4: Dimensional analysis | Course - StudyGenius | StudyGenius

Course Progress

Victories 0/39
Finished 0/39

StudyGenius Logo

4: Dimensional analysis

Play with a friend

Turn-based: you each do the quiz on your own time.

Play vs random

Face a random opponent that have completed this quiz before.

Choose your name

Choose your name

Play with a friend

Turn-based: you each do the quiz on your own time.

Play vs random

Face a random opponent that have completed this quiz before.

The quiz will be on the following text — learn it for the best chance to win.

Dimensional Analysis

Dimensional analysis is a powerful, fundamental tool in physics used to check the validity of equations, predict possible forms of physical relationships, convert units systematically, and reduce complex problems. It rests on the principle that physical quantities possess inherent dimensions (like length, mass, time), independent of the specific units used to measure them (like meters, kilograms, seconds).

The seven base quantities in the SI system and their dimensions are:

  • Length (L)
  • Mass (M)
  • Time (T)
  • Electric Current (I)
  • Thermodynamic Temperature (Θ\Theta)
  • Amount of Substance (N)
  • Luminous Intensity (J)

All other physical quantities are derived quantities, expressible as products of powers of these base dimensions. For example:

  • Velocity: [v]=LT1[v] = L T^{-1} (length per time)
  • Acceleration: [a]=LT2[a] = L T^{-2}
  • Force: [F]=MLT2[F] = M L T^{-2} (from Newton's Second Law, F=maF = ma)

The cornerstone is the Principle of Homogeneity (Dimensional Consistency): every term being added, subtracted, or compared in a physically meaningful equation must have identical dimensions. This is the primary method for checking equations. Consider the equation for position under constant acceleration: s=ut+12at2s = ut + \frac{1}{2}at^2.

  • [s]=L[s] = L
  • [ut]=[u][t]=(LT1)(T)=L[ut] = [u][t] = (L T^{-1})(T) = L
  • [12at2]=[a][t2]=(LT2)(T2)=L[\frac{1}{2}at^2] = [a][t^2] = (L T^{-2})(T^2) = L

All terms have dimension L, so the equation is dimensionally consistent. An equation like s=ut+ats = ut + at fails ([at]=LT1L[at] = L T^{-1} \neq L).

Beyond checking, dimensional analysis helps derive relationships. Suppose you suspect the period (T) of a simple pendulum depends on its length (l) and gravity (g), but not mass. Assuming TlagbT \propto l^a g^b:

  • [T]=T[T] = T
  • [la]=La[l^a] = L^a
  • [gb]=(LT2)b=LbT2b[g^b] = (L T^{-2})^b = L^b T^{-2b}
  • For consistency: La+bT2b=T1L0L^{a+b} T^{-2b} = T^1 L^0
  • Equate exponents:
    • For T: 2b=1b=1/2-2b = 1 \rightarrow b = -1/2
    • For L: a+b=0a+(1/2)=0a=1/2a + b = 0 \rightarrow a + (-1/2) = 0 \rightarrow a = 1/2

Thus, TlgT \propto \sqrt{\frac{l}{g}}. This illustrates the Buckingham Pi Theorem, which formalizes how dimensionless products (Pi terms) can be formed from variables to describe a system.

Dimensional analysis is also essential for unit conversions, ensuring quantities are expressed in compatible units before calculations.