In physics, dimensional analysis is a powerful tool for checking relationships between physical quantities. The foundation of the technique is consideration of the dimensions of physical quantities. What exactly do I mean by dimensions? Here the word does not mean the same thing as it does in Euclidean geometry. As a simple example lets consider the physical quantity of speed. As you drive down the highway your speedometer may report a speed of 55 miles/hour or equivalently 88.5 kilometers/hour. These two measurements of your vehicle’s speed have different units (miles/hour versus kilometers/hour) but they have equivalent dimensions, that is, both measurements are ratios of two base dimensions, distance (or length) over time. All physical quantities, if not already represented by base dimensions, are derived from a relatively small number of base dimensions. The table below summarized seven base dimensions encountered in physics, their common abbreviations, the SI units for physical quantities of each dimension, and some alternative non-SI units commonly encountered.

Dimension |
Dimension Symbol |
SI Unit |
Alternative Units |

Length | [L] | meter (m) | inch, mile, kilometer |

Time | [T] | second (s) | minute, hour, year |

Mass | [M] | kilogram (kg) | gram, slug |

Temperature | [Θ] | kelvin (K) | celsius, fahrenheit |

Current | [I] | ampere (A) | |

Luminous Intensity | [J] | candela (cd) | |

Amount of a Substance | [N] | mole (mol) |

In order to use dimensional analysis as a tool, it is important to understand the difference between the dimensions of a physical quantity and its units. Dimensions relate to the kind of physical quantity being measurement (speed, energy, force, etc.) while units communicate the system of measurement being used (e.g., kilometers/hour, Joules, Newtons, etc. in the SI system). To prevent confusion, dimensions are always represented using the bracket notation shown above. Take a minute to practice. Use the table above and your physics textbook to write the dimensions of the following quantities (answers at the bottom of the page).

**1) 9.8 m/s ^{2}**

**2) 100 N**

**3) 5 V**

Now that we have thoroughly defined what dimensions are, let’s illustrate how using dimensional analysis can help you succeed on a physics exam or the MCAT physical sciences exam. Consider the following MCAT-like question.

**4) A particularly dramatic form of mass wasting at Zion Canyon National Park occurs when strata of rock more resistant to erosion exist above strata more susceptible to erosion. The underlying rock erodes more rapidly leading to overhanging cliffs composed of the upper more resistant strata. The resistant rock eventually succumbs to gravity and calves away from the cliff falling freely to the valley floor below. Consider a boulder of mass m that falls a distance h from a cliff to the bottom of Zion Canyon. Which of the following most accurately describes the velocity of this boulder the instant before hitting the ground. The acceleration due to gravity is g.**

**A) (gh) ^{1/2}**

**B) 2gh**

**C) (2gh)**

^{1/2}**D) mgh**

Invoking conservation of mechanical energy, the kinetic energy of the boulder the instant before hitting the ground is equal to the gravitational potential energy of the boulder before it breaks away from the cliff, mgh = (1/2)mv^{2}. Therefore, after rearranging this equation we see that the velocity of the boulder the instant before hitting the ground is equal to option C. But what if you forgot the formulas for potential and kinetic energy? This is where dimensional analysis may help. First, determine the dimensions of the answer. Here the question concerns velocity which has dimensions of [L]/[T]. The key is recognizing that the dimensions of the correct answer must be the same. This is the essence of dimensional analysis as it applies to a question like this. Let’s determine the dimensions of the possible answers. Since g is an acceleration with dimensions of [L]/[T]^{2}, option A through D have dimensions of [L]/[T], [L]^{2}/[T]^{2}, [L]/[T], and [M][L]^{2}/[T]^{2}, respectively. Therefore, we can eliminate options B and D from consideration because they lack the required dimensions. In this case, dimensional analysis has not provided the solution but it has increased your chance of guessing the correct answer from 25% to 50%.

Another and perhaps more powerful use of dimensional analysis is helping you to remember formulas during the exam. Consider the following MCAT-like question.

**5) The capacitance (C) of an air filled parallel plate capacitor is related to the area of the plates (A) and the distance between the plates (d), C = ε _{0}(A/d), where ε_{0}=8.85 x 10^{-12} s^{4}A^{2}/(m^{3}kg) is the permittivity of free space. Consider a capacitor with plates of area 0.00001 m^{2} and separated by a distance of 0.02 m. When fully charged by a 9 V battery, how much charge is stored on each plate?**

Oh no, you can easily calculate the capacitance of the capacitor using the information given but you forget how to get charge from capacitance. You know it depends on voltage but was it Q=C/V? Q=V/C? Don’t panic, try using dimensional analysis. First use the relation giving the capacitance of a parallel plate capacitor to determine the dimensions of capacitance. The dimensions of ε_{0} are [T]^{4}[I]^{2}/([L]^{3}[M]), the dimensions of area are [L]^{2}, and the dimensions of distance are [L]. Put these all together and the dimensions of capacitance are [T]^{4}[I]^{2}/([L]^{2}[M]). From the definition of current (charge per unit time), charge has dimension of [I][T]. Voltage (see question 3 above) has dimensions of [M][L]^{2}/([I][T]^{3}). Low and behold, one gets the dimensions of charge when the dimensions of capacitance are multiplied by the dimensions of voltage. Therefore, the correct relationship between the charge on the capacitor, voltage, and capacitance is Q = CV. Now you can solve the problem.

This may all seem too complicated to rely on during an exam. But it isn’t, it just takes practice to achieve mastery, and mastery of dimensional analysis will pay huge dividends in ways not even mentioned here. When you start to think of physical quantities in terms of their underlying dimensions, you will start to understand that physical formulas summarize the linkages between physical quantities. Then, you will start to have better conceptual understanding of physics. If you are currently in a physics course, practice dimensional analysis when you study for exams and practice when you do your physics homework. In time, you will add a powerful new tool to your physics problem solving arsenal.

**Answers:**

1) [L]/[T]^{2}

2) A Newton is a kg m/s^{2} (refer to Newton’s Second Law, F=ma) so its dimensions are [M][L]/[T]^{2}

3) Voltage is electric potential energy per unit charge so 1 V = 1 J/C = 1 Nm/C = 1 kg m^{2} / (C s^{2}) = 1 kg m^{2} / (A s^{3}) so its dimensions are [M][L]^{2}/([I][T]^{3})