Suppose there are $N$ particles embedded in a rigid body which undergoes some random rotation such that:

$$ \overline{\overline {R}}_{ij} \otimes \vec{a}_{ij} = \vec{b}_{ij}$$


  • $i$ and $j$ are just two random particles out of the $N$ particles
  • $\vec{a}_{ij}$ is a vector connecting the $i^{th}$ and the $j^{th}$ particle
  • $\vec{b}_{ij}$ is the rotated version of $\vec{a}_{ij}$
  • $\overline{\overline {R}}_{ij}$ is the rotation tensor responsble for the rotaion of $\vec{a}_{ij}$ to $ \vec{b}_{ij}$

I intend to prove that $\overline{\overline {R}}_{ij}$ remains the same for all the combinations of $i$ and $j$ belonging to the set $[1,N]$ for $(i \neq j)$ using the most fundamental definition of a rigid body that it is a collection of particles which remains equidistant from each other.

An Update on my effort:

Using the definition of a rigid body - the distance between any two particles remains constant, we can also say that - the dot product of any two vectors joining the particles embedded in the rigid body remains the same before and after the rotation.

So, taking into account the $i^{th}, j^{th}, k^{th}$ and the $l^{th} $ particle, we can write -

the dot product of the vectors before the rotation = the dot product of the two vectors after rotation

$$ \vec{a}_{ji} \cdot \vec{a}_{lk}=(\overline{\overline {R}}_{ji} \otimes \vec{a}_{ji}) \cdot (\overline{\overline {R}}_{lk} \otimes \vec{a}_{lk})$$

Now, using the canonical isomorphism property of the tensor product: $$ \vec{x} \cdot (\overline{\overline Z} \otimes \vec{y}) = \vec{y} \cdot (\overline{\overline {Z}}^T \otimes \vec{x})$$

We have,

$$ \vec{a}_{ji} \cdot \vec{a}_{lk} = \vec{a}_{lk} \cdot [\overline{\overline {R}}^T_{lk} \otimes (\overline{\overline {R}}_{ji} \otimes \vec{a}_{ji}) ]$$

Using the associative property of tensor product:

$$\vec{a}_{ji} \cdot \vec{a}_{lk} = \vec{a}_{lk} \cdot (\overline{\overline {R}}^T_{lk} \otimes \overline{\overline {R}}_{ji}) \otimes \vec{a}_{ji} $$

Then again using the canonical isomorphism property, we have:

$$ \vec{a}_{ji} \cdot \vec{a}_{lk} = \vec{a}_{ji} \cdot (\overline{\overline {R}}_{lk} \otimes \overline{\overline {R}}^T_{ji}) \otimes \vec{a}_{lk} $$

Which implies, $$\vec{a}_{ji} \cdot \{ [\overline{\overline {1}} - (\overline{\overline {R}}_{lk} \otimes \overline{\overline {R}}^T_{ji})] \otimes \vec{a}_{lk} \}=0$$

Now, given the fact that the chosen vectors, $\vec{a}_{ji}$ and $\vec{a}_{lk}$, are non zero, the left hand side could only become zero under the following conditions:

  1. Either the last tensor product is zero
  2. Or the first dot product is zero
  3. Or the tensor, $[\overline{\overline {1}} - (\overline{\overline {R}}_{lk} \otimes \overline{\overline {R}}^T_{ji})]$, itself is zero
  4. Or any combination of the above three mentioned conditions are true

Now if somehow I prove that the first, the second and the fourth points are wrong, I'll be left with only one possibility that, $[\overline{\overline {1}} - (\overline{\overline {R}}_{lk} \otimes \overline{\overline {R}}^T_{ji})]=0$, proving which also happens to be my primary objective right now. And this is where I needed help. If what I asked for is proved, the rest of the task is pretty much tractable. Thank you.

  • $\begingroup$ Hint: Pick any third particle (k). Add the vectors between any two to yield the vector between 1st and 3rd. Now assume that rotation tensor is not unique for 1-3 vector. This should lead to something other than summation of all 3 vectors in rotated frame. Hence rotation matrix is unique. $\endgroup$ Jan 31, 2016 at 14:09
  • $\begingroup$ @ Sir Lewis Miller I will first write down what I think you meant. Correct me if I am wrong. Initially: $$ \vec{a}_{jk} + \vec{a}_{ij} = \vec{a}_{ki} $$ After rotation: $$ \overline{\overline {R}}_{jk} \otimes \vec{a}_{jk} + \overline{\overline {R}}_{ij} \otimes \vec{a}_{ij} = \overline{\overline {R}}_{ki} \otimes \vec{a}_{ki} $$ which also implies: $$ \vec{b}_{jk} + \vec{b}_{ij} = \vec{b}_{ki} $$ $\endgroup$
    – Dan
    Jan 31, 2016 at 18:39
  • $\begingroup$ But, sir, how do the above equations imply that: $\overline{\overline {R}}_{jk}=\overline{\overline {R}}_{ij}=\overline{\overline {R}}_{ki}$ ? I think, to prove all this, we'll also have to use the fact that the distance between the particles remains constant after rotation (to enforce the constraint imposed by the rigidity of the body). $\endgroup$
    – Dan
    Jan 31, 2016 at 18:39
  • $\begingroup$ The answer is elementary: by definition! A 3D rigid body is by definition a body such that there is a reference frame always at rest with it. If you describe the position of the body using another reference frame, at rest with the laboratory, the only rotation which enters the game is the one connecting the laboratory frame with the rest frame, only one rotation for all particles of the body. $\endgroup$ Jan 31, 2016 at 20:39
  • 1
    $\begingroup$ You are really asking, Why are rigid body rotations orthonormal such that $R^{-1} = R^\intercal$. This is the point you are stuck at, and you need to show this in the title, otherwise it confuses the reader. $\endgroup$ Feb 4, 2016 at 14:40

1 Answer 1


Your Problem

If you're trying to prove that all $R$ must be equal in order for all the distances to remain the same after rotation, you will fail, as it's not true. Rotation $R_{ij}$ could be replaced by any other rotation that only differs by a rotation about $a_{ij}$ for example say you have points $i=\langle 0,0,0 \rangle$ $j=\langle 1,0,0 \rangle$ and all $R$ were $\begin{bmatrix} 0 & 1 & 0 \\ -1 & 0 & 0 \\ 0 & 0 & 1 \end{bmatrix}$ then $R_{ij}$ could be any matrix such that $R_{ij} \otimes \langle 1,0,0 \rangle = \langle -1,0,0 \rangle$ and any matrix of the form $\begin{bmatrix} 0 & a & b \\ -1 & c & d \\ 0 & e & f \end{bmatrix}$ would suffice. This is a trivial example to demonstrate the point but the same concept it true in general.

What I think you're actually interested in

Prove that for a set of points with coordinates $p_i$ which are arbitrarily displaced to coordinates $p'_i$ that if the distances between the each pair of points remains constant across the displacement, that there exists a unique $R$ such that $R\otimes p_i=p'_i$

And to get you started with the proof you can use $|p_i-p_j|=|p'_i-p'_j|$ as the distance constraint, which can be rewritten as $(p_i-p_j)\cdot(p_i-p_j)=(p'_i-p'_j)\cdot(p'_i-p'_j)$

One thing that you will undoubtedly run into, is that the number of points matters. You must have at least $n\frac{n-1}2$ points, where n is the number of dimensions you have, for the unique constraint to hold true.

  • 1
    $\begingroup$ This question and this answer demonstrate that $n(n-1)/2$ is the optimal number. $\endgroup$
    – user154997
    Jun 17, 2017 at 23:08

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