Mixing
Angular velocity as eigenvector of rotation map
$begingroup$
In physics the angular velocity $omega$ usually is defined as follows:
For a map $B: mathbb{R} to SO(3)$ define $Omega(t):=dot{B}(t)B^t(t)$. This is a scew-symmetric matrix and hence there is a vector $omega(t) in mathbb{R}^3$ satisfying $Omega(t) x=omega(t) times x$ for $xinmathbb{R}^3$.
Does $omega$ always point in the direction of the "momentary axis of rotation of $B$"? That is, is $omega(t)$ an eigenvector of $B(t)$? I've tried to prove this by writing $omega_j(t) = frac{1}{2}sum_{k,l=1}^3epsilon_{jkl}Omega_{kl}(t)$ (were $epsilon_{jkl}$ is the Levi-Cita symbol). Then $$(B(t)omega(t))_j=-frac{1}{2}sum_{k,l,m=1}^3epsilon_{ikl}B_{ji}(t)epsilon_{ikl}Omega_{kl}(t)=-frac{1}{2}sum_{k,l,m,n=1}^3epsilon_{ikl}B_{ji}(t)dot{B}_{kn}(t)B_{ln}(t).$$ I don't know how to proceed further. Is this even true?
linear-algebra
asked Jan 1 at 22:02
Jannik PittJannik Pitt
374316
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add a comment |
$begingroup$
In physics the angular velocity $omega$ usually is defined as follows:
For a map $B: mathbb{R} to SO(3)$ define $Omega(t):=dot{B}(t)B^t(t)$. This is a scew-symmetric matrix and hence there is a vector $omega(t) in mathbb{R}^3$ satisfying $Omega(t) x=omega(t) times x$ for $xinmathbb{R}^3$.
Does $omega$ always point in the direction of the "momentary axis of rotation of $B$"? That is, is $omega(t)$ an eigenvector of $B(t)$? I've tried to prove this by writing $omega_j(t) = frac{1}{2}sum_{k,l=1}^3epsilon_{jkl}Omega_{kl}(t)$ (were $epsilon_{jkl}$ is the Levi-Cita symbol). Then $$(B(t)omega(t))_j=-frac{1}{2}sum_{k,l,m=1}^3epsilon_{ikl}B_{ji}(t)epsilon_{ikl}Omega_{kl}(t)=-frac{1}{2}sum_{k,l,m,n=1}^3epsilon_{ikl}B_{ji}(t)dot{B}_{kn}(t)B_{ln}(t).$$ I don't know how to proceed further. Is this even true?
linear-algebra
asked Jan 1 at 22:02
Jannik PittJannik Pitt
374316
$endgroup$
1
$begingroup$
lmao I always just said $leftlvertvecomegarightrvert = dfrac{dtheta}{dt}$
$endgroup$
– Chase Ryan Taylor
Jan 1 at 22:13
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@ChaseRyanTaylor What would $theta(t)$ be for the map $B$ as I defined?
$endgroup$
– Jannik Pitt
Jan 1 at 22:19
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I agree with Chase. Angular velocity in physics is usually defined through infinitesimal rotations. The OP's definition might be from some advanced mathematical perspective
$endgroup$
– Yuriy S
Jan 1 at 23:16
add a comment |
$begingroup$
In physics the angular velocity $omega$ usually is defined as follows:
For a map $B: mathbb{R} to SO(3)$ define $Omega(t):=dot{B}(t)B^t(t)$. This is a scew-symmetric matrix and hence there is a vector $omega(t) in mathbb{R}^3$ satisfying $Omega(t) x=omega(t) times x$ for $xinmathbb{R}^3$.
Does $omega$ always point in the direction of the "momentary axis of rotation of $B$"? That is, is $omega(t)$ an eigenvector of $B(t)$? I've tried to prove this by writing $omega_j(t) = frac{1}{2}sum_{k,l=1}^3epsilon_{jkl}Omega_{kl}(t)$ (were $epsilon_{jkl}$ is the Levi-Cita symbol). Then $$(B(t)omega(t))_j=-frac{1}{2}sum_{k,l,m=1}^3epsilon_{ikl}B_{ji}(t)epsilon_{ikl}Omega_{kl}(t)=-frac{1}{2}sum_{k,l,m,n=1}^3epsilon_{ikl}B_{ji}(t)dot{B}_{kn}(t)B_{ln}(t).$$ I don't know how to proceed further. Is this even true?
linear-algebra
asked Jan 1 at 22:02
Jannik PittJannik Pitt
374316
$endgroup$
In physics the angular velocity $omega$ usually is defined as follows:
For a map $B: mathbb{R} to SO(3)$ define $Omega(t):=dot{B}(t)B^t(t)$. This is a scew-symmetric matrix and hence there is a vector $omega(t) in mathbb{R}^3$ satisfying $Omega(t) x=omega(t) times x$ for $xinmathbb{R}^3$.
