Concepts: errors, dimensions and dimensional analysis

Units and Measurements24 min readLesson 3 of 5

3.1 Reporting an error The true value is never known, so a measurement is reported as a mean plus or minus an uncertainty. The absolute error of a single reading is the size of the difference from the mean. The mean absolute error of a set averages those differences and is quoted with the mean value, as in 3.62 plus or minus 0.05 s. The relative error divides the absolute error by the mean, and the percentage error multiplies that by 100. The relative error is the more useful number when two measurements are compared. An error of 0.01 g is 1 percent on a 1.02 g mass but only 0.1 percent on a 9.89 g mass, so the same instrument precision matters much more on the smaller quantity. That is why the same absolute uncertainty can be serious in one experiment and negligible in another.

Δ Z = Δ A + Δ B (Z = A \pm B), \frac{Δ Z}{Z} = \frac{Δ A}{A} + \frac{Δ B}{B} (Z = A B or Z = A / B)

Error combination rules for sums, differences, products and quotients

3.2 Combining errors in a calculation Add absolute errors when quantities add or subtract, and add relative errors when they multiply or divide. Do not mix the two rules: a sum answered with the multiplication rule, or the reverse, is a standard marking loss. For a power, multiply the relative error by the exponent, which is why a 2 percent error in the diameter of a sphere becomes a 6 percent error in its volume. Take a wire of length 150 cm measured with a scale of least count 0.1 cm, and diameter 0.08 cm measured with a screw gauge of least count 0.001 cm. The relative error in the length is 0.1/150, about 0.07 percent. The relative error in the diameter is 0.001/0.08, about 1.25 percent, and the cross sectional area, proportional to the square of the diameter, has twice that, about 2.5 percent. A result that divides by the area therefore carries the sum of the two, about 2.6 percent, even though the length was measured to much higher precision. When a formula is built from a product of powers, write it as a power law first and add the relative errors term by term. For a density found from a cylinder, rho = 4m / (pi d^2 l), the fractional errors are 0.02/97.42 for the mass, twice 0.02/20.20 for the diameter and 0.05/8.35 for the length. Convert to percentages and add: 0.02 + 0.20 + 0.60 gives about 0.82 percent. Round the final uncertainty to the same place as the value it accompanies.

3.3 Dimensions and dimensional formulae The dimensions of a quantity are the powers to which the base quantities are raised. In mechanics only mass, length and time appear, so every mechanical quantity can be written in the form M^a L^b T^c and recorded in square brackets. Velocity has dimensions of length divided by time, so its dimensional formula is [L T^-1]. Force is mass times acceleration, so [M L T^-2]. A dimensional equation sets a quantity equal to its dimensional formula. The dimensional formulae of volume, speed, force and density are [L^3], [L T^-1], [M L T^-2] and [M L^-3] when the zero powers are dropped. A quantity with no units, such as refractive index or relative density, is dimensionless and its formula is written [M^0 L^0 T^0]. The principle of homogeneity says that every term of a valid equation carries the same dimensions as every other term. Test each piece of x = x0 + v0 t + (1/2) a t^2. The position x and the initial position x0 are lengths. The term v0 t is a speed multiplied by a time, which is a length. The term (1/2) a t^2 is an acceleration multiplied by a time squared, which is again a length. All four terms are lengths, so the equation passes. A failing test is decisive. The kinetic energy formula K = m^2 v^3 has dimensions [M^2 L^3 T^-3], which is not energy, so it is wrong. The same method rejects K = ma, a force, and K = (1/2)mv^2 + ma, which adds incompatible terms. It cannot choose between (1/2)mv^2 and (3/16)mv^2, because both have the energy dimensions and only the definition decides the constant.

