Quaternions - Prove that two quaternions map to the same R












2












$begingroup$


I am given the following question: I need to prove that two quaternions map to the same Rotation Matrix in SO3 Space. It is demonstrated by this image:



enter image description here



Let w be v here.
I tried to work out the proof, but it isn't coming out correctly:



$Q=(cosfrac{theta}{2},w.sinfrac{theta}{2})$



$Q=cosfrac{theta}{2}+w_{1}.sinfrac{theta}{2}.mathbf{i}+w_{2}.sinfrac{theta}{2}.mathbf{j}+w_{3}.sinfrac{theta}{2}.mathbf{k}$



As per image, replace $theta$
with $2pi-theta$
and w with -w



$Q=cosfrac{2pi-theta}{2}+-w_{1}.sinfrac{2pi-theta}{2}.mathbf{i}+-w_{2}.sinfrac{2pi-theta}{2}.mathbf{j}+-w_{3}.sinfrac{2pi-theta}{2}.mathbf{k}$



$Q=-cosfrac{theta}{2}+w_{1}.sinfrac{theta}{2}.mathbf{i}+w_{2}.sinfrac{theta}{2}.mathbf{j}+w_{3}.sinfrac{theta}{2}.mathbf{k}$



It is not equal! I am left with $-cosfrac{theta}{2}$ which is not matching. Please tell me how to fix this






quaternions







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$endgroup$












  • $begingroup$
    You're told that there are two ways to encode the rotation, and you're trying to prove the two ways are one. What did you expect :)
    $endgroup$
    – rschwieb
    Jan 28 at 11:53










  • $begingroup$
    Related.
    $endgroup$
    – Jyrki Lahtonen
    Jan 31 at 18:48
















2












$begingroup$


I am given the following question: I need to prove that two quaternions map to the same Rotation Matrix in SO3 Space. It is demonstrated by this image:



enter image description here



Let w be v here.
I tried to work out the proof, but it isn't coming out correctly:



$Q=(cosfrac{theta}{2},w.sinfrac{theta}{2})$



$Q=cosfrac{theta}{2}+w_{1}.sinfrac{theta}{2}.mathbf{i}+w_{2}.sinfrac{theta}{2}.mathbf{j}+w_{3}.sinfrac{theta}{2}.mathbf{k}$



As per image, replace $theta$
with $2pi-theta$
and w with -w



$Q=cosfrac{2pi-theta}{2}+-w_{1}.sinfrac{2pi-theta}{2}.mathbf{i}+-w_{2}.sinfrac{2pi-theta}{2}.mathbf{j}+-w_{3}.sinfrac{2pi-theta}{2}.mathbf{k}$



$Q=-cosfrac{theta}{2}+w_{1}.sinfrac{theta}{2}.mathbf{i}+w_{2}.sinfrac{theta}{2}.mathbf{j}+w_{3}.sinfrac{theta}{2}.mathbf{k}$



It is not equal! I am left with $-cosfrac{theta}{2}$ which is not matching. Please tell me how to fix this






quaternions







share|cite|improve this question









$endgroup$












  • $begingroup$
    You're told that there are two ways to encode the rotation, and you're trying to prove the two ways are one. What did you expect :)
    $endgroup$
    – rschwieb
    Jan 28 at 11:53










  • $begingroup$
    Related.
    $endgroup$
    – Jyrki Lahtonen
    Jan 31 at 18:48














2












2








2





$begingroup$


I am given the following question: I need to prove that two quaternions map to the same Rotation Matrix in SO3 Space. It is demonstrated by this image:



enter image description here



Let w be v here.
I tried to work out the proof, but it isn't coming out correctly:



$Q=(cosfrac{theta}{2},w.sinfrac{theta}{2})$



$Q=cosfrac{theta}{2}+w_{1}.sinfrac{theta}{2}.mathbf{i}+w_{2}.sinfrac{theta}{2}.mathbf{j}+w_{3}.sinfrac{theta}{2}.mathbf{k}$



As per image, replace $theta$
with $2pi-theta$
and w with -w



$Q=cosfrac{2pi-theta}{2}+-w_{1}.sinfrac{2pi-theta}{2}.mathbf{i}+-w_{2}.sinfrac{2pi-theta}{2}.mathbf{j}+-w_{3}.sinfrac{2pi-theta}{2}.mathbf{k}$



