"You can't add apples and oranges," the old saying goes.
It reflects our experience that in applications the quantities
have units and keeping track of those units is worthwhile.
Everyone has done calculations such as this one
that use the units as a check.
However, the idea of including the units can be taken beyond bookkeeping.
It can be used to draw conclusions
about what relationships are possible among the physical quantities.
To start, consider the physics equation:
If the distance is in feet and the time is in seconds then this is a
true statement about falling bodies.
However it is not correct in other unit systems;
for instance, it is not correct in
the meter-second system.
We can fix that by making the a dimensional constant.
For instance, the above equation holds in the yard-second system.
So our first point is that
by "including the units" we mean that we are restricting our attention
to equations that use dimensional constants.
By using dimensional constants,
we can be vague about units
and say only that all quantities are measured in combinations
of some units of length , mass , and time .
We shall refer to these three
(these are the only three
dimensions that we shall need in this Topic).
For instance, velocity
could be measured in
or , but in all events it involves
some unit of length divided by some unit of time
so the dimensional formula of velocity is .
Similarly, the dimensional formula of density is .
We shall prefer using negative exponents over
the fraction bars and we shall include the dimensions with a
zero exponent, that is, we shall write the dimensional formula of
velocity as and that of density
In this context, "You can't add apples to oranges" becomes
the advice to check that all of an equation's terms have
the same dimensional formula.
An example is this version of the falling body equation: .
The dimensional formula of the term is .
For the other term, the dimensional formula of is
( is the dimensional constant given above as )
and the dimensional formula of is , so that of
the entire term is
Thus the two terms have the same dimensional formula.
An equation with this property is dimensionally homogeneous.
Quantities with dimensional formula are dimensionless.
For example, we measure an angle by taking
the ratio of the subtended arc to the radius
which is the ratio of a length to a length and
thus angles have the dimensional formula .
The classic example of using the units for more than bookkeeping, using them
to draw conclusions, considers the formula for the period of a pendulum.
The period is in units of time .
So the quantities on the other side of the equation must
have dimensional formulas that combine in such a way that
their 's and 's cancel and only a
The table on below has the quantities that
an experienced investigator would consider possibly relevant.
The only dimensional formulas involving are for
the length of the string and the acceleration due to gravity.
For the 's of these two to cancel, when they appear
in the equation they must be in ratio,
e.g., as , or as ,
or as .
Therefore the period is a function of .
This is a remarkable result: with a pencil and paper analysis,
before we ever took out the pendulum and made measurements,
we have determined something about the relationship
among the quantities.
To do dimensional analysis systematically, we need to know two things
(arguments for these are in (Bridgman 1931), Chapter II and IV).
The first is that each equation relating physical
quantities that we shall see involves a sum of terms,
where each term has the form
for numbers , ..., that measure the quantities.
For the second, observe that an easy way to construct a dimensionally
homogeneous expression is by taking a product of dimensionless quantities
or by adding such dimensionless terms.
Buckingham's Theorem states that any
complete relationship among quantities with
dimensional formulas can be algebraically manipulated into a form where
there is some function such that
for a complete set of dimensionless products.
(The first example below describes what makes a set of dimensionless
We usually want to express one of the quantities, for instance,
in terms of the others,
and for that we will assume that the above equality can be rewritten
where is dimensionless
and the products , ..., don't involve
(as with , here is just some function, this time
Thus, to do dimensional analysis we should
find which dimensionless products are possible.
For example, consider
again the formula for a pendulum's period.
length of string
mass of bob
acceleration due to gravity
arc of swing
By the first fact cited above, we expect the formula to
have (possibly sums of terms of) the form
To use the second fact, to find which combinations of the
powers , ...,
yield dimensionless products, consider this equation.
It gives three conditions on the powers.
Note that is and so the mass of the bob does not affect the period.
Gaussian reduction and parametrization of that system gives this
(we've taken as one of the parameters in order to express
the period in terms of the other quantities).
Here is the linear algebra.
The set of dimensionless products
contains all terms subject
to the conditions above.
This set forms a vector space under the
"" operation of multiplying two such products
and the "" operation of raising
such a product to the power of the scalar
(see Problem 5).
The term "complete set of dimensionless products"
in Buckingham's Theorem means a basis for this vector space.
We can get a basis
by first taking , and then , .
The associated dimensionless
products are and .
Because the set is complete,
Buckingham's Theorem says that
where is a function that we cannot determine from this analysis
(a first year physics text will show by other means
that for small angles it is approximately the constant
Thus, analysis of the relationships that are possible between the
quantities with the given dimensional formulas
has produced a fair amount of information: a pendulum's period does not
depend on the mass of the bob,
and it rises with the square root of the length of the string.
