It’s not widely appreciated how often physicists can guess the answer to a problem before they even start calculating. By combining a basic consistency requirement with scientific reasoning, they can often use a heuristic approach to solving problems that allows them to derive most of a formula without doing any work at all. This week I want to introduce this to you, and show you some of its power.
The trick, called “dimensional analysis” or “unit analysis” or “dimensional reasoning”, involves requiring consistency among units, sometimes called “dimensions.” For instance, the distance from the Earth to the Sun is, obviously, a length. We can state the length in kilometers, or in miles, or in inches; each is a unit of length. But for today’s purposes, it’s irrelevant which one we use. What’s important is this: the Earth-Sun distance has to be expressed in some unit of length, because, well, it’s a length! Or in physics-speak, it has the “dimensions of length.”
For any equation in physics of the form X = Y, the two sides of the equation have to be consistent with one another. If X has dimensions of length, then Y must also have dimensions of length. If X has dimensions of mass, then Y must also. Just as you can’t meaningfully say “I weigh twelve meters” or “I am seventy kilograms old”, physics equations have to make sense, relating weights to weights, or lengths to lengths, or energies to energies. If you see an equation X=Y where X is in meters and Y is in Joules (a measure of energy), then you know there’s a typo or a conceptual mistake in the equation.
In fact, looking for this type of inconsistency is a powerful tool, used by students and professionals alike, in checking calculations for errors. I use it both in my own research and when trying to figure out, when grading, where a student went wrong.
That’s nice, but why is it useful beyond checking for mistakes?
Sometimes, when you have a problem to solve involving a few physical quantities, there might be only one consistent equation relating them — only one way to set an X equal to a Y. And you can guess that equation without doing any work.
Well, that’s pretty abstract; let’s see how it works in a couple of examples.
Simple Example: Velocity, Radius and Period
First, a super-easy one just to illustrate the point. Suppose we want to find the velocity v of the Earth as it travels round the Sun. If we call the average radius of the Earth’s roughly circular orbit R, and we call T the time it takes to orbit the Sun, then what formula should we write down? Well, there’s only one possibility that’s consistent. Velocity v is a length traveled per time; it has dimensions of length over time. R is a length. T is a time. And so the equation that relates them must be of the form
- v = # R / T
where “#” is an unknown number that this argument doesn’t specify.
Since we don’t know this number, have we really learned anything? Yes we have! The formula cannot set v equal to R2/T, or T2/R, or R T, or R1.4 /T2.6. Any formula other than v = # R/T would relate a length per time to something that isn’t a length per time… and would therefore be nonsensical. Just by demanding sense, we have mostly solved the problem without doing any work at all, except for one unknown #.
If we want to be precise, we’ll still have to calculate the unknown #. If the orbit were circular that would be easy; # = 2π. For a realistic, elliptical orbit, you have to actually calculate it. Still, for a nearly-circular orbit like the Earth’s, this # it’s not going to be a billion for a nearly-circular orbit, nor is it going to be a billionth. It will be a number that’s not far from 2π, which in turn is not too, too far from 1. (2π is approximately 6.) So we can make an estimate without doing much, if any, calculation.
Interesting Example: Kepler’s Law, In Detail
Now let’s take a less trivial example, though still easy to do using other methods. Recently, using do-it-yourself techniques, I showed you how you yourself could derive Kepler’s third law, which relates the radius of a planet’s orbit R to its orbital period T, specifically that R3 is proportional to T2. We found this was true for objects that orbit the Sun. We also found it was true for objects that orbit the Earth, but with details that were different. Can we find a formula which is true both for the Sun and the Earth — one that explains the difference?
Well, under an assumption — that Newton’s gravity is involved somehow — we can. This is where physics reasoning and some experience comes in.
