Applications of dimensional analysis
The method of studying a physical phenomenon on the basis of dimensions is called dimensional analysis.
The three main uses of a dimensional analysis are described in detail in the following section
Checking the dimensional consistency of equation (Homogeneity of Dimensions in an Equation)
Every physical equation should be dimensionally balanced. This is called the principle of homogeneity. This principle states that, the dimensions of each term on both sides of an equation must be the same. On this basis, we can judge whether a given equation is correct or not. But a dimensionally correct equation may or may not be physically correct.
e.g. In the physical expression \(s=u t+\frac{1}{2} a t^2\), the dimensions of \(s, u t\) and \(\frac{1}{2} a t^2\) all are same.
Note: The physical quantities separated by the symbols,,\(+-=,\rangle,<\) etc., should have the same dimension.
For example, If the dimensions of all the terms are not same, the equation must be wrong. Let us check the equation
\(
x=u t+\frac{1}{2} a t^2
\)
for the dimensional homogeneity. Here \(x\) is the distance travelled by a particle in time \(t\) which starts at a speed \(u\) and has an acceleration \(a\) along the direction of motion.
\(
\begin{aligned}
{[x] } & =\mathrm{L} \\
{[u t] } & =\text { velocity × time }=\frac{\text { length }}{\text { time }} \times \text { time }=\mathrm{L} \\
{\left[\frac{1}{2} a t^2\right] } & =\left[\text { at }^2\right]=\text { acceleration } \times(\text { time })^2 \\
& =\frac{\text { velocity }}{\text { time }} \times(\text { time })^2=\frac{\text { length } / \text { time }}{\text { time }} \times(\text { time })^2=\mathrm{L}
\end{aligned}
\)
Thus, the equation is correct as far as the dimensions are concerned.
Note: that the dimension of \(\frac{1}{2} a t^2\) is same as that of \(a t^2\). Pure numbers are dimensionless. Dimension does not depend on the magnitude. Due to this reason the equation \(x=u t+a t^2\) is also dimensionally correct. Thus, a dimensionally correct equation need not be actually correct but a dimensionally wrong equation must be wrong.
Example 1: Test dimensionally if the formula \(t=2 \pi \sqrt{\frac{m}{F / x}}\) may be correct, where \(t\) is time period, \(m\) is mass, \(F\) is force and \(x\) is distance.
Solution: The dimension of force is \(\mathrm{MLT}^{-2}\). Thus, the dimension of the right-hand side is
\(
\sqrt{\frac{\mathrm{M}}{\mathrm{MLT}^{-2} / \mathrm{L}}}=\sqrt{\frac{1}{\mathrm{~T}^{-2}}}=\mathrm{T} .
\)
The left-hand side is time period and hence the dimension is T. The dimensions of both sides are equal and hence the formula may be correct.
Example 2: Write the dimensions of \(a\) and \(b\) in the relation,
\(
P=\frac{b-x^2}{a t}
\)
where, \(P\) is power, \(x\) the distance and \(t\) the time.
Solution: The given equation can be written as \(\mathrm{Pat}=b-x^2\)
Now, \([P a t]=[b]=\left[x^2\right] \quad\) or \([b]=\left[x^2\right]=\left[\mathrm{M}^0 \mathrm{~L}^2 \mathrm{~T}^0\right]\)
\(
[a]=\frac{\left[x^2\right]}{[P t]}=\frac{\left[\mathrm{L}^2\right]}{\left[\mathrm{ML}^2 \mathrm{~T}^{-3}\right][\mathrm{T}]}=\left[\mathrm{M}^{-1} \mathrm{~L}^0 \mathrm{~T}^2\right]
\)
Example 3: The velocity \(v\) of a particle depends upon the time \(t\) according to the equation \(v=a+b t+\frac{c}{d+t}\). Write the dimensions of \(a, b, c\) and \(d\).
