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Physics · Ch 2 — Units and Measurement

Deducing Relations Among Physical Quantities

2.6.2

Deducing Relations Among Physical Quantities

Deducing Relations Among Physical Quantities

The method of dimensions can sometimes be used to derive the form of a relationship between physical quantities, provided you know which quantities the dependent variable depends on. The method assumes a product-type dependence — that is, the dependent quantity can be expressed as a product of powers of the independent quantities, multiplied by a dimensionless constant.

To find the unknown exponents, suppose a physical quantity QQ depends on three independent quantities AA, BB, and CC. We write

Q=k AaBbCcQ = k\, A^a B^b C^c

where kk is a dimensionless constant and aa, bb, cc are exponents to be determined. Writing out the dimensions of each quantity on both sides and equating the powers of every base dimension present (mass, length, time, …) gives a set of simultaneous equations for aa, bb, and cc. Solving these equations pins down the exponents, and hence the form of the relationship — though never the value of kk itself.

General method: if QQ depends on AA, BB, CC as a product of powers,

Q=k AaBbCcQ = k\, A^a B^b C^c

equate the dimensions of QQ to [A]a[B]b[C]c[A]^a[B]^b[C]^c term by term for every base dimension present, and solve the resulting equations for aa, bb, cc.

As an illustration, consider the time period TT of a simple pendulum, which we expect to depend on its length ll, the mass of the bob mm, and the acceleration due to gravity gg. Writing T=k lxgymzT = k\, l^{x} g^{y} m^{z} and comparing dimensions:

[T]=[L]x[L T−2]y[M]z=Lx+yT−2yMz[T] = [L]^x [L\,T^{-2}]^y [M]^z = L^{x+y}T^{-2y}M^z

Equating powers of L, T, and M on both sides gives x+y=0x + y = 0, −2y=1-2y = 1, and z=0z = 0, so x=12x = \tfrac{1}{2}, y=−12y = -\tfrac{1}{2}, z=0z = 0. This gives

T=klgT = k\sqrt{\frac{l}{g}}

with no dependence on the mass of the bob at all. The method of dimensions cannot tell us the value of kk — that has to come from a full derivation or from experiment; it turns out k=2πk = 2\pi.

Watch out

Limitations of the method of dimensions. Powerful as it is, the method has real limits:

  • It cannot determine dimensionless constants (like the k=2πk = 2\pi above). These must be found by actual derivation, theory, or experiment.
  • It cannot be applied to relations involving trigonometric, exponential, or logarithmic functions, or to any relation that is a sum or difference of terms with different physical origins — the method only works when the relationship is a pure product of powers. …