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Concept 03 / 08Units and Measurements

Dimensional Analysis: Applications and Limitations

Using the principle of homogeneity to check equations, convert units between systems and derive physical relations by Rayleigh's method, plus a clear account of exactly where the method must fail.

Figure 1
The whole subject rests on one axiom, the principle of homogeneity: in any physically meaningful equation every additive term must have the same dimensions. The reason is not mathematical but semantic. Adding 3 kg to 5 s is not a wrong calculation, it is a meaningless statement, since the sum would change value under a change of the mass unit alone while the parts did not describe the same kind of thing. The figure above shows this for s=ut+12at2s = ut + \tfrac{1}{2}at^2: each of the three terms collapses to LL. Application 1, checking an equation. Test v2=u2+2asv^2 = u^2 + 2as: each term gives L2T−2L^2T^{-2}, so it passes. Test T=2πl/gT = 2\pi\sqrt{l/g}: LLT−2=T2=T\sqrt{\dfrac{L}{LT^{-2}}} = \sqrt{T^2} = T, so it passes. But note what "passes" means. A dimensionally correct equation may still be physically wrong, because dimensional analysis is blind to pure numbers. The equation s=ut+at2s = ut + at^2 passes the same test as s=ut+12at2s = ut + \tfrac{1}{2}at^2 and yet is false. Homogeneity is a necessary condition, never a sufficient one. The converse, however, is reliable: a dimensionally INCORRECT equation is certainly wrong, which makes this the single fastest way to kill wrong options in an exam. Application 2, converting units between systems. Since n1u1=n2u2n_1u_1 = n_2u_2 and u=MaLbTcu = M^aL^bT^c, n2=n1[M1M2]a[L1L2]b[T1T2]cn_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}. Convert 1 newton to dyne: force is MLT−2MLT^{-2}, so n2=1×[1 kg1 g]1[1 m1 cm]1[1 s1 s]−2=103×102=105n_2 = 1 \times \left[\dfrac{1\ \text{kg}}{1\ \text{g}}\right]^{1}\left[\dfrac{1\ \text{m}}{1\ \text{cm}}\right]^{1}\left[\dfrac{1\ \text{s}}{1\ \text{s}}\right]^{-2} = 10^3 \times 10^2 = 10^5. Hence 1 N =105= 10^5 dyne. Similarly 1 J =107= 10^7 erg, since ML2T−2ML^2T^{-2} gives 103×(102)2=10710^3 \times (10^2)^2 = 10^7. Application 3, deriving a relation (Rayleigh's method of indices). Assume the unknown quantity is a product of powers of the relevant variables times a dimensionless constant, then match exponents. Example: the time period of a simple pendulum, assumed to depend on length ll, mass mm and gg. Write T=k lxmygzT = k\,l^{x}m^{y}g^{z}. Then T=LxMy(LT−2)zT = L^{x}M^{y}(LT^{-2})^{z}, giving M:y=0M: y = 0; T:−2z=1⇒z=−12T: -2z = 1 \Rightarrow z = -\tfrac{1}{2}; L:x+z=0⇒x=12L: x + z = 0 \Rightarrow x = \tfrac{1}{2}. So T=kl/gT = k\sqrt{l/g}. Dimensional analysis has told us the period is independent of the bob's mass, a genuinely non-obvious physical result, but it cannot supply k=2πk = 2\pi. A more impressive example, the Planck length, built from GG, hh and cc alone. With [G]=M−1L3T−2[G] = M^{-1}L^3T^{-2}, [h]=ML2T−1[h] = ML^2T^{-1}, [c]=LT−1[c] = LT^{-1}, set LP=GαhβcγL_P = G^{\alpha}h^{\beta}c^{\gamma} and match: M:−α+β=0M: -\alpha + \beta = 0; T:−2α−β−γ=0T: -2\alpha - \beta - \gamma = 0; L:3α+2β+γ=1L: 3\alpha + 2\beta + \gamma = 1. From the first, β=α\beta = \alpha; from the second, γ=−3α\gamma = -3\alpha; substituting into the third, 3α+2α−3α=13\alpha + 2\alpha - 3\alpha = 1, so α=12\alpha = \tfrac{1}{2}, β=12\beta = \tfrac{1}{2}, γ=−32\gamma = -\tfrac{3}{2}. Hence LP=Ghc3≈(6.67×10−11)(6.63×10−34)2.7×1025≈4×10−35L_P = \sqrt{\dfrac{Gh}{c^3}} \approx \sqrt{\dfrac{(6.67\times10^{-11})(6.63\times10^{-34})}{2.7\times10^{25}}} \approx 4 \times 10^{-35} m. A few lines of bookkeeping produce the length scale at which quantum gravity must matter. Now the limitations, which JEE Advanced tests more often than the applications. (i) The dimensionless constant is undetermined. Dimensional analysis gives T∝l/gT \propto \sqrt{l/g} but never the 2π2\pi, and gives F∝ηrvF \propto \eta r v for Stokes drag but never the 6π6\pi. (ii) It cannot produce an equation containing a sum of terms. Each application of the method yields a single product of powers, so s=ut+12at2s = ut + \tfrac{1}{2}at^2 can never be derived, only checked term by term. (iii) It fails for relations involving transcendental functions: x=Asin⁡ωtx = A\sin\omega t, N=N0e−λtN = N_0e^{-\lambda t} and I=I0log⁡(...)I = I_0\log(...) are out of reach. (iv) It fails when the unknown depends on more independent quantities than there are relevant base dimensions. With only M, L, T available you get three equations, so four or more unknown exponents leave the system