Dimensional analysis is a mathematical tool often applied in physics, chemistry, and engineering to simplify a problem by reducing the number of variables to the smallest number of "essential" parameters. Systems which share these parameters are called similar and do not have to be studied separately.
The dimension of a physical quantity is the type of unit needed to express it. For instance, the dimension of a speed is distance/time and the dimension of a force is mass×distance/time². In mechanics, every dimension can be expressed in terms of distance (which physicists often call "length"), time, and mass, or alternatively in terms of force, length and mass. Depending on the problem, it may be advantageous to choose one or the other set of fundamental units. Every unit is a product of (possibly fractional) powers of the fundamental units, and the units form a group under multiplication.
In the most primitive form, dimensional analysis is used to check the correctness of algebraic derivations: in every physically meaningful expression, only quantities of the same dimension can be added or subtracted. The two sides of any equation must have the same dimensions. 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.
The above mentioned reduction of variables uses the Buckingham π-theorem as its central tool. This theorem describes how every physically meaningful equation involving n variables can be equivalently rewritten as an equation of n-m dimensionless parameters, where m is the number of fundamental units used. Furthermore, and most importantly, it provides a method for computing these dimensionless parameters from the given variables, even if the form of the equation is still unknown.
Two systems for which these parameters coincide are called similar; they are equivalent for the purposes of the equation, and the experimentalist who wants to determine the form of the equation can choose the most convenient one.
The π-theorem uses linear algebra: the space of all possible physical units can be seen as a vector space over the rational numbers if we represent a unit as the set of exponents needed for the fundamental units (with a power of zero if the particular fundamental unit is not present). Multiplication of physical units is then represented by vector addition within this vector space. The algorithm of the π-theorem is essentially a Gauss-Jordan elimination carried out in this vector space.
A typical application of dimensional analysis occurs in fluid dynamics. If a moving fluid meets an object, it exerts a force on the object, according to a complicated (and not completely understood) law. The variables involved are: the speed, density and viscosity of the fluid, the size of the body (expressed in terms of its frontal area A), and the force. Using the algorithm of the π-theorem, one can reduce these five variables to two dimensionless parameters: the drag coefficient and the Reynolds number.
That this is so becomes obvious when the force F is expressed as part of a function of the other variables in the problem:
- f(F,u,A,ρ,ν) = 0.
This rather odd form of expression is used because it does not assume a one-one relationship. Here, f is some function (as yet unknown) that takes five arguments. We note that the right hand side is zero in any system of units; so it should be possible to express the relationship described by f in terms of only dimensionless groups.
There are many ways of combining the five arguments of f to form dimensionless groups, but the Buckingham Pi theorem states that there will be two such groups. The most appropriate are the Reynolds number, given by
and the drag coefficient, given by
Thus the original law involving a function of five variables may be replaced by one involving only two:
where f is some function of two arguments. The original law is then reduced to a law involving only these two numbers.
Because the only unknown in the above equation is F, it is possible to express it as
- F = ρAu2f(Re).
Thus the force is simply ρAu2 times some (as yet unknown) function of the Reynolds number: a considerably simpler system than the original five-argument function given above.
Dimensional analysis thus makes a very very complex problem (trying to determine the behaviour of a function of five variables---essentially impossible in practice) a very much simpler one: the determination of a function of one variable, the Reynolds number.
The analysis also gives other information for free, so to speak. We know that, other things being equal, the drag force will be proportional to the frontal area of the body and to the density of the fluid. This kind of information often proves to be extremely valuable, especially in the early stages of a research project.
For everyday applications, the Reynolds number dependence is generally quite weak (for an exception, see the next paragraph) and if this is neglected (often a very reasonable proposition) the drag is proportional to the square of the speed. In practice, drag is very closely approximated by ignoring the Reynolds number dependence, ie
where k is some constant (physical analysis shows that k is one, to within an order of magnitude).
To empirically determine the Reynolds number dependence, instead of experimenting on huge bodies with fast flowing fluids (such as real-size airplanes in wind-tunnels), one may just as well experiment on small models with slow flowing, more viscous fluids, because these two systems are similar.
Consider Einstein's well-known equation E = mc². As stated above, the two sides of any equation must have the same dimensions. We can check this as follows.
- E is energy, which has units of mass × length² / time². (This is because energy = force × length, and force = mass × acceleration, and acceleration = length / time².)
- m is mass, which is a unit on its own.
- c is speed, which has units of length / time.
- The left-hand side, E, therefore has units of mass × length² / time².
- The right-hand side, mc², has units of mass × (length / time)² = mass × length² / time².
- The two sides therefore have the same dimensions.
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