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What is Dimensional Analysis?
What is Dimensional Analysis?
The recognition of concepts of dimensions, which guide the description of physical behaviour is of basic importance as only those physical quantities can be added or subtracted which have the same dimensions. A thorough understanding of dimensional analysis helps us in deducing certain relations among different physical quantities and checking the derivation, accuracy and dimensional consistency or homogeneity of various mathematical expressions. When magnitudes of two or more physical quantities are multiplied, their units should be treated in the same manner as ordinary algebraic symbols.We can cancel identical units in the numerator and denominator. The same is true for dimensions of a physical quantity. Similarly, physical quantities represented by symbols on both sides of a mathematical equation must have the same dimensions
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Checking the Dimensional Consistency of Equations
Checking the Dimensional Consistency of Equations
The magnitudes of physical quantities may be added together or subtracted from one another only if they have the same dimensions. In other words, we can add or subtract similar physical quantities. Thus, velocity cannot be added to force, or an electric current cannot be subtracted
from the thermodynamic temperature. This simple principle called the principle of homogeneity of dimensions in an equation is extremely useful in checking the correctness of an equation. If the dimensions of all the terms are not same, the equation is wrong. Hence, if we derive an expression for the length (or distance) of an object, regardless of the symbols appearing in the original mathematical relation, when all the individual dimensions are simplified, the remaining dimension must be that of length. Similarly, if we derive an equation of speed, the dimensions on both the sides of equation, when simplified, must be of length/ time, or [L T–1].
Dimensions are customarily used as a preliminary test of the consistency of an equation, when there is some doubt about the correctness of the equation. However, the dimensional consistency does not guarantee correct equations. It is uncertain to the extent of dimensionless quantities or functions. The arguments of special functions, such as the trigonometric, logarithmic and exponential functions must be dimensionless. A pure number, ratio of similar physical quantities, such as angle as the ratio (length/length), refractive index as the ratio (speed of light in vacuum/speed of light in medium) etc., has no dimensions.
Now we can test the dimensional consistency or homogeneity of the equation
\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9iaadIhadaWgaaWcbaGaaGimaaqabaGccqGHRaWkcaWG2bWaaSba % aSqaaiaaicdaaeqaaOGaamiDaiabgUcaRiaacIcacaaIXaGaai4lai % aaikdacaGGPaGaamyyaiaadshadaahaaWcbeqaaiaaikdaaaaaaa!44D9! x = {x_0} + {v_0}t + (1/2)a{t^2}\)
for the distance x travelled by a particle or body in time t which starts from the position x0 with an initial velocity \(v\)0 at time t = 0 and has uniform acceleration a along the of motion.
The dimensions of each term may be written as
[x]=[L]
[x0]=[L]
[ \(v\)0t]=[LT-1][T]
=[L]
[(1/2) at2]=[LT-2][T2]
=[L]
As each term on the right hand side of this equation has the same dimension,namely that of length, which is same as the dimension of left hand side of the equation, hence this equation is a dimensionally correct equation.It may be noted that a test of consistency of dimensions tells us no more and no less than a test of consistency of units, but has the advantage that we need not commit ourselves to a particular choice of units, and we need not worry about conversions among multiples and sub-multiples of the units. It may be borne in mind that if an equation fails this consistency test, it is proved wrong, but if it passes, it is not proved right. Thus, a dimensionally correct equation need not be actually an exact (correct) equation, but a dimensionally wrong (incorrect) or inconsistent equation must be wrong.
EXAMPLE 15
Let us consider an equation \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaWaaSaaaeaaca % aIXaaabaGaaGOmaaaacaWGTbGaamODamaaCaaaleqabaGaaGOmaaaa % kiabg2da9iaad2gacaWGNbGaamiAaaaa!3E2E! \frac{1}{2}m{v^2} = mgh\) where m is the mass of the body, \(v\) its velocity, g is the acceleration due to gravity and h is the height. Check whether this equation is dimensionally correct.
ANSWER
The dimensions of LHS are [M] [L T–1 ]2 = [M] [ L2 T–2]
= [M L2 T–2]
The dimensions of RHS are [M][L T–2] [L] = [M][L2 T–2]
= [M L2 T–2]
The dimensions of LHS and RHS are the same and hence the equation is dimensionally correct.
