One of the most important skills a physicist can have is to estimate the size of various effects without doing a detailed calculation. This basic skill is one of the hardest ones to learn, mostly because it is very easy to disbelieve: after all, you didn’t do a real calculation. But at the same time, it is one of the most important skills one can have. These estimates will tell you if you are going to be wasting your time doing a calculation or not. On the other hand, they migt give you an enormous amount of information without having a fundamental system of equations to solve: let us say, you don’t know if there is a relation between some set of phenomena, but order of magnitude estimates migt be able to do that for you. I will now describe various instances of these order of magnitude estimates to show you how powerful they are.
The first thing one learns in physics is that units are important. Physical quantities are not dimensionless numbers, they are measured and compared against other similar quantities, and if you decide that you have a standard object to compare it to, you call that a unit. For example, the meter was originally defined in terms of the circumference of the earth. But apart from the fact that we live here on the earth, the earth is just one more object that one could compare lengths to. In the old days, one could compare lengths to the length of the arm of the king or queen. Every time one had a new king or queen, one had to make a new set of units. I’m sure the merchants were not happy with this, which is why comparing to the earths circumference is a better option. At least it is more universal.
By the same token, a second derives from dividing a solar day into 86400 units of measurement, each of them called a second. The number 86400 is made of twenty four hours, each of which is divided into 60 minutes, each of which is divided into 60 seconds, and a second is not too different from the time that it takes you to say ‘one second’. Why did we choose to divide the day into 86400 pieces rather than 100000? After all, the second number is much nicer to do arithmetic with. However, you can instead blame the babylonians, because their numbering system was in base 60! This sounds bad by modern days, but 60 is exactly divisible by two, three, four, five and six, which makes divisions into equal parts with small integer denominators a lot easier. I’m sure babylonian accountants were very happy with that.
Finally, there is the unit of mass. In the MKS system, it is given in Kilograms, and a kilogram is the mass of one liter of water. A liter is a unit of volume. Why did we choose water, and not gold? Because most people have readily access to water and because water is liquid at room temperatures. That way the measurement of volumes by matter is more precise. That is why.
So all of these units are comparisons to randomly selected objects that we encounter every day and we consider important. Perhaps one of the most interesting statements one can make about physics is that after you have units for mass, time and length, the rest are derived concepts. That does not mean that other units are not useful. But it does mean that one can do an absolute comparison of properties to some controlled experiment.
As I discussed previously, once you find that the speed of light is the same for everyone, your units of time and length are related. Therefore, you can choose units where the speed of light is one. In this system of units, you measure lengths in light-seconds or light-years.
Similarly, once you discover quantum mechanics (I promise to talk about this in the future), you can see that energies can be related to inverse times. The constant responsible for that relation is Planck’s constant. Another pretty universal measurement.
In particle physics, it is customary to set
Where the h with the bar through it is Planck’s reduced constant: h divided by 2π.
A system of units with that property is called a system of natural units. The usual unit to compare to is GeV, a unit of energy whose numerical value is very close to the rest energy of a proton, or eV : the kinetic energy that one electron gets when moving through a difference of potential of one volt.
So now, let us talk about estimates. How do we put information together and make predictions?
Let us say that you have a gas, with some density ρ. The density is measured in . Let us say that you can also measure the pressure of the gas P. This is measured in units of Force per unit area, or . Now, we can play with units a little bit to try to eliminate some such units. Let us eliminate Kg. If you take P/ ρ, what units do you get? You get units of
These are units of velocity squared! So you would imagine that for the gas, the velocity unit given by
is rather important. But we know of some other velocities that are important in gases: the velocity of sound for example. If you don’t have anything else to go on, you could not be blamed for believing that the estimate v above might have something to do with the speed of sound. Indeed, when you estimate it, it differs from the true speed of sound by factors of order one (sometimes ). However, it is not exact. And the speed of sound is not just a multiple of v with a universal constant for all materials either. So you might imagine that the ratio of v to the velocity of sound might be important and that it might tell you something very important about the gas. Indeed, Laplace worked this out, and he found that the two numbers were related by the thermal properties of the gas in question. Indeed, one can use this equation to show that hot air transmits sound faster than cold air.
You might make similar estimates for a length of string under some tension. Indeed, that is how one tunes guitars: one varies the speed of waves on the string by modifying the tension.
Another favorite of mine is the following. Assume that someone gives you a cross section for absorption of light or some other particle by some dust, this is usually called σ. This cross section is measured in units of area. Let us also assume that you have some density of scatterers, this is measured in units of inverse volume and the usual symbol is n. If you take the cross section and divide by the density, you get a unit of length.
This unit of length is an estimate for the penetration thickness of the dust, or the mean free path before you have a collision. So if your dust or material is substantially longer than λ, you will absorb most particles. The penetration length of light in milk is rather small. This is why a glass of milk looks opaque.
Another fun example one can consider is the lifetime of the Z boson. I picked this one because this week I have been doing this calculation in class. The Z boson couples to fermions (matter), with a dimensionless coupling constant g, which is not too different from one tenth. If one is to estimate the decay rate of the Z boson, one would guess that it is estimated to be in natural units. The square uses a bit of quantum mechanics: the coupling of the Z particle above is related to the probability amplitude that a Z turns into a couple of fermions, and probabilities are measures by amplitudes squared. However, to make the whole thing have the right units (inverse time), you need the only time unit that is obvious in the problem: the one related to the mass of the Z particle. Lubos quite recently went at length on how to make some similar estimates in the case of evaporating black holes when they are Planck scale objects. It shows these principles at work.
Now, these estimates are important for various day to day problems. Let us say that someone tells you that they can deliver to you a home nuclear reactor, and that the failure rate is small and their lifetime is long: they don’t need work on them for 10-15 years and that after that you have to call the technicians to refresh the fuel. Now you have to imagine that there is a one in a million chance of someone who decides that they can fix their home nuclear reactor themselves by hitting it with a hammer. Thereby saving a pretty penny from not calling the technicians and paying for them. Would you grant the company that makes these reactors a permit to sell them?