Dimensional Analysis Study Notes

Chapter 1: Dimensional Analysis

Generalities

  • The term dimensional analysis is more comprehensive than just units of measurement. It encompasses fundamental concepts related to dimensions in physics and measurements.

Dimension Definition

  1. Dimension

    • A dimension is a fundamental property that describes physical quantities.

    • Mass (dimension) is distinct from its unit of measurement (kilogram).

    • Example: Measurement of distance utilizes various units (feet, meters, kilometers), each representing the dimension of length.

Dimensional Analysis

  1. Definition

    • Dimensional analysis is a method used in physics and sciences to verify relationships between physical quantities by determining their dimensions.

    • Key points:

      • Quantities on both sides of a mathematical equation must possess the same dimension.

      • All quantities must belong to the same system (base).

    • Important Note: Dimensional analysis does not guarantee the validity of physical laws; that is determined through mathematical analysis.

    • Example: The areas of a square and a triangle share the same dimension (𝐿²) but are not equal.

Fundamental Units

  1. Seven Fundamental Units

    • Dimensions are classified into seven fundamental categories:

      • Length (𝐿) = [𝑙]

      • Time (𝑇) = [𝑑]

      • Mass (𝑀) = [π‘š]

      • Temperature (πœƒ) = [πœƒ]

      • Electric current (𝐼) = [𝐼]

      • Amount of substance (𝑁) = [𝑛]

      • Luminous intensity (𝐽) = [𝐽]

Expressing Dimensions

  1. Expression of Physical Quantity

    • The dimension of a physical quantity (G) is expressed mathematically as:
      [𝐺]=πΏπ‘ŽM𝑏T𝑐I𝑑Jπ‘’πœƒπ‘“N𝑔[𝐺] = 𝐿^π‘Ž M^𝑏 T^𝑐 I^𝑑 J^𝑒 πœƒ^𝑓 N^𝑔

    • Variables (a, b, c, d, e, f, g) denote the powers associated with fundamental dimensions.

    • Dimensionless Quantities: Certain quantities, like plane angle and solid angle, are independent of basic dimensions and are considered dimensionless (e.g., sin x, cos x).

    • Notable Values:

      • [Numeric value] = 1

      • [angle] = 1

      • [cos Ξ±] = [sin Ξ±] = [tan Ξ±] = [cot Ξ±] = [ln x] = [e^x] = 1

      • Dimensionless quantities are referred to as pure numbers.

Properties of Dimensional Equations

  1. Properties

    • Avoid including a unit system when writing a dimensional equation.

    • If $[G] = 1$, then G is constant but could have a unit, such as $[2 ext{Ο€}] = 1$ (Radian or degrees).

    • Equations involving dimensions must satisfy the following:

      • If $G = A imes B$, then:
        [𝐺]=[𝐴][𝐡][𝐺] = [𝐴][𝐡]

      • If $G = A B$, then [𝐺]=[𝐴][𝐡][𝐺] = [𝐴][𝐡]

      • If $G = A^n$, then [𝐺]=[𝐴]n[𝐺] = [𝐴]ⁿ

      • Here, n is dimensionless.

      • Dimensions cannot be added or subtracted directly: $G = A Β± B
        ightarrow [G] = [A] = [B]$.

    • Important Notes:

      • A heterogeneous (non-homogeneous) equation is necessarily false.

      • A homogeneous equation is not necessarily true.

      • In mechanics, the primary quantities needed are mass (l) and time (t).

      • Denote quantities symbolically within square brackets (e.g., [𝐺]).

Examples of Dimensional Analysis

  1. Dimensional Equations

    • Example 1: For displacement ($s$), given by $s = l^2$, the dimension is:
      [s]=[l2]=[l]2=𝐿2[s] = [l²] = [l]^2 = 𝐿²

    • Example 2: For velocity ($v$), defined as $V = x/t$, its dimension becomes:
      [v]=[x][t]βˆ’1=LTβˆ’1[v] = [x][t]^{-1} = L T^{-1}

    • Example 3: For force ($F = ma$):
      [F]=[m][a]=MLTβˆ’2[F] = [m][a] = M L T^{-2}

Derived Quantities

  1. Categories of Physical Quantities

    • Physical Quantities: Any measurable property, like mass, time, area, volume, energy, density, and pressure.

