Proof of Logarithm map formulae from $SO(3)$ to $mathfrak {so}(3)$
According to exponential map, there also exist a logarithm map
$$log:SO(3) to mathfrak {so}(3).$$
Suppose a vector $t inmathfrak {so}(3)$ and $t=|t|w$, according to exponential map
$$R = cos|t|I+(1-cos |t|)ww^T+sin |t|w^{land}.$$
So the inverse(i.e. from rotation matrix $R$ to vector $t$) is
$$|t|=cos^{-1}left(frac{operatorname{Tr}(R)-1}{2}right)tag 1$$
$$w = frac{1}{2sin(|t|)}
begin{bmatrix} r_{32}-r_{23} \ r_{13}-r_{31} \ r_{21}-r_{12}
end{bmatrix} tag 2$$
where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?
linear-algebra lie-groups lie-algebras rotations
edited Dec 15 at 14:28
amWhy
191k28224439
asked Dec 9 at 4:16
Finley
373113
add a comment |
According to exponential map, there also exist a logarithm map
$$log:SO(3) to mathfrak {so}(3).$$
Suppose a vector $t inmathfrak {so}(3)$ and $t=|t|w$, according to exponential map
$$R = cos|t|I+(1-cos |t|)ww^T+sin |t|w^{land}.$$
So the inverse(i.e. from rotation matrix $R$ to vector $t$) is
$$|t|=cos^{-1}left(frac{operatorname{Tr}(R)-1}{2}right)tag 1$$
$$w = frac{1}{2sin(|t|)}
begin{bmatrix} r_{32}-r_{23} \ r_{13}-r_{31} \ r_{21}-r_{12}
end{bmatrix} tag 2$$
where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?
linear-algebra lie-groups lie-algebras rotations
edited Dec 15 at 14:28
amWhy
191k28224439
asked Dec 9 at 4:16
Finley
373113
WP not enough?
– Cosmas Zachos
Dec 10 at 23:14
add a comment |
According to exponential map, there also exist a logarithm map
$$log:SO(3) to mathfrak {so}(3).$$
Suppose a vector $t inmathfrak {so}(3)$ and $t=|t|w$, according to exponential map
$$R = cos|t|I+(1-cos |t|)ww^T+sin |t|w^{land}.$$
So the inverse(i.e. from rotation matrix $R$ to vector $t$) is
$$|t|=cos^{-1}left(frac{operatorname{Tr}(R)-1}{2}right)tag 1$$
$$w = frac{1}{2sin(|t|)}
begin{bmatrix} r_{32}-r_{23} \ r_{13}-r_{31} \ r_{21}-r_{12}
end{bmatrix} tag 2$$
where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?
linear-algebra lie-groups lie-algebras rotations
edited Dec 15 at 14:28
amWhy
191k28224439
asked Dec 9 at 4:16
Finley
373113
According to exponential map, there also exist a logarithm map
$$log:SO(3) to mathfrak {so}(3).$$
Suppose a vector $t inmathfrak {so}(3)$ and $t=|t|w$, according to exponential map
$$R = cos|t|I+(1-cos |t|)ww^T+sin |t|w^{land}.$$
So the inverse(i.e. from rotation matrix $R$ to vector $t$) is
$$|t|=cos^{-1}left(frac{operatorname{Tr}(R)-1}{2}right)tag 1$$
$$w = frac{1}{2sin(|t|)}
begin{bmatrix} r_{32}-r_{23} \ r_{13}-r_{31} \ r_{21}-r_{12}
end{bmatrix} tag 2$$
where $r_{ij}$ is elements of $R$. I can easily derive (1) while stuck in (2), can anyone help me out?
linear-algebra lie-groups lie-algebras rotations
linear-algebra lie-groups lie-algebras rotations
edited Dec 15 at 14:28
amWhy
191k28224439
asked Dec 9 at 4:16
Finley
373113
edited Dec 15 at 14:28
amWhy
191k28224439
asked Dec 9 at 4:16
Finley
373113
edited Dec 15 at 14:28
amWhy
191k28224439
edited Dec 15 at 14:28
amWhy
191k28224439
edited Dec 15 at 14:28
amWhy
191k28224439
191k28224439
asked Dec 9 at 4:16
Finley
373113
asked Dec 9 at 4:16
Finley
373113
asked Dec 9 at 4:16
Finley
373113
373113
WP not enough?
