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# Dimensions, dimensional homogeneity, and independent dimensions

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.

Next: Examples Up: Dimensional Analysis Previous: The SI system of

Vittorio Celli
Thu Jul 10 16:27:59 EDT 1997