 09 Apr 2007 08:27:53 +0100, mjb <mjb4567@yahoo.com> wrote:
People seem quite pedantic about using the term 'inertia tensor', rather
than say 'inertia matrix', why is this?
Is it because its form changes depending on the number of dimensions that we
are working in? For instance:
When rotating in 2 dimensions, then the torque (scalar) is related to the
angular acceleration (scalar) by a scalar (grade 0 tensor) second moment
of mass.
When rotating in 3 dimensions, then the torque bivector is related to the
angular acceleration bivector by a 3x3 matrix (grade 2 tensor)
So what happens in 4 dimensions? (I'm not necessarily talking about
space-time here) I'm just trying to extend the above sequence to 4 space
dimensions. As I understand it, both torque and angular acceleration would
be represented by a 6 dimensional bivector? So how do we relate these two
6D bivectors? Do we use a 6x6 matrix? Or if we follow the above sequence
then it would be some sort of grade 4 hypermatrix? But I can't imagine how
this fits into the equation T = I a?
So my best guess so far is that the inertia tensor is always a matrix? Is
there a scientific way to work this out? I would do an experiment but I
don't know how to apply a torque in 4D?
Is the importance of the term 'inertia tensor' more to do with the way that
it changes with a change of coordinates? For instance if we rotate the
coordinates with the rotation matrix (or is it a tensor) [R]? then:
T' = [R] T
a' = [R] a
No, T' = [R}^t T [R}
You must premultiply and postmultiply the inertia tensor T by the
rotation matrix and its inverse.
In 4 dimensons you'll find it hard to define a rotation matrix.
so:
T' = [R] I [R]^t a'
so
I' = [R] I [R]^t
I don't know much about tensors, do all tensors vary in this way with a
change of dimensions?
Does it make any difference that 'T' and 'a' are bivectors (as opposed to
vectors)?
Martin
John Polasek
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| User: "mjb" |
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| Title: Re: Inertia Tensor Fundamentals |
10 Apr 2007 02:37:46 AM |
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John C. Polasek wrote:
In 4 dimensons you'll find it hard to define a rotation matrix.
I read somewhere, I can't remember where and my memory may be faulty, that
one of the advantages of tensors is that equations can be defined in a way
that applies to any number of dimensions? Is this not true?
Is there a way to define the rotation matrix (or should that be rotation
tensor?) in terms of bivectors?
Thanks,
Martin
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| User: "John C. Polasek" |
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| Title: Re: Inertia Tensor Fundamentals |
10 Apr 2007 09:20:53 AM |
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On Mon, 09 Apr 2007 08:27:53 +0100, mjb <mjb4567@yahoo.com> wrote:
People seem quite pedantic about using the term 'inertia tensor', rather
than say 'inertia matrix', why is this?
Is it because its form changes depending on the number of dimensions that we
are working in? For instance:
When rotating in 2 dimensions, then the torque (scalar) is related to the
angular acceleration (scalar) by a scalar (grade 0 tensor) second moment
of mass.
When rotating in 3 dimensions, then the torque bivector is related to the
angular acceleration bivector by a 3x3 matrix (grade 2 tensor)
So what happens in 4 dimensions? (I'm not necessarily talking about
space-time here) I'm just trying to extend the above sequence to 4 space
dimensions. As I understand it, both torque and angular acceleration would
be represented by a 6 dimensional bivector? So how do we relate these two
6D bivectors? Do we use a 6x6 matrix? Or if we follow the above sequence
then it would be some sort of grade 4 hypermatrix? But I can't imagine how
this fits into the equation T = I a?
So my best guess so far is that the inertia tensor is always a matrix? Is
there a scientific way to work this out? I would do an experiment but I
don't know how to apply a torque in 4D?
Is the importance of the term 'inertia tensor' more to do with the way that
it changes with a change of coordinates? For instance if we rotate the
coordinates with the rotation matrix (or is it a tensor) [R]? then:
T' = [R] T
a' = [R] a
No, T' = [R}^t T [R}
You must premultiply and postmultiply the inertia tensor T by the
rotation matrix and its inverse.
