Derivative of quaternions
$begingroup$
I am trying to calculate the Jacobian of a function that has quaternions and 3D points in it. I refer to quaternions as $q$ and 3D points as $p$
$$h_1(q)=A C(q)p $$
$$h_2(q)=q_1otimes q otimes q_2 $$
where $Ain R^{3x3}$ and $C(q)$ is Direction cosine matrix.
I am using the Hamilton form for the quaternions.
I would like to calculate the following Jacobians:
$$H_1 = frac{partial h_1(q)}{partial q} $$
$$H_2 = frac{partial h_2(q)}{partial q} $$
Following Juao Sola's reference eq. 18 what I have is
$$H_1 = A^TC(q)^T[p]_x $$
$$H_2 = [q_1]_L[q_2]_R $$
Where $[q]_R$ and $[q]_L$ are the right and left handed conversion of quaternion to matrix form as defined in Juao Sola's reference eq. 18.
All rotations are body centric.
Is this correct?
Is there a better way to do this?
Can the expression be easily simplified?
derivatives quaternions
asked Jul 25 '16 at 7:42
Yonatan SimsonYonatan Simson
168
$endgroup$
add a comment |
$begingroup$
I am trying to calculate the Jacobian of a function that has quaternions and 3D points in it. I refer to quaternions as $q$ and 3D points as $p$
$$h_1(q)=A C(q)p $$
$$h_2(q)=q_1otimes q otimes q_2 $$
where $Ain R^{3x3}$ and $C(q)$ is Direction cosine matrix.
I am using the Hamilton form for the quaternions.
I would like to calculate the following Jacobians:
$$H_1 = frac{partial h_1(q)}{partial q} $$
$$H_2 = frac{partial h_2(q)}{partial q} $$
Following Juao Sola's reference eq. 18 what I have is
$$H_1 = A^TC(q)^T[p]_x $$
$$H_2 = [q_1]_L[q_2]_R $$
Where $[q]_R$ and $[q]_L$ are the right and left handed conversion of quaternion to matrix form as defined in Juao Sola's reference eq. 18.
All rotations are body centric.
Is this correct?
Is there a better way to do this?
Can the expression be easily simplified?
derivatives quaternions
asked Jul 25 '16 at 7:42
Yonatan SimsonYonatan Simson
168
$endgroup$
$begingroup$
I apologize for being so blunt, but your post is a major catastrophy because of your mistyping of the indices. In the first set of formulae, the second line should have $h_2$, not $h_1$. In the third set, in $H_1$ there should be $q$ instead of $q_1$, simply because there is no $q_1$ in $h_1$. Finally, in the same set, here should be a $q_2$ in $H_2$.
$endgroup$
– Alex M.
Jul 27 '16 at 12:56
$begingroup$
I edited my question and corrected the mistakes. Thanks for pointing out the errors
$endgroup$
– Yonatan Simson
Jul 27 '16 at 16:18
add a comment |
$begingroup$
I am trying to calculate the Jacobian of a function that has quaternions and 3D points in it. I refer to quaternions as $q$ and 3D points as $p$
$$h_1(q)=A C(q)p $$
$$h_2(q)=q_1otimes q otimes q_2 $$
where $Ain R^{3x3}$ and $C(q)$ is Direction cosine matrix.
I am using the Hamilton form for the quaternions.
I would like to calculate the following Jacobians:
$$H_1 = frac{partial h_1(q)}{partial q} $$
$$H_2 = frac{partial h_2(q)}{partial q} $$
Following Juao Sola's reference eq. 18 what I have is
$$H_1 = A^TC(q)^T[p]_x $$
$$H_2 = [q_1]_L[q_2]_R $$
Where $[q]_R$ and $[q]_L$ are the right and left handed conversion of quaternion to matrix form as defined in Juao Sola's reference eq. 18.
All rotations are body centric.
Is this correct?
Is there a better way to do this?
Can the expression be easily simplified?
derivatives quaternions
asked Jul 25 '16 at 7:42
Yonatan SimsonYonatan Simson
168
$endgroup$
I am trying to calculate the Jacobian of a function that has quaternions and 3D points in it. I refer to quaternions as $q$ and 3D points as $p$
$$h_1(q)=A C(q)p $$
$$h_2(q)=q_1otimes q otimes q_2 $$
where $Ain R^{3x3}$ and $C(q)$ is Direction cosine matrix.
I am using the Hamilton form for the quaternions.
I would like to calculate the following Jacobians:
$$H_1 = frac{partial h_1(q)}{partial q} $$
$$H_2 = frac{partial h_2(q)}{partial q} $$
Following Juao Sola's reference eq. 18 what I have is
$$H_1 = A^TC(q)^T[p]_x $$
$$H_2 = [q_1]_L[q_2]_R $$
Where $[q]_R$ and $[q]_L$ are the right and left handed conversion of quaternion to matrix form as defined in Juao Sola's reference eq. 18.
All rotations are body centric.
Is this correct?
Is there a better way to do this?
Can the expression be easily simplified?
derivatives quaternions
derivatives quaternions
asked Jul 25 '16 at 7:42
Yonatan SimsonYonatan Simson
168
asked Jul 25 '16 at 7:42
Yonatan SimsonYonatan Simson
168
edited Aug 1 '16 at 14:11
Yonatan Simson
asked Jul 25 '16 at 7:42
Yonatan SimsonYonatan Simson
168
asked Jul 25 '16 at 7:42
Yonatan SimsonYonatan Simson
168
asked Jul 25 '16 at 7:42
Yonatan SimsonYonatan Simson
168
168
$begingroup$
I apologize for being so blunt, but your post is a major catastrophy because of your mistyping of the indices. In the first set of formulae, the second line should have $h_2$, not $h_1$. In the third set, in $H_1$ there should be $q$ instead of $q_1$, simply because there is no $q_1$ in $h_1$. Finally, in the same set, here should be a $q_2$ in $H_2$.
$endgroup$
– Alex M.
Jul 27 '16 at 12:56
$begingroup$
I edited my question and corrected the mistakes. Thanks for pointing out the errors
$endgroup$
– Yonatan Simson
Jul 27 '16 at 16:18
add a comment |
$begingroup$
I apologize for being so blunt, but your post is a major catastrophy because of your mistyping of the indices. In the first set of formulae, the second line should have $h_2$, not $h_1$. In the third set, in $H_1$ there should be $q$ instead of $q_1$, simply because there is no $q_1$ in $h_1$. Finally, in the same set, here should be a $q_2$ in $H_2$.
$endgroup$
– Alex M.
