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How to Check the Correctness of Physical Equation by Dimensional Analysis?
Any theoretical probe involving the dimensions of different fundamental and derived physical quantities is called dimensional analysis. Usually, from this analysis,
- the dimensional correctness of a physical expression or equation can be checked;
- the value of a measured quantity can be converted from one system of units to another;
- relations among different physical quantities can be determined.
Dimensional analysis stands on the following basic principle:
The principle of dimensional homogeneity: This principle states that, in any mathematical expression or equation involving physical quantities, each term in the expression or each term on either side of the equation must have the same dimension.
For example, let us take an expression a + bc, or an equation d = a + bc. The principle asserts that, each term-a, bc and d-in the expression or the equation has the same dimension. This is obvious, because for example, some mass cannot be added with some length, or some work can never be equal to some density, etc.
Further, in a polynomial series of the form 1 + ax + bx2 + cx3 + ……….., the variable x must be dimensionless. Otherwise, the different terms of the expression would have different dimensions and they cannot be added. It is to be noted that functions like ex, sin x, cos x etc. are widely used in physics-each of them is actually a polynomial series; hence, x should be dimensionless.
Dimensional correctness of an equation
From the principle of dimensional homogeneity, by analysing the dimensions of both sides of an equation, the dimensional correctness of an equation may be checked.
Example: Let us check the dimensional correctness of the equation of motion, s = ut + \(\frac{1}{2}\) at2.
On LHS, [s] = L
On RHS, [u] = LT-1, [t] = T and [a] = LT-2
i.e., [RHS] = [ut] + [\(\frac{1}{2}\) at2]
Thus, dimension of physical quantities on both sides of the equation is L. So the equation is dimensionally correct.
Conversion between unit systems
The following examples are sufficient to illustrate the method:
i) Relation between joule and erg: 1 joule (J) and 1 erg are the SI and CGS units of work, respectively. Let, 1 J = n erg.
The dimension of work is ML2T-2. Using the SI and CGS fundamental units directly as per the dimension, we have
1 J = n erg or, 1 kg ᐧ m2 ᐧ s-2 = ng ᐧ cm2 ᐧ s-2
or, n = \(\frac{\mathrm{kg} \cdot \mathrm{m}^2 \cdot \mathrm{s}^{-2}}{\mathrm{~g} \cdot \mathrm{cm}^2 \cdot \mathrm{s}^{-2}}\) = \(\left(\frac{\mathrm{kg}}{\mathrm{g}}\right) \cdot\left(\frac{\mathrm{m}}{\mathrm{cm}}\right)^2\) = 1000 × (100)2
[As 1 kg = 1000 g and 1 m = 100 cm]
= 107
Thus, 1 J = 107erg, 1 erg = 10-7J.
ii) Conversion of the value of Young’s modulus from CGS system to SI: The value of Young’s modulus (Y) of iron is 2 × 1012 dyn ᐧ cm-2. The corresponding SI unit is N ᐧ m-2. Let, 2 × 1012 dyn ᐧ cm-2 = n N ᐧ m-2.
2 × 1012dyn ᐧ cm-2 = nN ᐧ m-2
or 2 × 1012g ᐧ cm-1 ᐧ s-2 = n kg ᐧ m-1 ᐧ s-2
or, n = 2 × 1012 × \(\frac{\mathrm{g} \cdot \mathrm{cm}^{-1} \cdot \mathrm{s}^{-2}}{\mathrm{~kg} \cdot \mathrm{m}^{-1} \cdot \mathrm{s}^{-2}}\)
= 2 × 1012 × \(\frac{1}{1000}\) × 100 = 2 × 1011
So, Y = 2 × 1011 N ᐧ m-2
Relationship among different physical quantities
i) Dependence of time period (T) of a simple pendulum on its effective length (l), acceleration due to gravity (g) and mass of the pendulum bob (m):
Let us assume,
T ∝ mx [when l and g are constant]
T ∝ ly [when m and g are constant]
T ∝ gz [when l and m are constant]
∴ T ∝ mxlygz
or, T = Kmxlygz, where k is a dimensionless constant of proportionality and x, y and z are numeric indices.
Now, [ml = M, [l] = L and [g] = LT-2.
Expressing LHS and RHS in terms of dimensions,
M0L0T = Mx × Ly × (LT-2)z
or, M0L0T = MxLy+zT-2z
Equating powers of the same bases from both sides we get,
x = 0
y + z = 0 or, y = -z
-2z = 1 or z = –\(\frac{1}{2}\)
∴ y = +\(\frac{1}{2}\) ∴ T = k\(\sqrt{\frac{l}{g}}\)
We cannot determine the value of k from this analysis. It is also important to note that, x = 0 means the time period does not depend on the mass of the pendulum.
ii) Dependence of frequency of vibration (n) of a stretched string on its length (l), mass per unit length (μ) and tension in the string (T):
Let us assume,
n ∝ lx [when T and μ are constant]
n ∝ Ty [when l and μ are constant]
n ∝ μz [when l and T are constant]
∴ n ∝ lxTyμz
or, n = klxTyμz, where k is a dimensionless constant of proportionality and x, y, z are numeric indices.
Here, [n] = T-1, [l] = L, [T] = MLT-2 and [μ] = ML-1.