In 4 dimensons you'll find it hard to define a rotation matrix.
so:
T' = [R] I [R]^t a'
so
I' = [R] I [R]^t
I don't know much about tensors, do all tensors vary in this way with a
change of dimensions?
Does it make any difference that 'T' and 'a' are bivectors (as opposed to
vectors)?
Martin
John Polasek
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| User: "John C. Polasek" |
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| Title: Re: Inertia Tensor Fundamentals |
11 Apr 2007 02:43:27 PM |
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On Mon, 09 Apr 2007 08:27:53 +0100, mjb <mjb4567@yahoo.com> wrote:
People seem quite pedantic about using the term 'inertia tensor', rather
than say 'inertia matrix', why is this?
Because it is most easily manipulated as a matrix.
Is it because its form changes depending on the number of dimensions that we
are working in? For instance:
When rotating in 2 dimensions, then the torque (scalar) is related to the
angular acceleration (scalar) by a scalar (grade 0 tensor) second moment
of mass.
When rotating in 3 dimensions, then the torque bivector is related to the
angular acceleration bivector by a 3x3 matrix (grade 2 tensor)
So what happens in 4 dimensions? (I'm not necessarily talking about
space-time here) I'm just trying to extend the above sequence to 4 space
dimensions. As I understand it, both torque and angular acceleration would
be represented by a 6 dimensional bivector? So how do we relate these two
6D bivectors? Do we use a 6x6 matrix? Or if we follow the above sequence
then it would be some sort of grade 4 hypermatrix? But I can't imagine how
this fits into the equation T = I a?
So my best guess so far is that the inertia tensor is always a matrix? Is
there a scientific way to work this out? I would do an experiment but I
don't know how to apply a torque in 4D?
Is the importance of the term 'inertia tensor' more to do with the way that
it changes with a change of coordinates? For instance if we rotate the
coordinates with the rotation matrix (or is it a tensor) [R]? then:
T' = [R] T
No, T' = [R}^t T [R}
You must premultiply and postmultiply the inertia tensor T by the
rotation matrix and its inverse when they are written as matrices.
In 4 dimensons you'll find it hard to define a rotation matrix.
a' = [R] a
No, Torque = RTR'a with column vectors or
Torque = aR'TR with row vectors
where a = domega/dt
Can you see the logic in the last equation? Even if we dont know how
to transforma a 2d rank tensor, we leave it alone, and take a into R'
to rotate a backwards, run it through the tensor, then rotate it
forward with R. It's a tensor sandwich.
T' = [R] I [R]^t a'
The last is nonsense, says T' = a' since the two R's equal identity
I' = [R] I [R]^t
I don't know much about tensors, do all tensors vary in this way with a
change of dimensions?
T is a tensor because it transforms like its vectors do, and the
operations can be carried out as matrices or as subscripted tensor
components.
Just because a matrix is declared a tensor doesn't make it so. The
metric tensor of relativity is really a mirror matrix because it
doesn't transform like its vectors and because its determinant is -1.
Forget this mapping fol de rol. In physics a 2d rank tensor represents
the characteristics of a he-man object.
Does it make any difference that 'T' and 'a' are bivectors (as opposed to
vectors)?
You could do with a little study, but pick your text carefully. If
it's by a full-fledged mathematician you'll be lost shortly. .
Martin
John Polasek
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| User: "Androcles" |
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| Title: Re: Inertia Tensor Fundamentals |
13 Apr 2007 12:49:54 AM |
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"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message =
news:tfdq13l3i4dkshksohorracfhs2hg0mtbg@4ax.com...
On Mon, 09 Apr 2007 08:27:53 +0100, mjb <mjb4567@yahoo.com> wrote:
=20
People seem quite pedantic about using the term 'inertia tensor', =
rather
than say 'inertia matrix', why is this?
=20
Because it is most easily manipulated as a matrix.
People seem quite pedantic about using the term 'sexual intercourse', =
rather
than say '*****', why is this?
