You cannot remember a formula, yet by looking only at the dimensions of its quantities you can nearly rebuild it, all but the constant out in front. Dimensional analysis does three jobs: it converts units, checks equations and builds them. And in eight places it gives up.
1. From one system to another
If a quantity is known in one system, dimensional analysis gives its value in another. When the dimension of the quantity is [MˣLʸTᶻ], the conversion formula is:
n₂ = n₁ [M₁/M₂]ˣ [L₁/L₂]ʸ [T₁/T₂]ᶻ
Here n₁ is the value in the first system, M₁, L₁, T₁ are its base units, and M₂, L₂, T₂ belong to the second system.
Watch the formula working. Change the quantity, change the value: the powers in the dimension decide the factor.
converterpick a quantity
SI to CGS, by dimension
Now, how many dynes is 1 N? Force has dimension [MLT⁻²], so x = 1, y = 1, z = −2. SI uses kg, m, s and CGS uses g, cm, s. Putting them in, n = (10³ g/1 g)(10² cm/1 cm)(1 s/1 s)⁻² = 10⁵. So 1 N = 10⁵ dyne, and the conversion factor is 10⁵ dyne per N. Put the steps in order.
in orderHint: dimension first.
Order the steps: 1 N to dyne
What must you know first? Which base quantities force is built from.
Right. Next step?
Not yet.
Right order. 1 N = 10⁵ dyne.
Example 1.3(1): how many dynes is 50 N? Start from 50 instead of 1 and use the same factor: 50 N = 50 × 1000 × 100 × 1 = 5×10⁶ dyne.
in orderHint: dimension first.
Order the steps: 50 N to dyne
Begin with what is given.
Yes. Then?
Not yet.
Right. 50 N = 5×10⁶ dyne.
The method can also find a unit. Example 1.3(2): if the unit of acceleration is 9.8 ms⁻² and of velocity 3×10⁸ ms⁻¹, what is the unit of time? Acceleration is velocity over time, so T = [v]/[a] = 3×10⁸/9.8 = 3.06×10⁷ s.
2. Is the equation correct?
To check an equation, find the dimensions of both sides separately, then compare. If they match, the equation is dimensionally correct.
The balance from the last note again, this time with this note's equations.
balancepick an equation
Put it on the balance
Take s = ut + ½at². [s] = L, [ut] = LT⁻¹ × T = L, [½at²] = LT⁻² × T² = L. Every term has dimension L and both sides agree, so the equation is dimensionally correct. Now judge six equations yourself.
sort themHint: look at each term's dimension.
Right or wrong?
Pick an equation and drop it in a box. Find the dimension of every term first.
Right. Next one.
Not yet. Write the two sides' dimensions separately and compare.
All in place. v = u + at² is wrong: v and u are both LT⁻¹ but at² is L. E = mc is wrong: E is [ML²T⁻²] while mc is [MLT⁻¹].
3. Building an equation
Write the quantity you want as a product of the quantities it depends on, each with an unknown power: x, y, z. Then equate the dimensions of the two sides. Each of M, L and T gives one equation, and the powers fall out of them.
For a simple pendulum take T = K mˣ lʸ gᶻ. Putting in dimensions, M: x = 0, L: y + z = 0, T: −2z = 1. So x = 0, z = −1/2, y = 1/2, and T = K√(l/g). Since x = 0 the period does not depend on mass; T ∝ √l and T ∝ 1/√g. The value of K, however, cannot be found by dimensional analysis.
Let the speed of sound v in a gas depend on the density ρ and the elasticity E: v ∝ ρˣEʸ. Solving gives x = −1/2 and y = 1/2, so v = K√(E/ρ). Experiment gives K = 1, so v = √(E/ρ).
Let the frequency n of a stretched string depend on the tension T, the length l and the mass per unit length m: n ∝ Tˣ lʸ mᶻ. Here T is tension, not time, with dimension [MLT⁻²]. Solving gives x = 1/2, y = −1, z = −1/2, so n = K(1/l)√(T/m). Experiment gives K = 1/2, so n = (1/2l)√(T/m).
Now derive all three yourself. Enter the powers and see which line fails to match.
calculatorHint: one equation per base dimension.
Find the powers
You enter the powers; the tool only compares dimensions. Nothing goes anywhere.
The same move works when the base quantities change. Take velocity (v), time (T) and force (F) as base quantities and the dimension of mass is [m] = [F][v]⁻¹[T] = Fv⁻¹T.