Jul 27 '16 at 12:56
$begingroup$
I edited my question and corrected the mistakes. Thanks for pointing out the errors
$endgroup$
– Yonatan Simson
Jul 27 '16 at 16:18
$begingroup$
I apologize for being so blunt, but your post is a major catastrophy because of your mistyping of the indices. In the first set of formulae, the second line should have $h_2$, not $h_1$. In the third set, in $H_1$ there should be $q$ instead of $q_1$, simply because there is no $q_1$ in $h_1$. Finally, in the same set, here should be a $q_2$ in $H_2$.
$endgroup$
– Alex M.
Jul 27 '16 at 12:56
$begingroup$
I apologize for being so blunt, but your post is a major catastrophy because of your mistyping of the indices. In the first set of formulae, the second line should have $h_2$, not $h_1$. In the third set, in $H_1$ there should be $q$ instead of $q_1$, simply because there is no $q_1$ in $h_1$. Finally, in the same set, here should be a $q_2$ in $H_2$.
$endgroup$
– Alex M.
Jul 27 '16 at 12:56
$begingroup$
I edited my question and corrected the mistakes. Thanks for pointing out the errors
$endgroup$
– Yonatan Simson
Jul 27 '16 at 16:18
$begingroup$
I edited my question and corrected the mistakes. Thanks for pointing out the errors
$endgroup$
– Yonatan Simson
Jul 27 '16 at 16:18
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
There are different ways to answer your question, but you probably want one of these two:
You want the derivative with respect to the 4 components of the quaternion q=w+ix+iy+iz, that is, with respect to a 4 vector $v_q=(w,x,y,z)^topin R^4$.
We have the derivative of the rotation wrt this vector q as:
$$
frac{partial qotimes p otimes q^*}{partial v_q} = 2[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
where:
$I$ is the 3x3 identity matrix.
$[p]_times$ is the skew symmetric matrix fromed from $p$.
$times$ is the cross product
$otimes$ is the quaternion product. This sign can be omitted since it's a product and write simply $qpq^*$.
So assuming A is a constant matrix, and knowing that $C(q)p = qpq^*$, your derivative is
$$
frac{partial (AC(q)p)}{partial v_q} = 2A[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
This is the derivative that e.g. Ceres is going to compute should you use automatic differentiation of your function using#include Jet.h.
You want the derivative with respect to the rotation itself seen as a 3-vector of the Lie algebra of the rotation group. The Lie-theory defines two Jacobians, left and right, for this, depending on whether you perturb the rotation on the right, $tilde R=Rexp([theta]_times)$, or on the left, $tilde R=exp([theta]_times)R$.
The right Jacobian of the rotation is:
$$
frac{partial C(q)p}{partial C} = -C(q)[p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -AC(q)[p]_times in R^{3times 3}
$$
The left Jacobian of the rotation is
$$
frac{partial C(q)p}{partial C} = -[C(q)p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -A[C(q)p]_times in R^{3times 3}
$$
answered Dec 28 '18 at 23:09
Joan SolàJoan Solà
1215
$endgroup$
add a comment |
$begingroup$
Quaternion derivatives are not that straight forward. Usually a direction is required (see here).
I'm assuming your thought process behind $H_2$ was something like
$$ h_2 = q_1 otimes q otimes q_2 = q_1 otimes (q otimes q_2) = [q_1]_L (q otimes q_2) = [q_1]_L [q_2]_R q$$
$$ H_2 = frac{partial h_2}{partial q} = [q_1]_L [q_2]_R $$
where I assigned L,R indices as in your reference. But that is the opposite indices of what you wrote. This also has other issues which I will get to in a bit.
For $H_1$, you have the direction cosine matrix as a function of $q$ but you don't appear to have taken the derivative of $C(q)$ at all. So I'm not following the logic there.
Looking at the definition of the direction cosine matrix in terms of the components of $q$, I feel this is venturing into the territory of abusing quaternions as a container of 4 variables.
There are a lot of things going on here, so let me try to untangle them a bit and note on them separately.
Representation of quaternions
If you wish to use linear algebra, there is already a real valued matrix representation of quaternions. I would suggest just using that representation if you want to clarify the content of some equations.
Instead, here quaternion are treated as a column vector (eq 7 of your reference), which leads to you then using two additional representations of quaternions to fit them into a linear algebra setting to represent multiplication. This confuses the mathematical content.
Using the definitions in your reference:
$$ [a + x i + y j + z k]_L =
begin{bmatrix}
a & -x & -y & -z \
x & a & -z & y \
y & z & a & -x \
z & -y & x & a \
end{bmatrix}$$
$$ [a + x i + y j + z k]_R =
begin{bmatrix}
a & -x & -y & -z \
x & a & z & -y \
y & -z & a & x \
z & y & -x & a \
end{bmatrix}$$
That then means:
$$[zk]_L [yj]_R =
begin{bmatrix}
0 & 0 & 0 & -z \
0 & 0 & -z & 0 \
0 & z & 0 & 0 \
z & 0 & 0 & 0
end{bmatrix}
begin{bmatrix}
0 & 0 & -y & 0 \
0 & 0 & 0 & -y \
y & 0 & 0 & 0 \
0 & y & 0 & 0
end{bmatrix}
=
begin{bmatrix}
0 & -yz & 0 & 0 \
-yz & 0 & 0 & 0 \
0 & 0 & 0 & -yz \
0 & 0 & -yz & 0
end{bmatrix}
$$
Which doesn't fit the form for any of the previous representations of a quaternion.
Therefore saying $H_2 = [q_1]_R [q_2]_L$, is stating the concept which is used here for a derivative of a quaternion function, can result in something outside of the set of functions of quaternions. Is this what you intended? How do you intend to interpret this object?
Representing multiple things with the same object
We can use a real value to represent temperature or distance, but these are distinct things so somehow these 'type labels' ('units') must be carried around to remind us of this.
Similarly, trying to represent both rotations and positions with quaternions can confuse things if one isn't careful to carry around these 'type labels'. For while it may make some notation closer to how you'd write the calculation in code, these objects (positions and rotations) are not on the same footing and their components transform differently if we change basis.
Even the underlying choice to represent a rotation with a quaternion (so that one can write $x'=q otimes x otimes q^*$ as done on pg 3 of your reference), doesn't actually relate to the underlying mathematical structures as cleanly as one might imagine. (For example, why do we need to use quaternion multiplication twice? Why is the angle off by a factor of 2? See this classic paper on
rotations and quaternions)
functions of quaternions
Not every "left linear" equation in quaterions
$$f(q) = a q + b$$
can be rewritten as a "right linear" equation
$$f(q) = q c + d$$
If we consider a linear equation of all combinations of multiplication from each side:
$$g(q) = a q b + c q + q d + e$$
Then how should one interpret the derivative of this function? Even for linear equations we are forced to be careful with our expectations.