– Cosmas Zachos
Dec 10 at 23:14
add a comment |
WP not enough?
– Cosmas Zachos
Dec 10 at 23:14
WP not enough?
– Cosmas Zachos
Dec 10 at 23:14
WP not enough?
– Cosmas Zachos
Dec 10 at 23:14
add a comment |
1 Answer
1
active
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This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
$$
Kequiv
left[begin{array}{ccc}
0 & -t_3 & t_2 \
t_3 & 0 & -t_1 \
-t_2 & t_1 & 0
end{array}right] ~,
$$
so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .
Now the crucial thing is to re-express the Rodrigues formula in this notation,
$$
R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
$$
Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).
So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.
Now antisymmetrize R,
$$
frac{R-R^T}{2}= sin|t| ~ hat K ,
$$
since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).
answered Dec 11 at 0:50
Cosmas Zachos
1,518520
add a comment |
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1 Answer
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votes
This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
$$
Kequiv
left[begin{array}{ccc}
0 & -t_3 & t_2 \
t_3 & 0 & -t_1 \
-t_2 & t_1 & 0
end{array}right] ~,
$$
so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .
Now the crucial thing is to re-express the Rodrigues formula in this notation,
$$
R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
$$
Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).
So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.
Now antisymmetrize R,
$$
frac{R-R^T}{2}= sin|t| ~ hat K ,
$$
since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).
answered Dec 11 at 0:50
Cosmas Zachos
1,518520
add a comment |
This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
$$
Kequiv
left[begin{array}{ccc}
0 & -t_3 & t_2 \
t_3 & 0 & -t_1 \
-t_2 & t_1 & 0
end{array}right] ~,
$$
so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .
Now the crucial thing is to re-express the Rodrigues formula in this notation,
$$
R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
$$
Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).
So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.
Now antisymmetrize R,
$$
frac{R-R^T}{2}= sin|t| ~ hat K ,
$$
since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).
answered Dec 11 at 0:50
Cosmas Zachos
1,518520
add a comment |
This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
$$
Kequiv
left[begin{array}{ccc}
0 & -t_3 & t_2 \
t_3 & 0 & -t_1 \
-t_2 & t_1 & 0
end{array}right] ~,
$$
so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .
Now the crucial thing is to re-express the Rodrigues formula in this notation,
$$
R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
$$
Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).
So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.
Now antisymmetrize R,
$$
frac{R-R^T}{2}= sin|t| ~ hat K ,
$$
since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).
answered Dec 11 at 0:50
Cosmas Zachos
1,518520
This is a standard formula accompanying the Rodrigues rotation formula, obvious if only you use the right notation! The important thing is is to streamline your orthogonal matrix R. Its logarithm is the antisymmetric matrix K, defined by the linear cross product at the end of your expansion of R,
$$
Kequiv
left[begin{array}{ccc}
0 & -t_3 & t_2 \
t_3 & 0 & -t_1 \
-t_2 & t_1 & 0
end{array}right] ~,
$$
so that $| t|^2= -frac{operatorname {Tr} (K^2)}{2}equiv | K|^2$; hence $K=|t| hat K$ with $| hat K|=1$ .
Now the crucial thing is to re-express the Rodrigues formula in this notation,
$$
R=(R^T)^{-1}= e^K= {mathbb 1} + sin |t| ~ hat K + (1-cos|t|) hat K ^2 .
$$
Note how different the coefficient of the identity looks, contrasted to your vector expression: it is not a mistake (read up on the point in WP.... well-wishers frequently "correct" it! Of course, this expression follows directly from the characteristic equation for K, not Rodrigues' original argument).
So, tracing this expression, and the fact that K is traceless since antisymmetric, your (1) follows.
Now antisymmetrize R,
$$
frac{R-R^T}{2}= sin|t| ~ hat K ,
$$
since the other two terms are symmetric matrices. Convert your "unit" antisymmetric matrix to your unit vector w and you have (2).
answered Dec 11 at 0:50
Cosmas Zachos
1,518520
edited Dec 11 at 19:36
answered Dec 11 at 0:50
Cosmas Zachos
1,518520
answered Dec 11 at 0:50
Cosmas Zachos
1,518520
answered Dec 11 at 0:50
Cosmas Zachos
1,518520
1,518520
add a comment |
add a comment |
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WP not enough?
– Cosmas Zachos
Dec 10 at 23:14