It's important when learning physics to get an idea of how big different things are. How big is an atomic nucleus? An atom? A cell? A planet? A galaxy? Often it's good enough to just have a rough idea.
If you are thinking about small things, its good to know about 4 important units of distance: the Bohr radius of the hydrogen atom, the Compton wavelength of the electron, the classical electron wavelength, and the Planck length.
If you want to know how big atoms and molecules are, you should start by understanding the Bohr radius. The Bohr radius r is the approximate size of a hydrogen atom in its ground state. It's only approximate, because the electron in a hydrogen atom is a fuzzy probability cloud, not a pellet whizzing around in a circular orbit! But that's okay.
I'd like to explain how the Bohr radius depends on the electron mass, the electron charge, and Planck's constant. In particular I want you to see why the Bohr radius is inversely proportional to the mass of the electron. This is an example of a general phenomenon in particle physics: a mass scale sets an inverse length scale.
Unfortunately, I forget what the Bohr radius equals, so I'll have to work it out.
First let's cheat and figure it out using dimensional analysis. (I want to prove to the real physicists there that I can do this stuff that mathematicians are usually scared to do.) We'll just get an order-of-magnitude estimate of the Bohr radius, modulo factors of 2, π, and the like.
So what could the hydrogen atom's radius depend on? It's two particles of equal charge attracted by the electric field, and it is kept from collapsing by quantum effects, namely the uncertainty principle. Since it involves a charged electron whizzing around, it'll depend on the electron charge e and the electron mass m. Since the electron is in a fuzzed-out orbit because of quantum effects, it'll involve Planck's constant ℏ. Since the proton is 1836 times as heavy as the electron, one can be approximate it quite well by being fixed, so the Bohr radius won't depend much on the proton mass. (We are making an approximation here; a better approach uses the center of mass frame and our approach will be off by terms proportional to the mass of the electron over the mass of the proton.) This is not a relativistic calculation so we won't use c.
In short, the answer will probably involve:
and nothing else. Okay. So using dimensional analysis, what can we do with these to get a quantity with dimensions of length? e has dimensions of charge. If we absorb Coulomb's constant into the definition of charge, charge has dimensions of length times force^{1/2}, since
force = charge^{2}/distance^{2},
by Coulomb's law.
Thus
To combine these to get a length, we fiddle around. Note that ℏ^{2}/e^{2} has dimensions ML, so ℏ^{2}/me^{2} has dimensions L.
Therefore the Bohr radius is
r = ℏ^{2}/me^{2}
times some dimensionless constant. Cheating, I peek in my uncle's old physics text, The New College Physics - A Spiral Approach, since right now I just happen to have packed up all my other books. The dimensionless constant turns out to be 1, and the Bohr radius works out to be 5 × 10^{-11} meters. Half an angstrom, in other words. Angstroms are the length scale of atomic physics.
Note that
r = ℏ^{2}/me^{2}
makes some sense in that it gets bigger when ℏgets bigger (more "quantum fuzziness" makes a bigger atom) and gets smaller when the electron's charge is bigger (more attractive pull). It also gets smaller when the mass of the electron gets bigger - here we see that bigger mass scales go with smaller length scales! But we want some understanding at a gut level why making the electron heavier would make the hydrogen atom smaller! It's known that this is true, by the way, because one can take a muon, which is just like an electron but 206.77 times heavier (and decays rapidly), and make muonic hydrogen which is about 206.77 times smaller. But why?
Here's a very shoddy "derivation" of the Bohr radius that makes it clear why a big electron mass would give small atoms! The uncertainty principle says that the product of the uncertainty of momentum and the uncertainty of position can't be too small:
Δp Δx ≥ ℏ
(leaving out factors of 2 etc.). Now in the hydrogen atom Δ x is just the Bohr radius and we can write Δ p, the uncertainty in momentum, as m Δv, where v is velocity. So we get
r = ℏ/m Δv. 1)
Note that I have replaced an inequality by an equality! That is because this whole argument is so bogus that it doesn't matter - all the = signs below should be read as "is approximately equal to [hopefully]". Mathematicians should learn to respect this sort of vague argument that physicists are so good at - one can often squeeze the right answer out of very sketchy reasoning!
