Dimensions

New Physics from Dimensional Analysis

Since physical equations express relations among physical measurements, if we understand scaling relations (see the discussion of considering change) and have some physical insight into how things behave, we can sometimes generate hypotheses for physical results.

An Example: Throwing a Ball

To see how this works, let's consider the problem of deciding how high a ball will go when we throw it up in the air. We want as our result, the maximum height, h, the ball reaches. To apply dimensional analysis, we consider the system and consider what dimensioned physical parameters we know that might have some relevance to the height.

We can begin by being rather complete. We have the ball, its properties (R, m, color, deformability,...), and its kinematic descriptive parameters (position = (x, y, z), velocity = (vx, vy, vz), acceleration = ...). We have the systems external to the ball acting on it (the earth's gravitational field = g, the density of the air = ρ,...).

What's the physics?

For a complex system, we may not be able to get an answer by dimensional analysis, since there may be many different ways of generating a length and once you have more than one, they may combine in complex (even in non-analytic) ways. So let's start with the simplest system: ignore the air and treat the earth as flat so g = const. With these assumptions, we know the size and shape of the ball will not matter, so the logical parameters for us to consider are:

Clearly, the three dimensions are going to be a problem. (We can improve things by using the vector character and writing vector-correct equations.) For now, let's simplify by assuming we are only working with one dimension -- up and down (typically, y). This then reduces us to 4 dimensioned parameters, y, vy, m, and g. It is clear that the initial position is going to matter, but it should come in easily. The height it goes after we release it should just be measured from our starting point. Changing our starting point will just change the result in a simple linear way. So let's take the initial value of y = 0 at our release point.

We are now left with three dimensioned variables.

From these, how can we construct a distance?

Dimensional Equations

We can't combine these quantities additively, since they have different dimensions. (See complex dimensions.) So we have to multiply them. A general way to combine them is to raise them each to an arbitrary power.

ma (vy)b gc.

Since [m] M, [v] = L/T and [g] = L/T2, this combination has the dimensions:

[ma (vy)b gc] = (M)a (L/T)b (L/T2)2 = MaLb+c T-b-2c.

If we want this to be a length ([h] = L) we must satisfy the equations:

a = 0
b+c = 1
-b-2c = 0.

These are easily solved to find:

a = 0
b = 2
c = -1.

This gives the resultL

h ~ (vy)2/g.

We use the "~" instead of "=" since there may be a dimensionless constant in front. In fact, there is: a "1/2" that we cannot derive using dimensional analysis. But we get pretty close without actually solving for any motion.

In general, if we want to combine dimensioned quantities to get a quantity of a particular dimension we can use a procedure like this. Notice, however, how much physics we put into the argument. Using dimensional analysis to create a solution is an art, relying heavily on one's understanding of the physics. It's not an algorithm.

RETURNS

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This page prepared by

Edward F. Redish
Department of Physics
University of Maryland
College Park, MD 20742
Phone: (301) 405-6120
Email: redish@umd.edu

Last revision 11. September, 2005.