Dimensional Formulae and Dimensional Equations

You already know that every physical quantity carries dimensions, and that those dimensions tell you the nature of the quantity rather than its size. The next step is learning a compact, standardised way to write those dimensions down and to use them in equations. That is what dimensional formulae and dimensional equations are all about, and mastering them opens the door to one of the most powerful tools in physics: dimensional analysis.

Writing a Dimensional Formula

A dimensional formula is the expression that shows exactly which base quantities go into a physical quantity, and what power each one carries. You write the base-dimension symbols inside square brackets, with their exponents, to build the formula.

Take volume as a first example. Volume measures the amount of space an object occupies. It equals length multiplied by breadth multiplied by height, and each of those three measurements is a length. So volume involves length raised to the third power, and neither mass nor time plays any part:

$[M^0 \, L^3 \, T^0]$

The zero exponents on $M$ and $T$ simply confirm that volume has nothing to do with mass or time. Any base dimension raised to zero equals one, so it contributes nothing to the expression. Including the zeros is optional, but it keeps things explicit, especially when you are just starting out.

Here are a few more dimensional formulae, each built directly from the definition of the quantity:

• Speed or velocity is distance divided by time. Distance is a length, so the formula is:

$[M^0 \, L \, T^{-1}]$

• Acceleration measures how quickly velocity changes over time. Since velocity already has $[L \, T^{-1}]$, dividing it by one more factor of time gives:

$[M^0 \, L \, T^{-2}]$

• Mass density is mass per unit volume. Replace mass with $[M]$ and volume with $[L^3]$:

$[M \, L^{-3} \, T^0]$

Notice the pattern. Each formula is built by starting from the definition of the quantity and then replacing every factor with its base dimension. No numerical coefficients appear because dimensional formulae capture the type of a quantity, not its value.

From Formula to Equation: Dimensional Equations

When you take a dimensional formula and write an equals sign between the quantity’s symbol (in square brackets) and the formula itself, you create a dimensional equation. It is simply a formal statement that says “this quantity has these dimensions.”

Here are four key dimensional equations:

$[V] = [M^0 \, L^3 \, T^0]$

$[v] = [M^0 \, L \, T^{-1}]$

$[F] = [M \, L \, T^{-2}]$

$[\rho] = [M \, L^{-3} \, T^0]$

Each equation reads naturally. The first one says “the dimensions of volume equal zero powers of mass, three powers of length, and zero powers of time.” The last one says “the dimensions of mass density are one power of mass, negative three powers of length, and zero powers of time.”

The distinction between a dimensional formula and a dimensional equation is small but worth noting. The formula is the right-hand side alone, for example $[M \, L \, T^{-2}]$. The equation is the complete statement, $[F] = [M \, L \, T^{-2}]$, which ties that formula to a specific physical quantity (force, in this case).

Deriving Dimensional Equations from Physical Relationships

You do not need to memorise a long list of dimensional formulae. Instead, you can derive them on the spot from the equation that defines the quantity. The process is straightforward:

1. Start with the defining equation of the quantity.
2. Replace each quantity on the right-hand side with its known dimensional formula.
3. Carry out the multiplication and division of the base dimensions.
4. Write the result as a dimensional equation.

Example: Force

The defining relationship is $F = m \times a$ (force equals mass times acceleration).

• Mass contributes $[M]$.
• Acceleration, as you worked out above, has $[L \, T^{-2}]$.

Multiplying these together:

$[F] = [M] \times [L \, T^{-2}] = [M \, L \, T^{-2}]$

Example: Mass density

Density is mass divided by volume: $\rho = m / V$.

• Mass contributes $[M]$.
• Volume contributes $[L^3]$.

Dividing:

$[\rho] = \frac{[M]}{[L^3]} = [M \, L^{-3} \, T^0]$

This approach works for any derived quantity, no matter how complex. You just keep substituting base dimensions until every factor is expressed in terms of $[M]$, $[L]$, $[T]$ (or the other base dimensions if the quantity involves electric current, temperature, and so on). A large collection of dimensional formulae for common physical quantities, derived in exactly this way, is listed in Appendix 9 of the NCERT textbook for quick reference.

The Foundation of Dimensional Analysis

Now that you can write dimensional formulae and equations, a powerful principle follows. It has to do with a simple but far-reaching idea: only physical quantities that share the same dimensions can be added or subtracted.

Think about what it would mean to add a length to a time. You would be combining metres with seconds, which is physically nonsensical. The result would have no clear meaning. This is not just a practical inconvenience; it reflects a deep constraint on how the physical world works. Quantities with different dimensions describe fundamentally different aspects of reality, and mixing them in a sum or difference produces nonsense.

This principle extends to entire equations. In any correct physical equation, every term on both sides must carry identical dimensions. If you find that one side of an equation has the dimensions of velocity while the other has the dimensions of acceleration, something has gone wrong in the derivation. This simple check, called verifying dimensional consistency (or dimensional homogeneity), is one of the most useful tools you will ever have for catching algebraic mistakes.

There is one more useful rule to keep in mind. When you multiply or divide physical quantities, their units and dimensions follow exactly the same rules as ordinary algebra. You can cancel identical units or dimensions that appear in both the numerator and the denominator, just as you would cancel common factors in a fraction. Similarly, the dimensions of every symbol on both sides of a mathematical equation must match. These ideas form the backbone of dimensional analysis, a technique you will explore in detail in the topics that follow.