Approximations & Small-Angle Expansions

The lie that makes physics solvable

A pendulum swinging through a small arc keeps almost perfect time. That fact runs every grandfather clock ever built, and it is not obvious: the equation that governs a pendulum, written honestly, cannot be solved with any function you know. Write down Newton's second law along the arc and you get $\ddot\theta = -(g/L)\sin\theta$. That $\sin$ is fatal. There is no formula for $\theta(t)$ in terms of sines, exponentials, logs, or anything else with a name.

So physicists tell a lie. For small $\theta$, they write $\sin\theta \approx \theta$, the equation becomes $\ddot\theta = -(g/L)\theta$, and out drops $T = 2\pi\sqrt{L/g}$. The lie is tiny (at $10^\circ$ it is wrong by less than a fifth of a percent) and without it there is no pendulum formula at all.

This lesson is about telling that kind of lie on purpose, and knowing exactly how big it is. You have just spent two lessons on derivatives and integrals. The derivative is about to earn its keep: it is the machine that generates every approximation below.

Near a point, every curve is a polynomial

Start with what a derivative means. If you know $f(0)$ and you know the slope $f'(0)$, then moving a short distance $x$ away from zero changes $f$ by about slope times distance:

$f(x) \approx f(0) + f'(0)\,x$

That is the tangent line, and it is the first approximation. It is exact at $x = 0$ and drifts away as you leave. Which direction does it drift? That depends on the curvature, which is $f''$. Adding a term that has the right second derivative and does not disturb the first two gives

$f(x) \approx f(0) + f'(0)\,x + \tfrac{1}{2}f''(0)\,x^2$

Keep going and you get the Taylor series: the $n$th term is $f^{(n)}(0)x^n/n!$. You will almost never need more than two terms, because each term is smaller than the one before by another factor of $x$, and the whole game is played where $x$ is small.

Run the machine on $\sin x$. You have $\sin 0 = 0$, and the derivative of $\sin$ is $\cos$, so $f'(0) = \cos 0 = 1$. The second derivative is $-\sin$, which is $0$ at the origin, so the $x^2$ term vanishes. The third derivative is $-\cos$, giving $-1$ at the origin, so the cubic term is $-x^3/3! = -x^3/6$. That is where the standard results come from:

$e^x \approx 1 + x, \qquad \ln(1+x) \approx x, \qquad (1+x)^n \approx 1 + nx$

Every one of these is exact at $x = 0$ and good for $|x| \ll 1$.

The binomial approximation, and the percent rule

Of the expansions above, $(1+x)^n \approx 1 + nx$ is the one you will use most, so derive it once and trust it forever. Let $f(x) = (1+x)^n$. Then $f(0) = 1$, and $f'(x) = n(1+x)^{n-1}$, so $f'(0) = n$. Tangent line: $1 + nx$. Done.

What makes it a workhorse is that $n$ can be anything. Negative: $\dfrac{1}{1+x} = (1+x)^{-1} \approx 1 - x$. Fractional: $\sqrt{1+x} \approx 1 + \tfrac{1}{2}x$. Both at once: $\dfrac{1}{\sqrt{1-x}} \approx 1 + \tfrac{1}{2}x$, which is the whole of special relativity at low speed.

Now the most useful consequence. Suppose some quantity is built from others by powers, like $T = 2\pi\sqrt{L/g} = 2\pi L^{1/2} g^{-1/2}$. Change $L$ by a small fraction $\delta L/L$ and hold $g$ fixed. Then

so $\dfrac{\Delta T}{T} \approx \dfrac{1}{2}\dfrac{\Delta L}{L}$. In words: multiply the fractional change by the exponent. A quantity that enters squared doubles its fractional change; one under a square root halves it; one in the denominator flips the sign.

That single sentence answers a whole family of problems without any calculus on the page. Lengthen a pendulum by $0.60\%$ and the period grows by $\tfrac{1}{2}(0.60\%) = 0.30\%$. Weaken $g$ by $0.30\%$ and, since $g$ carries exponent $-\tfrac{1}{2}$, the period grows by $0.15\%$.

Worked example 1: how much weaker is gravity in orbit

The field strength at height $h$ above the surface is

Pulling $R$ out of the bracket is the move that makes the whole thing work: it turns the expression into $(1 + \text{small})^n$, which is the only form the binomial rule accepts. Here the small thing is $x = h/R$ and the exponent is $n = -2$.

With $h = 250$ km $= 2.50\times10^5$ m and $R = 6.37\times10^6$ m,

$x = \frac{2.50\times10^5}{6.37\times10^6} = 0.03925$

That is comfortably less than one, so first order will do:

$g \approx g_0(1 - 2x) = g_0(1 - 0.0785)$

The field drops by $7.85\%$. Worth pausing on: the International Space Station orbits at about this height, and gravity there is $92\%$ of its surface value. Astronauts float because they are falling, not because gravity ran out.

Check the size of the error you accepted. The next term in the expansion of $(1+x)^{-2}$ is $\tfrac{1}{2}n(n-1)x^2 = 3x^2 = 3(0.03925)^2 = 0.0046$, so the true answer differs from yours in the fourth decimal place. First order was plenty.

Worked example 2: the tidal force, where the leading terms cancel

The Moon pulls on the near side of Earth harder than on the far side, and that difference raises the tides. Put the Moon at distance $d$ from Earth's center, and ask for the acceleration at the sublunar point (the spot on Earth directly under the Moon, a distance $R$ closer) compared with the acceleration of Earth as a whole.

$a_{\text{tide}} = \frac{GM}{(d-R)^2} - \frac{GM}{d^2}$

Here is the trap. Both terms are about $3.3\times10^{-5}\ \mathrm{m/s^2}$, and the answer is around $10^{-6}$. If you round either term to three digits and subtract, you get garbage. You must expand.