Potential "abuse" of quaternions
Now, it is possible that what you are trying to do has little to do with these complications. Somewhere along the way, ideas from quaternions get used for writing code in engineering, physics, graphics applications, but the equations eventually carry around so much specialized 'type information' that it really begins to feel more like linear algebra notation used merely to succinctly represent what the code is doing. In such cases quaternions eventually become abused as more of a way to carry around 4 parameters than anything else (which is how I'd describe many of the things at quaternions.com). In which case these linear algebra short hands can just be viewed as defining a calculation for a real valued function in four real variables. In which case you can discuss the partial derivatives of this with respect to any of the variables without the issues above.
So at the end of the day, if you are just using quaternions to calculate some transformation (likely including rotations without being forced to choose a basis for Euler angles), and you'd like to know the partial derivative of some transformation with respect to some parameter, you can always just expand out the transformation. It may not have some nice "quaternion" looking form, but that is just because you were actually just manipulating linear equations and nothing fundamentally "quaternion" in the first place.
answered Jul 31 '16 at 21:36
JustThinkingJustThinking
1708
$endgroup$
$begingroup$
You are right about the error in[q_1]_L[q_2]_R.
$endgroup$
– Yonatan Simson
Aug 1 '16 at 14:18
add a comment |
$begingroup$
Eq (163) in your notes has the really amazing equation:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial mathbf{q}} =
frac{partial(mathbf{R} mathbf{a})}{partial mathbf{q}} =
2 big[
w mathbf{a} +
mathbf{v} times mathbf{a} +
mathbf{v}^top mathbf{a} mathbf{I} +
mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big] $$
These engineering symbols for quaternions are quite fascinating. I have not digested all the symbols. In math we simply note that $x mapsto q x q^{-1}$ (which might be called $qast$) is a Rotation.
answered Aug 3 '16 at 3:34
cactus314cactus314
15.4k42269
$endgroup$
$begingroup$
It's worse than that. The reference is stating it equals $$2 big[ w mathbf{a} + mathbf{v} times mathbf{a} big| mathbf{v}^top mathbf{a} mathbf{I} + mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big]$$, which in their notation is indicating a quaternion with the "real part" three vector valued, and the "imaginary part" 3x3 matrix valued. This really is just abusing "quaternions" as an object that carries four components. It doesn't actually have anything to do with derivatives of quaterionic functions or quaternion analysis.
$endgroup$
– JustThinking
Aug 3 '16 at 6:19
add a comment |
$begingroup$
I needed to solve an easier problem then what I posted.
If $mathbf{a}=(0,a_x,a_y,a_z)^T$ and the differential is by
$bar{mathbf{a}}=(a_x,a_y,a_z)^T$ the answer simply becomes:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial bar{mathbf{a}}} =
frac{partial(mathbf{R} bar{mathbf{a}})}{partial bar{mathbf{a}}} = mathbf{R}$$
Where $mathbf{R}$ is the rotation matrix derived from $mathbf{q}$
answered Nov 23 '17 at 13:15
Yonatan SimsonYonatan Simson
168
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1870305%2fderivative-of-quaternions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
There are different ways to answer your question, but you probably want one of these two:
You want the derivative with respect to the 4 components of the quaternion q=w+ix+iy+iz, that is, with respect to a 4 vector $v_q=(w,x,y,z)^topin R^4$.
We have the derivative of the rotation wrt this vector q as:
$$
frac{partial qotimes p otimes q^*}{partial v_q} = 2[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
where:
$I$ is the 3x3 identity matrix.
$[p]_times$ is the skew symmetric matrix fromed from $p$.
$times$ is the cross product
$otimes$ is the quaternion product. This sign can be omitted since it's a product and write simply $qpq^*$.
So assuming A is a constant matrix, and knowing that $C(q)p = qpq^*$, your derivative is
$$
frac{partial (AC(q)p)}{partial v_q} = 2A[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
This is the derivative that e.g. Ceres is going to compute should you use automatic differentiation of your function using#include Jet.h.
You want the derivative with respect to the rotation itself seen as a 3-vector of the Lie algebra of the rotation group. The Lie-theory defines two Jacobians, left and right, for this, depending on whether you perturb the rotation on the right, $tilde R=Rexp([theta]_times)$, or on the left, $tilde R=exp([theta]_times)R$.
The right Jacobian of the rotation is:
$$
frac{partial C(q)p}{partial C} = -C(q)[p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -AC(q)[p]_times in R^{3times 3}
$$
The left Jacobian of the rotation is
$$
frac{partial C(q)p}{partial C} = -[C(q)p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -A[C(q)p]_times in R^{3times 3}
$$
answered Dec 28 '18 at 23:09
Joan SolàJoan Solà
1215
$endgroup$
add a comment |
$begingroup$
There are different ways to answer your question, but you probably want one of these two:
You want the derivative with respect to the 4 components of the quaternion q=w+ix+iy+iz, that is, with respect to a 4 vector $v_q=(w,x,y,z)^topin R^4$.
We have the derivative of the rotation wrt this vector q as:
$$
frac{partial qotimes p otimes q^*}{partial v_q} = 2[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
where:
$I$ is the 3x3 identity matrix.
$[p]_times$ is the skew symmetric matrix fromed from $p$.
$times$ is the cross product
$otimes$ is the quaternion product. This sign can be omitted since it's a product and write simply $qpq^*$.
So assuming A is a constant matrix, and knowing that $C(q)p = qpq^*$, your derivative is
$$
frac{partial (AC(q)p)}{partial v_q} = 2A[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
This is the derivative that e.g. Ceres is going to compute should you use automatic differentiation of your function using#include Jet.h.
You want the derivative with respect to the rotation itself seen as a 3-vector of the Lie algebra of the rotation group. The Lie-theory defines two Jacobians, left and right, for this, depending on whether you perturb the rotation on the right, $tilde R=Rexp([theta]_times)$, or on the left, $tilde R=exp([theta]_times)R$.
The right Jacobian of the rotation is:
$$
frac{partial C(q)p}{partial C} = -C(q)[p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -AC(q)[p]_times in R^{3times 3}
$$
The left Jacobian of the rotation is
$$
frac{partial C(q)p}{partial C} = -[C(q)p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -A[C(q)p]_times in R^{3times 3}
$$
answered Dec 28 '18 at 23:09
Joan SolàJoan Solà
1215
$endgroup$
add a comment |
$begingroup$
There are different ways to answer your question, but you probably want one of these two:
You want the derivative with respect to the 4 components of the quaternion q=w+ix+iy+iz, that is, with respect to a 4 vector $v_q=(w,x,y,z)^topin R^4$.
We have the derivative of the rotation wrt this vector q as:
$$
frac{partial qotimes p otimes q^*}{partial v_q} = 2[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
where:
$I$ is the 3x3 identity matrix.
$[p]_times$ is the skew symmetric matrix fromed from $p$.
