Dimensional Analysis

Basics – Units vs dimensions – Analysis

Basics

In Physics, the dimension of a quantity is the physical property it measures. Dimensions of length, mass and time are often written as \(L\), \(M\) and \(T\). Dimensionless quantities have dimension \(1\).

If the height of a building is \(h\), we say \(h\) has dimensions of length:

\[[h] = L\]

The square brackets can be read as “dimensions of”.

A speed, like \(v = 55\) miles per hour, has dimensions of length over time:

\[[v] = \frac{L}{T}\]

The volume of a sphere is \(V = \frac{4}{3}\pi r^3\). Here, the quantities \(\frac{4}{3}\) and \(\pi\) are dimensionless. So we write

\[[V] = \left [\frac{4}{3}\pi r^3 \right ] = \left [\frac{4}{3}\pi \right ] \left [ r^3 \right ] = 1 \cdot L^3 = L^3 \]

In fact any volume (sphere, cube, cylinder, blob) has dimensions of \(L^3\) — that’s what “volume” means. Density, which is mass per volume, has dimensions \(M L^{-3}\).

Example

The Hubble Constant, \(H_0\), is a measure of how fast the universe is expanding. It has been measured to be \(71.9\) kilometers per second per megaparsec. That means a galaxy \(1\) Mpc (one megaparsec) away is receding at speed of \(71.9\) km/s. A parsec (abbreviated pc) is an astronomical unit of distance equal to \(3.09 \times 10^{16}\) m.

\[ H_0 = 71.9 \; \frac{\rm{km/s}}{\rm{Mpc}} = 71.9 \; \frac{\rm{km}}{\rm{s}} \frac{1}{\rm{Mpc}} \]

The dimensions of \(H_0\) are

\[\displaystyle \left[ H_0 \right] = \frac{L}{T} \frac{1}{L} = T^{-1}.\]

Some rules

• Only quantities with the same dimension may be added or subtracted.
  • You can add \(5\) meters plus \(1\) inch, but you may not add \(5\) meters plus \(1\) hour.
• Multiplying (or dividing) dimensionful quantities will change the dimension of the product.
  • \(55\) miles divided by \(1\) hour will have dimension \(L/T\).
• Arguments of transcendental (ie, non-polynomial) functions must be dimensionless. These include sine, cosine, square-root, exponentials, logarithms, etc.
  • Example: if \(x(t) = x_0 e^{\omega t}\) then we must have \([\omega ] = T^{-1}\) so that \(\omega t\) is dimensionless.
  • If the height of a wave is (improperly) expressed as \(y = A \cos x\), where \(x\) is a location in meters, then the author really means that \(y = A \cos kx\), where \(k = 1\;\textup{m}^{-1}\). Then \(kx\) is dimensionless, \(\cos kx\) is dimensionless, and \([y] = [A]\).
  • Logarithms are slightly special, since adding (subtracting) logs is the same as multiplying (dividing) their arguments.
    • Example: in kinematic problem 5 we get the equation
      \(\displaystyle t = \int_{0}^{s} \frac{ds}{(v_0 - ks)} = \left[ -\frac{1}{k} \ln (v_0 - ks) \right]_0^s = -\frac{1}{k} \ln \frac{v_0-ks}{v_0}\).
      The middle expression appears to take the log of something with dimensions of velocity, but it’s ok since we’re really dividing two velocities when subtracting two logs.
• When using calculus, note the following:
  • Infinitesimal quantities (such as \(dx\) or \(dt\)) have dimensions of the variable used (just ignore the \(d\)’s).
    • \([\frac{dx}{dt}] = \frac{L}{T}\)
    • This is because \(dx\) is really a tiny \(\Delta x = x_f - x_i\), which has the same dimension as \(x\).
    • For second derivatives, just ignore the \(d\)’s: \([\frac{d^2x}{dt^2}] = \frac{L}{T^2}\).
  • An integral sign represents a summation (a Riemann sum), and so, like addition, it does not change the dimension.
    • \(\left[ \int v \; dt \right] = [v] \cdot [dt] = \frac{L}{T} \cdot T = L\)

Units vs dimensions

The concepts of “units” and “dimensions” are similar but not identical.