$Q=-cosfrac{theta}{2}+w_{1}.sinfrac{theta}{2}.mathbf{i}+w_{2}.sinfrac{theta}{2}.mathbf{j}+w_{3}.sinfrac{theta}{2}.mathbf{k}$



It is not equal! I am left with $-cosfrac{theta}{2}$ which is not matching. Please tell me how to fix this






quaternions







share|cite|improve this question









$endgroup$




I am given the following question: I need to prove that two quaternions map to the same Rotation Matrix in SO3 Space. It is demonstrated by this image:



enter image description here



Let w be v here.
I tried to work out the proof, but it isn't coming out correctly:



$Q=(cosfrac{theta}{2},w.sinfrac{theta}{2})$



$Q=cosfrac{theta}{2}+w_{1}.sinfrac{theta}{2}.mathbf{i}+w_{2}.sinfrac{theta}{2}.mathbf{j}+w_{3}.sinfrac{theta}{2}.mathbf{k}$



As per image, replace $theta$
with $2pi-theta$
and w with -w



$Q=cosfrac{2pi-theta}{2}+-w_{1}.sinfrac{2pi-theta}{2}.mathbf{i}+-w_{2}.sinfrac{2pi-theta}{2}.mathbf{j}+-w_{3}.sinfrac{2pi-theta}{2}.mathbf{k}$



$Q=-cosfrac{theta}{2}+w_{1}.sinfrac{theta}{2}.mathbf{i}+w_{2}.sinfrac{theta}{2}.mathbf{j}+w_{3}.sinfrac{theta}{2}.mathbf{k}$



It is not equal! I am left with $-cosfrac{theta}{2}$ which is not matching. Please tell me how to fix this






quaternions




quaternions






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Jan 28 at 3:13









user1436508user1436508

1688




1688












  • $begingroup$
    You're told that there are two ways to encode the rotation, and you're trying to prove the two ways are one. What did you expect :)
    $endgroup$
    – rschwieb
    Jan 28 at 11:53










  • $begingroup$
    Related.
    $endgroup$
    – Jyrki Lahtonen
    Jan 31 at 18:48


















  • $begingroup$
    You're told that there are two ways to encode the rotation, and you're trying to prove the two ways are one. What did you expect :)
    $endgroup$
    – rschwieb
    Jan 28 at 11:53










  • $begingroup$
    Related.
    $endgroup$
    – Jyrki Lahtonen
    Jan 31 at 18:48
















$begingroup$
You're told that there are two ways to encode the rotation, and you're trying to prove the two ways are one. What did you expect :)
$endgroup$
– rschwieb
Jan 28 at 11:53




$begingroup$
You're told that there are two ways to encode the rotation, and you're trying to prove the two ways are one. What did you expect :)
$endgroup$
– rschwieb
Jan 28 at 11:53












$begingroup$
Related.
$endgroup$
– Jyrki Lahtonen
Jan 31 at 18:48




$begingroup$
Related.
$endgroup$
– Jyrki Lahtonen
Jan 31 at 18:48










1 Answer
1






active

oldest

votes


















2












$begingroup$

Since $sin(pi-theta/2)$ equals $sin(theta/2)$, not $-sin(theta/2)$, you're missing minus signs.



Note $cos(theta/2)+sin(theta/2)mathbf{w}$ is expressible as $Q=exp(frac{1}{2}thetamathbf{w})$ (assuming $|mathbf{w}|=1$).



Replacing $thetamapsto2pi-theta$ and $mathbf{w}mapsto -mathbf{w}$ yields ${bfcolor{Red}{-}}Q$, not $Q$; you shouldn't be expecting the original quaternion in the first place. Both unit quaternions $Q$ and $-Q$ represent the same 3D rotation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    youre right, so now I get -Q. But how do i actually show that Q and -Q represent the same rotation. i guess physically it works and i can visualize, but how to show mathematically
    $endgroup$
    – user1436508
    Jan 29 at 16:39