$times$ is the cross product
$otimes$ is the quaternion product. This sign can be omitted since it's a product and write simply $qpq^*$.
So assuming A is a constant matrix, and knowing that $C(q)p = qpq^*$, your derivative is
$$
frac{partial (AC(q)p)}{partial v_q} = 2A[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
This is the derivative that e.g. Ceres is going to compute should you use automatic differentiation of your function using#include Jet.h.
You want the derivative with respect to the rotation itself seen as a 3-vector of the Lie algebra of the rotation group. The Lie-theory defines two Jacobians, left and right, for this, depending on whether you perturb the rotation on the right, $tilde R=Rexp([theta]_times)$, or on the left, $tilde R=exp([theta]_times)R$.
The right Jacobian of the rotation is:
$$
frac{partial C(q)p}{partial C} = -C(q)[p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -AC(q)[p]_times in R^{3times 3}
$$
The left Jacobian of the rotation is
$$
frac{partial C(q)p}{partial C} = -[C(q)p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -A[C(q)p]_times in R^{3times 3}
$$
answered Dec 28 '18 at 23:09
Joan SolàJoan Solà
1215
$endgroup$
There are different ways to answer your question, but you probably want one of these two:
You want the derivative with respect to the 4 components of the quaternion q=w+ix+iy+iz, that is, with respect to a 4 vector $v_q=(w,x,y,z)^topin R^4$.
We have the derivative of the rotation wrt this vector q as:
$$
frac{partial qotimes p otimes q^*}{partial v_q} = 2[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
where:
$I$ is the 3x3 identity matrix.
$[p]_times$ is the skew symmetric matrix fromed from $p$.
$times$ is the cross product
$otimes$ is the quaternion product. This sign can be omitted since it's a product and write simply $qpq^*$.
So assuming A is a constant matrix, and knowing that $C(q)p = qpq^*$, your derivative is
$$
frac{partial (AC(q)p)}{partial v_q} = 2A[wp+v×p , v^top p I + vp^top−pv^top−w[p]_times] in R^{3times4}
$$
This is the derivative that e.g. Ceres is going to compute should you use automatic differentiation of your function using#include Jet.h.
You want the derivative with respect to the rotation itself seen as a 3-vector of the Lie algebra of the rotation group. The Lie-theory defines two Jacobians, left and right, for this, depending on whether you perturb the rotation on the right, $tilde R=Rexp([theta]_times)$, or on the left, $tilde R=exp([theta]_times)R$.
The right Jacobian of the rotation is:
$$
frac{partial C(q)p}{partial C} = -C(q)[p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -AC(q)[p]_times in R^{3times 3}
$$
The left Jacobian of the rotation is
$$
frac{partial C(q)p}{partial C} = -[C(q)p]_times in R^{3times 3}
$$
and so your full Jacobian is
$$
frac{partial (AC(q)p)}{partial C} = -A[C(q)p]_times in R^{3times 3}
$$
answered Dec 28 '18 at 23:09
Joan SolàJoan Solà
1215
edited Jan 9 at 22:54
answered Dec 28 '18 at 23:09
Joan SolàJoan Solà
1215
answered Dec 28 '18 at 23:09
Joan SolàJoan Solà
1215
answered Dec 28 '18 at 23:09
Joan SolàJoan Solà
1215
1215
add a comment |
add a comment |
$begingroup$
Quaternion derivatives are not that straight forward. Usually a direction is required (see here).
I'm assuming your thought process behind $H_2$ was something like
$$ h_2 = q_1 otimes q otimes q_2 = q_1 otimes (q otimes q_2) = [q_1]_L (q otimes q_2) = [q_1]_L [q_2]_R q$$
$$ H_2 = frac{partial h_2}{partial q} = [q_1]_L [q_2]_R $$
where I assigned L,R indices as in your reference. But that is the opposite indices of what you wrote. This also has other issues which I will get to in a bit.
For $H_1$, you have the direction cosine matrix as a function of $q$ but you don't appear to have taken the derivative of $C(q)$ at all. So I'm not following the logic there.
Looking at the definition of the direction cosine matrix in terms of the components of $q$, I feel this is venturing into the territory of abusing quaternions as a container of 4 variables.
There are a lot of things going on here, so let me try to untangle them a bit and note on them separately.
Representation of quaternions
If you wish to use linear algebra, there is already a real valued matrix representation of quaternions. I would suggest just using that representation if you want to clarify the content of some equations.
Instead, here quaternion are treated as a column vector (eq 7 of your reference), which leads to you then using two additional representations of quaternions to fit them into a linear algebra setting to represent multiplication. This confuses the mathematical content.
Using the definitions in your reference:
$$ [a + x i + y j + z k]_L =
begin{bmatrix}
a & -x & -y & -z \
x & a & -z & y \
y & z & a & -x \
z & -y & x & a \
end{bmatrix}$$
$$ [a + x i + y j + z k]_R =
begin{bmatrix}
a & -x & -y & -z \
x & a & z & -y \
y & -z & a & x \
z & y & -x & a \
end{bmatrix}$$
That then means:
$$[zk]_L [yj]_R =
begin{bmatrix}
0 & 0 & 0 & -z \
0 & 0 & -z & 0 \
0 & z & 0 & 0 \
z & 0 & 0 & 0
end{bmatrix}
begin{bmatrix}
0 & 0 & -y & 0 \
0 & 0 & 0 & -y \
y & 0 & 0 & 0 \
0 & y & 0 & 0
end{bmatrix}
=
begin{bmatrix}
0 & -yz & 0 & 0 \
-yz & 0 & 0 & 0 \
0 & 0 & 0 & -yz \
0 & 0 & -yz & 0
end{bmatrix}
$$
Which doesn't fit the form for any of the previous representations of a quaternion.
Therefore saying $H_2 = [q_1]_R [q_2]_L$, is stating the concept which is used here for a derivative of a quaternion function, can result in something outside of the set of functions of quaternions. Is this what you intended? How do you intend to interpret this object?
Representing multiple things with the same object
We can use a real value to represent temperature or distance, but these are distinct things so somehow these 'type labels' ('units') must be carried around to remind us of this.
Similarly, trying to represent both rotations and positions with quaternions can confuse things if one isn't careful to carry around these 'type labels'. For while it may make some notation closer to how you'd write the calculation in code, these objects (positions and rotations) are not on the same footing and their components transform differently if we change basis.
