On 23 September 1999, after a journey of 670 million kilometres and 286 days, NASA's Mars Climate Orbiter fired its engine to slip into orbit around Mars — and was never heard from again. The spacecraft had dipped far too deep into the atmosphere and burned up. The cause was not a software crash or a hardware fault. One team had supplied a thruster figure in pound-force seconds while the flight software expected newton seconds. Nobody had checked that the units matched, and a 125-million-dollar probe paid the price.

The tool that would have caught it in seconds is the humblest and most powerful sanity check in all of physics: dimensional analysis. The idea is a single sentence, worth reading twice: every physical quantity carries a "kind" — a dimension — and you may only add, subtract, or equate quantities of the same kind. You cannot add a length to a time any more than you can add three apples to two Tuesdays. From that one rule flows a surprising amount: you can catch mistakes in an equation without solving it, and — astonishingly — you can guess the form of a physical law before you know any of the physics behind it.

The base dimensions: M, L, T

A dimension is the kind of a quantity, stripped of any particular unit. A length is a length whether you measure it in metres, miles or light-years — its dimension is [\text{L}], and we write the square brackets to mean "the dimension of". Mechanics needs just three base dimensions, from which every other mechanical quantity is built:

Everything mechanical is a product of powers of these. A velocity is a length per time, [\text{L}][\text{T}]^{-1}; an acceleration is [\text{L}][\text{T}]^{-2}; and, through F = ma, a force is

Build a few more the same way and a table of the mechanical world appears:

• Energy / work (Fd): [\text{M}][\text{L}]^{2}[\text{T}]^{-2};
• Pressure (force over area): [\text{M}][\text{L}]^{-1}[\text{T}]^{-2};
• Power (energy per time): [\text{M}][\text{L}]^{2}[\text{T}]^{-3};
• Momentum (mv): [\text{M}][\text{L}][\text{T}]^{-1};
• Density (mass per volume): [\text{M}][\text{L}]^{-3}.

Beyond mechanics the SI system adds four more base dimensions so that all of physics can be covered: electric current [\text{I}], temperature [\Theta], amount of substance [\text{N}], and luminous intensity [\text{J}]. Seven base dimensions in all — but for the mechanics on this page, [\text{M}], [\text{L}] and [\text{T}] are the whole story.

Dimensional homogeneity: the golden rule

Here is the rule that catches the Mars Orbiter mistake. In any equation that is physically meaningful, every term that is added, subtracted, or set equal must have exactly the same dimensions. This is called dimensional homogeneity. The two sides of an equals sign, and every separate term joined by a plus or minus, must all be "the same kind of thing".

Watch it work on a real kinematics formula. The velocity of an accelerating object is v = u + at, where u is the starting velocity, a the acceleration and t the time. Check each term:

• [v] = [\text{L}][\text{T}]^{-1}
• [u] = [\text{L}][\text{T}]^{-1}
• [at] = [\text{L}][\text{T}]^{-2}\cdot[\text{T}] = [\text{L}][\text{T}]^{-1}

All three terms are [\text{L}][\text{T}]^{-1} — the equation is homogeneous, and passes the check. Now sabotage it. Suppose a tired student writes v = u + at^{2} instead. The last term becomes

a plain length — and you cannot add a length [\text{L}] to a velocity [\text{L}][\text{T}]^{-1}. The equation is dimensionally inhomogeneous, so it is guaranteed wrong, and we knew it without plugging in a single number. That is the everyday power of the method: a fast, cheap check on every formula you write. Notice too that pure numbers — the \tfrac{1}{2} in s = ut + \tfrac{1}{2}at^{2}, or a 2\pi — are dimensionless; they carry no [\text{M}], [\text{L}] or [\text{T}] and never affect a dimensional check.

Deriving a law: the pendulum's period

The real magic of dimensional analysis is that it can hand you the shape of a physical law before you have written down a single equation of motion. The classic example is the simple pendulum: a bob swinging on a light string. What sets its period T — the time for one full swing?

