Dimensions of Physical Quantities, Dimensional Analysis and its Applications

In physics, we often deal with quantities that can be measured. These quantities have dimensions, which describe their fundamental nature. For example, length, mass, and time are fundamental dimensions. All other physical quantities can be expressed in terms of these fundamental dimensions. Understanding dimensions is crucial for building a strong foundation in physics.

Understanding Fundamental and Derived Quantities

Physical quantities are broadly classified into two types:

• Fundamental Quantities: These are the basic quantities that cannot be defined or expressed in terms of other physical quantities. The most common fundamental quantities in mechanics are length, mass, and time. Other fundamental quantities recognized in physics include temperature, electric current, luminous intensity, and amount of substance.
• Derived Quantities: These quantities are defined and expressed in terms of fundamental quantities. For example, velocity is defined as the rate of change of displacement with respect to time. Displacement has the dimension of length, and time has the dimension of time. Therefore, velocity is a derived quantity.

The Concept of Dimensions

Dimensions refer to the fundamental physical quantities that make up a derived physical quantity. We use symbols to represent these fundamental dimensions. The standard symbols are:

• Mass: M
• Length: L
• Time: T
• Temperature: $\Theta$ (Theta)
• Electric Current: A
• Amount of Substance: N
• Luminous Intensity: J

When expressing the dimensions of a derived quantity, we write them in terms of these symbols. For example, if a quantity's dimensions depend on mass raised to the power of 'a', length raised to the power of 'b', and time raised to the power of 'c', we write its dimensions as [M^a L^b T^c]. The square brackets [ ] denote "the dimensions of".

Dimensions of Common Physical Quantities

Let's determine the dimensions of some common physical quantities. This exercise will help you grasp the concept.

1. Area

Area = Length × Width. Both length and width are measures of length. So, Area = Length × Length = L × L = L². The dimensions of Area are [M⁰ L² T⁰]. We often omit dimensions with a power of zero.

2. Volume

Volume = Length × Width × Height. All are measures of length. So, Volume = L × L × L = L³. The dimensions of Volume are [M⁰ L³ T⁰].

3. Velocity

Velocity = Displacement / Time. Displacement has dimensions of length (L), and time has dimensions of time (T). So, Velocity = L / T = L T^-1. The dimensions of Velocity are [M⁰ L¹ T^-1].

4. Acceleration

Acceleration = Change in Velocity / Time. We know the dimensions of velocity are [L T^-1]. Time is [T]. So, Acceleration = (L T^-1) / T = L T^-2. The dimensions of Acceleration are [M⁰ L¹ T^-2].

5. Force

Force = Mass × Acceleration. Mass has dimensions [M], and acceleration has dimensions [L T^-2]. So, Force = M × (L T^-2) = M L T^-2. The dimensions of Force are [M¹ L¹ T^-2].

6. Density

Density = Mass / Volume. Mass has dimensions [M], and volume has dimensions [L³]. So, Density = M / L³ = M L^-3. The dimensions of Density are [M¹ L^-3 T⁰].

7. Momentum

Momentum = Mass × Velocity. Mass has dimensions [M], and velocity has dimensions [L T^-1]. So, Momentum = M × (L T^-1) = M L T^-1. The dimensions of Momentum are [M¹ L¹ T^-1].

8. Pressure

Pressure = Force / Area. Force has dimensions [M L T^-2], and area has dimensions [L²]. So, Pressure = (M L T^-2) / L² = M L^-1 T^-2. The dimensions of Pressure are [M¹ L^-1 T^-2].

9. Energy (Work)

Work = Force × Distance. Force has dimensions [M L T^-2], and distance has dimensions [L]. So, Work = (M L T^-2) × L = M L² T^-2. The dimensions of Energy are [M¹ L² T^-2].

10. Power

Power = Work / Time. Work has dimensions [M L² T^-2], and time has dimensions [T]. So, Power = (M L² T^-2) / T = M L² T^-3. The dimensions of Power are [M¹ L² T^-3].

Dimensional Analysis

Dimensional analysis is a powerful technique used in physics to check the correctness of equations, derive relationships between physical quantities, and convert units. It is based on the principle of homogeneity of dimensions.

Principle of Homogeneity of Dimensions

This principle states that an equation is dimensionally correct only if the dimensions of all the terms on both sides of the equation are the same. In simpler terms, you can only add or subtract quantities that have the same dimensions. For example, you can add lengths to lengths, but you cannot add length to mass.

Consider an equation: A + B = C. For this equation to be dimensionally valid, the dimensions of A, B, and C must all be identical. For example, in the equation of motion: $s = ut + \frac{1}{2}at^2$ Let's check the dimensions of each term:

• s (displacement): [L]
• ut: (u is velocity [LT^-1]) × (t is time [T]) = [LT^-1] × [T] = [L]
• $\frac{1}{2}at^2$: ( $\frac{1}{2}$ is a dimensionless constant) × (a is acceleration [LT^-2]) × (t² is time squared [T²]) = [LT^-2] × [T²] = [L]

Since the dimensions of all terms are [L], the equation is dimensionally correct.

Applications of Dimensional Analysis

Dimensional analysis has several important applications in physics:

1. Checking the Dimensional Consistency of Equations

As shown with the equation of motion example, we can use dimensional analysis to verify if an equation is physically plausible. If an equation is not dimensionally consistent, it is definitely incorrect. However, if it is dimensionally consistent, it does not guarantee it is correct (as dimensionless constants are ignored).