Even the underlying choice to represent a rotation with a quaternion (so that one can write $x'=q otimes x otimes q^*$ as done on pg 3 of your reference), doesn't actually relate to the underlying mathematical structures as cleanly as one might imagine. (For example, why do we need to use quaternion multiplication twice? Why is the angle off by a factor of 2? See this classic paper on
rotations and quaternions)
functions of quaternions
Not every "left linear" equation in quaterions
$$f(q) = a q + b$$
can be rewritten as a "right linear" equation
$$f(q) = q c + d$$
If we consider a linear equation of all combinations of multiplication from each side:
$$g(q) = a q b + c q + q d + e$$
Then how should one interpret the derivative of this function? Even for linear equations we are forced to be careful with our expectations.
Potential "abuse" of quaternions
Now, it is possible that what you are trying to do has little to do with these complications. Somewhere along the way, ideas from quaternions get used for writing code in engineering, physics, graphics applications, but the equations eventually carry around so much specialized 'type information' that it really begins to feel more like linear algebra notation used merely to succinctly represent what the code is doing. In such cases quaternions eventually become abused as more of a way to carry around 4 parameters than anything else (which is how I'd describe many of the things at quaternions.com). In which case these linear algebra short hands can just be viewed as defining a calculation for a real valued function in four real variables. In which case you can discuss the partial derivatives of this with respect to any of the variables without the issues above.
So at the end of the day, if you are just using quaternions to calculate some transformation (likely including rotations without being forced to choose a basis for Euler angles), and you'd like to know the partial derivative of some transformation with respect to some parameter, you can always just expand out the transformation. It may not have some nice "quaternion" looking form, but that is just because you were actually just manipulating linear equations and nothing fundamentally "quaternion" in the first place.
answered Jul 31 '16 at 21:36
JustThinkingJustThinking
1708
$endgroup$
$begingroup$
You are right about the error in[q_1]_L[q_2]_R.
$endgroup$
– Yonatan Simson
Aug 1 '16 at 14:18
add a comment |
$begingroup$
Quaternion derivatives are not that straight forward. Usually a direction is required (see here).
I'm assuming your thought process behind $H_2$ was something like
$$ h_2 = q_1 otimes q otimes q_2 = q_1 otimes (q otimes q_2) = [q_1]_L (q otimes q_2) = [q_1]_L [q_2]_R q$$
$$ H_2 = frac{partial h_2}{partial q} = [q_1]_L [q_2]_R $$
where I assigned L,R indices as in your reference. But that is the opposite indices of what you wrote. This also has other issues which I will get to in a bit.
For $H_1$, you have the direction cosine matrix as a function of $q$ but you don't appear to have taken the derivative of $C(q)$ at all. So I'm not following the logic there.
Looking at the definition of the direction cosine matrix in terms of the components of $q$, I feel this is venturing into the territory of abusing quaternions as a container of 4 variables.
There are a lot of things going on here, so let me try to untangle them a bit and note on them separately.
Representation of quaternions
If you wish to use linear algebra, there is already a real valued matrix representation of quaternions. I would suggest just using that representation if you want to clarify the content of some equations.
Instead, here quaternion are treated as a column vector (eq 7 of your reference), which leads to you then using two additional representations of quaternions to fit them into a linear algebra setting to represent multiplication. This confuses the mathematical content.
Using the definitions in your reference:
$$ [a + x i + y j + z k]_L =
begin{bmatrix}
a & -x & -y & -z \
x & a & -z & y \
y & z & a & -x \
z & -y & x & a \
end{bmatrix}$$
$$ [a + x i + y j + z k]_R =
begin{bmatrix}
a & -x & -y & -z \
x & a & z & -y \
y & -z & a & x \
z & y & -x & a \
end{bmatrix}$$
That then means:
$$[zk]_L [yj]_R =
begin{bmatrix}
0 & 0 & 0 & -z \
0 & 0 & -z & 0 \
0 & z & 0 & 0 \
z & 0 & 0 & 0
end{bmatrix}
begin{bmatrix}
0 & 0 & -y & 0 \
0 & 0 & 0 & -y \
y & 0 & 0 & 0 \
0 & y & 0 & 0
end{bmatrix}
=
begin{bmatrix}
0 & -yz & 0 & 0 \
-yz & 0 & 0 & 0 \
0 & 0 & 0 & -yz \
0 & 0 & -yz & 0
end{bmatrix}
$$
Which doesn't fit the form for any of the previous representations of a quaternion.
Therefore saying $H_2 = [q_1]_R [q_2]_L$, is stating the concept which is used here for a derivative of a quaternion function, can result in something outside of the set of functions of quaternions. Is this what you intended? How do you intend to interpret this object?
Representing multiple things with the same object
We can use a real value to represent temperature or distance, but these are distinct things so somehow these 'type labels' ('units') must be carried around to remind us of this.
Similarly, trying to represent both rotations and positions with quaternions can confuse things if one isn't careful to carry around these 'type labels'. For while it may make some notation closer to how you'd write the calculation in code, these objects (positions and rotations) are not on the same footing and their components transform differently if we change basis.
Even the underlying choice to represent a rotation with a quaternion (so that one can write $x'=q otimes x otimes q^*$ as done on pg 3 of your reference), doesn't actually relate to the underlying mathematical structures as cleanly as one might imagine. (For example, why do we need to use quaternion multiplication twice? Why is the angle off by a factor of 2? See this classic paper on
rotations and quaternions)
functions of quaternions
Not every "left linear" equation in quaterions
$$f(q) = a q + b$$
can be rewritten as a "right linear" equation
$$f(q) = q c + d$$
If we consider a linear equation of all combinations of multiplication from each side:
$$g(q) = a q b + c q + q d + e$$
Then how should one interpret the derivative of this function? Even for linear equations we are forced to be careful with our expectations.
Potential "abuse" of quaternions
Now, it is possible that what you are trying to do has little to do with these complications. Somewhere along the way, ideas from quaternions get used for writing code in engineering, physics, graphics applications, but the equations eventually carry around so much specialized 'type information' that it really begins to feel more like linear algebra notation used merely to succinctly represent what the code is doing. In such cases quaternions eventually become abused as more of a way to carry around 4 parameters than anything else (which is how I'd describe many of the things at quaternions.com). In which case these linear algebra short hands can just be viewed as defining a calculation for a real valued function in four real variables. In which case you can discuss the partial derivatives of this with respect to any of the variables without the issues above.
So at the end of the day, if you are just using quaternions to calculate some transformation (likely including rotations without being forced to choose a basis for Euler angles), and you'd like to know the partial derivative of some transformation with respect to some parameter, you can always just expand out the transformation. It may not have some nice "quaternion" looking form, but that is just because you were actually just manipulating linear equations and nothing fundamentally "quaternion" in the first place.
answered Jul 31 '16 at 21:36
JustThinkingJustThinking
1708
$endgroup$
$begingroup$
You are right about the error in[q_1]_L[q_2]_R.
$endgroup$
– Yonatan Simson
Aug 1 '16 at 14:18
add a comment |
$begingroup$
Quaternion derivatives are not that straight forward. Usually a direction is required (see here).