We see here why bigger m give smaller Bohr radius r: the uncertainty principle allows us to "squish up" the wavefunction of a more massive particle in a smaller region of space with the same Δv. But how big is Δv? To complete this shoddy "derivation" of Bohr's radius, note that Δv is probably about equal to the velocity of a classical electron going around a proton in an orbit the size of the Bohr radius. How do we relate the velocity of an orbiting electron and the radius of its orbit? We can get an approximate answer by simply equating the kinetic and potential energy
m(Δv)^{2}/2 = e^{2}/r 2)
since the virial theorem says kinetic and potential energies are about the same magnitude. Now squaring 1) we get
(Δv)^{2} = ℏ^{2}/m^{2} r^{2} 3)
and dividing 2) by 3) we get
m/2 = e^{2} m^{2} r/ℏ^{2}
or ignoring the factor of 2, just
r = ℏ^{2}/me^{2}!
I have emphasized by the use of words such as "shoddy" that this is a rough heuristic argument and not at all a mathematical derivation of the Bohr radius. Indeed, no matter how sloppy our argument was it was bound to give the same answer as long as it was dimensionally correct. This should make one distrust the argument! But one can make it utterly precise, as Schrödinger did. We see at any rate that the Bohr radius can be guessed by essentially classical reasoning together with the uncertainty principle, and this length scale is naturally proportional to an inverse mass scale - the inverse of the electron mass!
The Compton wavelength of a particle, roughly speaking, is the length scale at which relativistic quantum field theory becomes crucial for its accurate description. A simple way to think of it is this. Trying to localize an electron to within less than its Compton wavelength makes its momentum so uncertain that it can have an energy large enough to make an extra electron-positron pair! This is the length scale at which quantum field theory, which describes particle creation, becomes REALLY important for describing electrons. The Compton wavelength of the electron is the characteristic length scale of QED (quantum electrodynamics).
It's easy to guess how big the Compton wavelength is using the knowledge that it depends only on the mass of the electron, relativity and quantum mechanics. Mass has dimension M. Length has dimension L. Time has dimension T. In relativity we have a constant, the speed of light, with dimensions L/T, and in quantum mechanics we have a constant, Planck's constant, with dimensions ML^{2}/T = energy times time = momentum times position. These two constants enable us to express units of mass in terms of dimensions of inverse length. I.e.:
M = (ML^{2}/T)(T/L)1/L = ℏ/c 1/L.
So in particular the Compton wavelength should be about
L_{Compton} = ℏ/mc.
This is about 4 × 10^{-13} meters.
In fact, this is usually called the "reduced" Compton wavelength. What people usually call the Compton wavelength is 2π times as big, about 2 × 10^{-12} meters. That's because the wavelength of a wave is really not the reciprocal of its frequency: it's 2π divided by the frequency. But I'm not worrying much about factors of 2π.
We can also derive the Compton wavelength in a slightly more enlightening manner as follows. The energy of an electron at rest is
mc^{2}
as someone once noted. Say we try to confine an electron in a region of size L. Then the uncertainty principle says that its momentum can only be known up to an error Δ p, where
L Δp ≥ ℏ.
If we make L small enough Δ p will be so big the electron may have a kinetic energy bigger than twice the rest energy mc^{2}. This would be enough to form an electron-positron pair! This effect -- the creation of new particles while trying to determine the position of old ones -- will kick in at the Compton wavelength
L_{Compton} p = ℏ 4)
where here p is the momentum at which the kinetic energy of an electron is about mc^{2} (forgetting the factor of 2, naturally). Recall the souped-up version of E = mc^{2} good for moving particles:
E^{2} = m^{2}c^{4} + p^{2}c^{2},
and set the "kinetic energy squared" p^{2}c^{2} equal to the rest energy squared, (mc^{2})^{2}. We get
p^{2}c^{2} = m^{2}c^{4} or p = mc!
