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I am reading a book by Arfken and Weber (Mathematical methods for physicists), in the section regarding rotations in $\mathbb{R}^3$. They express the elements of a rotation matrix in Cartesian coordinates, in terms of unit vectors $ S_{\mu\nu} = \hat{e^\prime}_\mu \cdot \hat{e}_{\nu} $. It says that because the dot product is a projection of the prime coordinate into the unprimed, it is therefore the change in $x_\nu$ produced by a change in $x^\prime_\mu$, so $$\hat{e^\prime}_\mu \cdot \hat{e}_{\nu} = \frac{\partial x_\nu}{\partial x_\mu'}$$

I am really struggling to see why they are equal in a more intuitive way, and with certainty.

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    $\begingroup$ expressed as a differential form A differential form is not the same thing as a partial derivative. $\endgroup$
    – Ghoster
    May 24, 2023 at 17:42
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    $\begingroup$ Shouldn’t there be a prime on one of the $x$’s in the partial derivative? $\endgroup$
    – Ghoster
    May 24, 2023 at 17:44

2 Answers 2

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I would imagine that the simplest way to show this is to note that the position vector $\mathbf x$ can be expressed in either basis:

$$x'^j \hat e_j' = \mathbf x = x^i \hat e_i$$ A given set of Cartesian basis vectors don't change with position, and they're orthonormal. Differentiate both sides with respect to $x^k$ and dot both sides with $\hat e'_\ell$ and the result follows immediately.

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First observe, that for any matrix, we can pick up its $i,j$ entry by applying it first to a column vector that is zero apart from $1$ at its $j$'th position and then dotting it into a similar vector that has $1$ only at its $i$'th position (equivalently, applying from the left such a row vector). More explicitly, if we have a square $n \times n$ matrix $\mathbf{M}$ and two $1 \times n$ vectors:

$$\hat{e_i} = (0,\dots,0,\underbrace{1}_{\text{i'th position}},0,\dots,0)$$ $$\hat{e_j} = (0,\dots,0,\underbrace{1}_{\text{j'th position}},0,\dots,0)$$

Then:

$$M_{ij} = \hat{e_i}\mathbf{M}\hat{e_j}{}^T$$


The transformation that relates an unprimed coordinate system $(x_1,x_2,x_3)$ with a primed one that's rotated w.r.t it by angle $\phi$ is:

\begin{align*} x_1'(x_1,x_2,x_3) &= x_1\cos\phi -x_2\sin\phi\tag{1} \\ x_2'(x_1,x_2,x_3) &= x_1\sin\phi + x_2\cos\phi\tag{2} \\ x_3'(x_1,x_2,x_3) &= x_3 \tag{3}\end{align*}

We can also represent it via a $3\times 3$ matrix:

$$\mathbf{R}(\phi) = \left[\begin{array}{2}\cos\phi&-\sin\phi&0\\\sin\phi&\cos\phi&0\\0&0&1\end{array}\right]$$

Now if we have the usual unit basis vectors for the unprimed coordinates:

\begin{align*} \hat{e_1} &= (1,0,0) \\ \hat{e_2} &= (0,1,0) \\ \hat{e_3} &= (0,0,1) \end{align*}

We can apply the above transformation $\mathbf{R}$ to each of them, to find the corresponding $\hat{e_1}'\ , \hat{e_2}'\ , \hat{e_3}'$, and we will get:

\begin{align*} \hat{e_1}' &= \mathbf{R}(\phi)\hat{e_1} = (\cos\phi,\sin\phi,0) \\ \hat{e_2}' &= \mathbf{R}(\phi)\hat{e_2} = (-\sin\phi,\cos\phi,0) \\ \hat{e_3}' &= \mathbf{R}(\phi)\hat{e_3} = (0,0,1) \end{align*}

Considering our initial observation, we also see then that indeed, if we dot between any $\hat{e_j}'$ and any $\hat{e_i}$ we get:

$$\hat{e_i} \cdot \hat{e_j}' = \hat{e_i} \cdot (\mathbf{R}(\phi)\hat{e_j}) = R_{ij}$$

In addition, the statement that:

$$\hat{e_i} \cdot \hat{e_j}' = \frac{\partial{x_i'}}{\partial{x_j}}$$

Can be explained by the fact that the transformation is linear, hence, when you partially derive any of the relations $(1),(2),(3)$ with respect to one of their dependent variables $x_1,x_2,x_3$, you pick up only the coefficient of this variable. This characterizes all linear transformations.

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