I'm assuming your thought process behind $H_2$ was something like
$$ h_2 = q_1 otimes q otimes q_2 = q_1 otimes (q otimes q_2) = [q_1]_L (q otimes q_2) = [q_1]_L [q_2]_R q$$
$$ H_2 = frac{partial h_2}{partial q} = [q_1]_L [q_2]_R $$
where I assigned L,R indices as in your reference. But that is the opposite indices of what you wrote. This also has other issues which I will get to in a bit.
For $H_1$, you have the direction cosine matrix as a function of $q$ but you don't appear to have taken the derivative of $C(q)$ at all. So I'm not following the logic there.
Looking at the definition of the direction cosine matrix in terms of the components of $q$, I feel this is venturing into the territory of abusing quaternions as a container of 4 variables.
There are a lot of things going on here, so let me try to untangle them a bit and note on them separately.
Representation of quaternions
If you wish to use linear algebra, there is already a real valued matrix representation of quaternions. I would suggest just using that representation if you want to clarify the content of some equations.
Instead, here quaternion are treated as a column vector (eq 7 of your reference), which leads to you then using two additional representations of quaternions to fit them into a linear algebra setting to represent multiplication. This confuses the mathematical content.
Using the definitions in your reference:
$$ [a + x i + y j + z k]_L =
begin{bmatrix}
a & -x & -y & -z \
x & a & -z & y \
y & z & a & -x \
z & -y & x & a \
end{bmatrix}$$
$$ [a + x i + y j + z k]_R =
begin{bmatrix}
a & -x & -y & -z \
x & a & z & -y \
y & -z & a & x \
z & y & -x & a \
end{bmatrix}$$
That then means:
$$[zk]_L [yj]_R =
begin{bmatrix}
0 & 0 & 0 & -z \
0 & 0 & -z & 0 \
0 & z & 0 & 0 \
z & 0 & 0 & 0
end{bmatrix}
begin{bmatrix}
0 & 0 & -y & 0 \
0 & 0 & 0 & -y \
y & 0 & 0 & 0 \
0 & y & 0 & 0
end{bmatrix}
=
begin{bmatrix}
0 & -yz & 0 & 0 \
-yz & 0 & 0 & 0 \
0 & 0 & 0 & -yz \
0 & 0 & -yz & 0
end{bmatrix}
$$
Which doesn't fit the form for any of the previous representations of a quaternion.
Therefore saying $H_2 = [q_1]_R [q_2]_L$, is stating the concept which is used here for a derivative of a quaternion function, can result in something outside of the set of functions of quaternions. Is this what you intended? How do you intend to interpret this object?
Representing multiple things with the same object
We can use a real value to represent temperature or distance, but these are distinct things so somehow these 'type labels' ('units') must be carried around to remind us of this.
Similarly, trying to represent both rotations and positions with quaternions can confuse things if one isn't careful to carry around these 'type labels'. For while it may make some notation closer to how you'd write the calculation in code, these objects (positions and rotations) are not on the same footing and their components transform differently if we change basis.
Even the underlying choice to represent a rotation with a quaternion (so that one can write $x'=q otimes x otimes q^*$ as done on pg 3 of your reference), doesn't actually relate to the underlying mathematical structures as cleanly as one might imagine. (For example, why do we need to use quaternion multiplication twice? Why is the angle off by a factor of 2? See this classic paper on
rotations and quaternions)
functions of quaternions
Not every "left linear" equation in quaterions
$$f(q) = a q + b$$
can be rewritten as a "right linear" equation
$$f(q) = q c + d$$
If we consider a linear equation of all combinations of multiplication from each side:
$$g(q) = a q b + c q + q d + e$$
Then how should one interpret the derivative of this function? Even for linear equations we are forced to be careful with our expectations.
Potential "abuse" of quaternions
Now, it is possible that what you are trying to do has little to do with these complications. Somewhere along the way, ideas from quaternions get used for writing code in engineering, physics, graphics applications, but the equations eventually carry around so much specialized 'type information' that it really begins to feel more like linear algebra notation used merely to succinctly represent what the code is doing. In such cases quaternions eventually become abused as more of a way to carry around 4 parameters than anything else (which is how I'd describe many of the things at quaternions.com). In which case these linear algebra short hands can just be viewed as defining a calculation for a real valued function in four real variables. In which case you can discuss the partial derivatives of this with respect to any of the variables without the issues above.
So at the end of the day, if you are just using quaternions to calculate some transformation (likely including rotations without being forced to choose a basis for Euler angles), and you'd like to know the partial derivative of some transformation with respect to some parameter, you can always just expand out the transformation. It may not have some nice "quaternion" looking form, but that is just because you were actually just manipulating linear equations and nothing fundamentally "quaternion" in the first place.
answered Jul 31 '16 at 21:36
JustThinkingJustThinking
1708
$endgroup$
Quaternion derivatives are not that straight forward. Usually a direction is required (see here).
I'm assuming your thought process behind $H_2$ was something like
$$ h_2 = q_1 otimes q otimes q_2 = q_1 otimes (q otimes q_2) = [q_1]_L (q otimes q_2) = [q_1]_L [q_2]_R q$$
$$ H_2 = frac{partial h_2}{partial q} = [q_1]_L [q_2]_R $$
where I assigned L,R indices as in your reference. But that is the opposite indices of what you wrote. This also has other issues which I will get to in a bit.
For $H_1$, you have the direction cosine matrix as a function of $q$ but you don't appear to have taken the derivative of $C(q)$ at all. So I'm not following the logic there.
Looking at the definition of the direction cosine matrix in terms of the components of $q$, I feel this is venturing into the territory of abusing quaternions as a container of 4 variables.
There are a lot of things going on here, so let me try to untangle them a bit and note on them separately.
Representation of quaternions
If you wish to use linear algebra, there is already a real valued matrix representation of quaternions. I would suggest just using that representation if you want to clarify the content of some equations.
Instead, here quaternion are treated as a column vector (eq 7 of your reference), which leads to you then using two additional representations of quaternions to fit them into a linear algebra setting to represent multiplication. This confuses the mathematical content.
Using the definitions in your reference:
$$ [a + x i + y j + z k]_L =
begin{bmatrix}
a & -x & -y & -z \
x & a & -z & y \
y & z & a & -x \
z & -y & x & a \
end{bmatrix}$$
$$ [a + x i + y j + z k]_R =
begin{bmatrix}
a & -x & -y & -z \
x & a & z & -y \
y & -z & a & x \
z & y & -x & a \
end{bmatrix}$$
That then means:
$$[zk]_L [yj]_R =
begin{bmatrix}
0 & 0 & 0 & -z \
0 & 0 & -z & 0 \
0 & z & 0 & 0 \
z & 0 & 0 & 0
end{bmatrix}
begin{bmatrix}
0 & 0 & -y & 0 \
0 & 0 & 0 & -y \
y & 0 & 0 & 0 \
0 & y & 0 & 0
end{bmatrix}
=
begin{bmatrix}
0 & -yz & 0 & 0 \
-yz & 0 & 0 & 0 \
0 & 0 & 0 & -yz \
0 & 0 & -yz & 0
end{bmatrix}
$$
Which doesn't fit the form for any of the previous representations of a quaternion.