EXAMPLE 16
The SI unit of energy is J = kg m2 s–2; that of speed \(v\) is m s–1 and of acceleration a is m s–2. Which of the formulae for kinetic energy (K) given below can you rule out on the basis of dimensional arguments (m stands for the mass of the body) :
(a) K = m2 \(v\)3
(b) K = (1/2)m\(v\)2
(c) K = ma
(d) K = (3/16)m\(v\)2
(e) K = (1/2)m\(v\)2 + ma
ANSWER
Every correct formula or equation must have the same dimensions on both sides of the equation. Also, only quantities with the same physical dimensions can be added or subtracted. The dimensions of the quantity on (b) and (d); [M L T–2] for (c). The quantity on the right side of (e) has no proper dimensions since two quantities of different dimensions have been added. Since the kinetic energy K has the dimensions of [M L2 T–2], formulas (a), (c) and (e) are ruled out. Note that dimensional arguments cannot tell which of the two, (b) or (d), is the correct formula. For this, one must turn to the actual definition of kinetic energy (see Chapter 6). The correct formula for kinetic energy is given by (b).
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Deducing Relation among the Physical Quantities
Deducing Relation among the Physical Quantities
The method of dimensions can sometimes be used to deduce relation among the physical quantities. For this we should know the dependence of the physical quantity on other quantities (upto three physical quantities or linearly independent variables) and consider it as a product type of the dependence. Let us take an example.
EXAMPLE 17
Consider a simple pendulum, having a bob attached to a string, that oscillates under the action of the force of gravity. Suppose that the period of oscillation of the simple pendulum depends on its length (l), mass of the bob (m) and acceleration due to gravity (g). Derive the expression for its time period using method of dimensions.
ANSWER
The dependence of time period T on the quantities l, g and m as a product may be written as :
T = k lx gy mz
where k is dimensionless constant and x, y
and z are the exponents.
By considering dimensions on both sides, we have
[Lo Mo T1] =[L1 ]x [L1 T –2 ]y [M1 ]z
= Lx+y T–2y Mz
On equating the dimensions on both sides, we have
x + y = 0; –2y = 1; and z = 0
So that,\(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamiEaiabg2 % da9maalaaabaGaaGymaaqaaiaaikdaaaGaaiilaiaadMhacqGH9aqp % cqGHsisldaWcaaqaaiaaigdaaeaacaaIYaaaaiaacYcacaWG6bGaey % ypa0JaaGimaaaa!4217! x = \frac{1}{2},y = - \frac{1}{2},z = 0\)
Then \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamivaiabg2 % da9iaadUgacaWGSbWaaWbaaSqabeaacaaIXaGaai4laiaaikdaaaGc % caWGNbWaaWbaaSqabeaacqGHsislcaaIXaGaai4laiaaikdaaaGcca % WGVbGaamOCaiaacYcacaWGubGaeyypa0Jaam4AamaakaaabaWaaSaa % aeaacaWGSbaabaGaam4zaaaaaSqabaaaaa!47C3! T = k{l^{1/2}}{g^{ - 1/2}}or,T = k\sqrt {\frac{l}{g}} \)
Note that value of constant k can not be obtained by the method of dimensions. Here it does not matter if some number multiplies the right side of this formula, because that does not affect its dimensions.
Actually, k = 2\(\pi\) so that T = \(% MathType!MTEF!2!1!+- % feaagKart1ev2aaatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaaGOmaiabec % 8aWnaakaaabaWaaSaaaeaacaWGSbaabaGaam4zaaaaaSqabaaaaa!3A77! 2\pi \sqrt {\frac{l}{g}} \)
Dimensional analysis is very useful in deducing relations among the interdependent physical quantities. However, dimensionless constants cannot be obtained by this method. The method of dimensions can only test the dimensional validity, but not the exact relationship between physical quantities in any equation. It does not distinguish between the physical quantities having same dimensions.
A number of exercises at the end of this chapter will help you develop skill in dimensional analysis.