Converting units between systems

Physicists routinely need to translate a measurement from one system of units into another — a force measured in newtons in an SI lab must become dynes in an old CGS paper, an energy in joules must become ergs. Instead of memorising a separate conversion table for every possible quantity, dimensional analysis hands us one universal rule: since the numerical value of a quantity times its unit never changes, the dimensional formula of the quantity tells us exactly how its number must scale when the mass, length and time units change. This article builds that single conversion formula from first principles, applies it to force and energy, and flags the exponent mistakes that cost the most marks. It is the third and final syllabus use of dimensional analysis, after checking correctness and deriving relations — and it leans on the same dimensional formulae you have already learned to write down.

The core insight

A quantity's numerical value in a new unit system is found by scaling its old numerical value with the ratios of the two systems' mass, length and time units, each raised to the power given by that quantity's own dimensional formula.

The core idea

A physical quantity is the same physical amount no matter which unit system describes it — only the number attached to it changes when the unit changes. For a quantity $Q$ with dimensional formula $[M^aL^bT^c]$ , measured as $n_1$ in a first system (base units $M_1, L_1, T_1$ ) and as $n_2$ in a second system (base units $M_2, L_2, T_2$ ), this invariance is written $n_1u_1 = n_2u_2$ , where $u_1$ and $u_2$ are the two systems' units for $Q$ , built from their own base units raised to $a$ , $b$ , $c$ . Rearranging gives the working conversion formula:

n_2 = n_1\left(\frac{M_1}{M_2}\right)^{a}\left(\frac{L_1}{L_2}\right)^{b}\left(\frac{T_1}{T_2}\right)^{c}

where: $n_1, n_2$ are the numerical values of $Q$ in the old and new systems, $a, b, c$ are the exponents in $Q$ 's dimensional formula $[M^aL^bT^c]$ , and $M_1/M_2$ , $L_1/L_2$ , $T_1/T_2$ are the ratios of the two systems' mass, length and time units. Three things make this rule powerful. First, it needs only the dimensional formula of $Q$ — nothing else about how $Q$ is defined. Second, the same three ratios ( $M_1/M_2$ , $L_1/L_2$ , $T_1/T_2$ ) work for every mechanical quantity; only the exponents $a$ , $b$ , $c$ change. Third, it explains why physical laws themselves never need rewriting between CGS, FPS, MKS and SI — a law like $F = ma$ holds in every system; only the size of the "unit of force" differs, and that difference is exactly what the formula above computes.

Figure

Bar chart comparing the SI to CGS base unit ratios for mass length and time next to the resulting force and energy conversion factors

The general conversion formula for mechanical quantities

The goal is to derive $n_2 = n_1(M_1/M_2)^{a}(L_1/L_2)^{b}(T_1/T_2)^{c}$ from nothing more than the statement that a physical quantity's value does not depend on which unit system is used to describe it, so that the formula can then be applied to any mechanical quantity once its dimensional exponents $a$ , $b$ , $c$ are known.

Step 1. Start from invariance. The same physical quantity $Q$ , expressed in two systems, satisfies $n_1u_1 = n_2u_2$ — the number times the unit is a fixed physical amount, so a bigger unit must come with a smaller number and vice versa.

Step 2. Write each unit in terms of its system's base units. Since $Q$ has dimensional formula $[M^aL^bT^c]$ , its unit in system 1 is $u_1 = M_1^aL_1^bT_1^c$ and in system 2 is $u_2 = M_2^aL_2^bT_2^c$ , where $M_1, L_1, T_1$ and $M_2, L_2, T_2$ are the mass, length and time base units of the two systems.

Step 3. Substitute into the invariance equation. $n_1M_1^aL_1^bT_1^c = n_2M_2^aL_2^bT_2^c$ . This single line already contains the whole method: everything else is algebra to isolate $n_2$ .

Step 4. Solve for $n_2$ by dividing both sides by $M_2^aL_2^bT_2^c$ and grouping each base unit with its partner: $n_2 = n_1\left(\dfrac{M_1}{M_2}\right)^{a}\left(\dfrac{L_1}{L_2}\right)^{b}\left(\dfrac{T_1}{T_2}\right)^{c}$ . Each ratio is a pure number (a length divided by a length, and so on), so $n_2$ comes out as a pure number too — exactly the numerical value of $Q$ in the new system.

