# Example: Creating equations using dimensional analysis

## Understanding the situation

Since physical equations express relations among physical measurements, if we understand scaling relations (functional dependence) and have some physical insight into how things behave, we can sometimes generate hypotheses for physical results.

## Example: Throwing a ball straight up

To see how this works, let's consider the following problem.

Generate a formula that gives an estimate (up to a dimensionless constant) that shows how high a ball will go when we throw it up in the air and shows what that height depends on. Identify relevant parameters and build a formula for the maximum height, h, the ball reaches in terms of those parameters.

## Solving this problem

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 (radius $R$, mass $m$, color, deformability,...), and its kinematic descriptive parameters (position = ($x, y, z$), velocity = ($v_x, v_y, v_z$), 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 ways. So let's start with the simplest system: ignore the air and treat the earth as flat so g is a constant. With these assumptions, we know the size and shape of the ball will not matter, so the logical parameters for us to consider are:

• The ball's initial position, ($x, y, z$)
• The ball's initial velocity, ($v_x, v_y, v_z$)
• The ball's mass. $m$
• The earth's gravitational field, $g$

Clearly, moving in three dimensions is 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 parameters that have a dimensionality (are created from measurements): $y, v_y, m$, 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 additive way. So let's take the initial value of $y= 0$ as our release point.

We are now left with three dimensioned variables.

• The ball's initial upward velocity, ($v_y$)
• The ball's mass. $m$
• The earth's gravitational field, $g$

We'd like to know how high the ball will go. The best way to do that, of course, is to use physical principles (for example, Newton's 2nd law or The conservation of mechanical energy.) Interestingly enough, if we are primarily interested in questions of functional dependence and scaling, we can often answer without using any physical principles at all, other than the kinds of measurements that are involved!

This is a rather dramatic results. Suppose I know how high the ball goes when it goes up with a velocity $v_0$ and want to know, "How fast does it need to go in order to go twice as high?" Or, "I know how high one object goes with a velocity $v_0$. How high will an object of twice the mass go if it has that same velocity?"

To see how this works, let's see if  we can construct a combination of the input data of the problem that yields a distance as output. This in only a question of dimensional analysis.

#### Building a distance using dimensional equations

We can't combine these quantities additively since they have different dimensions. (See Complex Dimensions and Dimensional Analysis for details.) So we have to multiply them. A general way to combine them is to raise them each to an arbitrary power.

$$m^a (v_y)^b g^c$$

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

$$[m^a (v_y)^b g^c] = (\text{M})^a (\text{L/T})^b (\text{L/T}^2)^c = \text{M}^a \text{L}^{b-c} \text{T}^{-b-2c}$$

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, and c = -1 which gives the result:

$$h \sim {v_y}^2 /g.$$

We use the symbol "$\sim$" 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.

This tells us a number of important results:

1. In the absence of air, the height the ball rises does not depend on its mass, only on its initial velocity.
2. The height it rises depends on the square of the velocity with which it is thrown upward, so throwing it up twice as fast will make it go 4 times as high.

Neither of these results are immediately obvious from just thinking about throwing a ball.

## Putting the example in perspective

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 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.

Joe Redish 9/11/05

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