  • $begingroup$
    Do you know how quaternions are used to represent 3D rotations? Given a quaternion $q$, with polar form $q=exp(frac{1}{2}theta{bf w})$, and a 3D vector $bf x$ (in other words, a purely imaginary quaternion), the conjugation $q{bf x}q^{-1}$ has the effect of rotating $bf x$ around $bf w$ by the angle $theta$. Since $q{bf x}q^{-1}=(-q){bf x}(-q)^{-1}$ for all $bf x$, we conclude $q$ and $-q$ represent the same 3D rotation.
    $endgroup$
    – arctic tern
    Jan 31 at 2:41











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1 Answer
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1 Answer
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active

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active

oldest

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active

oldest

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2












$begingroup$

Since $sin(pi-theta/2)$ equals $sin(theta/2)$, not $-sin(theta/2)$, you're missing minus signs.



Note $cos(theta/2)+sin(theta/2)mathbf{w}$ is expressible as $Q=exp(frac{1}{2}thetamathbf{w})$ (assuming $|mathbf{w}|=1$).



Replacing $thetamapsto2pi-theta$ and $mathbf{w}mapsto -mathbf{w}$ yields ${bfcolor{Red}{-}}Q$, not $Q$; you shouldn't be expecting the original quaternion in the first place. Both unit quaternions $Q$ and $-Q$ represent the same 3D rotation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    youre right, so now I get -Q. But how do i actually show that Q and -Q represent the same rotation. i guess physically it works and i can visualize, but how to show mathematically
    $endgroup$
    – user1436508
    Jan 29 at 16:39












  • $begingroup$
    Do you know how quaternions are used to represent 3D rotations? Given a quaternion $q$, with polar form $q=exp(frac{1}{2}theta{bf w})$, and a 3D vector $bf x$ (in other words, a purely imaginary quaternion), the conjugation $q{bf x}q^{-1}$ has the effect of rotating $bf x$ around $bf w$ by the angle $theta$. Since $q{bf x}q^{-1}=(-q){bf x}(-q)^{-1}$ for all $bf x$, we conclude $q$ and $-q$ represent the same 3D rotation.
    $endgroup$
    – arctic tern
    Jan 31 at 2:41
















2












$begingroup$

Since $sin(pi-theta/2)$ equals $sin(theta/2)$, not $-sin(theta/2)$, you're missing minus signs.



Note $cos(theta/2)+sin(theta/2)mathbf{w}$ is expressible as $Q=exp(frac{1}{2}thetamathbf{w})$ (assuming $|mathbf{w}|=1$).



Replacing $thetamapsto2pi-theta$ and $mathbf{w}mapsto -mathbf{w}$ yields ${bfcolor{Red}{-}}Q$, not $Q$; you shouldn't be expecting the original quaternion in the first place. Both unit quaternions $Q$ and $-Q$ represent the same 3D rotation.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    youre right, so now I get -Q. But how do i actually show that Q and -Q represent the same rotation. i guess physically it works and i can visualize, but how to show mathematically
    $endgroup$
    – user1436508
    Jan 29 at 16:39












  • $begingroup$
    Do you know how quaternions are used to represent 3D rotations? Given a quaternion $q$, with polar form $q=exp(frac{1}{2}theta{bf w})$, and a 3D vector $bf x$ (in other words, a purely imaginary quaternion), the conjugation $q{bf x}q^{-1}$ has the effect of rotating $bf x$ around $bf w$ by the angle $theta$. Since $q{bf x}q^{-1}=(-q){bf x}(-q)^{-1}$ for all $bf x$, we conclude $q$ and $-q$ represent the same 3D rotation.
    $endgroup$
    – arctic tern
    Jan 31 at 2:41














2












2








2





$begingroup$

Since $sin(pi-theta/2)$ equals $sin(theta/2)$, not $-sin(theta/2)$, you're missing minus signs.



Note $cos(theta/2)+sin(theta/2)mathbf{w}$ is expressible as $Q=exp(frac{1}{2}thetamathbf{w})$ (assuming $|mathbf{w}|=1$).



Replacing $thetamapsto2pi-theta$ and $mathbf{w}mapsto -mathbf{w}$ yields ${bfcolor{Red}{-}}Q$, not $Q$; you shouldn't be expecting the original quaternion in the first place. Both unit quaternions $Q$ and $-Q$ represent the same 3D rotation.






share|cite|improve this answer









$endgroup$



Since $sin(pi-theta/2)$ equals $sin(theta/2)$, not $-sin(theta/2)$, you're missing minus signs.