For the next example we try to determine the period of revolution of
two bodies in space orbiting each other under mutual gravitational attraction.
An experienced investigator could expect that these are the relevant
To get the complete set of dimensionless products we consider the
which results in a system
with this solution.
As earlier, the linear algebra here is that
the set of dimensionless products of these quantities forms a vector space,
and we want to produce a basis for that space,
a "complete" set of dimensionless products.
One such set, gotten from setting and ,
and also setting and is
With that, Buckingham's Theorem says that
any complete relationship among these quantities is stateable this form.
An important application of the prior formula is
when is the mass of the sun and is the mass
of a planet.
Because is very much greater than ,
the argument to is approximately ,
and we can wonder whether this part of the formula remains approximately
constant as varies. One way to see that it does is this.
The sun is so much larger than the planet that the mutual rotation is
approximately about the sun's center. If we vary the planet's mass
by a factor of (e.g., Venus's mass is
times Earth's mass), then the force of attraction
is multiplied by , and times the force
acting on times the mass gives, since
, the same acceleration, about the same center
(approximately). Hence, the orbit will be the same and so its period
will be the same, and thus the right side of the above equation also
remains unchanged (approximately). Therefore,
is approximately constant as
varies. This is Kepler's Third Law: the square of the
period of a planet is proportional to the cube of the mean radius of
its orbit about the sun.
The final example was one of the first explicit applications of
dimensional analysis. Lord Raleigh considered the speed of a wave in
deep water and suggested these as the relevant quantities.
velocity of the wave
density of the water
acceleration due to gravity
gives this system
with this solution space
(as in the pendulum example, one of the quantities turns out not to be
involved in the relationship).
There is one dimensionless product, ,
and so is times a constant
( is constant since it is a function of no arguments).
As the three examples above show, dimensional analysis
can bring us far toward expressing the relationship among the quantities.
For further reading,
the classic reference is (Bridgman 1931)—this brief book is delightful.
Another source is (Giordano, Wells & Wilde 1987)..
A description of dimensional analysis's place in
modeling is in (Giordano, Jaye & Weir 1986)..
Consider a projectile, launched with initial velocity , at an angle . An investigation of this motion might start with the guess that these are the relevant quantities. (de Mestre 1990)
angle of launch
acceleration due to gravity
Show that is a complete set of dimensionless products. (Hint. This can be done by finding the appropriate free variables in the linear system that arises, but there is a shortcut that uses the properties of a basis.)
These two equations of motion for projectiles are familiar: and . Manipulate each to rewrite it as a relationship among the dimensionless products of the prior item.
Einstein (Einstein 1911) conjectured that the infrared characteristic frequencies of a solid may be determined by the same forces between atoms as determine the solid's ordanary elastic behavior. The relevant quantities are
number of atoms per cubic cm
mass of an atom
Show that there is one dimensionless product. Conclude that, in any complete relationship among quantities with these dimensional formulas, is a constant times . This conclusion played an important role in the early study of quantum phenomena.
The torque produced by an engine has dimensional formula
. We may first guess that it depends on the
engine's rotation rate (with dimensional formula
), and the volume of air displaced (with
dimensional formula ) (Giordano, Wells & Wilde 1987).
Try to find a complete set of dimensionless products. What goes wrong?
Adjust the guess by adding the density of the air (with dimensional formula ). Now find a complete set of dimensionless products.
Dominoes falling make a wave. We may conjecture that the wave speed depends on the spacing between the dominoes, the height of each domino, and the acceleration due to gravity . (Tilley)
Find the dimensional formula for each of the four quantities.
Show that is a complete set of dimensionless products.
Show that if is fixed then the propagation speed is proportional to the square root of .
Prove that the dimensionless products form a vector space under the operation of multiplying two such products and the operation of raising such the product to the power of the scalar. (The vector arrows are a precaution against confusion.) That is, prove that, for any particular homogeneous system, this set of products of powers of , ...,
is a vector space under:
(assume that all variables represent real numbers).
The advice about apples and oranges is not right. Consider the familiar equations for a circle and .
Check that and have different dimensional formulas.
Produce an equation that is not dimensionally homogeneous (i.e., it adds apples and oranges) but is nonetheless true of any circle.
The prior item asks for an equation that is complete but not dimensionally homogeneous. Produce an equation that is dimensionally homogeneous but not complete.
(Just because the old saying isn't strictly right, doesn't keep it from being a useful strategy. Dimensional homogeneity is often used as a check on the plausibility of equations used in models. For an argument that any complete equation can easily be made dimensionally homogeneous, see (Bridgman 1931, Chapter I, especially page 15.)