General Physics Topics

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Applications of Dimensions


Broadly speaking, dimension is the nature of a Physical quantity. Understanding of this nature helps us in many ways.

Following are some of the applications of the theory of dimensional analysis in Physics: 

(i) To find the unit of a given physical quantity in a given system of units:

By expressing a physical quantity in terms of basic quantity we find its dimensions. In the dimensional formula replacing M, L, T by the fundamental units of the required system, we get the unit of physical quantity. However, sometimes we assign a specific name to this unit. 

Illustration:

              Force is numerically equal to the product of mass and acceleration 

                                 i.e. Force = mass x acceleration 

          or [F] = mass x velocity/time= mass x displacement/(time)2 ) = mass x
                             
                                        length/(time)2)

                               = [M] x [LT-2] = [MLT-2]

Its unit in SI system will be Kgms-2 which is given a specific name “newton (N)”. 
Similarly, its unit in CGS system will be gmcms-2 which is called “dyne”.

(ii) To find dimensions of physical constants or coefficients:

The dimension of a physical quantity is unique because it is the nature of the physical quantity and the nature does not change. If we write any formula or equation incorporating the given physical constant, we can find the dimensions of the required constant or co-efficient.

Illustration:

      From Newton’s law of Gravitation, the exerted by one mass upon another is

                          F=G (m1 m2)/r2 or G=(Fr2)/(m1 m2 )

                         or [G] = ([MLT]2][L-2]) / ([M][M]) = [M-1 L3 T-2 ]

                       We can find its SI unit which is m3/Kgs2.

(iii) To convert a physical quantity from one system of units to another:

This is based on the fact that for a given physical quantity, magnitude x unit = constant 
So, when unit changes, magnitude will also change. 

Illustration:

                          Convert one Newton into dyne

Solution: 

Dimensional formula for Newton = [MLT-2

               Or 1 N = 1 Kg m/s2 ; But 1 kg = 103 g and 1 m = 102 cm

                 Therefore 1 N = ((103 g)(102 cm))/s2 = 105 g cm/s2 = 105 dyne

(iv) To check the dimensional correctness of a given physical relation:

This is based on the principle that the dimensions of the terms on both sides on an equation must be same. This is known as the ‘principle of homogeneity’. If the dimensions of the terms on both sides are same, the equation is dimensionally correct, otherwise not. 

Caution: It is not necessary that a dimensionally correct equation is also physically correct but a physically correct equation has to be dimensionally correct. 

Illustration: 

(i) Consider the formula, T=2√(l/g)

Where T is the time period of oscillation of a simple pendulum in a simple harmonic motion, l and g are the length of the pendulum and gravitational constants respectively. Check this formula, whether it is correct or not, using the concept of dimension. 

As we know [g] = [LT2

                     Therefore  [T] = √(([L])/([LT-2])) = [T] s

Thus the above equation is dimensionally correct (homogenous) and later you will come to know that it is physically also correct.

(ii) Consider the formula s=ut -1/3 at2. Check this formula whether it is correct or not, using the concept of dimension.

Dimensionally 

                            [L] = [LT-1] [L] – [LT-2] [T2

                                   => [L] = [L] – [L] 

In this case also the formula is dimensionally correct but, you know that it is physically incorrect as the correct formula is given by
 
                                   S = ut + 1/3at2 

(v) As a research tool to derive new relations: 

One of the aims of scientific research is to discover new laws relating different physical quantities. The theory of dimensions (in the light of principal of homogeneity) provides us with a powerful tool of research in the preliminary stages of investigation [It must be again emphasized that mere dimensional correctness of an equation does not ensure its physical correctness]

Limitations of the theory of dimensions



The limitations are as follows:

(i) If dimensions are given, physical quantity may not be unique as many physical quantities have the same dimension. For example, if the dimensional formula of a physical quantity is [ML2T-2] it may be work or energy or even moment of force. 

(ii) Numerical constants, having no dimensions, cannot be deduced by using the concepts of dimensions. 

(iii) The method of dimensions cannot be used to derive relations other than product of power functions. Again, expressions containing trigonometric or logarithmic functions also cannot be derived using dimensional analysis, e.g. 

              s = ut + 1/3at2    or   y = a sincot    or    P= P0e (–Mgh)/RT 

    cannot be derived. However, their dimensional correctness can be verified. 

(iv) If a physical quantity depends on more than three physical quantities, method of dimensions cannot be used to derive its formula. For such equations, only the dimensional correctness can be checked. For example, the time period of a physical pendulum of moment of inertia I, mass m and length l is given by the following equation.

T = 2√(I/mgl) (I is known as the moment of Inertia with dimensions of [ML2] through dimensional analysis), though we can still check the dimensional correctness of the equation (Try to check it as an exercise).

(v) Even if a physical quantity depends on three Physical quantities, out of which two have the same dimensions, the formula cannot be derived by theory of dimensions, and only its correctness can be checked e.g. we cannot derive the equation.

 

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