Because it is most easily manipulated as a snob.
.
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| User: "Dirk Van de moortel" |
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| Title: Re: Inertia Tensor Fundamentals |
13 Apr 2007 03:42:06 PM |
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"Androcles" <Engineer@hogwarts.physics.co.uk> wrote in message news:6SETh.13626$aB1.6638@fe3.news.blueyonder.co.uk...
"John C. Polasek" <jpolasek@cfl.rr.com> wrote in message
news:tfdq13l3i4dkshksohorracfhs2hg0mtbg@4ax.com...
On Mon, 09 Apr 2007 08:27:53 +0100, mjb <mjb4567@yahoo.com> wrote:
People seem quite pedantic about using the term 'inertia tensor', rather
than say 'inertia matrix', why is this?
Because it is most easily manipulated as a matrix.
People seem quite pedantic about using the term 'sexual intercourse', rather
than say '*****', why is this?
because most people aren't uneducated sick frustrated old
bed-tied bastards like you?
Dirk Vdm
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| User: "Uncle Al" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 10:06:42 AM |
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mjb wrote:
People seem quite pedantic about using the term 'inertia tensor', rather
than say 'inertia matrix', why is this?
Is it because its form changes depending on the number of dimensions that we
are working in? For instance:
When rotating in 2 dimensions, then the torque (scalar) is related to the
angular acceleration (scalar) by a scalar (grade 0 tensor) second moment
of mass.
When rotating in 3 dimensions, then the torque bivector is related to the
angular acceleration bivector by a 3x3 matrix (grade 2 tensor)
So what happens in 4 dimensions? (I'm not necessarily talking about
space-time here) I'm just trying to extend the above sequence to 4 space
dimensions. As I understand it, both torque and angular acceleration would
be represented by a 6 dimensional bivector? So how do we relate these two
6D bivectors? Do we use a 6x6 matrix? Or if we follow the above sequence
then it would be some sort of grade 4 hypermatrix? But I can't imagine how
this fits into the equation T = I a?
There are no rotation axes in 4-D.
So my best guess so far is that the inertia tensor is always a matrix? Is
there a scientific way to work this out? I would do an experiment but I
don't know how to apply a torque in 4D?
Is the importance of the term 'inertia tensor' more to do with the way that
it changes with a change of coordinates? For instance if we rotate the
coordinates with the rotation matrix (or is it a tensor) [R]? then:
T' = [R] T
a' = [R] a
so:
T' = [R] I [R]^t a'
so
I' = [R] I [R]^t
I don't know much about tensors, do all tensors vary in this way with a
change of dimensions?
Does it make any difference that 'T' and 'a' are bivectors (as opposed to
vectors)?
Martin
--
Uncle Al
http://www.mazepath.com/uncleal/
(Toxic URL! Unsafe for children and most mammals)
http://www.mazepath.com/uncleal/lajos.htm#a2
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| User: "Androcles" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 11:49:28 AM |
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"Uncle Al" <UncleAl0@hate.spam.net> wrote in message =
news:461A5682.86353CDD@hate.spam.net...
[snip river of *****]
10) Imbecile
9) Cretin
8) Moron
7) *****
6) Idiot
5) Schmuck
4) Turkey
3) FUCKHEAD
2) Beta cannot be derived
1) GPS works.
0) http://www.androcles01.pwp.blueyonder.co.uk/GPS/GPS.htm
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| User: "mjb" |
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| Title: Re: Inertia Tensor Fundamentals |
09 Apr 2007 10:44:15 AM |
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Uncle Al wrote:
There are no rotation axes in 4-D.
I don't think I used the word 'axes' did I?
I thought (from reading Feynmans lectures) that rotation in 'n' dimensions
happens in a 'directed plane' which is represented by a bivector.
So in 4D there are 6 combinations of 2 from 4 so this bivector is
represented by 6 scalar values.
As I understand it 3D is a special case where a vector is the dual of the
bivector.
What are you trying to say here? are you telling me I'm wrong about this?
Martin
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