## Saturday, March 18, 2006

### Dimensional Analysis

What is Dimensional Analysis, and what is it for?

Dimensional analysis is a technique used by physicists (and other scientists) to check the validity of equations. It is often also used to arrive at the equations in the first place! The basic principle of Dimensional analysis is the common sense principle that you need to compare apples with apples. In physics, this translates to “for two quantities to be equal, they must have the same dimensions.” That is, if the object on the left hand side of an equals sign must have the same dimensions as the quantity on the right hand side.

But, what are dimensions?

The dimension of an object tells you what sort of quantity it is. There are four basic dimensions, Length (L), Mass (M), Time (T) and Electric Charge (Q), corresponding to the four basic quantities. The dimensions of more complicated quantities can be expressed as powers of these four. A useful property of dimension is that the dimension of a product, is the product of the dimensions, that is the dimension of x*y is the dimension of x times the dimension of y. Furthermore, the arguments to exponential, trigonometric and logarithmic functions must be dimensionless numbers, which is often achieved by multiplying a certain physical quantity by a suitable constant of the inverse dimension.

For example, you might want to know the dimensions of a force: From Newton’s Laws, we know that force is mass times acceleration, and as the SI units for acceleration are m/s2, we know that the dimensions of force must be the product of the dimensions of mass (M) and acceleration (L/T2). That is, the dimensions of force (usually written [force]) are ML/T2. Similarly, [velocity] is L/T.

How do I use this to find equations?

Say you’re in a basic motion exam, and you are trying to solve a problem like

“a 10kg ball is thrown upwards from ground level on the moon (where gravity is one sixth that of earth) with an initial speed of 15 m/s. How high will it be after 2 seconds, 5 seconds and 10 seconds? When will it bounce?”

Now, you remember that there was a formula in your notes that told you the displacement (s) of a mass after some time (t), given its initial velocity (u) and the constant acceleration (a), but you can’t quite remember it – was it

“s=ut2+1/2 a2t” or “s=ut+1/2 at”?

you’re not sure about the exponents (the “squares”), but you are pretty sure that you’ve got the symbols in the right places, and you know that that ½ is definitely there – how do you go about checking the formula?

Let’s start with the first guess, and compare the dimensions of the terms:

[s]=L as displacement is a length, so we need the dimensions of the other two terms to L also. [ut^2] = [u][t]^2=L/T*T2=LT! Clearly this term is incorrect, but let’s check the second term as well:

[1/2 at]=[1/2][a][t], but ½ is just a number – or in the terminology of physics: it is dimensionless, so we can say [1/2]=1. Then [1/2 at]=1*L2/T4*T=L2/T3! Not only does this term not match the displacement, it doesn’t even match the ut2 term it is added to! Just as we cannot compare apples and oranges, we can’t add apples to oranges!

If you have a look at the second guess, you’ll see that it is also wrong, so we need another guess – but, if we look at our first guess, the first term had am extra T, so if we remove one of the times from it, it will have the correct dimension so it must be ut. The second term is wrong by a factor of L/T3, which could be explained by having one too many accelerations and one too few times – so it should be ½ at2. No doubt you’ve now seen one of the shortcomings of dimensional analysis – at no point in our calculation, have we been able to make any conclusion about whether the ½ was correct – all we know is that there is some dimensionless object in front of each term – it could be one, it could be ½ or it could be p – and there could be a different one in front of the ut term. Unfortunately, dimensional analysis doesn’t tell us anything about particular dimensionless constants, those you will have to remember (or work out some other way – if you know calculus, the laws for constant acceleration come from integrating Newton’s Second Law – twice in this case).

So know we know that “s=ut+1/2 at2” with out having had to remember it exactly. This is a useful exam time saver – if you’re struggling to recall a formula, you can at least rule out dimensionally inconsistent guesses quickly, leaving more time to find the correct option.

Can it do more?

Of course it can – with Dimensional analysis, you can make a pretty good guess as to the structure of equations you have never been shown.

How can I possibly guess equations I’ve never been taught?

Well, first of all, we have to go back to the principle of comparing apples to apples. So, if say you want to come up with the formula for the vibration frequency of a mass on a spring, you know that what ever the formula you’re looking for is, it has to have the same dimensions as frequency. As frequency is a measure of how many times something happens per unit of time, you can see that [frequency] is T-1.

The second step is to identify all the things that the frequency could depend upon. If you’ve studied springs at all, you will know that they obey Hooke’s Law: the restoring force (F) pulling (or pushing) a stretched (or compressed) spring back to its equilibrium point is proportional to the distance (s) the spring has been stretched or compressed. The constant of proportionality, traditionally called “k”, completely describes how stiff the spring is (so you don’t need to worry about the properties of the metal the spring is made from or the physical size and shape of the spring). Also, it seems clear that heavy masses and light masses will move at different speeds, so the only terms we need to worry about are the spring constant (k) and the mass (m).

So, we guess that the frequency will have some sort of power relationship to the mass and spring constant:

“f=mx*ky

where x and y are constants that are to be determined. Now, we apply the apples to apple rule and choose x and y so that the dimensions on the left and right hand side match.

[f]=1/T

[m]x=Mx

[k]y=??

To evaluate the dimensions of k, we need to go back to our definitions:

Hookes Law: F=-ks (the minus sign indicates that the force is in the opposite direction to the stretching)

So, from Hooke’s Law, we know that [k]=[F]/[s]=[F]/L.

Newtons Second Law: F=ma thus, [F]=[m][a]=ML/T2.

Therefore, [k]=M/T2.

So then

[mxky]=Mx*(M/T2)y=M(x+y)/T2y

so, if this is to be equal to 1/T, we need to satisfy

x+y=0 and 2y =1

so x = -½ and y = ½.

Thus, we have shown that:

f a (k/m)1/2 (that is, the frequency is proportional to the square root of the spring constant divided by the mass on the end of the spring).

In fact, I know that f is exactly equal to (k/m)1/2, but the dimensional analysis technique cannot show that. For that, we need to use some calculus, which will be the subject of another article.