Can someone explain how these angles are equivalent?

Question

Linked here is a great tutorial I've been reading on contravariant and covariant representations of vectors. I'm following along well up until page 5 and 6. Here, I'd like to know why in Figure 4.7(a), the angle created between $\bf r$ and the new axis $x_1$ is equivalent to $\alpha$.

Further, on page 6, it's stated that, from Figure 4.7, $x_1 = x^1 + x^2\cos\alpha$ and $x_2 = x^2 + x^1\cos\alpha$. Can someone break those assumptions down?

Answer 1

I'd like to know why in Figure 4.7(a), the angle created between $\bf r$ and the new axis $x_1$ is equivalent to $\alpha$.

It hinges on the fact that ${\bf a}^1 \perp {\bf a}_2$. If you know that, then all you need to know is that all the angles in a triangle sum to $180º$ in order to conclude that the two angles $\alpha$ are in fact congruent.

Remember that the goal was to devise a coordinate system in which ${\bf e}^1\perp{\bf a}_2$ (as stated in the first sentence on p. 4-5). Looking at the diagram, it appears that the coordinate system was designed such that that ${\bf a}^1\parallel{\bf e}^1$, which therefore implies that ${\bf a}^1 \perp {\bf a}_2$.

In addition, Holo describes in the comment section a way to soundly conclude that ${\bf a}^1 \perp {\bf a}_2$:

We know that ${\bf a}^1 \perp {\bf a}_2$. Look at the right part of the image, we know that $\bf{L_2e^2}\perp \bf x^1\implies {\bf a}^2 \perp {\bf a}_1$. We can show that this is the same from the other side of $\bf r$ and conclude that ${\bf a}^1 \perp {\bf a}_2$.

The diagram isn’t labeled for a geometer, so I will try my best to reference the image. For this, let $\rm O$ be the origin, let hats indicate angles and let me label “points” with the nearest symbol:

Standard trig gives us these:

$$\widehat{\mathrm{O}x_1\mathbf{L}_1} = 90º$$ $$\widehat{\mathrm{O}\mathbf{L}_1x_1} = \alpha$$ $$\widehat{\mathbf{a}_1\mathrm{O}\mathbf{a}^1} = 90º-\alpha$$

Our conclusion that ${\bf a}^1 \perp {\bf a}_2$ gives us

$$\widehat{\mathbf{a}^1\mathrm{O}\mathbf{a}_2}=90º$$

We are simply left to deduce that

$$\widehat{\mathbf{a}_1\mathrm{O}\mathbf{a}_2} = \widehat{\mathbf{a}^1\mathrm{O}\mathbf{a}_2} - \widehat{\mathbf{a}_1\mathrm{O}\mathbf{a}^1} = 90º - \left(90º-\alpha\right) = \alpha$$

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Further, on page 6, it's stated that, from Figure 4.7, $x_1 = x^1 + x^2\cos\alpha$ and $x_2 = x^2 + x^1\cos\alpha$.

In regard to your second concern, redirect your attention to figure 4.4:

You can definitely see how, looking at the bottom-right right triangle and remembering that we’re dealing with parallel lines,

$$\begin{align} (\text{adjacent leg}) &= (\text{hypotenuse})\cos\alpha \\ &= x^2 \cos \alpha \end{align}$$

which you then add to $x^1$ to get $x_1$. The same logic applies in calculating $x_2$.

Does that clear things up for you?