underdetermined. The honest fix is Buckingham's pi theorem, which says the answer is a function of the dimensionless groups you can form, and that unknown function is precisely what dimensions cannot supply. (v) It cannot distinguish quantities with identical dimensions (work vs torque), cannot distinguish scalars from vectors, and gives no information about a quantity that enters only through a dimensionless combination. Key Takeaways: - Homogeneity: every additive term has the same dimensions. Necessary, not sufficient. - Conversion: n2=n1[M1/M2]a[L1/L2]b[T1/T2]cn_2 = n_1[M_1/M_2]^{a}[L_1/L_2]^{b}[T_1/T_2]^{c}. Remember 1 N =105= 10^5 dyne, 1 J =107= 10^7 erg. - Rayleigh's method: write the unknown as k AxByCzk\,A^{x}B^{y}C^{z}, equate powers of M, L, T. - Useful results: T=2πl/gT = 2\pi\sqrt{l/g}, v=T/μv = \sqrt{T/\mu} for a string, vesc=2GM/Rv_{esc} = \sqrt{2GM/R}, LP=Gh/c3L_P = \sqrt{Gh/c^3}. - Pitfall: declaring a dimensionally correct equation "proved". Numerical factors are invisible. - Pitfall: concluding a quantity is irrelevant because it dropped out; it may enter through a dimensionless ratio. - Pitfall: using the method when four or more quantities are involved and expecting a unique answer. Brain-Teaser 1: The frequency ν\nu of a vibrating stretched string depends on its length ll, the tension FF and the mass per unit length μ\mu. Derive the relation, and then state what the derivation cannot tell you. Solution: Put ν=k lxFyμz\nu = k\,l^{x}F^{y}\mu^{z}. Dimensions: T−1=Lx(MLT−2)y(ML−1)zT^{-1} = L^{x}(MLT^{-2})^{y}(ML^{-1})^{z}. Matching M: y+z=0y + z = 0. Matching T: −2y=−1-2y = -1, so y=12y = \tfrac{1}{2} and therefore z=−12z = -\tfrac{1}{2}. Matching L: x+y−z=0x + y - z = 0, so x=z−y=−12−12=−1x = z - y = -\tfrac{1}{2} - \tfrac{1}{2} = -1. Hence ν=klFμ\nu = \dfrac{k}{l}\sqrt{\dfrac{F}{\mu}}. Check the inner quantity: MLT−2ML−1=L2T−2\dfrac{MLT^{-2}}{ML^{-1}} = L^2T^{-2}, whose square root is LT−1LT^{-1}, a speed, divided by ll gives T−1T^{-1}. Correct. What it cannot tell you is the value of kk, which is 12\tfrac{1}{2} for the fundamental mode and n2\tfrac{n}{2} for the nn-th harmonic. So dimensional analysis delivers the entire physics of how frequency scales, and is silent about the discrete mode structure, which comes from boundary conditions. Brain-Teaser 2: The period of a simple pendulum actually does depend on the amplitude for large swings, with T≈2πl/g(1+θ0216)T \approx 2\pi\sqrt{l/g}\left(1 + \dfrac{\theta_0^2}{16}\right) for moderate θ0\theta_0. Yet the dimensional derivation above concluded T=kl/gT = k\sqrt{l/g} with no amplitude in sight. Is the derivation wrong? Solution: No, it is incomplete in a way that is entirely predictable. The amplitude θ0\theta_0 is an angle, hence dimensionless, hence completely invisible to a dimensional balance. Repeating the derivation with θ0\theta_0 included changes nothing: the exponent of a dimensionless quantity is never fixed by matching powers of M, L and T. The correct general statement, which is what Buckingham's pi theorem gives, is T=l/g  f(θ0)T = \sqrt{l/g}\;f(\theta_0), where ff is an unknown dimensionless function. The elementary result is the special case f(θ0)→2πf(\theta_0) \to 2\pi as θ0→0\theta_0 \to 0. The lesson generalises far beyond pendulums: whenever a problem contains a dimensionless ratio (an angle, a Reynolds number, a ratio of two lengths, a refractive index), dimensional analysis can only determine the answer up to an arbitrary function of that ratio. A student who claims "dimensional analysis proves the period is amplitude independent" has mistaken a blind spot for a theorem. Brain-Teaser 3: In a hypothetical system of units, the unit of force is 100 N, the unit of length is 10 m and the unit of time is 100 s. What is the unit of mass in this system? Solution: Do not convert the given numbers; instead express mass in terms of the chosen base quantities. Since [F]=MLT−2[F] = MLT^{-2}, we have [M]=[F][T]2[L]−1[M] = [F][T]^2[L]^{-1}. Therefore the unit of mass is (100 N)(100 s)210 m=100×10410\dfrac{(100\ \text{N})(100\ \text{s})^2}{10\ \text{m}} = \dfrac{100 \times 10^4}{10} kg =105= 10^5 kg. The routine error is to treat force, length and time as though they were still derived from a kilogram; here force is itself a base unit of the new system, and mass is the derived one. A quick sanity check: the new force unit is 100 times bigger and the new time unit is 100 times bigger (so T2T^2 is 10410^4 times bigger), while the length unit is only 10 times bigger, so the mass unit must be enormous, which 10510^5 kg indeed is.

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