Why is it interesting?

Dimensional analysis is an extremely powerful tool that allows us to derive the solution for many complicated systems or equations, without doing any actual calculations.

For example, complicated Gaussian integrals can be "calculated" using dimensional analysis, up to some constant, using solely dimensional analysis. Another example, is the speed of an apple at the ground after being dropped from some height, or the frequency of a pendulum. For more details, see the "Examples" tab.

In addition, dimensional analysis is one of the most important tools in quantum field theory.

Layman

Student

Looking at the Lagrangian density in (1), we can easily work out what the units of the constant e, $\mu$, G, etc., are. All terms in the Lagrangian density must have units [mass]$^4$, because length and time have units of inverse mass and the Lagrangian density integrated over spacetime must have no units. From the $m\bar{\Psi}\Psi$ term, we see that the electron field must have units [mass]$^{3/2}$, because $\frac{3}{2}+\frac{3}{2}+1 = 4$. The derivative operator (the rate of change operator) has units of [mass]$^1$, and so the photon field also has units [mass]$^1$. Now we can work out what the units of the coupling constants are. As I said before, the electric charge turns out to be a pure number, to have no units. But then as you add more and more powers of fields, more and more derivatives, you are adding more and more quantities that have units of positive powers of mass, and since the Lagrangian density has to have fixed units of [mass]$^4$, therefore the mass dimensions of the associated coupling constants must get lower and lower, until eventually you come to constants like $\mu$ and G which have negative units of mass. (Specifically, $\mu$ has the units of [mass]$^{-1}$, while G has the units [mass]$^{-2}$.) Such terms in (1) would completely spoil the agreement between theory and experiment for the magnetic moment of the electron, so experimentally we can say that they are not there to a fantastic order of precision, and for many years it seems that this could be explained by saying that such terms must be excluded because they would give infinite results, as in (4).

Of course, that is exactly what we are looking for: a theoretical framework based on quantum mechanics, and a few symmetry principles, in which the specific dynamical principle, the Lagrangian, is only mathematically consistent if it takes one particular form. At the end of the day, we want to have the feeling that "it could not have been any other way".

Towards the final laws of physics by Steven Weinberg

Recommended Resources:

• A brilliant discussion with many great examples can be found in chapter 1 of Mahajan's book "Stree-fighting Mathematics".
• https://particlephd.wordpress.com/2008/12/08/dimensional-analysis-for-animals/
• see http://math.ucr.edu/home/baez/renormalizability.html for how dimensional analysis is used in quantum field theory.

Researcher

Recommended Resources:

• Dimensional Analysis in field theory by P.M Stevenson

Common Question 1

Common Question 2

Examples

Speed of an apple after being dropped from some height h

Our task is to calculate the speed of an apple right before it hits the ground, after being dropped from some height $h$. The relevant quantities are the height $h$ with units [meters] and the gravitational acceleration $g$ with units [meters/seconds$^2$].

We search for a velocity, which has units [meters/seconds]. Hence, the solutions must be proportional to $\sqrt{gh}$, because this is the only possible combination with units [meters/seconds]. The full correct solution is indeed $v_f = - 2 \sqrt{gh}$.

This example is from page 4 of Mahajan's "Street-fighting Mathematics" book.

Frequency of a pendulum

Gaussian integral

Dimensional analysis is not only important in physics, but can be used in pure mathematics, too. Let's consider a general Gaussian integral

$$\int_{-\infty}^\infty e^{- \alpha x^2} dx = ???$$

No quantity here has units, because it's a pure math problem. However we are free, of course, to assign some as we wish.

First, we need to note that the argument of the exponential function is not allowed to have any units. This is necessary, because we can write the exponential function as a series (Taylor series): $e^y = 1 +y +y^2/2 + \ldots$. Thus, if $y$ had units, we would have to add objects with different units, which doesn't make sense. (Say $y$ had units $[m]$, then $y^2$ had units $[m^2]$ etc.)

With this in mind, we now assign dimensions to the quantities that appear in our integral. This is only a tool to analyse the integral and there is no physical meaning whatsoever involved here. You can assign any units that you want, as long as the argument of the exponential function has no units. A convenient choice for the units of $x$ is $[m]$. Then $\alpha$ mus have units $[m^{-2}]$, because then the exponent has no units.

Then $e^{- \alpha x^2}$ has no units. What about the rest of our expression? Well $\int_{-\infty}^\infty$ has no units, and $dx$ is simply a little bit of $x$ and hence has the same units as $x$. Therefore our solution on the right-hand side must have units $[m]$.

We integrate over $x$ and therefore, $x$ is not allowed to appear in the solution. Thus, we know immediately that our solution must be proportional to $1 / \sqrt{\alpha}$, because $\alpha$ has units $[m^{-2}]$ and therefore $1 / \sqrt{\alpha}$ has units $[m]$ as needed. Therefore:

$$\int_{-\infty}^\infty e^{- \alpha x^2} dx \sim 1 / \sqrt{\alpha} .$$

We can find the missing constant, by considering a simple case like $\alpha = 1$. This well known integral is

$$\int_{-\infty}^\infty e^{- x^2} dx = \sqrt{\pi} .$$

Our reasoning using dimensional analysis is of course valid for all values of $\alpha$ and thus also for $\alpha =1$. Thus, we can conclude that the full solution is

$$\int_{-\infty}^\infty e^{- \alpha x^2} dx = \sqrt{\pi / \alpha} .$$

This example is from page 8 of Mahajan's "Street-fighting Mathematics" book.

(Take note that integrals are not always giving an area as a result. This really depends on the function $f(x)$ that appears in the integral $\int_{-\infty}^\infty f(x) dx$. If $f(x)$ has units $[m]$, we get indeed an area. But if $f(x)$ has no units, as in the example above, we don't get an area as a result.)

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