Worked example

Problem: Convert a force of $1$ newton (SI) into dynes (CGS) using dimensional analysis, given $M_1 = 1\ \text{kg} = 1000\ \text{g}$ , $L_1 = 1\ \text{m} = 100\ \text{cm}$ , $T_1 = 1\ \text{s}$ in SI, and $M_2 = 1\ \text{g}$ , $L_2 = 1\ \text{cm}$ , $T_2 = 1\ \text{s}$ in CGS.

Step 1. Write the dimensional formula of force: $[F] = [M^1L^1T^{-2}]$ , so $a = 1$ , $b = 1$ , $c = -2$ .

Step 2. Apply the general formula with $n_1 = 1$ : $n_2 = 1\left(\dfrac{M_1}{M_2}\right)^{1}\left(\dfrac{L_1}{L_2}\right)^{1}\left(\dfrac{T_1}{T_2}\right)^{-2}$ .

Step 3. Substitute the ratios: $M_1/M_2 = 1000\ \text{g}/1\ \text{g} = 1000$ , $L_1/L_2 = 100\ \text{cm}/1\ \text{cm} = 100$ , $T_1/T_2 = 1\ \text{s}/1\ \text{s} = 1$ .

Step 4. Compute: $n_2 = 1000^1 \times 100^1 \times 1^{-2} = 1000 \times 100 = 10^5$ . So $1\ \text{N} = 10^5\ \text{dyne}$ , matching the standard conversion.

Competitive stretch

The same method extends immediately to energy, which will crop up wherever CGS and SI results must be compared side by side. Energy has $[E] = [M^1L^2T^{-2}]$ , so $a = 1$ , $b = 2$ , $c = -2$ ; using the same ratios, $n_2 = 1000^1 \times 100^2 \times 1^{-2} = 1000 \times 10000 = 10^7$ , giving $1\ \text{J} = 10^7\ \text{erg}$ . Notice only the $L$ exponent changed — from $1$ to $2$ — yet it moved the answer by two more powers of ten, which is exactly why getting each exponent right individually matters so much.

Common traps

Using a wrong or misremembered dimensional formula for $Q$ — every exponent error propagates straight into a wrong conversion factor.
Applying the same power to $M$ , $L$ and $T$ ratios instead of each quantity's own exponents $a$ , $b$ , $c$ — a very common slip when $b \ne 1$ , as in the energy example above.
Mixing base units of different systems inside one ratio, such as comparing kilograms to centimetres, instead of comparing like to like — mass to mass, length to length, time to time — between the two systems.
Assuming the $M$ , $L$ , $T$ method extends untouched to quantities with electrical or thermal base units — for the purely mechanical CGS, FPS, MKS and SI conversions in this syllabus, only mass, length and time are needed.

Ask the tutor: Want to try converting pressure or momentum between CGS and SI using this same method, or need a refresher on setting up the ratios correctly?

Why does only the numerical value change between unit systems, and not the physical quantity itself?

The physical quantity — a certain force, a certain energy — is a fixed amount of "stuff," independent of how humans choose to label it. The unit is just the labelling scale, like measuring the same rope with a metre stick or a foot ruler. Because $n \times u$ must stay the same fixed amount, a smaller unit (like a dyne compared with a newton) forces a bigger number, and a bigger unit forces a smaller number. Dimensional analysis simply tracks this trade-off systematically using the quantity's $M$ , $L$ , $T$ exponents.

Can this conversion method be used for any physical quantity at all?

Only for quantities built purely from mass, length and time — the mechanical quantities covered by this syllabus, such as force, energy, pressure, momentum and velocity. It assumes the quantity's defining relation is a simple product of powers of $M$ , $L$ , $T$ , which is exactly what a dimensional formula captures. Quantities involving electric charge, current or temperature need extra base units beyond $M$ , $L$ , $T$ , and are outside this method as taught here.

What is the fastest way to avoid the exponent mistakes mentioned in the common traps?

Write out the dimensional formula of the quantity first, explicitly labelling $a$ , $b$ and $c$ before touching any numbers — this is the same discipline used when checking dimensional correctness or deriving relations earlier in the chapter. Then substitute the $M_1/M_2$ , $L_1/L_2$ , $T_1/T_2$ ratios one at a time next to their matching exponent, rather than multiplying everything together in your head. Slowing down at this labelling step is what prevents mixing up which ratio gets which power.