In the question defined in the following post:
i.e. the set of linear equations $\bar{x}=5x−2y$ and $\bar{y}=3x+2y$ which define a linear transformation, how can I understand the fact that the distance in the unbarred system between $P=(0,−1)$ and $Q=(2,1)$ is $\sqrt{8}$, and the distance between these same two points in the barred system is also $\sqrt{8}$ (using the formula with the $g_{ij}$, yet, when I plot $P$ and $Q$ in the barred system, the distance (via simple Pythagoras) is $\sqrt{136}$, not $\sqrt{8}$. This is called the "invariant" distance, so according to the numerical answer, the distance is the same in both coordinate systems ($\sqrt{8}$), but according to the points I plot in the two coordinate systems, i.e. the $xy$ graphs, the distance is different ($\sqrt{8}$ in the unbarred, and $\sqrt{136}$ in the barred).
This example comes from Schaum's Outlines for Tensor Calculus by David C. Kay, 1988, pages 11-12. He then has a similar "linear transformation" example on page 18, and when I plot the two points in that example in the barred and unbarred systems, the Pythagoras distance in both cases is $\sqrt{2}$. So both the "graphical method" and the formula with the $g_{ij}$ give the same answers.
Is the page 18 example just lucky? And I should not expect to be able to find the distance using Pythagoras in the barred coordinate system?
I wouldn't be so confused if both examples told me the same thing, i.e. that these distances are different in the two coordinate systems (when I try to plot the points in the barred system and use Pythagoras to "check the answer"), or that they're the same (plotting the points and using Pythagoras).
thanks, js.
We must be careful about what we mean when we call something "invariant". You could ask two of your friends to measure the height of your door. One friend could come back to you with the number $80$ while the other says $203$. Obviously one friend used imperial units while the other used SI (or something to this effect). Both measured a distance with the same physical meaning and thus want this quantity to be "invariant" or "intrinsic to the space".
Think about the above example like this, friend $\# 1$ broke the door down into a square grid where each "tick" represents an inch. Friend $\# 2$ broke the door down into a different square grid where the "ticks" are now centimeters and thus much closer together. We model this as the coordinate change
$$ \bar{x} = 2.54 x \text{ }$$ $$\bar{y} = 2.54 y\text{ .}$$
Simply counting the number of "ticks" we pass from one end of the door to the other clearly won't give us an invariant measure of the distance. This directly implies that the Pythagorean method (which is based on counting "ticks") will not work either for the case when distance doesn't follow coordinate curves. What you are doing is a slightly more generalized version of what went wrong in this simplified example. We will first fix this issue for a general Riemannian manifold and then apply this to your problem.
Given any metric $g_{ij}$ for a Riemannian manifold, we can ask what the distance is between any two simply-connected points $P$ and $Q$. Being mathematicians, we would first ask the question "what do we mean by distance?" We can clearly take any one of an infinite number of paths between $P$ and $Q$ and track the length of the path. The smart way (and potentially most obvious way) would be to compute the shortest distance between the points. We know that the shortest path is given parametrically by the solution to the geodesic equation
$$ \ddot{x}^k + \Gamma_{ij}^k \: \dot{x}^i \: \dot{x}^j = 0 \text{ .}$$
Now, all we have to do is take the solution to the above and compute the path length
$$S = \int_P^Q \sqrt{g_{ij} \: \dot{x}^i \: \dot{x}^j} \text{ d}t \text{ .}$$
Technically speaking, the geodesic equation can be very, very, very hard to solve for a general manifold. You however are working in an affine coordinate system so we can make some massively impactful simplifications. The metric we use for working in barred coordinates
$$ \bar{g}_{ij} = \frac{1}{16^2}\begin{bmatrix} 13&-11\\-11&29 \end{bmatrix} $$
is constant. We can immediately say that the Christoffel symbols vanish and that the geodesic equation
$$\begin{align} \ddot{\bar{x}} = 0\\\ddot{\bar{y}} = 0 \end{align}$$
has the straight-line solution
$$\begin{align} \bar{x} = b_1 + b_2 t&\\\bar{y} = c_1 + c_2 t &\text{ .} \end{align}$$
Simplifying further that $P$ corresponds with $t=0$ and $Q$ corresponds with $t=1$ we can say that
$$\begin{align} \bar{x} &= 2 + 6 t\\\bar{y} &= -2 + 10 t \text{ .} \end{align}$$
We can go further and say that $\dot{\bar{x}} = 6$ and $\dot{\bar{y}} = 10$ are both constant which implies that the path length $S$ is
$$ S = \int_0^1 \sqrt{g_{ij} \: \dot{\bar{x}}{}^i \: \dot{\bar{x}} {}^j} \text{ d}t = \sqrt{g_{ij} \: \dot{\bar{x}}{}^i \: \dot{\bar{x}} {}^j}\int_0^1 \text{ d}t=\sqrt{g_{ij} \: \dot{\bar{x}}{}^i \: \dot{\bar{x}} {}^j} \text{ .} $$
Doing the double sum, we see that $$\sqrt{g_{ij} \: \dot{\bar{x}}{}^i \: \dot{\bar{x}} {}^j} = \sqrt{8} \text{ .}$$
If you prefer matrix notation, we have
$$\sqrt{\mathbf{V}^\top \bar{\mathbf{g}}\mathbf{V}} = \sqrt{8}$$
$$\text{where}$$
$$\mathbf{V} = \begin{bmatrix} \dot{\bar{x}}\\\dot{\bar{y}} \end{bmatrix} = \begin{bmatrix} 6\\10 \end{bmatrix} \text{ .}$$
$${}$$
Please let me know if anything is unclear or if I made any glaring errors.