So by 4) we have
L_{Compton} = ℏ/mc
as before. Again, a mass scale sets a length scale.
I strongly advise the reader to take the ratio of the Bohr radius by the Compton wavelength to see how much bigger the atomic length scale is than the length scale at which quantum field theory becomes really important. It's about 137 if I didn't screw up. Of course this is a famous number, one over the fine structure constant. I let the reader work out what it is in terms of ℏ, e, and c. It's good to keep this in mind for a gut-level understanding of microphysics: quantum field theory effects start really mattering for electrons on a distance scale 1/137 the size of the hydrogen atom. This is why people were able to notice these field-theoretic effects and develop QED not too long after they came up with a quantum-mechanical description of the hydrogen atom.
Another characteristic length scale is the length scale at which renormalization becomes really important. Renormalization is an aspect of field theory which deals with such issues as the fact that the electromagnetic field produced by an electron has energy and thus should be counted as part of the mass of the electron! The length scale at which these effects become really important is called the classical electron radius. It's important to note that it is really classical, not quantum mechanical, because it only depends on classical electromagneticsm, which doesn't involve ℏ, and the formula for the rest energy of an electron, which involves c but not ℏ. Indeed, renormalization was an issue in classical field theory before quantum field theory came along.
So the classical electron radius should just depend on the mass of the electron, its charge, and the speed of light. Recall these have units
so to get a length out of these we should form e^{2}/mc^{2}. So, without doing any real work, we can guess
r_{e} = e^{2}/mc^{2}.
We can derive the classical electron radius by working out the electric field outside of a ball having charge equal to that of the electron, e, and radius L, then working out the energy of this electric field, and then setting that energy equal to the electron mass m. Solving for L we get a formula for the electron radius r_{e}. In other words, the classical electron radius is the radius the electron would have to have for all of its mass to be due to the electric field it produced, assuming it was a charged shell. Up to miscellaneous factors we get
r_{e} = e^{2}/mc^{2},
of course; since the actual calculation is not very exciting I'll skip it.
It's worth noting that the classical electron radius is 1/137 as big as the Compton wavelength of the electron - the all-important fine structure constant again! So we have 3 length scales:
e^{2}/ℏc.
It's a dimensionless constant depending only on ℏ, e, and c. In this respect it's more fundamental than any of the length scales mentioned, because all the length scales mentioned involve the electron mass, and one could work them out for particles other than the electron, whereas
e^{2}/ℏc
is truly universal, once you remember that the "electron charge" is nothing specific to the electron but is a basic aspect of electromagnetism that applies to all charged particles. (Yes, quarks apparently have charge 1/3, but that doesn't really affect my point.) In other words, the fine structure constant is a dimensionless measure of how strong the electromagnetic force is, and we have seen that it sets the ratio of 3 important length scales.
Now for one final length scale - still smaller. This is the length scale at which quantum gravity should become important - the Planck length l. On the scale of the Planck length, it's possible that the structure of spacetime becomes quite different from the four-dimensional manifold we know and love. Spacetime itself becomes a foam (according to Wheeler) or a bucket of dust (according to Wheeler) or a bubbling sea of virtual black holes (according to Hawking) or a weave of knots or tangles (according to Ashtekar, Rovelli, and Smolin). In short, it's weird, but beyond that nobody really knows. To be more precise, the Planck length is the length scale at which quantum mechanics, gravity and relativity all interact very strongly. Thus it depends on ℏ, c, and Newton's gravitational constant G. These have dimensions
so to get a length we have to use
L_{Planck} = (ℏG/c^{3})^{1/2},
This makes some sense because the bigger ℏis, the more "quantum" the universe is, so the bigger the length is at which quantum gravity matters. Also, the bigger G is, the stronger gravity is, so the bigger the length is at which quantum gravity matters. The bigger c is, the less "relativistic" the universe is, so the smaller the Planck length is. Of course, Planck's constant and the gravitational constant are actually very small, so the Planck length is really small. The Planck length is about 1.6 × 10^{-35} meters. This is much smaller than the length scales I was talking about before - ridiculously small. That's why we haven't seen any (obvious) signs of quantum gravity effects, and why it will be so hard to do any quantum gravity experiments.
