Understanding the cosine as a partial derivative.

Question

From the first chapter of Arfken's Mathematical Methods for physicists (rotation of the coordinate axis):

To go on to three and, later, four dimensions, we find it convenient to use a more compact notation. Let \begin{equation} x → x_1, \textbf{ } y → x_2 \end{equation}

\begin{equation} a_{11} = cos\phi,\textbf{ } a_{12} = sin\phi \end{equation}

\begin{equation} a_{21} = −sin\phi, \textbf{ } a_{22} = cos\phi \end{equation}

Then Eqs. become

\begin{equation} x′_1 = a_{11}x_1 + a_{12}x_2 \end{equation}

\begin{equation} x′_2 = a_{21}x_1 + a_{22}x_2. \end{equation}

The coefficient $a_{ij}$ may be interpreted as a direction cosine, the cosine of the angle between $x'_i$ and $x_j$ ; that is,

This is all good. Later, the book states:

From the definition of $a_{ij}$ as the cosine of the angle between the positive $x′_i$ direction and the positive $x_j$ direction we may write (Cartesian coordinates):

\begin{equation} a_{ij}=\frac{\partial x'_i}{\partial x_j} \end{equation}

Where did that come from? I can't find anything about the cosine as a partial derivative, and I don't see how that works.

Answer 1

You can verify-

$\frac{\partial x_1'}{\partial x_1}=a_{11},\frac{\partial x_1'}{\partial x_2}=a_{12},\frac{\partial x_2'}{\partial x_1}=a_{21}\ and\ \frac{\partial x_2'}{\partial x_2}=a_{22}$

Answer 2

When you choose to represent the system of equations as a sum, it looks like this:

$$ x^{'}_i = \sum_{j=1}^N a_{ij}x_j $$

For example, in the two-dimensional case (N=2), it follows that:

$$ x^{'}_1 = a_{11}x_1 + a_{12}x_2 $$ $$ x^{'}_2 = a_{21}x_1 + a_{22}x_2 $$

As you can see, both systems are functions of $x_1$ and $x_2$, so we can differentiate each one of those in order to obtain the required $a_{ij}$. Lets see one of them:

$$ \frac{\partial x^{'}_1}{\partial x_1} = \frac{\partial (a_{11}x_1 + a_{12}x_2)}{\partial x_1} = a_{11} $$

The same procedure is done for the rest of the $a_{ij}$ required cosines. Trying to generalize this, we can differentiate the sum with respect of $x_j$, it then looks as follow:

$$ \frac{\partial x^{'}_i}{\partial x_j} = a_{ij} $$

That's where it comes from :)