Returning to the discussion above, recall that if the units of length are
changed by a factor of L, and the units of time are changed by a factor of
T , then the units of velocity change by a factor of 
. We call the dimensions
of the velocity; it tells
us the factor by which the numerical value of the velocity changes under a
change in the units (within the LMT-class). Following a convention
suggested by Maxwell, we denote the dimensions of a physical quantity 
by 
; thus, 
. A dimensionless quantity would have 
; i.e., its numerical value is the same in all systems of units
within a given class. What about more complicated quantities such as force?
From Newton's second law, F=ma, so that 
. Proceeding
in this way, we can easily construct the dimensions of any physical
quantity; some of the more commonly encountered quantities are included in
Table 1.1.
| [L] | Length | L |
| Mass | M | |
| Time | T | |
| Velocity | | |
| Acceleration |  | |
| Force |  | |
 | Mass density |  |
| Pressure |  | |
 | Angle | 1 |
| Energy |  | |
 | Temperature |  |
| Entropy |  | |
| Electric current | I | |
| Electric charge | I T | |
| Electric field |  | |
| Magnetic field |  |
We see that all of the dimensions in the Table are power law monomials, of the form (in the LMT-class)
where C and (a,b,c) are constants. In fact, this is a general result
which can be proven mathematically; see Sec. 1.4 of Barenblatt's book. This
property is often called dimensional homogeneity,
and is really the key to dimensional analysis. To see why this
is useful, consider again the determination of the period of a point
pendulum, in a more abstract form. We have for the dimensions 
, 
, [l]=L, and [m]=M. If 
is a
function of (g,l,m), then its dimensions must be a power-law monomial of
the dimensions of these quantities. We then have
with a, b, and c constants which are determined by comparing the dimensions on both sides of the equation. We see that
The solution is then a=1/2, b=-1/2, c=0, and we recover Eq. (1.2).
A set of quantities 
is said to have independent dimensions
if none of these quantities
have dimensions which can be represented as a product of powers of the
dimensions of the remaining quantities. As an example, the density ( 
), the velocity ( ), and the force ( 
) have independent dimensions, so that there is no product of powers of these quantities which is dimensionless (note 1)
On the other
hand, the density, velocity, and pressure ( 
) are not
independent, for we can write 
; i.e., 
is a
dimensionless quantity.
Now suppose we have a relationship between a quantity a which is being
determined in some experiment (which we will refer to as the governed
parameter), and a set of quantities 
which are under
experimental control (the governing parameters). This relationship is
of the general form
where have independent dimensions. This means that
the dimensions of the governed parameter a are determined by the
dimensions of , while all of the 
's with s>k
have dimensions that can be written as products of powers of of the
dimensions of ; e.g., 
, would be dimensionless, with 
an appropriately
chosen set of constants. With this set of definitions, it is possible to
prove that Eq. (1.7) can be written as
with 
some function of dimensionless quantities only. The great
simplification is that while the function f in Eq. (1.7) was a
function of n variables, the function in Eq. (1.8) is
only a function of n-k variables. Eq. (1.8) is a mathematical
statement of Buckingham's 
-Theorem,
which is the central result of dimensional analysis. The formal
proof can again be found in Barenblatt's book. Dimensional analysis cannot
supply us with the dimensionless function - we need a real theory
for that.
As a simple example of how this works, let's return to the pendulum, but
this time we'll assume that the mass can be distributed, so that we relax
the condition of the mass being concentrated at a point. The governed
parameter is the frequency ; the governing parameters are g, l
(which we can interpret as the distance between the pivot point and the
center of mass), m, and the moment of inertia about the pivot point, I.
Since 
, the set (g,m,l,I) is not independent; we can choose as
our independent parameters (g,m,l) as before, with 
a
dimensionless parameter. In the notation developed above, n=4 and k=3.
Therefore, dimensional analysis tells us that
with some function which cannot be determined from dimensional
analysis alone; we need a theory in order to determine it.
Footnote:
(1) Prove this formally by writing [F]a[v]b
[
]c = 1,
and then show that the only solution is a = b = c = 0.
Alternatively, show that it is impossible to write [F] =
[]a
[v]b for any a,b.
-class.