Note that in all the previous three examples a length scale was proportional on the inverse of a mass - in particular, the electron mass. The Planck length is peculiar in that it does not depend on a mass in this way. Of course, it depends on the gravitational constant, which has a lot to do with mass! In fact, the combination of gravity, relativity and quantum mechanics sets a mass scale - the Planck mass - as well as a length scale. The Planck mass is huge (by particle physics standards) so the Planck length is puny.
A rough way of understanding the Planck length is as follows. Every mass determines a Schwarzschild radius - that is, the radius of the event horizon of a black hole having that mass. Now this is curious, in that I've been saying repeatedly that a mass scale sets an inverse length scale, but the radius of a black hole is proportional to its mass. Of course, this is dimensionally possible in that the gravitational constant involves units of mass. We'll work out the Schwarzschild radius of a given mass in a minute. But also every mass determines a Compton wavelength, as I explained earlier:
L_{Compton} = ℏ/mc 5)
We can then work out how big a black hole we need for its Compton wavelength to equal its Schwarzschild radius! This sort of black hole will have mass about equal to the Planck mass, and radius about equal to the Planck length.
What does this mean? Well, remember that the Compton wavelength of a particle is the length scale at which quantum field theory becomes very important in describing it. So the Planck length is the size of a black hole for which quantum field theory becomes very important. Hawking has predicted that black holes of any size emit radiation due to quantum-field-theoretic effects - the bigger the black hole, the less radiation. His calculations treat the black hole classically and only use quantum field theory in treating the electromagnetic radiation. For a black hole about as big as the Planck length one would expect this approximation to break down drastically.
To be picturesque, we can say that if we have a black hole about the size of the Planck length, and we try to locate it to an accuracy equal to its radius, the Heisenberg uncertainty principle makes the the momentum of the black hole so poorly known that there may be enough energy around to create another black hole of that size! I warn the reader to take this with a massive grain of salt, since there is no good theory of this sort of thing yet - much less any experimental evidence. But people have sharpened this sort of thought experiment and seen that things get awfully funny at the Planck length. By analogy with particle physics, one might expect processes involving virtual black holes to be very important at this length scale. Hawking and others have written interesting papers on reactions induced by virtual black holes... but I would not take these predictions too seriously yet.
Okay - let's start with Newtonian gravity:
force = -Gm_{1}m_{2}/r^{2}
and remember that we can write the potential energy as
V = -Gm_{1}m_{2}/r
Now let's say we have a little particle with mass m_{2} in the field of a big thing with mass m_{1}, and let's compute its escape velocity. That's the velocity for which its kinetic energy plus the potential energy above is zero, i.e.:
m_{2}v^{2}/2 = Gm_{1}m_{2}/r
Calling m_{1} simply m, we get
v = (2Gm/r)^{1/2}
Black holes were in fact predicted before general relativity simply by noting that the escape velocity can become larger than c, so that light cannot escape! Since we are being deliberately sloppy in these articles, let's use that idea to guess the Schwarzschild radius of a black hole of mass m. We get
c = (2Gm/r)^{1/2}
or
r = 2Gm/c^{2} 6)
I don't have any books at hand at the moment, but I think this is close to the actual Schwarzschild radius. It could be a bit off, due the nonlinearity of general relativity. It's just one of those constant factors that we are blithely ignoring here. I'm shocked that even I kept the "2" around above!
Okay, so now set the Schwarzschild radius - 6) - equal to the Compton wavelength - 5) - and forget that darn "2", getting
m^{2} = ℏc/G
for the Planck mass. Plugging that into formula 1), we get the Planck length:
L_{Planck} = (ℏG/c^{3})^{1/2}
as expected.