[K] = [M L^{2} T^{- 2}], [m^{2} v^{3}] = [M^{2} L^{3} T^{- 3}], [ma] = [M L T^{- 2}]

Dimensional test of three proposed kinetic energy formulae

3.4 Deducing a relation from dimensions When a quantity is known to depend on a few others as a product, the dimensions can fix the exponents up to a dimensionless constant. For a simple pendulum, assume the period T depends on the length l, the mass m and the acceleration due to gravity g. Write T = k l^x g^y m^z. The left side has dimensions [T]. The right side has dimensions L^(x+y) T^(-2y) M^z. Matching powers gives x + y = 0, -2y = 1 and z = 0, so y = -1/2, x = 1/2 and z = 0. The period is proportional to the square root of l divided by g, and the mass drops out. The constant k cannot be found this way, and in the exact result it is 2 pi. The method also fails to distinguish quantities with the same dimensions, such as work and torque or frequency and angular velocity, so it can never establish a relation on its own. It is a check, and the exact coefficient needs theory or experiment.

T = k l^{x} g^{y} m^{z} \Rightarrow [T] = [L^{x + y} T^{- 2 y} M^{z}] \Rightarrow y = - \frac{1}{2}, x = \frac{1}{2}, z = 0

Dimensional determination of the pendulum exponents

3.5 Exercise practice: the NCERT set in short Exercises 1.1 and 1.2 ask for unit conversions: 1 cm^3 = 10^-6 m^3, the cylinder surface area 2 pi r h + 2 pi r^2 in mm^2, 18 km h^-1 = 5 m s^-1, and G = 6.67 x 10^-11 in CGS units. Exercise 1.3 repeats the calorie conversion with new units of mass, length and time, giving 4.2 alpha^-1 beta^-2 gamma^2. Exercise 1.4 asks you to attach a reference to each statement, such as an atom compared with a cricket ball, or the speed of a jet with the speed of sound. Exercise 1.5 makes the speed of light the unit, so light travelling for 8 min 20 s covers 500 such units. Exercise 1.6 compares precision through least count: the optical instrument that reads to a wavelength is the finest, about 6 x 10^-7 m. Exercise 1.7 gives the hair thickness as 3.5 mm / 100 = 0.035 mm. Exercise 1.8 explains the screw gauge as pitch divided by the number of divisions, and says the accuracy cannot grow without limit because the scale cannot be read beyond its least count. Exercise 1.9 finds the linear magnification from the ratio of areas, the square root of 1.55 m^2 divided by 1.75 cm^2, which is about 94. Exercise 1.10 counts significant figures in six values, from two for 0.007 m^2 to four for 0.2370 g cm^-3. Exercise 1.11 gives the sheet area to three significant figures and its volume to three. Exercise 1.12 keeps decimal places in the addition and in the difference of the two gold pieces. Exercise 1.13 spots the missing c inside the Lorentz factor: m = m0 / sqrt(1 - v^2/c^2). Exercise 1.14 uses the given radius to get the volume occupied by one mole of hydrogen atoms, and exercise 1.15 compares that with the molar volume of the gas, a ratio that comes out large because a gas is mostly empty space. Exercise 1.16 explains the motion of nearby objects seen from a train as an effect of the viewing angle, and exercise 1.17 checks the mean density of the Sun, near 1.4 x 10^3 kg m^-3, which lies between the density of a solid and a gas.

Exam

In a dimensional analysis answer, show the dimensional equation for each term before you conclude. In an error answer, show the fractional errors term by term and only then add them; a bare percentage with no working rarely earns full marks.

in hinglish

Error report karte waqt best value plus uncertainty likhte hain. Absolute error single reading ki deviation hai, relative error usko mean se divide karta hai, aur percentage error 100 se multiply. Addition aur subtraction mein absolute errors add hote hain, product aur quotient mein relative errors, aur power mein relative error power se multiply hota hai, isliye diameter ka 2 percent error sphere ke volume mein 6 percent ban jaata hai. Dimensions powers hote hain base quantities ke, aur mechanics mein sirf M, L, T kaafi hain. Principle of homogeneity kehta hai equation ke har term ki dimensions same hongi; mismatch mila to equation pakka galat hai, lekin match hona sahi hone ki guarantee nahi. Simple pendulum wale example mein T = k l^x g^y m^z se x = 1/2, y = -1/2, z = 0 milta hai, constant 2 pi dimensions se nahi nikalta. Exercise practice mein unit conversion, least count se precision, significant figures aur Sun ki density wale questions aa jaate hain, inhe formula ke saath karna.