Ages and ages ago, Jennifer Ouellette commented on the start of the Basic Concepts series with a list of topics she'd like to see done. One of these was "Size and Scaling:"

First, let's tackle the jargon problem: Just what the heck is an order of magnitude? I use the phrase all the time now, after years of hanging around physicists, but as a budding science writer, I found the term a bit opaque, and I'd wager the average person on the street is a bit unclear on the specifics, too. Second, this is one of those areas where a picture really can be worth a thousand words -- or, barring that, it helps to paint a word picture. Many science museums use the "powers of ten" approach when discussing various size scales in the universe, from the subatomic level to the farthest reaches of the universe. It's been around since at least the 1960s, and it endures because it's effective. But it's just a start. The laws of physics actually start to change as one approaches the subatomic level, and a clear explication of how size and scaling can change a system's behavior would help with the public's chronic confusion about a number of things

This has been saved in my RSS reader for months, but I keep not getting around to it. This is one of those ideas that exists in a really tricky grey area: It's sufficiently basic that it's actually hard to articuate what's going on, but it's still abstract enough that it's hard to see why it matters. Even when we try to explicitly teach students about order-of-magnitude estimates, they don't really see the point, and I admit, I didn't either, until I spent a few years working in the field.

I was reminded of this topic while reading a book for an upcoming review, and I figured I really ought to take a crack at it. This will be somewhat less focussed than some of the previous Basic Concepts entries, and I'm going to lump together two slightly different ideas: Order-of-Magnitude Estimation and Dimensional Analysis, both of which are an important part of the physicist's mental toolkit.

The term **order of magnitude**, as Jennifer notes, is casually thrown around by physicists and astronomers all the time. In casual use, it means something like "roughly equal," but it has a fairly precise technical meaning, based on scientific notation. If you recall your middle school math, or have one of those calculators that display in scientific notation, you'll remember that any number can be written as a decimal multiplied by ten to some power. Thus, we can write 3,141.59 as:

3.14159 x 10

^{3}

In this notation, "order of magnitude" means "just the 10^{3} part." 3,141.59 is of the same order of magnitude as 1,500 or 4,567.8. There's a little uncertainty about whether you round up or not-- most people would probably say that 9,000 is the same order of magnitude as 10,000, but it's not clear whether 5,123 would be of order 1,000 or 10,000-- I've seen both, and used both, in different circumstances.

Either way, "order of magnitude" applied to an individual number just means the appropriate power of ten. Two numbers that are of the same order are within a factor of 10 of each other, and two numbers separated by an order of magnitude differ by a factor of ten-- 150 is an order of magnitude smaller than 3,141.59, and 45,678 is an order of magnitude larger.

This is an extremely coarse way of looking at numbers (though it might pass for precision measurement in astronomy), but it's actually more useful than you might think. Physicists frequently do order-of-magnitude calculations as a first step in solving problems.

For example, let's say that you want to find the velocity of a beam of electrons that have been accelerated through some potential difference (something that has come up several times in my class this term). To do this, you set the change in potential energy of the charges equal to the kinetic energy they acquire, and get something like:

qV = 1/2 mv

^{2}

where q is the charge, V is the electric potential difference (in volts), m is the mass, and v is the speed we're looking for. If you re-arrange this, you find an equation for the speed:

v = (2qV/m)

^{1/2)}

An electron has a charge of 1.602 x 10^{-19} coulombs and a mass of 9.109 x 10^{-31} kg, and let's say we're talking about a voltage difference of 250V. Getting an exact answer to this requires a fair bit of button-punching on a calculator, but you can do an order of magntiude estimate in seconds by stripping each number down to just an order of magnitude, and putting that in.

The charge is just 10^{-19} C, the mass is 10^{-30} kg, and the voltage is 10^{2} V, so this becomes:

v ≈ (10

^{-19}10^{3}/ 10^{-30})^{(1/2)}= (10^{13})^{(1/2)}= 10^{6.5}

So, we expect the speed of the electrons to be somewhere between 10^{6} and 10^{7} m/s, depending on the exact values. In fact, it's 9.377 x 10^{6}, using the numbers I gave above.

Why is this useful? I mean, to get the correct answer, you're going to need to go back and plug all these numbers into the calculator anyway, so why not just do that from the beginning?

Well, in this case, it's useful because it gives us a quick check on the validity of our assumptions. When we wrote down that first equation, we used the classical expression for the kinetic energy. If the electrons turn out to be moving close to the speed of light, that expression will give results that are badly wrong, and would need to be replaced with the full relativistic expression. Of course, the relativistic formula is a pain in the ass to work with, and we'd rather not use it if we don't have to.