Solution: From principle of homogeneity,
\(
\begin{aligned}
& {[a]=[v] \text { or }[a]=\left[\mathrm{LT}^{-1}\right]} \\
& {[b t]=[v]}
\end{aligned}
\)
\(
[b]=\frac{[v]}{[t]}=\frac{\left[\mathrm{LT}^{-1}\right]}{[\mathrm{T}]} \text { or }[b]=\left[\mathrm{LT}^{-2}\right]
\)
Similarly,
\(
[d]=[t]=[T]
\)
Further,
\(
\frac{[c]}{[d+t]}=[v] \text { or }[c]=[v][d+t]
\)
\(
[c]=\left[\mathrm{LT}^{-1}\right][\mathrm{T}]
\)
\(
[c]=[\mathrm{L}]
\)
∴ Dimensions of \(a=\left[\mathrm{LT}^{-1}\right]\)
Dimensions of \(b=\left[\mathrm{LT}^{-2}\right]\)
Dimension of \(c=[\mathrm{L}]\)
Dimension of \(d=[\mathrm{T}]\)
Example 4: The following equation gives a relation between the mass \(m_1\), kept on a surface of area \(A\) and the pressure \(p\) exerted on this area
\(
p=\frac{\left(m_1+m_2\right) x}{A}
\)
What must be the dimensions of the quantities \(x\) and \(m_2\) ?
Solution: Since, all the terms of a mathematical equation should have the same dimensions.
Therefore,
\(
[p]=\left[\frac{\left(m_1+m_2\right) x}{A}\right] \dots(i)
\)
Only the quantities having same dimensions and nature can be added to each other.
Here, \(m_2\) is added to mass \(m_1\).
Hence, \(\left[m_2\right]=\left[m_1\right]=[\mathrm{M}]\)
Also, the quantity obtained by the addition of \(m_1\) and \(m_2\) would have the same dimensions as that of mass.
\(
\therefore \quad\left[m_1+m_2\right]=[\mathrm{M}]
\)
Now, going back to Eq. (i),
\(
\begin{aligned}
& {[p]=\frac{\left[m_1+m_2\right][x]}{[A]} } \\
\Rightarrow & {\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\frac{[\mathrm{M}][x]}{\left[\mathrm{L}^2\right]} } \\
\Rightarrow & {\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]=\left[\mathrm{ML}^{-2}\right][x] } \\
\Rightarrow & \frac{\left[\mathrm{ML}^{-1} \mathrm{~T}^{-2}\right]}{\left[\mathrm{ML}^{-2}\right]}=[x] \Rightarrow[x]=\left[\mathrm{LT}^{-2}\right]
\end{aligned}
\)
Hence, the quantity \(x\) represents acceleration. In this example, it is the acceleration due to gravity \(g \cdot\left(m_1+m_2\right) g\) represents the weight exerted by two masses \(m_1, m_2\) on the area \(A\).
To convert a physical quantity from one system of units to other system of units
This is based on the fact that the product of the numerical value \((n)\) and its corresponding unit \((u)\) is a constant, i.e.
\(
n(u)=\text { constant } \text { or } n_1\left[u_1\right]=n_2\left[u_2\right]
\)
Suppose the dimensions of a physical quantity are \(a\) in mass, \(b\) in length and \(c\) in time. If the fundamental units in one system are \(\mathrm{M}_1, \mathrm{~L}_1\) and \(\mathrm{T}_1\) and in the other system are \(\mathrm{M}_2, \mathrm{~L}_2\) and \(\mathrm{T}_2\), respectively. Then, we can write
\(
n_1\left[\mathrm{M}_1^a \mathrm{~L}_1^b \mathrm{~T}_1^c\right]=n_2\left[\mathrm{M}_2^a \mathrm{~L}_2^b \mathrm{~T}_2^c\right] \dots(i)
\)
\(
n_2=n_1 \frac{u_1}{u_2}=n_1\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]^a\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_2}\right]^b\left[\frac{\mathrm{~T}_1}{\mathrm{~T}_2}\right]^c
\)
Here, \(n_1\) and \(n_2\) are the numerical values in two system of units, respectively. Using Eq. (i), we can convert the numerical value of a physical quantity from one system of units into the other system.