Therefore saying $H_2 = [q_1]_R [q_2]_L$, is stating the concept which is used here for a derivative of a quaternion function, can result in something outside of the set of functions of quaternions. Is this what you intended? How do you intend to interpret this object?
Representing multiple things with the same object
We can use a real value to represent temperature or distance, but these are distinct things so somehow these 'type labels' ('units') must be carried around to remind us of this.
Similarly, trying to represent both rotations and positions with quaternions can confuse things if one isn't careful to carry around these 'type labels'. For while it may make some notation closer to how you'd write the calculation in code, these objects (positions and rotations) are not on the same footing and their components transform differently if we change basis.
Even the underlying choice to represent a rotation with a quaternion (so that one can write $x'=q otimes x otimes q^*$ as done on pg 3 of your reference), doesn't actually relate to the underlying mathematical structures as cleanly as one might imagine. (For example, why do we need to use quaternion multiplication twice? Why is the angle off by a factor of 2? See this classic paper on
rotations and quaternions)
functions of quaternions
Not every "left linear" equation in quaterions
$$f(q) = a q + b$$
can be rewritten as a "right linear" equation
$$f(q) = q c + d$$
If we consider a linear equation of all combinations of multiplication from each side:
$$g(q) = a q b + c q + q d + e$$
Then how should one interpret the derivative of this function? Even for linear equations we are forced to be careful with our expectations.
Potential "abuse" of quaternions
Now, it is possible that what you are trying to do has little to do with these complications. Somewhere along the way, ideas from quaternions get used for writing code in engineering, physics, graphics applications, but the equations eventually carry around so much specialized 'type information' that it really begins to feel more like linear algebra notation used merely to succinctly represent what the code is doing. In such cases quaternions eventually become abused as more of a way to carry around 4 parameters than anything else (which is how I'd describe many of the things at quaternions.com). In which case these linear algebra short hands can just be viewed as defining a calculation for a real valued function in four real variables. In which case you can discuss the partial derivatives of this with respect to any of the variables without the issues above.
So at the end of the day, if you are just using quaternions to calculate some transformation (likely including rotations without being forced to choose a basis for Euler angles), and you'd like to know the partial derivative of some transformation with respect to some parameter, you can always just expand out the transformation. It may not have some nice "quaternion" looking form, but that is just because you were actually just manipulating linear equations and nothing fundamentally "quaternion" in the first place.
answered Jul 31 '16 at 21:36
JustThinkingJustThinking
1708
edited Jul 31 '16 at 21:48
answered Jul 31 '16 at 21:36
JustThinkingJustThinking
1708
answered Jul 31 '16 at 21:36
JustThinkingJustThinking
1708
answered Jul 31 '16 at 21:36
JustThinkingJustThinking
1708
1708
$begingroup$
You are right about the error in[q_1]_L[q_2]_R.
$endgroup$
– Yonatan Simson
Aug 1 '16 at 14:18
add a comment |
$begingroup$
You are right about the error in[q_1]_L[q_2]_R.
$endgroup$
– Yonatan Simson
Aug 1 '16 at 14:18
$begingroup$
You are right about the error in
[q_1]_L[q_2]_R.$endgroup$
– Yonatan Simson
Aug 1 '16 at 14:18
$begingroup$
You are right about the error in
[q_1]_L[q_2]_R.$endgroup$
– Yonatan Simson
Aug 1 '16 at 14:18
add a comment |
$begingroup$
Eq (163) in your notes has the really amazing equation:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial mathbf{q}} =
frac{partial(mathbf{R} mathbf{a})}{partial mathbf{q}} =
2 big[
w mathbf{a} +
mathbf{v} times mathbf{a} +
mathbf{v}^top mathbf{a} mathbf{I} +
mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big] $$
These engineering symbols for quaternions are quite fascinating. I have not digested all the symbols. In math we simply note that $x mapsto q x q^{-1}$ (which might be called $qast$) is a Rotation.
answered Aug 3 '16 at 3:34
cactus314cactus314
15.4k42269
$endgroup$
$begingroup$
It's worse than that. The reference is stating it equals $$2 big[ w mathbf{a} + mathbf{v} times mathbf{a} big| mathbf{v}^top mathbf{a} mathbf{I} + mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big]$$, which in their notation is indicating a quaternion with the "real part" three vector valued, and the "imaginary part" 3x3 matrix valued. This really is just abusing "quaternions" as an object that carries four components. It doesn't actually have anything to do with derivatives of quaterionic functions or quaternion analysis.
$endgroup$
– JustThinking
Aug 3 '16 at 6:19
add a comment |
$begingroup$
Eq (163) in your notes has the really amazing equation:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial mathbf{q}} =
frac{partial(mathbf{R} mathbf{a})}{partial mathbf{q}} =
2 big[
w mathbf{a} +
mathbf{v} times mathbf{a} +
mathbf{v}^top mathbf{a} mathbf{I} +
mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big] $$
These engineering symbols for quaternions are quite fascinating. I have not digested all the symbols. In math we simply note that $x mapsto q x q^{-1}$ (which might be called $qast$) is a Rotation.
answered Aug 3 '16 at 3:34
cactus314cactus314
15.4k42269
$endgroup$
$begingroup$
It's worse than that. The reference is stating it equals $$2 big[ w mathbf{a} + mathbf{v} times mathbf{a} big| mathbf{v}^top mathbf{a} mathbf{I} + mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big]$$, which in their notation is indicating a quaternion with the "real part" three vector valued, and the "imaginary part" 3x3 matrix valued. This really is just abusing "quaternions" as an object that carries four components. It doesn't actually have anything to do with derivatives of quaterionic functions or quaternion analysis.
