On the orthogonality of the rotation matrix

[brotherbobby]

Homework Statement

Show how the rotation matrix is orthogonal in three dimensional Euclidean space $E_3$ when it acts on vectors. Remember that rotation should preserve the length of the vector.

Relevant Equations

If the vector $\mathbf x = x_i \hat e_i$ is acted on by a rotation matrix $\mathbb {R}$, we obtain a different (rotated) vector $\mathbf x' = x'_i \hat e_i$, where $\boxed{x'_i = R_{ij} x_j}$, $R_{ij}$'s being the components of the rotation matrix $\mathbb {R}$.

The length of a vector $|\mathbf{x}|^2 = x_i x_i$.

Let me start with the rotated vector components : $x'_i = R_{ij} x_j$. The length of the rotated vector squared : $x'_i x'_i = R_{ij} x_j R_{ik} x_k$. For this (squared) length to be invariant, we must have $R_{ij} x_j R_{ik} x_k = R_{ij} R_{ik} x_j x_k = x_l x_l$.

If the rotation matrix components supported the relation $\boxed{R_{ij} R_{ik} = \delta_{jk}}$, we find that the above equation would hold good, $l$ being a dummy variable which can be replaced by $j$ or $k$.

However, I have proved sufficiency : Given that $R_{ij} R_{ik} = \delta_{jk}$, the length of a vector remains unchanged.

I am stuck as to the necessity : If the length of a vector is given to be unchanged, show how $\boxed{R_{ij} R_{ik} = \delta_{jk}}$.

A help as to prove the necessary condition would be welcome.

 

[George Jones]

we must have $R_{ij} x_j R_{ik} x_k = R_{ij} R_{ik} x_j x_k = x_l x_l$.

And, e.g., $x_l = \delta_{jl} x_j$.

 

[vanhees71]

For necessity note that the condition that $\vec{x}^2$ is unschanged must hold for all vectors $\vec{x}$. What can you conclude for $(\vec{x}+\vec{y})^2$ where $\vec{x}$ and $\vec{y}$ are arbitrary vectors?

 

[brotherbobby]

For necessity note that the condition that $\vec{x}^2$ is unschanged must hold for all vectors $\vec{x}$. What can you conclude for $(\vec{x}+\vec{y})^2$ where $\vec{x}$ and $\vec{y}$ are arbitrary vectors?

Let me write out the equations as you put it.

As the arbitrary vector $\vec x$ would have its (squared) length unchanged, we can say that $\left( \vec x + \vec y \right)^2 = \left( \vec x + \vec y \right)^2 = \left( \vec x' + \vec y' \right)^2 = \left( \rm R \vec x + \rm R \vec y \right)^2 \Rightarrow \vec x^2 + \vec y^2 + 2 \vec x \cdot \vec y = (\rm R \vec x)^2 + (\rm R \vec y)^2 + 2 R_{ij} R_{ik} x_i x_j$.

For this to be valid, seeing the last term, we have $\boxed{R_{ij} R_{ik} = \delta_{jk}}$.

Thank you very much.

Please let me know if I have been correct when you have the time.

 

[vanhees71]

Of course the last term in your long equation should be $2 R_{ij} R_{ik} x_i y_j$. Then, since $(R \vec{x})^2=\vec{x}^2$ and $(R\vec{y})^2=\vec{y}^2$ you have from your equation necessarily $R_{ij} R_{ik} x_j y_k=\delta_{jk} x_j x_k$ which means, since this has to hold for any $\vec{x}$ and $\vec{y}$ that $R_{ij} R_{ik}=\delta_{jk}$. In matrix notation this reads $R^{\text{T}} R=1$, where $R^{\text{T}}$ is the transposed matrix, i.e., writing the columns of $R$ as the rows of $R^{\text{T}}$.