For example, Let us first write the dimensional formula of pressure.
We have \(P=\frac{F}{A}\)
Thus, \([P]=\frac{[F]}{[A]}=\frac{\mathrm{MLT}^{-2}}{\mathrm{~L}^2}=\mathrm{ML}^{-1} \mathrm{~T}^{-2}\)
so, \(1 \operatorname{pascal}=(1 \mathrm{~kg})(1 \mathrm{~m})^{-1}(1 \mathrm{~s})^{-2}\)
\(1 \mathrm{CGS} \text { pressure }=(1 \mathrm{~g})(1 \mathrm{~cm})^{-1}(1 \mathrm{~s})^{-2}\)
Thus,
\(
\begin{aligned}
\frac{1 \text { pascal }}{1 \text { CGS pressure }} & =\left(\frac{1 \mathrm{~kg}}{1 \mathrm{~g}}\right)\left(\frac{1 \mathrm{~m}}{1 \mathrm{~cm}}\right)^{-1}\left(\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right)^{-2} \\
& =\left(10^3\right)\left(10^2\right)^{-1}=10
\end{aligned}
\)
or, \(\quad 1\) pascal \(=10\) CGS pressure.
Example 5: Find the value of \(100 J\) on a system which has \(20 \mathrm{~cm}, 250 \mathrm{~g}\) and half minute as fundamental units of length, mass and time.
Solution: The dimensional formula of work is \(=\left[\mathrm{ML}^2 \mathrm{~T}^{-2}\right]\)
To convert a physical quantity from one system of units to other system of units, we use the following formula
\(
\begin{aligned}
n_2 & =n_1\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]^a\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_2}\right]^b\left[\frac{\mathrm{~T}_1}{\mathrm{~T}_2}\right]^c \\
n_2 & =100\left[\frac{1 \mathrm{~kg}}{250 \mathrm{~g}}\right]^1\left[\frac{1 \mathrm{~m}}{20 \mathrm{~cm}}\right]^2\left[\frac{1 \mathrm{~s}}{0.5 \mathrm{~min}}\right]^{-2} \\
& =100\left[\frac{1000 \mathrm{~g}}{250 \mathrm{~g}}\right]^1\left[\frac{100 \mathrm{~cm}}{20 \mathrm{~cm}}\right]^2\left[\frac{1 \mathrm{~s}}{30 \mathrm{~s}}\right]^{-2} \\
& =100 \times 4 \times 25 \times 30 \times 30=9 \times 10^6 \text { new units }
\end{aligned}
\)
Example 6: The value of gravitational constant is \(G=6.67 \times 10^{-11} \mathrm{~N}-\mathrm{m}^2 / \mathrm{kg}^2\) in SI units. Convert it into CGS system of units.
Solution: The dimensional formula of \(G\) is \(\left[\mathrm{M}^{-1} \mathrm{~L}^3 \mathrm{~T}^{-2}\right]\).
To convert a physical quantity from one system of units to other system of units, we use the following formula
\(
\begin{aligned}
n_1\left[\mathrm{M}_1^{-1} \mathrm{~L}_1^3 \mathrm{~T}_1^{-2}\right] & =n_2\left[\mathrm{M}_2^{-1} \mathrm{~L}_2^3 \mathrm{~T}_2^{-2}\right] \\
n_2 & =n_1\left[\frac{\mathrm{M}_1}{\mathrm{M}_2}\right]^{-1}\left[\frac{\mathrm{~L}_1}{\mathrm{~L}_2}\right]^3\left[\frac{\mathrm{~T}_1}{\mathrm{~T}_2}\right]^{-2} \\
& =6.67 \times 10^{-11}\left[\frac{1 \mathrm{~kg}}{10^{-3} \mathrm{~kg}}\right]^{-1}\left[\frac{1 \mathrm{~m}}{10^{-2} \mathrm{~m}}\right]^3\left[\frac{1 \mathrm{~s}}{1 \mathrm{~s}}\right]^{-2}
\end{aligned}
\)
\(
n_2=6.67 \times 10^{-8}
\)
Thus, value of \(G\) in CGS system of units is \(6.67 \times 10^{-8}\) dyne \(\mathrm{cm}^2 / \mathrm{g}^2\).