$endgroup$
– JustThinking
Aug 3 '16 at 6:19
add a comment |
$begingroup$
Eq (163) in your notes has the really amazing equation:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial mathbf{q}} =
frac{partial(mathbf{R} mathbf{a})}{partial mathbf{q}} =
2 big[
w mathbf{a} +
mathbf{v} times mathbf{a} +
mathbf{v}^top mathbf{a} mathbf{I} +
mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big] $$
These engineering symbols for quaternions are quite fascinating. I have not digested all the symbols. In math we simply note that $x mapsto q x q^{-1}$ (which might be called $qast$) is a Rotation.
answered Aug 3 '16 at 3:34
cactus314cactus314
15.4k42269
$endgroup$
Eq (163) in your notes has the really amazing equation:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial mathbf{q}} =
frac{partial(mathbf{R} mathbf{a})}{partial mathbf{q}} =
2 big[
w mathbf{a} +
mathbf{v} times mathbf{a} +
mathbf{v}^top mathbf{a} mathbf{I} +
mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big] $$
These engineering symbols for quaternions are quite fascinating. I have not digested all the symbols. In math we simply note that $x mapsto q x q^{-1}$ (which might be called $qast$) is a Rotation.
answered Aug 3 '16 at 3:34
cactus314cactus314
15.4k42269
answered Aug 3 '16 at 3:34
cactus314cactus314
15.4k42269
answered Aug 3 '16 at 3:34
cactus314cactus314
15.4k42269
answered Aug 3 '16 at 3:34
cactus314cactus314
15.4k42269
15.4k42269
$begingroup$
It's worse than that. The reference is stating it equals $$2 big[ w mathbf{a} + mathbf{v} times mathbf{a} big| mathbf{v}^top mathbf{a} mathbf{I} + mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big]$$, which in their notation is indicating a quaternion with the "real part" three vector valued, and the "imaginary part" 3x3 matrix valued. This really is just abusing "quaternions" as an object that carries four components. It doesn't actually have anything to do with derivatives of quaterionic functions or quaternion analysis.
$endgroup$
– JustThinking
Aug 3 '16 at 6:19
add a comment |
$begingroup$
It's worse than that. The reference is stating it equals $$2 big[ w mathbf{a} + mathbf{v} times mathbf{a} big| mathbf{v}^top mathbf{a} mathbf{I} + mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big]$$, which in their notation is indicating a quaternion with the "real part" three vector valued, and the "imaginary part" 3x3 matrix valued. This really is just abusing "quaternions" as an object that carries four components. It doesn't actually have anything to do with derivatives of quaterionic functions or quaternion analysis.
$endgroup$
– JustThinking
Aug 3 '16 at 6:19
$begingroup$
It's worse than that. The reference is stating it equals $$2 big[ w mathbf{a} + mathbf{v} times mathbf{a} big| mathbf{v}^top mathbf{a} mathbf{I} + mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big]$$, which in their notation is indicating a quaternion with the "real part" three vector valued, and the "imaginary part" 3x3 matrix valued. This really is just abusing "quaternions" as an object that carries four components. It doesn't actually have anything to do with derivatives of quaterionic functions or quaternion analysis.
$endgroup$
– JustThinking
Aug 3 '16 at 6:19
$begingroup$
It's worse than that. The reference is stating it equals $$2 big[ w mathbf{a} + mathbf{v} times mathbf{a} big| mathbf{v}^top mathbf{a} mathbf{I} + mathbf{v}mathbf{a}^top - mathbf{a}mathbf{v}^top - w[mathbf{a}]_times big]$$, which in their notation is indicating a quaternion with the "real part" three vector valued, and the "imaginary part" 3x3 matrix valued. This really is just abusing "quaternions" as an object that carries four components. It doesn't actually have anything to do with derivatives of quaterionic functions or quaternion analysis.
$endgroup$
– JustThinking
Aug 3 '16 at 6:19
add a comment |
$begingroup$
I needed to solve an easier problem then what I posted.
If $mathbf{a}=(0,a_x,a_y,a_z)^T$ and the differential is by
$bar{mathbf{a}}=(a_x,a_y,a_z)^T$ the answer simply becomes:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial bar{mathbf{a}}} =
frac{partial(mathbf{R} bar{mathbf{a}})}{partial bar{mathbf{a}}} = mathbf{R}$$
Where $mathbf{R}$ is the rotation matrix derived from $mathbf{q}$
answered Nov 23 '17 at 13:15
Yonatan SimsonYonatan Simson
168
$endgroup$
add a comment |
$begingroup$
I needed to solve an easier problem then what I posted.
If $mathbf{a}=(0,a_x,a_y,a_z)^T$ and the differential is by
$bar{mathbf{a}}=(a_x,a_y,a_z)^T$ the answer simply becomes:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial bar{mathbf{a}}} =
frac{partial(mathbf{R} bar{mathbf{a}})}{partial bar{mathbf{a}}} = mathbf{R}$$
Where $mathbf{R}$ is the rotation matrix derived from $mathbf{q}$
answered Nov 23 '17 at 13:15
Yonatan SimsonYonatan Simson
168
$endgroup$
add a comment |
$begingroup$
I needed to solve an easier problem then what I posted.
If $mathbf{a}=(0,a_x,a_y,a_z)^T$ and the differential is by
$bar{mathbf{a}}=(a_x,a_y,a_z)^T$ the answer simply becomes:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial bar{mathbf{a}}} =
frac{partial(mathbf{R} bar{mathbf{a}})}{partial bar{mathbf{a}}} = mathbf{R}$$
Where $mathbf{R}$ is the rotation matrix derived from $mathbf{q}$
answered Nov 23 '17 at 13:15
Yonatan SimsonYonatan Simson
168
$endgroup$
I needed to solve an easier problem then what I posted.
If $mathbf{a}=(0,a_x,a_y,a_z)^T$ and the differential is by
$bar{mathbf{a}}=(a_x,a_y,a_z)^T$ the answer simply becomes:
$$ frac{partial(mathbf{q} otimes mathbf{a} otimes mathbf{qast})}{partial bar{mathbf{a}}} =
frac{partial(mathbf{R} bar{mathbf{a}})}{partial bar{mathbf{a}}} = mathbf{R}$$
Where $mathbf{R}$ is the rotation matrix derived from $mathbf{q}$
answered Nov 23 '17 at 13:15
Yonatan SimsonYonatan Simson
168
answered Nov 23 '17 at 13:15
Yonatan SimsonYonatan Simson
168
answered Nov 23 '17 at 13:15
Yonatan SimsonYonatan Simson
168
answered Nov 23 '17 at 13:15
Yonatan SimsonYonatan Simson
168
168
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f1870305%2fderivative-of-quaternions%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
I apologize for being so blunt, but your post is a major catastrophy because of your mistyping of the indices. In the first set of formulae, the second line should have $h_2$, not $h_1$. In the third set, in $H_1$ there should be $q$ instead of $q_1$, simply because there is no $q_1$ in $h_1$. Finally, in the same set, here should be a $q_2$ in $H_2$.
$endgroup$
– Alex M.
Jul 27 '16 at 12:56
$begingroup$
I edited my question and corrected the mistakes. Thanks for pointing out the errors
$endgroup$
– Yonatan Simson
Jul 27 '16 at 16:18