Factor $d$ out of the first denominator, exactly as before:

Now subtract, and the $1$ cancels the second term exactly:

$a_{\text{tide}} \approx \frac{GM}{d^2}\cdot\frac{2R}{d} = \frac{2GMR}{d^3}$

Notice what just happened: the answer came entirely from the first-order term, because the zeroth-order pieces annihilated each other. When you are subtracting two nearly equal things, the term you would normally throw away is the entire answer.

Numbers, with $M = 7.35\times10^{22}$ kg, $R = 6.37\times10^{6}$ m, $d = 3.84\times10^{8}$ m, $G = 6.674\times10^{-11}$:

$GM = (6.674\times10^{-11})(7.35\times10^{22}) = 4.905\times10^{12}$ $2GMR = 2(4.905\times10^{12})(6.37\times10^{6}) = 6.25\times10^{19}$ $d^3 = (3.84\times10^{8})^3 = 5.66\times10^{25}$ $a_{\text{tide}} = \frac{6.25\times10^{19}}{5.66\times10^{25}} = 1.1\times10^{-6}\ \mathrm{m/s^2}$

About one ten-millionth of $g$, and it moves the oceans.

Worked example 3: how small does a small angle have to be

A pendulum swung to amplitude $\theta_0$ does not have period exactly $T_0 = 2\pi\sqrt{L/g}$. Carrying the expansion of $\sin\theta$ one term further leads to the correction

$T = T_0\left(1 + \frac{\theta_0^2}{16} + \dots\right)$

with $\theta_0$ in radians. Two questions follow immediately.

First: how bad is a sloppy clock? Suppose the clock is regulated on the assumption of $T_0$ but actually swings to $\theta_0 = 20^\circ$. Convert:

$\theta_0 = 20\times\frac{\pi}{180} = 0.3491\ \text{rad}$ $\frac{\theta_0^2}{16} = \frac{0.1219}{16} = 0.007615$

Each real swing takes $0.7615\%$ longer than the clock thinks. Over a day of $86400$ s that is

$86400 \times 0.007615 = 658\ \text{s}$

The clock loses about eleven minutes a day. Note that $L$ and $g$ never appeared: the fractional error does not care how long the pendulum is, which is exactly the kind of thing an expansion is good at telling you.

Second: run it backwards. What amplitude keeps the simple formula accurate to $0.10\%$? Set the correction equal to the tolerance:

$\theta_0 = 0.1265 \times \frac{180}{\pi} = 7.2^\circ$

So "small angle" means about seven degrees if you want three-digit accuracy, and more like twenty degrees if one percent is fine. There is no universal answer, and any problem that says "small" without a tolerance is telling you the tolerance is loose.

Choosing the small parameter

The algebra is easy. The judgement is the hard part, and it comes down to one question you must answer before writing anything: small compared with what?

An expansion parameter has to be dimensionless. "The height is small" is meaningless: small next to what? In the orbit problem, the answer was $h/R$. In the tide problem, $R/d$. In relativity, $v/c$. In a decay problem, $t/\tau$. Find the ratio, confirm it is much less than one, and expand in it. If you cannot name a dimensionless small quantity, you do not yet have an approximation, you have a hope.

Three more that appear constantly:

Short-time decay. A sample with half-life $t_{1/2}$ decays as $N = N_0 e^{-t/\tau}$ where $\tau = t_{1/2}/\ln 2$. Over a short interval, the fraction decayed is $1 - e^{-t/\tau} \approx t/\tau$. For a $12.0$ h half-life and $6.0$ minutes $= 0.10$ h: $\tau = 12.0/0.693 = 17.31$ h, so the fraction is $0.10/17.31 = 0.00578$. The small parameter is $t/\tau$, not $t$.

Slow relativity. $\gamma = (1 - v^2/c^2)^{-1/2} \approx 1 + \tfrac{1}{2}v^2/c^2$. At $v = 0.010c$ this gives $\gamma - 1 = \tfrac{1}{2}(10^{-4}) = 5\times10^{-5}$. The small parameter is $v^2/c^2$, which is already tiny when $v/c$ is merely small: that is why Newtonian mechanics survives so well.

Sagging a curve into a parabola. A circle of radius $R$ dips below its own tangent by $R - \sqrt{R^2 - r^2} = R\left(1 - \sqrt{1 - r^2/R^2}\right) \approx R\cdot\tfrac{1}{2}(r^2/R^2) = r^2/2R$. That is the air gap under a lens in Newton's rings, the shape of a shallow dish, and the reason small sections of spheres act like parabolas. Small parameter: $r^2/R^2$.

How to recognise that this is the technique

The cue is almost always a word, not an equation. Reach for an expansion when a problem says:

• "small oscillations", "small amplitude", "slightly displaced" (expand the potential or the force about the equilibrium point, and you will land on simple harmonic motion.
• "to first order in", "estimate", "approximately") the problem is telling you outright.
• "for $r \gg d$", "far from", "thin", "shallow" (form the ratio of the small length to the big one and expand in it.
• a percent change, or an uncertainty, in a formula built from powers) use the exponent rule and skip the calculus entirely.
• a difference of two nearly equal quantities: tides, dipoles, beat frequencies, clock corrections. Here the expansion is not a convenience, it is the only way to get a nonzero answer without twelve digits of precision.

And one negative cue worth internalising. If a problem hands you a number like $\theta_0 = 20^\circ$ and asks "how much does the simple formula get wrong", it wants the first correction, not the leading term. The leading term is what you already assumed. The question is always about the next one down.

You now have the whole toolkit: vectors, units, derivatives, integrals, and the expansions that make hard functions behave. What is left is knowing which tool to pick up and in what order, which is the subject of Problem-Solving Strategy.