This used to be on the MEI mechanics A-level syllabus. The chemistry department at PMC used to ask the Maths dept. to teach it in year ten so as to make their work easier. The result was, quite often, that the Physics dept. moaned that we were teaching them how to find answers without doing any Physics. Clearly miscommuni-cation.

Think of this as a good way to do mathematical modelling.

Here’s an example first, so as to set the scene:

I want a formula for the period of a pendulum. I think (these are assumptions) that the formula depends upon mass, gravity and pendulum length. So I’m guessing that there is a formula of the form Period=k (mass)^{a }(gravity)^{b} (length)^{c}.

Dimensionally, using square brackets from convention and M,L,T for mass length and time, then we have [T]1=[M]^{a} [LT-2]^{b} [L]^{c} which is [T]^{1}=M^{a}L^{b+c}T^{-2b} . Equating dimensions rapidly gives us a=0 —there’s no mass on the left side and a appears only once in the last equation, i.e. the period of a pendulum is independent of mass; a little work shows that b=-½ and c=+½ , i.e. that T=k(Ɩ/g)^{1/2}. Subsequent experiment might show that k = (2π)^{-½}. Or, if you prefer, T²= Ɩ / 2πg

In general, you assume that there is a formula of the general form A^{a}B^{b}C^{c} = X where all the capital letters represent physical properties in their usual units. Expressing these as mass length and time produces the indices for the units and leaves unknown a dimensionless (unit free) constant of proportionality. It is great to show that some factor you think maybe important is irrelevant, such as in that first example.

Suppose you think air resistance of cars depends upon the speed of the car, the density of the air and the length of the car. Air resistance would be a force, so measured in newtons, MLT^{-2}. Density is ML^{-3} and speed is LT^{-1}. Therefore we have MLT^{-2} = [LT^{-1}]^{a }[ML^{-3}]^{b} [L]^{c} = M^{b}L^{a-3b+c}T^{-a}

So M¹=M^{b}, L^{1}= L^{a-3b+c }and T^{-2 }= T^{-a}, all of which says a=2, b=1 and c=2,

Air resistance = k (speed)²(air density)¹(length)². So longer cars have lower air resistance and the air resistance depends upon the square of the speed. I have ignored other factors such as air viscosity—the inability of a fluid to move out of the way, a sort of fluid friction. There are two reasons for this; (i) you probably haven’t met the idea yet and (ii) this method only copes with three terms in a formula. I also haven’t included the cross-sectional area of the car as it will be confused by the length aspect.

Viscosity of the dynamic variety is measured in pascal seconds, which are newton seconds per square metre, or ML^{1-2}T^{-2+1}=ML^{-1}T^{-1}. Classically, consider a fluid flow at velocity u between two plates of area A and separation y and this viscosity thing, µ. Then the force, F, (from our point of view, some unidentified force, but the one resisting motion) can be found using dimensional analysis:

F=u^{a}A^{b}y^{c}µ^{d} =[LT^{-1}]^{a }[L²]^{b} [L]^{c} [ML^{-1}T^{-1}]^{d} => [MLT^{-2}]=M^{d}L^{a+2b+c-d}T^{-a-d}^{ }

so we know d=a=1 and that 2b+c=1, so we could support b=1, c=-1 i.e. **F= µuA/y,**

but at the moment we would say F= µu A^{b}y^{1-2b} or F= µuy (A/y²)^{b}

This exhibits the problem I was avoiding in looking at air resistance, in mixing two units that are only in length. It also exhibits some of the problem that occurs when we have more than three variables to handle—we don’t have enough equations to produce a definitive answer and so the result can at best be expressed in some mixture of those components.

If we suppress the area of the plates either side of tehg flow and keep the separation,

then F=u^{a }y^{b} µ^{c}

so [MLT^{-2}]=[LT^{-1}]^{a }[L]^{b} [ML^{-1}T^{-1}]^{c } => a=b=c=1 and F=uyµ but that implies that separation of the plates is directly proportional to the force experienced, which is not true (it is inversely so). What is true is that the force is directly proportional to *some* length component, but we have not yet ascertained what that component is (or, if you like, how to measure it).

DJS 20130530

...more to come if someone asks...

Exercise:

1. Develop a formula for an oscillating mass on the end of a spring (spring constant

dim(k)=MT^{-2}) in a gravity field.

2. The wikipedia article suggests you might consider the energy in a vibrating wire. Assume that the relevant features are length and amplitude (Ɩ and A, both L), tension (s, a force) and linear density, dim(ρ)=ML^{-1}. Show that no experiments are required to show that energy is directly proportional to the tension, nor any to show that energy is inversely proportional to the linear density.

3. Find the horizontal range of a projectile of initial velocity u and gravity g with vertical and horizontal components of velocity Vy and Vx respectively. Separate L into X and Y so that dim(Vy)=YT^{-1}.

DJS 20130530

1 T²=C m/k for some constant C that might be 2π. Note g is irrelevant - or at the very least apparently so.

2 dim(E)=ML^{2}T^{-2}=dim(l^{a}A^{b}s^{c}ρ^{d})=[L]^{a+b} [MLT^{-2}]^{c} [ML^{-1}]^{d} = M^{c +d} L^{a+b+c-d }T^{-2c} => c=1, d=-1; 0=a+b+1 so E = l^{a} /A^{a+1} s /ρ.

3 X=dim(V_{x}^{a}.V_{y}^{b}g^{c})= [XT^{-1}]^{a} [YT^{-1}]^{b} [YT^{-2}]^{c} = X^{a} Y^{b+c} T^{-(a+b+2c}) =>a=b=1, c=-1. So Range = k Vx.Vy /g and we might guess that the dot product is u², being scalar. However, wikipedia and others point out that vector products are not coped with well; if you’re interested, look up Siano.