Limitations of Dimensional Analysis

Dimensional analysis is powerful, but it isn't magic. Knowing what it cannot do is as important as knowing what it can.

Concept

The method works because physically meaningful equations must be dimensionally homogeneous. But it has clear limitations:

1. It Cannot Determine Dimensionless Constants

If a quantity depends on others via $Q = k\, X^a Y^b\,\ldots,$ dimensional analysis fixes the exponents $a, b, \ldots$ but not the constant $k$. For example, for a simple pendulum we get $T \propto \sqrt{l/g}$ — but the $2\pi$ is invisible to dimensions.

2. It Cannot Handle More Variables Than Independent Dimensions

If a quantity depends on more variables than the number of independent dimensions in the problem, the method gives a family of solutions, not a unique one.

Example: Range of a projectile depends on $u$, $g$, $\theta$. Since $\theta$ is dimensionless, dimensional analysis only gives $R \propto u^2/g$ — it cannot recover $\sin 2\theta$.

3. It Cannot Distinguish Quantities With the Same Dimensions

Work and torque both have dimensions $ML^2T^{-2}$, yet they are physically different (scalar vs vector, energy vs rotational tendency). Similarly, energy and moment of a couple share dimensions but are not the same thing.

4. It Cannot Handle Trig, Log, or Exponential Functions

The form $\sin(\omega t)$, $e^{-\lambda t}$, or $\ln(x/x_0)$ contains a dimensionless argument, but dimensional analysis cannot tell you which functional form is correct. It can't distinguish $\sin\theta$ from $\theta - \theta^3/6$ etc.

5. Cannot Be Applied to Sums of Different-Power Terms in the Same Variable

Equations like $x = at + bt^2$ work term-by-term but can't be guessed a priori by dimensional analysis if multiple terms with different powers of the same variable exist.

6. Fails for Equations Containing Truly Dimensional Constants

Some constants (like $g$, $G$, $\hbar$) carry hidden dimensions; you must include them as variables. If you omit them, the analysis will fail or give wrong relations.

Worked Example

Q: Show that dimensional analysis cannot determine the angular dependence of the range of a projectile.

Solution: Let $R = k\, u^a g^b \theta^c$ where $\theta$ is dimensionless.

• $[R] = L$, $[u] = LT^{-1}$, $[g] = LT^{-2}$, $[\theta] = 1$.

Equating: $L = L^a T^{-a} L^b T^{-2b} = L^{a+b} T^{-a-2b}$.

So $a + b = 1$, $-a - 2b = 0 \Rightarrow a = 2$, $b = -1$.

Therefore $R = k\, u^2/g\, \cdot f(\theta)$. The function $f(\theta)$ is undetermined by dimensions — we know from full kinematics it's $\sin(2\theta)$, but dimensions can't tell us that.

Common Confusions

• Dimensional consistency is necessary, not sufficient. $E = mv$ is dimensionally wrong; $E = mv^2$ is dimensionally right but missing the factor $1/2$.
• Two quantities with identical dimensions need not be interchangeable (e.g., work vs torque).
• A "dimensionless constant" still represents a physical idea (like Reynolds number) — being dimensionless doesn't mean unimportant.

Key Takeaways

• Dimensional analysis cannot find dimensionless multipliers like $\pi$ or $1/2$.
• It fails when more variables than independent dimensions are involved.
• It cannot determine pure functions (trig, log, exp) of dimensionless arguments.
• Same dimensions $\ne$ same physical quantity.
• Use it as a quick sanity check, not as a substitute for derivation.