What Are Dimensional Formulae and Dimensional Equations?

Physical quantities in physics are classified as either fundamental or derived, and their quantitative relationships are expressed using specific mathematical structures based on their dependence on base units. This relationship is formalised using the concepts of dimensional formulae and dimensional equations, which allow mathematical analysis of physical laws and conversions between units.

Formal Statement of the Dimensional Formula of a Physical Quantity

If a derived physical quantity $Q$ depends on the fundamental quantities of mass ($M$), length ($L$), time ($T$), electric current ($I$), temperature ($\Theta$), amount of substance ($N$), and luminous intensity ($J$), its dimensional formula is expressed as:

$Q = M^a L^b T^c I^d \Theta^e N^f J^g$

The exponents $a$, $b$, $c$, $d$, $e$, $f$, $g$ may be positive, negative, or zero, according to the dimensional dependence of $Q$ on each base quantity.

The dimensional formula of a physical quantity is the expression representing those base quantities with their respective powers in square brackets. Formally, the dimensional formula of $Q$ is written as $[M^a L^b T^c I^d \Theta^e N^f J^g]$.

Dimensional Formulae of Selected Common Quantities

For a quantity such as area $A$, which is defined physically as the product of two lengths, the relationship is $A = \text{length} \times \text{breadth}$. Substituting both as $L$ yields $A = L \times L = L^2$. Therefore, the dimensional formula is:

$[A] = [L^2] = [M^0 L^2 T^0]$

For velocity $v$, defined as displacement divided by time, $v = \dfrac{L}{T}$, so the dimensional formula is:

$[v] = \left[\dfrac{L}{T}\right] = [L^1 T^{-1}] = [M^0 L^1 T^{-1}]$

For force $F$, defined as mass multiplied by acceleration, $F = M \times a$, and since acceleration is $\text{velocity} / \text{time} = \dfrac{L}{T^2}$, the dimensional formula for $F$ is:

$[F] = [M] \times [a] = [M^1] \times [L^1 T^{-2}] = [M^1 L^1 T^{-2}]$

For additional dimensional formulae of physical quantities, refer to the Dimensional Formulae And Dimensional Equations page.

Mathematical Structure of Dimensional Equations

A dimensional equation is established by replacing each physical quantity in an equation with its respective dimensional formula. For a physical equation, validity requires that every term added or compared must possess identical dimensions, guaranteeing dimensional homogeneity.

For example, consider the kinematic equation $s = ut + \frac{1}{2} a t^2$, where $s$ is displacement, $u$ is initial velocity, $t$ is time, and $a$ is acceleration.

The dimensional formula for $s$ is $[L]$.

For $ut$: $[u][t] = [L T^{-1}][T^1] = [L T^{-1 + 1}] = [L^1 T^0] = [L]$.

For $\displaystyle \frac{1}{2} a t^2$: $[a][t^2] = [L T^{-2}][T^2] = [L T^{-2 + 2}] = [L T^0] = [L]$.

As each term yields $[L]$, the equation is dimensionally consistent.

A dimensional equation for force is constructed as $[F] = [M L T^{-2}]$ as previously derived.

For further clarity on foundational dimensional principles, refer to the Introduction To Dimensions page.

Systematic Derivation of a Dimensional Formula from a Physical Law

To obtain the dimensional formula of a derived quantity, each variable is replaced sequentially with its dimensional formula, followed by stepwise algebraic simplification. Consider work, defined as $W = F \times s$ where $F$ is force and $s$ is displacement:

Step 1. Substitute the dimensional formula for force: $[F] = [M L T^{-2}]$.

Step 2. Substitute the dimensional formula for displacement: $[s] = [L]$.

Step 3. Multiply dimensions: $[W] = [M L T^{-2}] \cdot [L]$.

Step 4. Apply the law of exponents: $[W] = [M^1 L^{1+1} T^{-2}] = [M^1 L^2 T^{-2}]$.

This yields the dimensional formula of work (and energy) as $[M L^2 T^{-2}]$.

