## The Physics of Scaling Laws and Dimensional Analysis

I’ve been puzzling over dimensional analysis and scaling laws lately. Believe it or not, professional physicists sometimes use dimensional analysis to do a quick back of the envelope calculation and usually get the correct answer to within an order of magnitude or so. In this post I’m going to give three examples of how the “magic” of dimensional analysis can be used to guess the answer.  I’ll start with a simple example from introductory physics  and move to more advanced examples.

1)  If a baseball is dropped from a height ‘h’ above the ground, how long until it hits the ground neglecting air resistance?   We know that since air resistance is neglected the size of the baseball is irrelevant, and Galileo tells us the mass of the baseball is also unimportant.  That leaves the height from which the ball is dropped, h, which is in units of length [L], and the acceleration due to gravity, g, which is in units of length per time squared [L/T^2].  So, arranging these two quantities to give the dimension of time is simple:

$T=L^a\left(\frac{L}{T^2}\right)^b$

Which gives,

$T=L^{a+b} T^{-2b}$

We need time  so the power for length must be zero, so a = -b, and b=-1/2.  Thus a = 1/2 and b = -1/2 and voila:

$\Delta t = \sqrt{\frac{h}{g}}$

And this answer is only off by dimensionless factor of ~1.4 (root 2)!

2) As it turns out you can do precisely the same thing to figure out how large craters will be when you drop rocks into the sand (this was the topic of my previous post). The critical assumption you have to make is in deciding which variables are important. Mainly, you have to decide if density is the only important quantity for the sand.  (It turns out that for other materials there are other dimensional quantities you must include.)  Then you assume the only other important quantity is the kinetic energy of the ball.  Finally, you do dimensional analysis exactly as I did above and you will find the same scaling law I’ve found in the last post for relating “meteor” kinetic energy (KE) to crater diameter (D):
$D \propto KE^{1/4}$

Notice how you can get the same answer as I did in the last post without thinking about all the physics and kinematics equations.

3) Now here’s a truly bizarre application of dimensional analysis to the Schrodinger equation that I saw on the first day of a quantum mechanics class. Let’s say that you’re tired of solving partial differential equations in three dimensions and you just want to guess what the ground state energy of an electron is in a hydrogen atom.  You bypass the Schrodinger equation in all its glory and simply write the (classical) energy equation for an electron:
$E= \frac{p^2}{2m}-\frac{e^2}{4 \pi \epsilon_0 r}$
Now you think, in quantum mechanics you always have to throw in Planck’s constant, so how do you do that? Well we have momentum and position…mmm…momentum and position, then BAM!, you remember that planck’s contant (h-bar) has units of position times momentum because you remember the Heisenberg uncertainty principle.  So let’s just postulate through dimensional analysisthat:
$p r = \hbar$
and substitute that into the energy equation, eliminating position:
$E= \frac{p^2}{2m}-\frac{e^2 p}{4 \pi \epsilon_0 \hbar}$
Now we want to find the lowest energy level, so we do the usual calculus to minimize this function, E(p), and find:
$p_{min}=\frac{e^2 m}{4 \pi \epsilon_0 \hbar}$
Which tells us that:
$E_{min}= -\frac{e^4m}{2(4 \pi \epsilon_0)^2 \hbar^2} \sim -13.6eV$
And out of shear dumb luck, that is not just the approximate answer, it’s the exact answer to the full solution of the Schrodinger equation! Before I celebrate too much, I should admit that I could have used h instead of h-bar (which is just h/2pi). In that case I would have gotten -0.34 eV, which is definitely in the right ball park, but not nearly as impressive as getting the exact answer.

Of course you could just use dimensional on all the relevant constants in the equation.  You know that, being quantum mechanics you have include Planck’s constant.  Obviously the electric force is involved so you have to include the fundamental unit of charge and the permittivity of free space in the following way:

$\frac{e^2}{4 \pi \epsilon_0}$

And finally you include the mass of the electron.  If you have some experience with these types of problems you will know that you don’t have to include the proton mass because it’s so much heavier than the electron.  Then you have to solve the following equation for a, b and c.

$ML^2 T^{-2}=\left[\frac{e^2}{4 \pi \epsilon_0}\right]^a \left[m_e\right]^b \left[\hbar\right]^c$

Where the brackets [] mean the units of the constant and the right hand side of the equation are the units of energy.  When you do this you will find that a = 4, b = 1, and c = -2 which gives:

$\frac{m_e e^4}{(4 \pi \epsilon_0)^2\hbar^2} \sim 27.2 eV$

This result is only off by a factor of two!  Also, you should know that a negative sign belongs there as we’re talking about bound energies.  So, if the above were all you knew about the hydrogen atom, you would be able to rightly conclude that all the bound electronic energy levels are between 0 eV (which is to say the electron is free) and negative several tens of eV.  And this is totally correct!  Then when you have more time you can show that all the bound levels are between 0 eV and -13.6 eV.

I know it sounds crazy, just cobbling together relevant quantites to get important results, but it works.  If you understand the physics of the problem well enough, and you can guess what the relevant parameters are, then you’re ready to pull out the envelope and start calculating.