Does $omega$ always point in the direction of the "momentary axis of rotation of $B$"? That is, is $omega(t)$ an eigenvector of $B(t)$? I've tried to prove this by writing $omega_j(t) = frac{1}{2}sum_{k,l=1}^3epsilon_{jkl}Omega_{kl}(t)$ (were $epsilon_{jkl}$ is the Levi-Cita symbol). Then $$(B(t)omega(t))_j=-frac{1}{2}sum_{k,l,m=1}^3epsilon_{ikl}B_{ji}(t)epsilon_{ikl}Omega_{kl}(t)=-frac{1}{2}sum_{k,l,m,n=1}^3epsilon_{ikl}B_{ji}(t)dot{B}_{kn}(t)B_{ln}(t).$$ I don't know how to proceed further. Is this even true?
linear-algebra
linear-algebra
asked Jan 1 at 22:02
Jannik PittJannik Pitt
374316
asked Jan 1 at 22:02
Jannik PittJannik Pitt
374316
asked Jan 1 at 22:02
Jannik PittJannik Pitt
374316
asked Jan 1 at 22:02
Jannik PittJannik Pitt
374316
asked Jan 1 at 22:02
Jannik PittJannik Pitt
374316
374316
1
$begingroup$
lmao I always just said $leftlvertvecomegarightrvert = dfrac{dtheta}{dt}$
$endgroup$
– Chase Ryan Taylor
Jan 1 at 22:13
$begingroup$
@ChaseRyanTaylor What would $theta(t)$ be for the map $B$ as I defined?
$endgroup$
– Jannik Pitt
Jan 1 at 22:19
$begingroup$
I agree with Chase. Angular velocity in physics is usually defined through infinitesimal rotations. The OP's definition might be from some advanced mathematical perspective
$endgroup$
– Yuriy S
Jan 1 at 23:16
add a comment |
1
$begingroup$
lmao I always just said $leftlvertvecomegarightrvert = dfrac{dtheta}{dt}$
$endgroup$
– Chase Ryan Taylor
Jan 1 at 22:13
$begingroup$
@ChaseRyanTaylor What would $theta(t)$ be for the map $B$ as I defined?
$endgroup$
– Jannik Pitt
Jan 1 at 22:19
$begingroup$
I agree with Chase. Angular velocity in physics is usually defined through infinitesimal rotations. The OP's definition might be from some advanced mathematical perspective
$endgroup$
– Yuriy S
Jan 1 at 23:16
1
1
$begingroup$
lmao I always just said $leftlvertvecomegarightrvert = dfrac{dtheta}{dt}$
$endgroup$
– Chase Ryan Taylor
Jan 1 at 22:13
$begingroup$
lmao I always just said $leftlvertvecomegarightrvert = dfrac{dtheta}{dt}$
$endgroup$
– Chase Ryan Taylor
Jan 1 at 22:13
$begingroup$
@ChaseRyanTaylor What would $theta(t)$ be for the map $B$ as I defined?
$endgroup$
– Jannik Pitt
Jan 1 at 22:19
$begingroup$
@ChaseRyanTaylor What would $theta(t)$ be for the map $B$ as I defined?
$endgroup$
– Jannik Pitt
Jan 1 at 22:19
$begingroup$
I agree with Chase. Angular velocity in physics is usually defined through infinitesimal rotations. The OP's definition might be from some advanced mathematical perspective
$endgroup$
– Yuriy S
Jan 1 at 23:16
$begingroup$
I agree with Chase. Angular velocity in physics is usually defined through infinitesimal rotations. The OP's definition might be from some advanced mathematical perspective
$endgroup$
– Yuriy S
Jan 1 at 23:16
add a comment |
1 Answer
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No this is not true in general. Take for example
$$ B = begin{pmatrix} 0 & -sinphi & -cosphi\ 0 & cosphi & -sinphi\ 1 & 0 &0 end{pmatrix}$$
Then
$$dot B = dotphibegin{pmatrix} 0 & -cosphi & sinphi \ 0 & -sinphi & -cosphi \ 1 & 0 & 0end{pmatrix}
quadtext{and}quad Omega = dotphibegin{pmatrix} 0 & -1& 0 \
1 & 0 & 0 \
0 & 0 & 0 end{pmatrix}, omega = begin{pmatrix}0\0\-dotphiend{pmatrix} $$
However, if $dotphi(0) neq 0$ then $omega(0)$ is not an eigenvector of $B(0)$.
There also is no reason why this should be true. The matrix $B$ represents an orientation, not a rotation. It's Eigenvectors are therefore meaningless. The "momentary rotation" of the object if given by $exp(Omega)$.
And indeed, since $Omega omega = omega times omega = 0$ we have $exp(Omega)omega = omega$.
answered Jan 1 at 23:05
0x5390x539
1,067317
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add a comment |
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$begingroup$
No this is not true in general. Take for example
$$ B = begin{pmatrix} 0 & -sinphi & -cosphi\ 0 & cosphi & -sinphi\ 1 & 0 &0 end{pmatrix}$$
Then
$$dot B = dotphibegin{pmatrix} 0 & -cosphi & sinphi \ 0 & -sinphi & -cosphi \ 1 & 0 & 0end{pmatrix}
quadtext{and}quad Omega = dotphibegin{pmatrix} 0 & -1& 0 \
1 & 0 & 0 \
0 & 0 & 0 end{pmatrix}, omega = begin{pmatrix}0\0\-dotphiend{pmatrix} $$
However, if $dotphi(0) neq 0$ then $omega(0)$ is not an eigenvector of $B(0)$.