The order-of-magntiude calculation tells us quickly that we're ok using the classical expression. The speed of light is of order 10^{8} m/s, so our answer of 10^{6}-10^{7} m/s is between 1% and 10% of the speed of light. The relativistic formula gives only a 1% correction when v is 10% of the speed of light (a result I know from another easy order-of-magnitude calculation), so for the precision we've got here, we're not going to notice the errors caused by using the classical formula.

"Well, fine," you say, "But I still have to put the full numbers in, so I haven't really saved any work. But it's easy to check an order-of-magntiude calculation, where it's hard to know if you've gone wrong with a full calculation. And with the order-of-magnitude estimate in hand, you can also know when you've done something horribly wrong in the full calculation.

Order of magntiude estimates are more important in physics than most other sciences because physics necessarily spans a hunge number of orders of magnitude. You can see that from this example-- the numbers in this problem range from 10^{-30} on up to 10^{2}, a span of 32 orders of magntiude. Even if you restrict the possible operations to addition, subtraction, multiplication, and division, that means the correct answer might be anywhere between 10^{-30} and 10^{+30}, which gives you absolutely no way to judge whether the number that comes out of the calculator is right. An order-of-magnitude calculation lets you restrict the possibilities to the point where you can hope to say something sensible about whether you've done things correctly.

These estimates are also key to understanding the scaling problem that Jennifer mentions. You can use order-of-magnitude calculations to show when odd effects start to have a significant effect-- for example, the statement above that relativistic effects only start to become significant when the speed gets above about 10% of the speed of light. Orders of magnitude are crucial to keeping straight what physical effects we need to consider-- you wouldn't use string theory to calculate the force needed to accelerate a car, because the order of magnitude is all wrong.

Closely related to the idea of order-of-magnitude estimation is the idea of **dimensional analysis**. It works pretty much the same way that order-of-magnitude estimation does, but in this case, you throw away the numbers entirely, and just work with the units. So, our speed equation becomes:

v ≈ (C V/ kg)

^{(1/2)}

That doesn't seem to have helped, but I know from basic E&M that one volt per meter is equal to one newton per coulomb, and from basic mechanics I know that one newton is one kilogram meter per second squared, so:

v ≈ (C V/ kg)

^{(1/2)}= ((~~C~~(~~kg~~m/s^{2})/~~C~~)m/~~kg~~)^{(1/2)}= (m^{2}/s^{2})^{(1/2)}

This means that the big bundle of stuff on the right hand side of that equation ends up giving you a number with the units of velocity. Which, in turn, means that we most likely did the original calculation correctly, and can have some faith that the quantity we've calculated is the thing we're looking for.

Neither of these methods will tell you whether you've really gotten the right answer in the full calculation, but they'll help catch the most egregious sorts of mistakes that get made in the course of doing basic physics problems. It's hard to get undergraduates to understand this, but I spent hours as a graduate student making a crude simulation of some atomic collision processes, and dimensional analysis saved me more than once.

Dimensional analysis is not just about checking the answers to problems you already know how to do, though. If you understand dimensional analysis well enough, you can actually use it to make new discoveries.

A nice example of this comes from my recent bedtime reading, Götz Hoeppe's Why the Sky Is Blue, which describes how Lord Rayleigh determined the wavelength dependence of the scattering process that bears his name using nothing but dimensional analysis. He recognized that there are only a handful of things that can possibly affect the intensity of the light scattered from particles in the air: the wavelength of the light, the size of the particles, the density of the particles, the density of the surrounding medium, and the distance between the scattering particles and the observer. Using simple dimensional considerations, he was able to deduce that the scattering intensity has to be inversely proportional to the fourth power of the wavelength, which is the correct dependence. And, incidentally, explains why the sky is blue.

- Log in to post comments

Another example of dimensional analysis that I like:

In 1947, details of atomic bombs were still highly classified, including their yield. However, a time lapse series of photos of a nuclear explosion was published in newspapers and magazines. G. I. Taylor applied dimensional analysis to this "blast radius as a function of time" data and accurately derived the bomb yield (which did not make certain parties in the government and military happy).

See here and here.

I'm old enough to have used slide rules as my first calculating devices. Finding the order of magnitude of the solution was a pre-check for where the decimal point would be located when the answer was finally found. After a little experience with this practice, we learned that sizing things up early would tell us if the thing we're looking for might be too small to matter, which made this a useful way to eliminate false leads.

Similarly, dimensional analysis was presented as way of finding when we've started off on the wrong foot. Decades later, there was a Scientific American Library book on the uses of dimensional analysis, and a fine example was the derivation of a means for estimating the energy of an atomic bomb from the diameter of the fireball in two closely (and precisely) spaced photographs.

