Dimensional Analysis
The study of the relationship between physical quantities with the help of dimensions and units of measurement is termed dimensional analysis. Dimensional analysis is essential because it keeps the units the same, helping us perform mathematical calculations smoothly.
Unit Conversion and Dimensional Analysis
Dimensional analysis use conversion factors to get the same units. To help you understand the stated better, let’s say you want to know how many metres make 3 km?
We know that 1000 metres make 1 km,
Therefore,
3 km = 3 × 1000 metres = 3000 metres
Here, the conversion factor is 1000 metres.
Using Dimensional Analysis to Check the Correctness of Physical Equation
Let’s say that you don’t remember whether
- time = speed/distance, or
- time = distance/speed
We can check this by making sure the dimensions on each side of the equations match.
Reducing both the equations to its fundamental units on each side of the equation, we get
- \(\begin{array}{l}[T]=\frac{[L][T]^{-1}}{L}=[T]^{-1}\,\,(Wrong) \end{array} \)
- \(\begin{array}{l}[T]=\frac{[L]}{[L][T]^{-1}}=[T]\,\,(Right) \end{array} \)
However, it should be kept in mind that dimensional analysis cannot help you determine any dimensionless constants in the equation.
Read more like this:
Homogeneity Principle of Dimensional Analysis
Principle of Homogeneity states that dimensions of each of the terms of a dimensional equation on both sides should be the same. This principle is helpful because it helps us convert the units from one form to another. To better understand the principle, let us consider the following example:
Example 1: Check the correctness of physical equation s = ut + ½ at2. In the equation, s is the displacement, u is the initial velocity, v is the final velocity, a is the acceleration and t is the time in which change occurs.
Solution:
We know that L.H.S = s and R.H.S = ut + 1/2at2
The dimensional formula for the L.H.S can be written as s = [L1M0T0] ………..(1)
We know that R.H.S is ut + ½ at2 , simplifying we can write R.H.S as [u][t] + [a] [t]2
[L1M0T-1][L0M0T-1] +[L1M0T-2][L0M0T0]
=[L1M0T0]………..(2)
From (1) and (2), we have [L.H.S] = [R.H.S]
Hence, by the principle of homogeneity, the given equation is dimensionally correct.
Applications of Dimensional Analysis
Dimensional analysis is a fundamental aspect of measurement and is applied in real-life physics. We make use of dimensional analysis for three prominent reasons:
- To check the consistency of a dimensional equation
- To derive the relation between physical quantities in physical phenomena
- To change units from one system to another
Limitations of Dimensional Analysis
Some limitations of dimensional analysis are:
- It doesn’t give information about the dimensional constant.
- The formula containing trigonometric function, exponential functions, logarithmic function, etc. cannot be derived.
- It gives no information about whether a physical quantity is a scalar or vector.
Additional Solved Problems
1. Check the correctness of the physical equation v2 = u2 + 2as2.
Solution:
The computations made on the L.H.S and R.H.S are as follows:
L.H.S: v2 = [v2] = [ L1M0T–1]2 = [ L1M0T–2] ……………(1)
R.H.S: u2 + 2as2
Hence, [R.H.S] = [u]2 + 2[a][s]2
[R.H.S] = [L1M0T–1]2 + [L1M0T–2][L1M0T0]2
[R.H.S] = [L2M0T–2] + [L1M0T–2][L2M0T0]
[R.H.S] = [L2M0T–2] + [L1M0T–2][L2M0T0]
[R.H.S] = [L2M0T–2] + [L3M0T–2]…………………(2)
From (1) and (2), we have [L.H.S] ≠ [R.H.S]
Hence, by the principle of homogeneity, the equation is not dimensionally correct.
2. Evaluate the homogeneity of the equation when the rate flow of a liquid has a coefficient of viscosity η through a capillary tube of length ‘l’ and radius ‘a’ under pressure head ‘p’ given as
\(\begin{array}{l}\frac{dV}{dt}=\frac{\pi p a^4}{8l\eta}\end{array} \)
Solution:
\(\begin{array}{l} \frac{ dV }{ dt } = \frac{ \pi p ^ { 4 }}{ 8 l \eta } \end{array} \)
\(\begin{array}{l} [\textup{L.H.S}] = \frac{ [dV] }{[dt]} = \frac{[M^{0} L^{3} T^ {0}]}{[M^{0} T^{0} T^{1}]} = [M^{0}L^{3}T^{-1}] \textup{ …..(1)} \end{array} \)
\(\begin{array}{l} [\textup{R.H.S}] = \frac{[p] [a] ^ {4}}{[l] [\eta]} \end{array} \)
\(\begin{array}{l} \therefore [\textup{R.H.S}] = \frac{ [M ^ { 1 } L ^ { -1 } T ^ { -2 }] [M ^ { 0 } L ^ { 1 } T ^ { 0 }]^ { 4 }} { [M ^ { 0 } L ^ { 1 } T ^ { 0 }] [M ^ { 1 } L ^ { -1 } T ^{ -1 }] } \end{array} \)
\(\begin{array}{l} = \frac{ [M^{ 1 } L ^ { -1 } T ^ { -2 }] [M ^ { 0 }L ^ { 4 } T ^ { 0 } ] }{ [M^{ 1 }L^ { 0 }T^ { -1 }] } \end{array} \)
\(\begin{array}{l} = \frac{ [M^ { 1 } L ^ { 3 } T^ { 2 }] }{ [M^ { 1 } L^ { 0 } T ^ { -1 }] } = [M^ { 0 }L ^ { 3 }T^{ -1 }] \textup{ …….(2)} \end{array} \)
From (1) and (2), we have [L.H.S] = [R.H.S]
Hence, by the principle of homogeneity, the given equation is homogenous.
Frequently Asked Questions – FAQs
What is dimensional analysis?
Dimensional analysis is the study of the relationship between physical quantities with the help of dimensions and units of measurement.
State true or false: Dimensional analysis can not be used to find dimensionless constants.
State the principle of homogeneity of dimensions
The principle of homogeneity of dimensions states that an equation is dimensionally correct if the dimensions of the various terms on either side of the equation are the same.
Why do we use dimensional analysis?
- To check the consistency of a dimensional equation
- To derive the relation between physical quantities in physical phenomena
- To change units from one system to another.
What are the limitations of dimensional analysis?
- It doesn’t give information about the dimensional constant.
- The formula containing trigonometric function, exponential functions, logarithmic function, etc. cannot be derived.
- It gives no information about whether a physical quantity is a scalar or vector.