There also is no reason why this should be true. The matrix $B$ represents an orientation, not a rotation. It's Eigenvectors are therefore meaningless. The "momentary rotation" of the object if given by $exp(Omega)$.
And indeed, since $Omega omega = omega times omega = 0$ we have $exp(Omega)omega = omega$.
answered Jan 1 at 23:05
0x5390x539
1,067317
$endgroup$
add a comment |
$begingroup$
No this is not true in general. Take for example
$$ B = begin{pmatrix} 0 & -sinphi & -cosphi\ 0 & cosphi & -sinphi\ 1 & 0 &0 end{pmatrix}$$
Then
$$dot B = dotphibegin{pmatrix} 0 & -cosphi & sinphi \ 0 & -sinphi & -cosphi \ 1 & 0 & 0end{pmatrix}
quadtext{and}quad Omega = dotphibegin{pmatrix} 0 & -1& 0 \
1 & 0 & 0 \
0 & 0 & 0 end{pmatrix}, omega = begin{pmatrix}0\0\-dotphiend{pmatrix} $$
However, if $dotphi(0) neq 0$ then $omega(0)$ is not an eigenvector of $B(0)$.
There also is no reason why this should be true. The matrix $B$ represents an orientation, not a rotation. It's Eigenvectors are therefore meaningless. The "momentary rotation" of the object if given by $exp(Omega)$.
And indeed, since $Omega omega = omega times omega = 0$ we have $exp(Omega)omega = omega$.
answered Jan 1 at 23:05
0x5390x539
1,067317
$endgroup$
add a comment |
$begingroup$
No this is not true in general. Take for example
$$ B = begin{pmatrix} 0 & -sinphi & -cosphi\ 0 & cosphi & -sinphi\ 1 & 0 &0 end{pmatrix}$$
Then
$$dot B = dotphibegin{pmatrix} 0 & -cosphi & sinphi \ 0 & -sinphi & -cosphi \ 1 & 0 & 0end{pmatrix}
quadtext{and}quad Omega = dotphibegin{pmatrix} 0 & -1& 0 \
1 & 0 & 0 \
0 & 0 & 0 end{pmatrix}, omega = begin{pmatrix}0\0\-dotphiend{pmatrix} $$
However, if $dotphi(0) neq 0$ then $omega(0)$ is not an eigenvector of $B(0)$.
There also is no reason why this should be true. The matrix $B$ represents an orientation, not a rotation. It's Eigenvectors are therefore meaningless. The "momentary rotation" of the object if given by $exp(Omega)$.
And indeed, since $Omega omega = omega times omega = 0$ we have $exp(Omega)omega = omega$.
answered Jan 1 at 23:05
0x5390x539
1,067317
$endgroup$
No this is not true in general. Take for example
$$ B = begin{pmatrix} 0 & -sinphi & -cosphi\ 0 & cosphi & -sinphi\ 1 & 0 &0 end{pmatrix}$$
Then
$$dot B = dotphibegin{pmatrix} 0 & -cosphi & sinphi \ 0 & -sinphi & -cosphi \ 1 & 0 & 0end{pmatrix}
quadtext{and}quad Omega = dotphibegin{pmatrix} 0 & -1& 0 \
1 & 0 & 0 \
0 & 0 & 0 end{pmatrix}, omega = begin{pmatrix}0\0\-dotphiend{pmatrix} $$
However, if $dotphi(0) neq 0$ then $omega(0)$ is not an eigenvector of $B(0)$.
There also is no reason why this should be true. The matrix $B$ represents an orientation, not a rotation. It's Eigenvectors are therefore meaningless. The "momentary rotation" of the object if given by $exp(Omega)$.
And indeed, since $Omega omega = omega times omega = 0$ we have $exp(Omega)omega = omega$.
answered Jan 1 at 23:05
0x5390x539
1,067317
edited Jan 1 at 23:24
answered Jan 1 at 23:05
0x5390x539
1,067317
answered Jan 1 at 23:05
0x5390x539
1,067317
answered Jan 1 at 23:05
0x5390x539
1,067317
1,067317
add a comment |
add a comment |
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$begingroup$
lmao I always just said $leftlvertvecomegarightrvert = dfrac{dtheta}{dt}$
$endgroup$
– Chase Ryan Taylor
Jan 1 at 22:13
$begingroup$
@ChaseRyanTaylor What would $theta(t)$ be for the map $B$ as I defined?
$endgroup$
– Jannik Pitt
Jan 1 at 22:19
$begingroup$
I agree with Chase. Angular velocity in physics is usually defined through infinitesimal rotations. The OP's definition might be from some advanced mathematical perspective
$endgroup$
– Yuriy S
Jan 1 at 23:16