First, if gravity is at work, an experienced physicist knows that Newton’s constant G always appears, because this constant characterizes the overall strength of gravity. The dimensions of G have to be consistent with Newton’s gravitational force equation
- F = G M m / r2
which gives the force of gravity between two objects of mass M and m that are separated by a distance r. Rearranging for convenience, we can write this as
- G = F r2 /(M m)
In first-year physics we learn that force has dimensions of mass times length divided by time squared. M and m have dimensions of mass, and r has dimensions of length. From the above equation G = F r2 /(M m), we find the dimensions of G itself:
- dimensions of G = dimensions of F r2/(M m) [for consistency!]
- = (dimensions of F) * (dimensions of r2)/(dimensions of M m)
- = (mass * length / time2) * (length2)/(mass2) = (length3/time2/mass)
Moreover, under our assumption that gravity is at work, and since we are considering objects that all orbit the same central body, such as the Sun, we can guess that the mass of that central body comes in somehow. Let’s refer to that mass as “M”.
So now to Kepler’s law: might there be an equation that relates gravity’s ever-present constant, G, the mass of a central body, M, the period T of an object orbiting that central body, and the radius R of that orbit? Well, how about
- G / M = # R4/T2
- G M5 = # R9 T2
- G M3/2 = # T5/R7/3 ?
No Way! Each of these possible equations is nonsense! because the dimensions of the left hand side are not equal to the dimensions of the right hand side!
But there is in fact (as I’ll convince you in a moment) one and only one possible answer that could make sense! That’s this one:
- G M = # R3/T2
And this confirms that for a particular central object of mass M, all objects that orbit it have the same ratio for R3 to T2. In other words, you can guess Kepler’s third law of orbits simply by using dimensional analysis. No complicated equations are required.
Again, # is an unknown number that we would have to calculate. But even though we don’t know it yet, we’re most of the way to finding the formula we want, and we haven’t done any work other than checking dimensions!
Why is this the only possible formula? One can be systematic about it, but here’s a quick way to see it. R and T have no units of mass, but G and M do. So to relate G and M to R and T, there must be some combination of G and M in which the dimensions of mass cancel. Since, as we just saw above, G has dimensions of something divided by mass, G M is the only combination where the dimensions of mass cancel, leaving only dimensions of length and time! In fact GM has dimensions of length3 divided by time2 — and that means G M can only be related to R3/T2. That’s all there is to it!
A little calculation shows that the unknown # is (2π)2, approximately 39.5, which is not too, too far from 1. (It’s not that close, admittedly. But remember that this unknown # could have been 483,248,342,198 or 0.000000000000932 — and so, relative to what it could have been, it’s still pretty close to 1.) This tendency for these unknown #’s to be not to so far from 1 is one we need to keep an eye on.
Yet we don’t even need to know the unknown # to learn something extremely important! Suppose we have studied R3/T2 for the Moon and satellites moving around the Earth, and R3/T2 for the planets orbiting the Sun. We have
- G Msun = # R3/T2 for objects orbiting the Sun
For instance we could focus on the Earth’s orbit, so we would take R to be the Earth-Sun distance RES and T to be one Earth year TE. Meanwhile,
- G Mearth = # R3/T2 for objects orbiting the Earth
Here we could focus on the Moon’s orbit, and take R to be the Moon-Earth distance RME and T to be one Moon month TM. Now we can take the ratio of these two expressions! The G cancels, and so does the unknown #! That leaves us with
- Msun/Mearth = (R3/T2)earth_around_sun/(R3/T2)moon_around_earth = (RES3/RME3)(TM2/TE2)
Since the Earth-Sun distance RES is about 388 times the Moon-Earth distance, and the Earth-Year is about 13.4 times the Moon-Month, we find
- Msun/Mearth = (388)3/(13.4)2 = 325,000
which is correct, to within a few percent. Look at that! We calculated the ratio of the Sun’s mass to the Earth’s mass just using dimensional analysis! All we needed to know was the distances and times relevant to the orbits of the Earth and Moon.
We never had to solve a gravity equation to figure this out! We just had to assume that gravity was involved somehow.
I hope that this convinces you that if you use this reasoning well, it can be immensely powerful, with the potential to simplify difficult problems dramatically. Next time we’ll look at how dimensional analysis can be (and was) used to learn things about relativity and black holes, and then we’ll look at atomic physics and beyond.