Note $cos(theta/2)+sin(theta/2)mathbf{w}$ is expressible as $Q=exp(frac{1}{2}thetamathbf{w})$ (assuming $|mathbf{w}|=1$).



Replacing $thetamapsto2pi-theta$ and $mathbf{w}mapsto -mathbf{w}$ yields ${bfcolor{Red}{-}}Q$, not $Q$; you shouldn't be expecting the original quaternion in the first place. Both unit quaternions $Q$ and $-Q$ represent the same 3D rotation.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 28 at 4:21









arctic ternarctic tern

12k31536




12k31536












  • $begingroup$
    youre right, so now I get -Q. But how do i actually show that Q and -Q represent the same rotation. i guess physically it works and i can visualize, but how to show mathematically
    $endgroup$
    – user1436508
    Jan 29 at 16:39












  • $begingroup$
    Do you know how quaternions are used to represent 3D rotations? Given a quaternion $q$, with polar form $q=exp(frac{1}{2}theta{bf w})$, and a 3D vector $bf x$ (in other words, a purely imaginary quaternion), the conjugation $q{bf x}q^{-1}$ has the effect of rotating $bf x$ around $bf w$ by the angle $theta$. Since $q{bf x}q^{-1}=(-q){bf x}(-q)^{-1}$ for all $bf x$, we conclude $q$ and $-q$ represent the same 3D rotation.
    $endgroup$
    – arctic tern
    Jan 31 at 2:41


















  • $begingroup$
    youre right, so now I get -Q. But how do i actually show that Q and -Q represent the same rotation. i guess physically it works and i can visualize, but how to show mathematically
    $endgroup$
    – user1436508
    Jan 29 at 16:39












  • $begingroup$
    Do you know how quaternions are used to represent 3D rotations? Given a quaternion $q$, with polar form $q=exp(frac{1}{2}theta{bf w})$, and a 3D vector $bf x$ (in other words, a purely imaginary quaternion), the conjugation $q{bf x}q^{-1}$ has the effect of rotating $bf x$ around $bf w$ by the angle $theta$. Since $q{bf x}q^{-1}=(-q){bf x}(-q)^{-1}$ for all $bf x$, we conclude $q$ and $-q$ represent the same 3D rotation.
    $endgroup$
    – arctic tern
    Jan 31 at 2:41
















$begingroup$
youre right, so now I get -Q. But how do i actually show that Q and -Q represent the same rotation. i guess physically it works and i can visualize, but how to show mathematically
$endgroup$
– user1436508
Jan 29 at 16:39






$begingroup$
youre right, so now I get -Q. But how do i actually show that Q and -Q represent the same rotation. i guess physically it works and i can visualize, but how to show mathematically
$endgroup$
– user1436508
Jan 29 at 16:39














$begingroup$
Do you know how quaternions are used to represent 3D rotations? Given a quaternion $q$, with polar form $q=exp(frac{1}{2}theta{bf w})$, and a 3D vector $bf x$ (in other words, a purely imaginary quaternion), the conjugation $q{bf x}q^{-1}$ has the effect of rotating $bf x$ around $bf w$ by the angle $theta$. Since $q{bf x}q^{-1}=(-q){bf x}(-q)^{-1}$ for all $bf x$, we conclude $q$ and $-q$ represent the same 3D rotation.
$endgroup$
– arctic tern
Jan 31 at 2:41




$begingroup$
Do you know how quaternions are used to represent 3D rotations? Given a quaternion $q$, with polar form $q=exp(frac{1}{2}theta{bf w})$, and a 3D vector $bf x$ (in other words, a purely imaginary quaternion), the conjugation $q{bf x}q^{-1}$ has the effect of rotating $bf x$ around $bf w$ by the angle $theta$. Since $q{bf x}q^{-1}=(-q){bf x}(-q)^{-1}$ for all $bf x$, we conclude $q$ and $-q$ represent the same 3D rotation.
$endgroup$
– arctic tern
Jan 31 at 2:41


















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