    • Types of Physical Quantities:

    1. Scalar Quantities: Described by magnitude only (e.g., mass, temperature).

    2. Vectorial Quantities: Described by both magnitude and direction (e.g., velocity, force).

Measurement Methods

  1. Measuring Physical Processes

    • Two primary methods exist:

      • Direct Measurement (e.g., using a ruler for distance).

      • Calculation using mathematical relationships (e.g., calculating speed using $Ξ½ = rac{x}{T}$).

International System of Units

  1. Unit Standards

    • The value of a physical quantity is expressed in relation to standard units known as units.

    • The MKSA International System was adopted in 1946, establishing fundamental units:

      • Length = Meter (m)

      • Mass = Kilogram (kg)

      • Time = Second (s)

      • Current Intensity = Ampere (A)

      • Temperature = Kelvin (K)

      • Amount of Substance = Mole (Mol)

      • Luminous Intensity = Candela (Cd)

    • Derived Units: Area (mΒ²), Velocity (m/s), Force (kg m/sΒ²).

Alternative Unit Systems

  1. CGS System

    • The CGS system (Centimeter, Gram, Second) is less common, established in 1847, where:

      • Distance = Centimeters (cm)

      • Force = Dynes (Dyne)

      • Energy = Ergs (Erg).

Dimensional Consistency Examples

  1. Example Analysis

    • Velocity:
      [v]=[x][t]βˆ’1=LTβˆ’1[v] = [x][t]^{-1} = L T^{-1}

    • Acceleration:
      [a]=[v][t]βˆ’1=LTβˆ’2[a] = [v][t]^{-1} = L T^{-2}

    • Force:
      [F]=[m][a]=MLTβˆ’2[F] = [m][a] = M L T^{-2}

    • Charge Quantity:
      [Q]=[i][t]=IT[Q] = [i][t] = IT

    • Kinetic Energy:
      E=rac12mv2<br>ightarrow[E]=ML2Tβˆ’2E = rac{1}{2}mv^2 <br>ightarrow [E] = M L^{2} T^{-2}

Exercises on Dimensional Consistency

  1. Exercise 01: Analyze the change in displacement with acceleration and time:
    x=rac12at2x = rac{1}{2} a t^2

    • Left side dimension:
      [x]=L[x] = L

    • Right side dimension:
      [rac12][a][t2]=LTβˆ’2T2=L[ rac{1}{2}][a][t^2] = L T^{-2} T^2 = L

    • The equation is dimensionally correct.

  2. Exercise 02: Deriving the period in relation to length and gravitational force:
    T=klxgyT = k l^x g^y

    • Identify:

      • Since $[T] = T$, $[l] = L$, and constant K is dimensionless:

      • $[T] = [k][l]^x[g]^y$ we express:

      • Given the relation for gravitational force ($
        ho = mg$):

      • Using the dimensions:

      • [g]=[<br>ho]m[g] = [<br>ho] m

      • Therefore, by substituting dimensions, we derive:
        [T]=Lx(LTβˆ’2)y[T] = L^x (LT^{-2})^y

    • Solving yields:

      • $x + y = 0$, and $-2y = 1$ leading to:

      • $y = - rac{1}{2}, ext{ and } x = rac{1}{2}$.

    • Finally, the relationship for the period is:
      T=kracl1/2g1/2T = k rac{l^{1/2}}{g^{1/2}}

Chapter 1: Dimensional Analysis - Conclusion

  • Comprehensive and exhaustive understanding of dimensional analysis, units, and derived quantities fosters a deeper appreciation for the mathematical relationships in physics and related fields.