Make a list of everything the period could plausibly depend on. Common sense suggests three candidates: the mass of the bob m, the length of the string L, and the strength of gravity g (an acceleration, [\text{L}][\text{T}]^{-2}). We guess the law is a product of powers of these, times some unknown dimensionless number k:

Now demand that the dimensions balance. The left side is a pure time, [\text{T}]. The right side, written out, is

For this to hold, the power of each base dimension must match on both sides. Reading off M, L and T in turn gives three little equations:

• [\text{M}]:  a = 0  — the mass drops out entirely!
• [\text{T}]:  -2c = 1 \;\Rightarrow\; c = -\tfrac{1}{2}
• [\text{L}]:  b + c = 0 \;\Rightarrow\; b = +\tfrac{1}{2}

Substituting back, the exponents assemble themselves into a single unavoidable form:

The one thing dimensional analysis cannot give you is the pure number k. A proper solution of the equation of motion (for small swings) reveals k = 2\pi, so the full result is T = 2\pi\sqrt{L/g}. But look how far we got with pure bookkeeping: we deduced that the period grows as the square root of the length, is independent of mass, and weakens with gravity — the entire structure of the law — without ever writing Newton's second law.

Seeing the √(L/g) law

The graph plots the pendulum's period T = 2\pi\sqrt{L/g} against string length L. Because the relationship is a square root, the curve rises steeply for short strings and then flattens — quadrupling the length only doubles the period. Use the slider to switch gravity between Earth, the Moon and Jupiter: on the low-gravity Moon the same pendulum swings lazily (a longer period), while Jupiter's strong gravity whips it back quickly.

A one-metre pendulum on Earth has a period of T = 2\pi\sqrt{1/9.81} \approx 2.0\ \text{s} — which is why a "seconds pendulum", ticking once per swing, is very close to a metre long. Take that same pendulum to the Moon and its period stretches to about 4.9\ \text{s}, because gravity there is roughly six times weaker.

More worked examples

Example 1 — the speed of a wave on a string. How fast does a wave travel along a taut string? Plausibly it depends on the tension F pulling the string (a force, [\text{M}][\text{L}][\text{T}]^{-2}) and on how heavy the string is per unit length, the linear density \mu ([\text{M}][\text{L}]^{-1}). Guess v = k\,F^{a}\,\mu^{b} and balance dimensions:

Matching each base dimension: from [\text{T}], -2a = -1 so a = \tfrac{1}{2}; from [\text{M}], a+b = 0 so b = -\tfrac{1}{2} (and the [\text{L}] equation, a-b = 1, checks out). Hence

and here the constant really is k = 1: the exact answer is v = \sqrt{F/\mu}.

Example 2 — checking an unfamiliar formula. A student claims the energy of a spring is E = \tfrac{1}{2}kx, where the spring constant k has dimensions [\text{M}][\text{T}]^{-2} (from F = kx) and x is an extension, [\text{L}]. Test it: [kx] = [\text{M}][\text{T}]^{-2}\cdot[\text{L}] = [\text{M}][\text{L}][\text{T}]^{-2} — that is a force, not an energy ([\text{M}][\text{L}]^{2}[\text{T}]^{-2}). The formula is wrong; it is missing a factor of x. The correct E = \tfrac{1}{2}kx^{2} restores the extra [\text{L}] and the dimensions balance.

Example 3 — spotting a dimensionless group. For a ball falling through air, the drag force is often written F_D = \tfrac{1}{2}C_D\,\rho\,A\,v^{2}. Check that the drag coefficient C_D must be a pure number. With density \rho = [\text{M}][\text{L}]^{-3}, area A = [\text{L}]^{2} and v^{2} = [\text{L}]^{2}[\text{T}]^{-2}:

which is already a force. So C_D (and the \tfrac{1}{2}) must be dimensionless for the equation to balance — exactly as it should be.

Counting the groups: a taste of Buckingham's π theorem

The pendulum trick worked because there was essentially one way to combine the ingredients into something dimensionless. When can we count on that, and when might there be more freedom? The answer is the Buckingham π theorem, one of the crown jewels of dimensional analysis.

Return to the pendulum. Its variables are T, L, g, m, so n = 4, and they involve the three base dimensions [\text{M}], [\text{L}], [\text{T}], so k = 3. The theorem promises p = 4 - 3 = 1 dimensionless group — and indeed the only one you can form (ignoring the inert mass) is

which must therefore equal a constant, giving gT^{2}/L = \text{const}, i.e. T \propto \sqrt{L/g} — the very result we derived. Because there is just one group, the law is pinned down up to a number. When p is larger, the theorem tells you how many independent dimensionless "knobs" the problem has, which is precisely why engineers can test a tiny model aircraft in a wind tunnel and trust the results for the full-size plane: match the dimensionless groups (like the Reynolds number) and the physics scales.