Applications of Dimension

A clear 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 different physical expressions. We evaluate the size and state of things utilizing Dimensional Analysis. It assists us with concentrating on the idea of items numerically. It includes lengths and points along with mathematical properties like uniforms and straightness. The essential idea of dimension is that we can add and take away just those amounts that have similar dimensions. Essentially, two actual amounts are equivalent on the off chance that they have similar dimensions.

The four basic applications of dimension are as follows:

1. To check the correctness of physical relation.
2. To derive the relation between various physical quantities.
3. To convert the value of physical quantities from one system of unit into another system of units.
4. To find the dimensions of constants in the given equation.

1. Correctness of Physical Relation

One of the most common uses of dimension is that we can examine the correctness of physical relations.

For example: check the correctness of the following formula \(v^2=u^2+2aS\)

Solution:

The dimension on the left-hand side of the equation is

\(\left[v^2\right]=\left[LT^{-1}\right]^2=\left[L^2T^{-2}\right]\)

The dimension of the right hand side of the equation is

\(\left[LT^{-1}\right]^2+2\left[LT^{-2}\right]\left[L\right]\)
= \(\left[L^2T^{-2}\right]+2\left[L^2T^{-2}\right]\)
= \(3\left[L^2T^{-2}\right]\)

Here, 3 is a dimensionless constant.

So, Dimension of LHS = Dimension of RHS

Hence, the given equation is dimensionally correct.

Here is an example of dimensionally incorrect expression.

2. Derive relation between Physical Quantities

Using the principle of homogeneity of dimensions, we can establish the relationship between various physical quantities. In order to deduce the relation between physical quantities using dimensions, we should know the dependence of the physical quantity with other physical quantities. To demonstrate how we can derive the relation let's carefully observe the following example.

Derive the formula for the time period of a simple pendulum using dimensional analysis.

Solution: The time period (t) of a simple pendulum depends upon the length (l) of the pendulum and acceleration due to gravity (g).

Therefore, we can write \(t\propto l^a\)

and, \(t\propto g^b\)

Combining the equations, we get

\(t\propto l^a\cdot g^b\)

or, \(t=k\cdot l^a\cdot g^b\) --------- (1), where k is the dimensionless proportionality constant and a and b are the indices that are to be determined.

Next, we compare the dimension of the left and right-hand sides of equation (1)

The dimension of t is [T]

The dimension of l is [L]

The dimension of g = [LT^-2]

Substituting the dimensional formula of physical quantities in equation (1), we get

[T] = [L]^a [LT^-2]^b

or, \(\left[M^0L^0T\right]=\left[M^0L^{a+b}T^{-2b}\right]\)

Equating the powers of similar physical quantities on both sides of equation, we get

a + b = 0 ------------- (2)

and, -2b = 1 or b = \(-\frac12\) ----------- (3)

Substituting the value of b from equation (3) in equation (2) we get,

a = -b = \(\frac12\)

Finally using the value of a and b in equation (1), we get,

\(t=k\cdot l^\frac12g^{-\frac12}=k\sqrt{\frac lg}\)

The value of k is calculated experimentally and is found to be 2π

∴ \(t=2\mathrm\pi\sqrt{\frac{\mathrm l}{\mathrm g}}\)

3. Convert the Unit from one to other System

The dimensional method is useful to convert one system of the unit to another system of the unit. The physical quantity (X) in the given (initial) system of the unit can be written as:

\(X=N_1\times U_1\)
Or, \(X=N_1\left[{M^a}_1{L^b}_1{T^c}_1\right]\) ------------- (1)

Where, a, b and, c are the dimensions of mass, length and the time respectively.

Again,

The same physical quantity (X) in another (final) system of unit

\(X=N_2\times U_2\)
Or, \(X=N_2\left[{M^a}_2{L^b}_2{T^c}_2\right]\) ------------- (2)

From equation (1) and (2), we get

\(N_2\left[{M^a}_2{L^b}_2{T^c}_2\right]=N_1\left[{M^a}_1{L^b}_1{T^c}_1\right]\)

\(\therefore N_2=N_1\left[\frac{M_1}{M_2}\right]^a\left[\frac{L_1}{L_2}\right]^b\left[\frac{T_1}{T_2}\right]^c\)

Using this dimensional relation we can convert unit from one system to another system of unit.

For example check this question (convert 20 dyne into Newton), or navigate to exercises section to find other examples

4. Determine the Dimensional Constant of Equation

Using dimensional analysis, we can determine the dimensions of constants appearing in the equation. Below is an example that illustrates the process.

Determine the dimension of 'a' and 'b' in the equation: \(\left(P+\frac a{V^2}\right)\left(V-b\right)=RT\).

According to the principle of homogeneity of dimensions, to add or subtract, each term should have the same dimensions.

Therefore, dimensions of P = dimensions of \(\frac a{V^2}\)

That is, \(\left[P\right]=\left[\frac a{V^2}\right]\)

Or, \(\left[ML^{-1}T^{-2}\right]=\frac{\left[a\right]}{\left[L^3\right]^2}=\frac{\left[a\right]}{\left[L^6\right]}\)

Or, \(\left[a\right]=\left[ML^5T^{-2}\right]\)

And,

\(\left[V\right]=\left[b\right]\)

\(\Rightarrow\left[b\right]=\left[L^3\right]\)

Thus, the dimensions of a are [M L⁵ T^-2] and the dimension of b is [^L3]

Limitations of Dimensional Analysis

Despite the usefulness of dimensional analysis, there are a few important limitations of it. Here are the five

1. The dimensional analysis does not give any information about dimensionless constants.
2. If the quantity depends on more than three other physical quantities having dimensions, the formula cannot be derived.
3. We cannot derive the formula containing trigonometric functions, logarithmic functions, exponential functions, etc. It is best suited for linear functions only.
4. The exact form of a relationship cannot be determined when there is more than one part in any relation.
5. It gives no information about the physical quantity, whether it is vector or scalar.

• The four applications of dimensional analysis are:
  • To check the correctness of physical relation.
  • To derive the relation between various physical quantities.
  • To convert the value of physical quantities from one system of units into another system of units.
  • To find the dimensions of constants in the given equation.
    
    
• The correctness of physical relation can be determined by comparing the dimension of the left-hand side (LHS) and the dimension of the right-hand side (RHS)
• We can use the formula, \(N_2=N_1\left[\frac{M_1}{M_2}\right]^a\left[\frac{L_1}{L_2}\right]^b\left[\frac{T_1}{T_2}\right]^c\)\) to convert unit from one to another system using dimension.
  
  
• Despite the usefulness of dimensions, there are some limitations. They are:
  • The dimensional analysis does not give any information about dimensionless constants.
  • If the quantity depends on more than three other physical quantities having dimensions, the formula cannot be derived.
  • We cannot derive the formula containing trigonometric functions, logarithmic functions, exponential functions, etc. It is best suited for linear functions only.
  • The exact form of a relationship cannot be determined when there is more than one part in any relationship.
  • It gives no information about the physical quantity,