Then, M0L0T-1 = Lxᐧ(MLT-2)y ᐧ (ML-1)z
or, M0L0T-1 = Lx+y-z ᐧ My+z ᐧ T-2y
Equating the powers of same bases,
x + y – z = 0, y + z = 0, -2y = -1
So, y = \(\frac{1}{2}\)
Then, z = –\(\frac{1}{2}\) and x = -1
Hence, n = \(\frac{k}{l} \sqrt{\frac{T}{\mu}}\)
iii) Dependence of viscous force (F) on the radius of a ball falling through a viscous fluid (a), coefficient of vis-cosity of the fluid (η) and terminal velocity (v) of the ball:
Let us assume,
F ∝ ax [when η and v are constant]
F ∝ ηy [when a and v are constant]
F ∝ vz [when a and η are constant]
∴ F ∝ axηy
or, F = kaxηyvz, where k is a dimensionless constant of proportionality and x, y, z are numeric indices.
Here, [F] = MLT-2, [a] = L, [η] = ML-1T-1 and [vl = LT-1.
So, MLT-2 = Lx ᐧ (ML-1T-1)J ᐧ (LT-1)z
or, MLT-2 = Lx-y+zᐧMyᐧT-y-z
Equating powers of same bases,
x – y + z = 1
y = 1
and -y – z = -2 or, y + z = 2
Then, z = 1 and x = l
∴ F = kaηv
Detailed dimensional analysis shows that F does not depend on the densities of the fluid and of the material of the ball. We omitted those details here.
Limitations of dimensional analysis
i) The value of a constant in an equation cannot be deter-mined.
Example: In a simple pendulum the time period of the bob depends on length and acceleration due to gravity.
The relation is T = k\(\sqrt{\frac{l}{g}}\). We cannot determine the value of constant k from dimensional analysis.
ii) No relation, containing a constant which is not dimen-sionless, can be established.
Example: We cannot determine the nature of dependence of force on mass and distance in Newton’s law of gravitation, as it contains a constant G which is not dimensionless.
iii) If any relation contains a dimensionless quantity, we cannot determine its nature of dependence with others present in the relation.
Example: If a body is displaced by a distance s when a force is acting on it, then work done by the force depends on the applied force, displacement of the body and the angle between the force and displacement. This is expressed by the relation W = Fscosθ, where θ is the angle between the direction of applied force and that of displacement. As θ is a dimensionless quantity we cannot determine the relation by dimensional analysis.
iv) If a physical quantity depends on different quantities having the same dimension, the relation among them cannot be obtained.
Example: The volume of a right cylinder depends on its radius and on its length. Here, the radius and the length are entirely different quantities related to the cylinder, but they have the same dimension—the dimension of length. So, the relation V ∝ r2l cannot be obtained from dimensional analysis.
Numerical Examples
Example 1.
What will be the conversion factor when you change a value expressed in newton to dyne?
Solution:
N and dyn are the units of force in SI and CGS sys-tems respectively.
The dimension of force, [F] = MLT-2
Let 1 N = n dyn
Then using the base units in the dimension,
1 kg ᐧ m ᐧ s-2 = ng ᐧ cm ᐧ s-2
or, n = \(\frac{1 \mathrm{~kg}}{1 \mathrm{~g}}\) × \(\frac{1 \mathrm{~m}}{1 \mathrm{~cm}}\) × 1 = 105
∴ 1N = 105 dyn
So, the conversion factor is 105.
Example 2.
Taking electric potential V as a fundamental quantity instead of electric current, find the dimension of electric current in terms of the dimension of the electric potential.
Solution:
Since V (electric potential) \(=\frac{P(\text { power })}{I(\text { electric current })}\)
[I] = \(\frac{[P]}{[V]}\) = \(\frac{\mathrm{ML}^2 \mathrm{~T}^{-3}}{\mathrm{~V}}\) = ML2T-3V-1.
Example 3.
Find the dimension of universal gravitational constant, taking the units of density D, velocity V and work W as base units instead of those of mass, distance and time.
Solution:
From Newton’s law of gravitation, F = G\(\frac{M_1 M_2}{r^2}\) we have
[G] = \(\frac{[F]\left[r^2\right]}{\left[m^2\right]}\) = \(\frac{\mathrm{MLT}^{-2} \times \mathrm{L}^2}{\mathrm{M}^2}\) = M-1L3T-2
Let M-1L3T-2 = DxVyWz
= (ML-3)x × (LT-1)y × (ML2T-2)z
= Mx+z × L-3x+y+2z × T-y-2z
Equating powers of same bases,
x + z = -1
-3x + y + 2z = 3 and
-y – 2z = -2
Solving for x, y, z, we get x = –\(\frac{1}{3}\), y = \(\frac{10}{3}\) and z = –\(\frac{2}{3}\)
∴ [G] = \(\mathrm{D}^{-\frac{1}{3}} \mathrm{v}^{\frac{10}{3}} \mathrm{w}^{-\frac{2}{3}}\)
Example 4.
In a system of units, the unit of length is defined as the distance travelled by light in space in 1 s and the unit of time as the time taken by the earth to revolve round the sun once. Find the velocity unit in CGS system.
Solution:
In this system, unit distance = 3 × 1010 cm and unit time = 365.25 d = 3.156 × 107 s
Velocity unit in this system
= \(\frac{3 \times 10^{10}}{3.156 \times 10^7}\) = 950.57 cm ᐧ s-1