4. Where the method gives up
There are eight limitations. One: the value of a constant cannot be found, such as the 2π in T = 2π√(l/g). Two: a dimensional constant, such as G, blocks the relation. Three: a dimensionless factor, such as cos θ in W = Fs cos θ, cannot be related to the rest.
Four: when more than L, M, T are involved, an equation cannot be built, as with heat and temperature. Five: the value of a dimensionless constant cannot be found. Six: with exponential, trigonometric or logarithmic quantities, such as sin θ, no relation can be established.
Seven: scalars and vectors cannot be told apart. Eight: equal dimensions do not mean the same quantity. Torque, work and energy all have dimension [ML²T⁻²], yet work and energy are scalars and torque is a vector. Pair each limit with its example.
match themHint: what kind of obstacle is the example?
Spot the limit
Pick up a limit, then drop it in its example's box.
Right. Next one.
Not there. Ask what the obstacle is in this example: an unknown constant, an extra quantity, or the identity of the quantity?
All seven pairs matched. The method checks and shapes, but it never fixes the number.
5. Check yourself
Three questions in the exam's style. Start with the first.
check yourselfpick one answer
Question 1
What is the dimension of √(L/g)?
T⁻² has the wrong sign and double the power; after the square root T to the first power remains.
T² is the dimension of L/g; the square root is still to be taken.
T⁻¹ is the dimension of frequency; the period of a pendulum has a positive power.
Right. L/g = L ÷ LT⁻² = T², so √(L/g) has dimension T.
Now the second.
check yourselfpick one answer
Question 2
In C.G.S units the value of the Boltzmann constant is how many times its value in S.I. units?
10⁻⁷ is inverted; the C.G.S unit (erg) is smaller, so the number is larger.
Right. The Boltzmann constant has dimension ML²T⁻²θ⁻¹; as 1 J = 10⁷ erg, the C.G.S value is 10⁷ times larger.
10⁻⁵ belongs to the newton-dyne conversion; here the unit of energy is involved.
10⁵ is the N-to-dyne factor; this is energy (joule to erg), so it is 10⁷.
In the last one, match the dimension of every term.
check yourselfpick one answer
Question 3
The force of 1 newton (1 N) is equal to how many dyne?
10 dyne is far too small; kg to g gives 10³ and m to cm gives 10².
100 dyne counts only the length factor and leaves out the 10³ from mass.
Right. Force is [MLT⁻²]: n = 10³ × 10² × 1 = 10⁵, so 1 N = 10⁵ dyne.
10⁴ dyne leaves out part of the mass or length factor.
The whole note on eight cards.
cut from the notetap to turn over
Eight cards
QuestionWhat is the formula for converting a quantity from one system to another?
Answern₂ = n₁ [M₁/M₂]ˣ [L₁/L₂]ʸ [T₁/T₂]ᶻ
QuestionShow that s = ut + ½at² is dimensionally correct.
Answer[s] = L; [ut] = L; [½at²] = LT⁻²×T² = L. All terms agree, so it is correct.
QuestionDerive the period of a simple pendulum by dimensional analysis.
AnswerFrom T = K mˣ lʸ gᶻ, x = 0, y = 1/2, z = −1/2; T = K√(l/g).
QuestionDerive the speed of sound in a gas by dimensional analysis.
AnswerFrom v ∝ ρˣEʸ, x = −1/2, y = 1/2; v = K√(E/ρ), with K = 1 by experiment.
QuestionDerive the frequency of a stretched string by dimensional analysis.
AnswerFrom n ∝ Tˣ lʸ mᶻ, x = 1/2, y = −1, z = −1/2; n = K(1/l)√(T/m), K = 1/2.
QuestionWhy can a dimensional analysis not give the value of a constant? Give an example.
AnswerDimensions fix only powers, not numbers, for example the 2π in T = 2π√(l/g).
QuestionWhich equations can dimensional analysis not handle?
AnswerThose with exponential, trigonometric or logarithmic quantities, and those with more than L, M, T.
QuestionTorque, work and energy have the same dimension; are they the same quantity?
AnswerNo. Work and energy are scalars, torque is a vector; dimensional analysis cannot tell.
Source: Physics 1st Paper, Classes 11-12 (NCTB), Chapter 1 “Physical World and Measurement”, section 1.7-1.9.