## Tuesday, December 14, 2010

### Three examples of dimensional analysis

Let's say I drop a bowling ball from a building.  How long does it take to hit the ground?

Probably many of you are now thinking about how much you hated your high school physics class.  What was that formula you used for falling bowling balls?  Geez, who wants to remember that equation and solve it?

But let's say that we just want an approximate answer.  In physics, there's a special method called dimensional analysis which is useful to find approximate answers to questions like this.

The first step is to figure out what quantities could be involved in the answer.  For the bowling ball, it could only depend on the mass, the height, and the strength of gravity.  Then we figure out the units of all these quantities, as well as the units of the answer.

 Quantity Units Mass of ball kilograms Height of ball meters Strength of gravity meters/second2 Time to fall seconds

How can we combine the three quantities to get the answer?  One thing's for sure, they'd have to be combined in a way that gives the correct units.  So here's a guess:

It's a pretty good guess too.  If you work out the problem using kinematics equations (which isn't really that hard), you get the same answer, but multiplied by the square root of two.  That means that the dimensional analysis was only off by 40%; it's almost correct by physics standards.  Dimensional analysis also successfully predicted that the time to fall does not depend on the mass of the bowling ball.

Math can be hard.  But in the end, the math just ends up with a number, like the square root of two.  Or maybe it ends up with one half.  Or 2*pi.  Or something.  It's highly unlikely that the math will end up with a factor of a thousand.  So whatever you get, you're probably at least within a factor of 10.

I've seen one exception where dimensional analysis is off by more than a factor of 10.  My last physics post discussed energy lost by radiation from a thermal pot.  According to the Stefan-Boltzmann law, the rate of energy loss per unit surface area is proportional to temperature to the fourth power.  What I didn't tell you is that the Stefan-Boltzmann law can almost be derived by dimensional analysis.

The first step is to figure out what quantities could be involved in the answer.  To do this, you need to know something about the physics, but not much.  It turns out that the answer depends only on the temperature and fundamental constants.

 Quantity Symbol Units Speed of light c meters/second Planck's constant h Joule seconds Boltzmann constant k Joules/Kelvin Temperature T Kelvin Power loss per unit area P/A Joules/(second meter2)

If you try dimensional analysis on this problem, you predict the following:
But it turns out that this answer is wrong.  The real answer is about 40 times bigger.  More precisely, it's bigger by a factor of 2*pi5/15. The reason for this is that there is some really nasty math involved.  And when I say the math is nasty, I'm not joking around.  At one point, you have to calculate the following:

I have no idea how to calculate that one.  I'd have to look it up (it's called the Riemann zeta function).

But the dimensional analysis wasn't a complete failure.  We at least showed that the power loss is proportional to temperature to the fourth power.  That's the most important result!

I have one last example of dimensional analysis used in particle physics.  One of the holy grails in particle physics is to figure out the correct way to combine General Relativity and Quantum Field Theory.  Now, nobody knows for sure the underlying physics of the situation.  However, we do know that we can combine fundamental constants to get a special length, called the Planck length.

The Planck length is calculated from Planck's constant (from quantum mechanics), the gravitational constant (from gravitational theory), and the speed of light (from relativity theory).  The result is a length that is mind-bogglingly small, 25 orders of magnitude smaller than an atom.  What does it mean?

The significance of the Planck length depends on what theory we're using.  String theorists think that the Planck length is about the length of a string.  If the universe has extra dimensions, perhaps these extra dimensions are Planck length in size. Some theorists think that space is quantized into lengths about the Planck length.

And since nobody really knows the underlying physics, nobody knows what mathematical factors may appear.  Maybe a factor of pi will show up.  At worst, a factor of 40 might appear.  But even if they're off by a lot, one thing's for sure: the Planck length is tiny!  If strings exist, strings will be tiny!  If space is quantized, it's quantized into really tiny pieces!  Dimensional analysis tells us that much.

I wonder how mathematicians and philosophers would react to the method of dimensional analysis.  I suspect a lot of head-banging would be involved.  How can those dang physicists be sure that this is sound reasoning?  Well, no one is really sure.  Luckily we can use experiments for external verification.

Larry Hamelin said...

We physics dummies need the intermediate steps. How do you get from mass, height & gravity to the square root of height/strength? And I'm completely lost in the temperature loss case. Can you work it out for us step by step?

miller said...

Abbreviations:
g = strength of gravity
h = height
t = time to fall
m = meters, units of length
s = seconds, units of time

Guess: t = g^a * h^b
...where a and b are some unknown exponents.

(Units of t) = s
(Units of g^a * h^b) = (m/s^2)^a * m^b = m^(a+b) * s^(-2a)

Therefore:
s = m^(a+b) * s^(-2a)
0 = a+b (powers of m)
1 = -2a (powers of s)

Solution:
a = -1/2, b = 1/2
t = g^(-1/2) * h^(1/2) = sqrt(h/g)

The Boltzmann Law involves more obscure units (What's a Joule? What's a Kelvin?), but the method is the same. It's not especially enlightening to show it out in detail.

Larry Hamelin said...

Guess: t = g^a * h^b

Why aren't you including mass? I mean, I know why: the gravitational force is proportional to mass, but the inertia is also proportional to mass, so they cancel out. But does this fall out of dimensional analysis or more fundamental considerations?

miller said...

You can take out mass, because that's the only quantity with units of kilograms. You could leave it in too, you'd get the same answer. schrodingasdawg said...

If you're interested in knowing how to calculate the zeta of four,

Write the Fourier series of a normalized triangle wave. Square the series and integrate over the period. Using the orthogonality of the sine functions, you'll get a series over (2n+1)^{-4} times some constant equals 1. Note that the zeta of four is just the sum over even n of n^{-4} and the sum over odd n of the same, and that the sum over even n is just 1/16 the zeta function itself. So zeta(4) is 16/15 times the sum of (2n+1)^{-4}.

As far as dimensional analysis goes, I think it's a need trick for deriving scaling laws and making rough estimates, but not much more. It won't work so well when you have multiple quantities with the same units, and I remain totally unconvinced that such things as Planck length, Planck mass etc. are physically meaningful. But then, I can be a pretty strong skeptic. schrodingasdawg said...

And by need trick I meant neat trick. What a silly typo...