The example showed it well, but its worth mentioning explicitly: the utility of an order of magnitude calculation is that you can use the rules of exponents (and logs if needed) to make the computation reasonable by hand.

I find it really helpful for catching bad units - if the answer if 10^4 to large, I probably forgot to change cm to m for an area.

"An order-of-magnitude calculation lets you restrict the possibilities to the point where you can hope to say something sensible about whether you've done things correctly."

I like this aspect. Though I'd say that one can use only negligible increased amount of work to use the order-of-magnitude concept for setting rough bounds, not just a rough estimate. Chad presented the values by didn't lay it out explicitly. Laying out numbers (as much for practice in my head as for anyone reading),

mass of electron -> take to be 10x10^-31 = 10^-30

electric potential difference:

V=250 V

upper bound:

magnitude of charge on electron -> take to be 2x10^-19 C

lower bound:

magnitude of charge on electron -> take to be 1x10^-19 C

For the upper bound, we get (ignoring the units)

2qV= 2 x 2x10^-19 x 250 = 1000 x 10^-19 =10^(3-19),

so the order-of-magnitude estimate is then the value of 10^7 Chad gave us: 10^((-19+30+3)/2) = 10^7

For the lower bound,

2qV= 2 x 1x10^-19 x 250 = 500 x 10^-19. Then we say "well, 500 is the same order of magnitude as 100=10^2". Then we replace 10^3 above with 10^2, giving an order-of-magnitude estimate is then the value of 10^((-19+30+2)/2) = 10^6.5

Hey, this was fun to practice!

As I argued on a "new tools in education" discussion, adopting new tools changes the skill set for better and for worse but

it becomes even more important to make estimations for checking.The last time I saw someone really goof with this was in a discussion about the size of the hominid fossil record. The commenter got his densities (confusing dm^3 with cm^3) and weights off. IIRC he thought he was able to stuff ~ 10 000 pulverized skeletons massing "50 kg" in a coffin.

An estimate would have given him a weight of 500 Mg, or 500 metric ton from bone dust in a small volume. Hmm...

This ask metafilter post asks What are the best rules, formulas and tricks in math?. It's got some nice tips for everyday back-of-the-envelope calculations.

1) Mentioning orders of magnitude reminded me Harlow Shapley's Flight From Chaos, a dated, but otherwise well written discussion of the orders of magnitude in the universe from subatomic particles to the intergalactic universe. Shapley was sort of the Carl Sagan of my parents'generation.

2) Why not combine slide rule approximation AND dimensional analysis? Have you ever seen the K&E Analon slide rule? Instead of numbers on its scales, it had physical units of distance, time, mass and charge. If you multiplied mass (on the C scale) by velocity (on the D scale), you get momentum!

I breifly taught MS/HS science at a very small school and constantly had problems with my students in chemistry and physics not having any idea what the answer should be and therefore not realizing they had made a mistake on the calculator. They had never been taught to estimate the answer first. I tried to teach them order of magnitude (and simple) estimation. I also tried to get them to use dimensional analysis to make sure they had put the correct given information in for the correct variables in a problem (Life is a word problem, kids - deal with it).

There is a good reason I didn't teach long. I never did understand how they got out of elementary school without understanding the usefulness of estimation. Thankfully, I will never again beat my head against the wall trying to understand why they think there are only 2 molecules in 32kg of water.

I just remembered an AJP article on the uses of dimensional analysis for estimating the magnitudes of neglected second-order effects:

http://dx.doi.org/10.1119/1.1574042

The Industrial Physicistalso had a nice primer on dimensional analysis:http://www.aip.org/tip/INPHFA/vol-1/iss-1/p21.pdf

Nice thread. Gave me some good ideas and exmaple to use in class. Here is a simple version of a joke that I've used. You guys might know a version of it. I always find it interesting to see that only a very small fraction of the students laugh, a few more smile, but most have a really empty expression on their faces.

Joke: A General has to capture and enemy fortress. He calls over a scout over an issues an order.

"Soldier. Go and find out how many men are defending that fortress!"

"Yes Sir"

The scout leaps on his horse and gallops off.

Less than 30 minutes later he is already back.

The general is quite suprised.

"SOLDIER! You are back already?"

"Yes Sir. The number of men in the fort is 2016! Sir!"

"Young man. No how in God's name did you come to find that out is such a short time?"

"Sir! Easy Sir! The fort has four towers. In each tower I counted four men. 4 times 4 is 16. Then the men inside the fort are about 2000. Sir!"