Deducing relation between the physical quantities
If we know the factors on which a given physical quantity depends, we can find a formula relating to those factors.
Example 7: The frequency ( \(f\) ) of a stretched string depends upon the tension \(F\) (dimensions of force), length \(l\) of the string and the mass per unit length \(\mu\) of string. Derive the formula for frequency.
Solution: Suppose, the frequency \(f\) depends on the tension raised to the power \(a\), length raised to the power \(b\) and mass per unit length raised to the power \(c\).
Then, \(f \propto(F)^a(l)^b(\mu)^c\)
\(
f=k(F)^a(l)^b(\mu)^c \dots(i)
\)
Here, \(k\) is a dimensionless constant of proportionality.
Thus, \([f]=[\mathrm{F}]^a[l]^b[\mu]^c\)
\(
\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]=\left[\mathrm{MLT}^{-2}\right]^a[\mathrm{~L}]^b\left[\mathrm{ML}^{-1}\right]^c
\)
\(
\left[\mathrm{M}^0 \mathrm{~L}^0 \mathrm{~T}^{-1}\right]=\left[\mathrm{M}^{a+c} \mathrm{~L}^{a+b-c} \mathrm{~T}^{-2 a}\right]
\)
For dimensional balance, the dimensions on both sides should be same.
Thus, \(a+c=0 \dots(ii)\)
and \(a+b-c=0 \dots(iii)\)
\(-2 a=-1 \dots(iv)\)
Solving these three equations, we get
\(
a=\frac{1}{2}, \quad c=-\frac{1}{2}
\)
and \(b=-1\)
Substituting these values in Eq. (i), we get
\(
f=k(F)^{1 / 2}(l)^{-1}(\mu)^{-1 / 2} \text { or } f=\frac{k}{l} \sqrt{\frac{F}{\mu}}
\)
Experimentally, the value of \(k\) is found to be \(\frac{1}{2}\).
Hence, \(f=\frac{1}{2 l} \sqrt{\frac{F}{\mu}}\)
Example 8: The centripetal force \(F\) acting on a particle moving uniformly in a circle may depend upon mass ( \(m\) ), velocity (\(v\)) and radius (\(r\)) of the circle. Derive the formula for \(F\) using the method of dimensions.
Solution: Let \(F=k(m)^x(v)^y(r)^z \dots(i)\)
Here, \(k\) is a dimensionless constant of proportionality.
Writing the dimensions of RHS and LHS in Eq. (i), we have
\(
\begin{aligned}
{\left[\mathrm{MLT}^{-2}\right] } & =[\mathrm{M}]^x\left[\mathrm{LT}^{-1}\right]^y[\mathrm{~L}]^z \\
& =\left[\mathrm{M}^x \mathrm{~L}^{y+z} \mathrm{~T}^{-y}\right]
\end{aligned}
\)
Equating the powers of \(\mathrm{M}, \mathrm{L}\) and \(T\) on both sides, we have
\(
\begin{aligned}
& x=1, y=2 \text { and } y+z=1 \\
& z=1-y=-1
\end{aligned}
\)
Putting the values in Eq. (i), we get
\(
\begin{array}{ll}
F=k m v^2 r^{-1}=k \frac{m v^2}{r} & \\
F=\frac{m v^2}{r} & (\text { where }, k=1)
\end{array}
\)
Defects or limitations of dimensional analysis
The method of dimensional analysis has the following limitations:
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