Detailed Derivation of the Dimensional Formula for Thermal Conductivity

Consider Fourier’s law for the rate of steady-state heat flow through a rod, where the relationship is:

$\dfrac{Q}{t} = k \dfrac{A (T_1-T_2)}{L}$

where $Q$ is heat energy transferred, $t$ is time, $A$ is cross-sectional area, $T_1-T_2$ is temperature difference, $L$ is the length, and $k$ is thermal conductivity.

Rearrange for $k$:

$k = \dfrac{Q}{t} \cdot \dfrac{L}{A (T_1-T_2)}$

Start with the dimensional formula for heat energy, $Q$; $[Q] = [\text{work}] = [M L^2 T^{-2}]$.

Dimensional formula for time: $[t] = [T]$.

Therefore, $[Q/t] = [M L^2 T^{-2}] / [T] = [M L^2 T^{-2}][T^{-1}] = [M L^2 T^{-3}]$.

Dimensional formula for length: $[L] = [L]$.

Dimensional formula for area: $[A] = [L^2]$.

Dimensional formula for temperature difference: $[T_1 - T_2] = [\Theta]$.

Compute the denominator:

$[A(T_1-T_2)] = [L^2][\Theta] = [L^2 \Theta^1]$.

Therefore, $[k] = [M L^2 T^{-3}] \cdot [L] / [L^2 \Theta^1]$.

Multiply numerator: $[M L^2 T^{-3}] \cdot [L] = [M L^{2+1} T^{-3}] = [M L^3 T^{-3}]$.

Now divide by $[L^2 \Theta]$:

$[k] = \dfrac{[M L^3 T^{-3}]}{[L^2 \Theta]} = [M L^{3-2} T^{-3} \Theta^{-1}] = [M L^1 T^{-3} \Theta^{-1}]$.

Result: The dimensional formula for thermal conductivity $k$ is $[M^1 L^1 T^{-3} \Theta^{-1}]$.

To deepen understanding of derivational approaches, see also the Area And Perimeter Formula page.

Verifying Dimensional Homogeneity in a Differential Equation

Consider a mechanical damped system governed by $m u'' + b u' + k u = 0$, where $m$ is mass, $u$ is displacement, $b$ is damping coefficient, and $k$ is spring constant. The derivative $u' = \dfrac{du}{dt}$ and $u'' = \dfrac{d^2u}{dt^2}$.

$[u] = [L]$.

$[u'] = [u]/[t] = [L][T^{-1}] = [L T^{-1}]$.

$[u''] = [u']/[t] = [L T^{-1}] / [T] = [L T^{-2}]$.

$m u''$ thus has dimensions $[M][L T^{-2}] = [M L T^{-2}]$.

$b u'$ yields $[b][L T^{-1}]$. To match dimensionality with $m u''$, $[b][L T^{-1}] = [M L T^{-2}]$, so $[b] = [M T^{-1}]$.

$k u$ gives $[k][L] = [M L T^{-2}]$, hence $[k] = [M T^{-2}]$.

Since every term in the differential equation possesses the same dimension $[M L T^{-2}]$, the equation is dimensionally homogeneous.

Principal Use Cases of Dimensional Formulae and Equations

Dimensional analysis provides a robust mathematical check by ensuring dimensional homogeneity, facilitating unit conversions, deducing relations for unknown formulae, and establishing the dimensional dependence of physical constants in equations. However, dimensional methods have limitations in capturing dimensionless constants and cannot deduce equations containing trigonometric, exponential, or logarithmic functions, or relations involving more than three independent physical quantities.

For a comparison with geometric measure theory, refer to the Difference Between Length And Height page.

Limitations in the Application of Dimensional Equations

Dimensional equations cannot determine dimensionless constants of proportionality, nor can they derive formulae involving summation of unlike dimensions, or formulae where quantities are related through non-algebraic (e.g., trigonometric, logarithmic) relations. Additionally, dimensional analysis cannot uniquely derive a formula if the corresponding physical quantity depends on more than three variables with independent dimensions.