• All dimensionful quantities have units.
  • For velocity, \([v] = \frac{L}{T}\). Units could be m/s or mi/hr.
• Some dimensionless quantities have units.
  • Angles are dimensionless: \([\theta] = 1\), but units could be radians or degrees.
    • For radians this is clearly true since an angle in radians is defined as an arc length over radius: \([\theta] = \left[ \frac{s}{r} \right] = \frac{L}{L} = 1\)
  • Number counts are dimensionless. “Five apples” is dimensionless, but could be said to have “units” of apples.
• Variable quantities have dimension, but undefined units.
  • For position equation \(x=vt\), both sides have dimension \(L\). But the units could be meters or nm or light-years. You would not write “\(x=vt\) m”, because \(x=vt\) is true in any units.
• Unit conversion ratios are dimensionless.
  • A ratio with the same dimensions in the numerator and denominator must be dimensionless. This is how we convert units without changing dimension.
  • To convert a pressure, say \(14.7\) lb/in², into metric units of N/m², you multiply by ratios that change the units but not the dimension:

\[ P = 14.7 \; \frac{\textup{lb}}{\textup{in}^2} \cdot \left( \frac{4.45\;\textup{N}}{1 \; \textup{lb}} \right) \cdot \left( \frac{1 \; \textup{in}}{2.54 \;\textup{cm}} \right)^2 \cdot \left( \frac{100 \;\textup{cm}}{1 \; \textup{m}} \right)^2 = 1.01 \times 10^5 \; \frac{\textup{N}}{\textup{m}^2} \]

Analysis

For any equation, both sides must have the same dimension, and those dimensions can be treated as algebraic quantities. This technique of dimensional analysis can be used to

• double-check equations you have derived, and
• deduce the Physics of a situation with little knowledge.

Example: double-check your algebra

If you make an algebra mistake, there is a good chance the dimensions got messed up. An occasional check of dimensional consistency is a good way to catch errors early.

Observe Alice’s solution to a certain dynamics problem. At some point she obtains an equation for acceleration:

\[ a = g (\sin \theta - \mu \cos \theta). \]

Since she knows that \(\mu\) is a dimensionless quantity, a quick check shows that her acceleration equation is dimensionally correct. The right-hand side is proportional to \(g\), the free-fall acceleration.

Also, her velocity at the end of the problem is

\[ v = \sqrt{ 2 gh \left( 1 - \frac{\mu}{\tan \theta} \right) } \]

The part in parentheses is clearly dimensionless. So checking the dimension of the rest:

\[ \left[ \sqrt{ 2 gh } \right] = \sqrt{ [ gh ] } = \sqrt{ \frac{L}{T^2} \cdot L } = \sqrt{ \frac{L^2}{T^2} } = \frac{L}{T} \]

These are dimensions of velocity, which is what we were hoping. Dimensional consistency is not a guarantee that the equation is correct, but it increases one’s confidence significantly.

Example: deduce the Physics

• When the concept of energy is developed in first-semester Physics, an object’s kinetic energy is shown to be \(K = \tfrac{1}{2} m v^2\). Thus the dimension of energy is equal to the dimension of mass times velocity-squared, \([m] \cdot [v]^2\). Based on dimensional analysis alone, with no knowledge of Special Relativity, one may wonder what this would mean if the velocity were \(c\), the speed of light. This speculation leads one to consider \(E=mc^2\), the important relation of mass-energy equivalence.
• Walter Lewin (MIT) shows with dimensional analysis that free-fall time must be proportional to the square-root of the drop height.
• If you’ve taken Archie’s Physics 1 class, you’ve seen how a UK physicist was able, with dimensional analysis alone, to determine the highly-classified energy of the Trinity nuclear detonation.

Last modified: October, 2025