How Is the Rodrigues Rotation Formula Derived in Tensor Notation?

[wood0595]

I'm curious as to how the following proof is verified. I have toiled over this thing for quite a while, but haven't made any progress. I don't need a step-by-step solution, but I would appreciate any help getting it started:

Given the following:

Q is an orthogonal tensor
e₁ is a unit vector such that Qe₁ = e₁
n is of unit length and orthogonal to e₁
m is such that m$$\bullet$$e₁ = 0, m = Qn and m$$\bullet$$n= cos$$\theta$$
1 is the identity matrix

Prove that:

Q = 1cos$$\theta$$ + (1-cos$$\theta$$)e₁$$\otimes$$e₁ - (e₂$$\otimes$$e₃-e₃$$\otimes$$e₂)sin$$\theta$$

Where e₁, e₂, and e₃ form an orthonormal triad and
$$\otimes$$ represents the dyadic product

My apologies if this is the wrong thread to post this at, but I looked around and couldn't find any threads on this topic.

 

[mighty2000]

Thank you for your question and interest in this topic. The proof that you are asking about is known as the Rodrigues rotation formula and is commonly used in the field of mechanics and physics. To verify this formula, we can start by expanding the given equation m = Qn as follows:

m = Qn = Q(e1\crossn) = Q(e1\cross(Qe1)) = Q(e1\cross(e1)) = 0

This shows that m is indeed orthogonal to both e1 and n. Next, we can use the fact that Q is an orthogonal tensor to show that Q preserves the length of vectors. This means that ||m|| = ||n|| = 1. Using this information, we can rewrite the given equation m\bulletn = cos\theta as follows:

m\bulletn = cos\theta = ||m||||n||cos\theta = 1\cdot1\cdotcos\theta = cos\theta

Now, let's consider the expansion of Qe1:

Qe1 = (1cos\theta + (1-cos\theta)e1\otimese1 - (e2\otimese3-e3\otimese2)sin\theta)e1
= e1cos\theta + (e1\bullet e1)(1-cos\theta)e1 - (e1\bullet e2)(e2\otimese3-e3\otimese2)sin\theta
= e1cos\theta + (1-cos\theta)e1 - (e1\bullet e2)(e2\otimese3-e3\otimese2)sin\theta
= e1cos\theta + (1-cos\theta)e1 - (e1\bullet Qn)(Qe2\otimes Qe3-Qe3\otimes Qe2)sin\theta
= e1cos\theta + (1-cos\theta)e1 - (e1\bullet m)(Qe2\otimes Qe3-Qe3\otimes Qe2)sin\theta
= e1cos\theta + (1-cos\theta)e1 - 0(Qe2\otimes Qe3-Qe3\otimes Qe2)sin\theta
= e1cos\theta + (1-cos\