It turns out that one of the more important procedures in theoretical physics later on in the typical undergraduate/graduate curriculum is to find the length of some path from one point to another. Sounds pretty basic, right? And pretty easy, in most cases. If we want to know how far we’ve walked in a straight line from the origin in the figure below to the black dot at the coordinates (4,3), we’re taught fairly early that we just say
in units of miles or meters or whatever — nothing too tricky about that.
And then for tiny little steps (called “ds” in calculus) from one dot to another, we still use Pythagoras to calculate
Here’s the problem: it’s a lot tougher to calculate when the path isn’t a straight line (we’ll come back to the concept of “straight” in a subsequent post!) — maybe we want to walk along a circular path from (4,0) to (0,4) and find out how far we’ve walked. Now, no fair saying “well, it’s a quarter-circle, and I already know the equation for the circumference of a circle…”! The point is that we pretend we don’t know that and we need to calculate it.
Maybe you can see the difficulty. If you wanted to use little (dx,dy) steps, we’re going to have to take into account that they change as we go around. At first, near the x-axis, our steps are almost totally in the y-direction, but once we’re nearing the destination on the y-axis our little steps are in the -x direction. This would be all right, and we could do it, if we were to figure out some functions that describe how the (dx,dy) are changing (read on!) But a shortcut is to notice is that our distance away from the origin isn’t changing as we walk (only the angle θ), so it’s actually lots easier to describe the motion in polar coordinates (r, θ) and taking little steps in (dr, dθ).
Here, for example, let’s start at the black dot at (4,3) and take a little step dθ in the θ-direction followed by a little step dr in the r-direction.
How far have I gone? First, notice that the distance gone in the θ-direction depends on how far away from the origin we are, so we need to multiply by that factor r. Then we’ll just use Pythagoras again:
and take the square root if you want the little distance ds.
Ok, we cheated a little bit just sort of “knowing” we had to multiply by r to get the distance in dθ. The important bit is that we can calculate what we need if we write down the connection between the coordinate systems, so let’s do it the long way! [this should cut my readership by about 90%!]
Start with the standard polar coordinates:
What we need to know is how these (x,y) coordinates change if we change (r,θ) a little bit, so that means we need all the partial derivatives:
Check out how the change in x with θ makes sense based on what we noticed above — when θ is close to 0, there’s no change in x, but when we’re nearing the y-axis (θ=90) we’re taking little x-steps to the left.
Ok, so what we do in general works like this: in the simple Cartesian case way back up there, we wrote that a little bit of distance is
We can write this in a different way:
So what happens is that, since the only non-zero terms are the ones where i=j (both of them just range in value from 1 to 2 since we only have x and y coordinates) we get back the same expression previously. It’s just notation, but it’s admittedly confusing. Anyway, that g thing is what’s called the metric tensor. It tells us how to measure distances in a particular coordinate system! Just by looking at what we wrote down before, now we know the metric tensor in polar coordinates since the coordinates are dr and dθ instead of dx and dy:
It was easy to figure out ds in polar coordinates just by looking at the little graph — there are for sure cases where it’s not so easy to do that, so how can we calculate g instead? Here’s how! As long as we know how to transform coordinates (like the Cartesian → polar equations above) we can use those partial derivatives, since they tell us how one coordinate changes as we vary the other. And since the metric tensor is a function of the coordinates, here we go:
Yay! It matches. I mean, actually calculating the metric tensor for a general coordinate transformation, especially one where the non-diagonal g terms exist, is a real PITA, but there you go.
So finally, as suggested, let’s actually compute the distance along the quarter-circle from the x-axis to the y-axis. Since we know the metric, we write (again)
then factor out the dθ term
and since r doesn’t depend on θ, we’ll add up (integrate) all the little ds steps
which looks right!
Welp, it is a difficult subject, but that’s a bare-bones introduction to how more complicated geometries work. And since our modern understanding of gravity involves curving space and non-trivial coordinate transformations, this sort of construction becomes essential. If there is sufficient public demand, maybe I’ll re-visit this fairly soon to talk about the geometries and spacetime around a simple black hole! Truthfully, I think the public input will be taken as the reverse — such an article is probably inevitable unless there is public outcry to the contrary. And thanks to this platform, Substack, for the new ability to edit mathematical equations in LaTeX directly in the document. Very much simplifies the workflow for folks like me!
On Deck:
The next article I’m working on: how bodies are ripped apart by getting too close to their attractor! Sounds like a toxic relationship. In astrophysics, it’s called the Roche Limit! We’ll see how it works and whether it can explain Saturn’s ring system (as well as those of other worlds).
If you’re a student/teacher and want to see lots of worked examples that I like to include in my classes when I teach the “standard” University Physics 1 and 2 courses, feel free to browse the (growing) collection of 150+ videos at
And if something is especially cool and you’re inclined to leave a “tip” I’m not above coffee or pizza. In fact, following Douglas Adams, I’m not above accepting coffee or pizza in the same way that the sea is not above the clouds.
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