How Do You Multiply Quaternions Correctly?

Click For Summary
SUMMARY

The discussion focuses on the correct multiplication of quaternions, specifically the expression exp(iψ/2)exp(kθ/2)exp(iф/2). The user attempts to demonstrate that this expression simplifies to cos(θ/2)exp(i[ψ+ф]/2) + ksin(θ/2)exp(i[ψ-ф]/2). Through detailed calculations, the user confirms that the final term should indeed be ψ-ф, contrary to the paper's assertion. The conversation emphasizes the importance of understanding quaternion algebra and the commutative properties of scalar terms.

PREREQUISITES
  • Understanding of quaternion algebra
  • Familiarity with exponential functions in complex numbers
  • Knowledge of trigonometric identities
  • Experience with mathematical proofs and simplifications
NEXT STEPS
  • Study quaternion multiplication rules and properties
  • Learn about the geometric interpretation of quaternions
  • Explore the applications of quaternions in 3D graphics and rotations
  • Investigate the relationship between quaternions and Euler angles
USEFUL FOR

Mathematicians, computer graphics developers, and anyone interested in advanced algebraic structures and their applications in 3D modeling and physics simulations.

amiras
Messages
64
Reaction score
0

Homework Statement


Its not really homework problem, and you may find it silly because its only multiplication problem, but I cannot get the right answer by multiplying quaternions.

Basically this is what i want to show:

exp(iψ/2)exp(kθ/2)exp(iф/2) = cos(θ/2)exp(i[ψ+ф]/2) + ksin(θ/2)exp(i[ψ-ф]/2)


Homework Equations





The Attempt at a Solution



I begin writing:
exp(kθ/2) = cos(θ/2) + ksin(θ/2)

Then multiplying:
exp(iψ/2)*[cos(θ/2) + ksin(θ/2)] = exp(iψ/2)cos(θ/2) + exp(iψ/2)*ksin(θ/2) =
= cos(θ/2)exp(iψ/2) + sin(θ/2) exp(iψ/2)*k

Since scalar terms can are commutative in quaternions algebra.

Finally multiplying answer above with the final exp(iф/2)

[cos(θ/2)exp(iψ/2) + sin(θ/2) exp(iψ/2)*k] * exp(iф/2) =
= cos(θ/2)exp(iψ/2)exp(iф/2) + sin(θ/2)exp(iψ/2)*k*exp(iф/2) =
= cos(θ/2)exp(i(ψ+ф)/2) + sin(θ/2)*k*exp(-iψ/2)*exp(iф/2) =
= cos(θ/2)exp(i(ψ+ф)/2) + k*sin(θ/2)*exp(i(ф-ψ)/2)

Here I used that exp(iψ/2)*k = k*exp(-iψ/2)

And no matter how I do it I always get the same answer, with the last exponential term having ф-ψ, and the paper says it should be ψ-ф
 
Physics news on Phys.org
hi amiras! :smile:

yours looks correct to me …

exp(iψ/2)exp(kθ/2)exp(iф/2)

= exp(iψ/2)cos(θ/2)exp(iф/2) + exp(iψ/2)ksin(θ/2)exp(iф/2)

= cos(θ/2)exp(i(ψ+ф)/2) + kexp(-iψ/2)sin(θ/2)exp(iф/2)
 
Thanks for confirming me, from now on I'l stop blindly following and start thinking by myself. :)
 

Similar threads

  • · Replies 13 ·
Replies
13
Views
13K
  • · Replies 3 ·
Replies
3
Views
2K
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 18 ·
Replies
18
Views
2K
  • · Replies 9 ·
Replies
9
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 1 ·
Replies
1
Views
6K