2. Deriving Relationships Between Physical Quantities

We can use dimensional analysis to find an equation relating different physical quantities, provided we know the dependence of one quantity on others. This is particularly useful when the exact form of the equation is unknown or difficult to derive through conventional methods.

Example: Deriving the formula for the time period of a simple pendulum. Let's assume the time period (T) of a simple pendulum depends on its length (l), the mass of the bob (m), and the acceleration due to gravity (g). We can write this relationship as: T $\propto$ l^a m^b g^c Introducing a constant of proportionality k: T = k l^a m^b g^c Now, let's write the dimensions of each quantity:

Substituting these dimensions into the equation: [T] = [L]^a [M]^b [LT^-2]^c [M⁰ L⁰ T¹] = [L^a] [M^b] [L^c T^-2c] [M⁰ L⁰ T¹] = [M^b L^a+c T^-2c]

Now, we equate the powers of M, L, and T on both sides:

• For M: 0 = b
• For L: 0 = a + c
• For T: 1 = -2c

From the equation for T, we get c = -1/2. From the equation for L, we get a = -c = -(-1/2) = 1/2. From the equation for M, we get b = 0.

Substituting these values back into the assumed relation: T = k l^1/2 m⁰ g^-1/2 T = k l^1/2 / g^1/2 T = k $\sqrt{l/g}$

Dimensional analysis gives us the form of the equation. The actual value of the constant k (which is 2$\pi$ in this case) cannot be determined by this method.

3. Converting Units from One System to Another

Dimensional analysis can be used to convert a physical quantity from one system of units to another. For example, converting Joules (SI unit of energy) to ergs (CGS unit of energy).

Example: Convert 1 Newton to dynes. We know that Force = Mass × Acceleration. In the SI system: 1 N = 1 kg × 1 m/s² In the CGS system: 1 dyne = 1 g × 1 cm/s² The dimensions of force are [M L T^-2]. Let's say we want to convert a quantity $Q$ from system 1 to system 2. $Q_1 = n_1 [M_1 L_1 T_1^{-2}]$ (where $n_1$ is the numerical value and units are represented by M, L, T) $Q_2 = n_2 [M_2 L_2 T_2^{-2}]$ Since the quantity is the same, $Q_1 = Q_2$. $n_1 [M_1 L_1 T_1^{-2}] = n_2 [M_2 L_2 T_2^{-2}]$ We want to find $n_2$ in terms of $n_1$. $n_2 = n_1 \left( \frac{M_1}{M_2} \right) \left( \frac{L_1}{L_2} \right) \left( \frac{T_1}{T_2} \right)^{-2}$

For converting Newtons (SI) to dynes (CGS): System 1 (SI): $M_1 = 1$ kg, $L_1 = 1$ m, $T_1 = 1$ s System 2 (CGS): $M_2 = 1$ g, $L_2 = 1$ cm, $T_2 = 1$ s We know 1 kg = 1000 g and 1 m = 100 cm. $n_1 = 1$ (for 1 Newton) $n_2 = 1 \left( \frac{1 \text{ kg}}{1 \text{ g}} \right) \left( \frac{1 \text{ m}}{1 \text{ cm}} \right) \left( \frac{1 \text{ s}}{1 \text{ s}} \right)^{-2}$ $n_2 = 1 \left( \frac{1000 \text{ g}}{1 \text{ g}} \right) \left( \frac{100 \text{ cm}}{1 \text{ cm}} \right) (1)^{-2}$ $n_2 = 1 \times 1000 \times 100 \times 1$ $n_2 = 100,000$ So, 1 Newton = 100,000 dynes.

Limitations of Dimensional Analysis

While powerful, dimensional analysis has certain limitations:

• It cannot determine dimensionless constants of proportionality (like 2$\pi$ or 1/2).
• It cannot be used to derive equations involving more than three physical quantities if the dimensions of the quantities are not independent (e.g., if a quantity depends on A, B, and C, but the dimensions of A, B, and C are linearly dependent).
• It cannot distinguish between quantities with the same dimensions (e.g., Work and Torque both have dimensions of [M L² T^-2]).
• It cannot handle equations involving trigonometric, exponential, or logarithmic functions, as these are not expressible in terms of M, L, and T alone.

Dimensional Formulae for Other Fundamental Quantities

Let's extend our understanding to other fundamental quantities beyond M, L, T.

Temperature

The fundamental dimension for temperature is $\Theta$ (Theta). Example: Coefficient of thermal expansion has dimensions of [$\Theta^{-1}$].

Electric Current

The fundamental dimension for electric current is A. Example: Charge (Q) = Current (I) × Time (t). Dimensions of Charge: [A T].

Electric Potential (Voltage)

Potential = Work / Charge. Work has dimensions [M L² T^-2]. Charge has dimensions [A T]. So, Electric Potential = (M L² T^-2) / (A T) = M L² T^-3 A^-1. Dimensions of Electric Potential: [M¹ L² T^-3 A^-1].

Resistance

Resistance (R) = Voltage (V) / Current (I). Voltage has dimensions [M L² T^-3 A^-1]. Current has dimensions [A]. So, Resistance = (M L² T^-3 A^-1) / A = M L² T^-3 A^-2. Dimensions of Resistance: [M¹ L² T^-3 A^-2].

Summary Table of Dimensions

It's beneficial to have a quick reference for the dimensions of frequently encountered quantities.

Mastering the dimensions of these quantities and the principles of dimensional analysis will significantly aid in your understanding and problem-solving in physics. Practice deriving dimensions for new quantities and applying the principles to check equations.