Topic 5 of 6

Binomial Series & Relativity

The binomial series $(1+x)^\alpha = 1 + \alpha x + \ldots$ is the most powerful single-series approximation in physics. It contains square roots, reciprocals, and — most famously — the Lorentz factor. Expanding relativity in powers of $v/c$ literally recovers Newtonian mechanics as the leading term.

Key Concepts

Binomial Approximation

For

|x| \ll 1

(1+x)^\alpha \approx 1 + \alpha x

. The key is identifying

x

and

\alpha

correctly for the expression at hand.

Lorentz Factor

\gamma = (1-\beta^2)^{-1/2}

where

\beta = v/c

. For

\beta \ll 1

\gamma \approx 1 + \beta^2/2 + 3\beta^4/8 + \ldots

— a binomial series with

\alpha = -1/2

and

x = -\beta^2

Relativistic KE

K = (\gamma-1)mc^2

. Expanding:

K = \frac{1}{2}mv^2 + \frac{3mv^4}{8c^2} + \ldots

The leading term is exactly classical KE; the next term is the first relativistic correction.

Newtonian Limit

Every relativistic formula reduces to its Newtonian counterpart in the limit

v/c \to 0

. This is guaranteed by the binomial expansion — Newtonian mechanics is the

v/c \to 0

Taylor series of special relativity.

Key Equations

Binomial Series

(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2!}x^2 + \frac{\alpha(\alpha-1)(\alpha-2)}{3!}x^3 + \cdots

Valid for

|x| < 1

and any real

\alpha

Lorentz Factor Expansion

\gamma = \frac{1}{\sqrt{1-\beta^2}} = 1 + \frac{\beta^2}{2} + \frac{3\beta^4}{8} + \frac{5\beta^6}{16} + \cdots

Binomial series with

\alpha = -1/2

x = -\beta^2

. Converges for

\beta < 1

Relativistic KE Expansion

K = (\gamma-1)mc^2 = \frac{1}{2}mv^2 + \frac{3mv^4}{8c^2} + \cdots

Classical KE emerges as the leading term.

Worked Example

Recovering Classical KE from Relativity

Problem

Show that the relativistic kinetic energy $K = (\gamma - 1)mc^2$ reduces to $\frac{1}{2}mv^2$ for $v \ll c$ .

Solution

Expand $\gamma = (1-\beta^2)^{-1/2}$ using the binomial series with $\alpha = -1/2$ , $x = -\beta^2$ :

\gamma = 1 + \frac{1}{2}\beta^2 + \frac{3}{8}\beta^4 + \cdots

Subtract 1 and multiply by $mc^2$ :

K = mc^2\!\left(\frac{\beta^2}{2} + \frac{3\beta^4}{8} + \cdots\right) = \frac{mc^2\beta^2}{2} + \cdots

Since $\beta = v/c$ , we have $mc^2\beta^2/2 = mc^2 v^2/(2c^2) = mv^2/2$ — exactly classical kinetic energy.

Answer

K \approx \frac{1}{2}mv^2

to leading order in

v/c

. Classical KE is the zeroth-order binomial term.

Practice

Exercises

6 problems

1 of 6

Using $(1+x)^{1/3} \approx 1 + x/3$ for small $x$ , approximate $(1.09)^{1/3}$ .

Your answer

2 of 6

For $v = 0.1c$ , compute the Lorentz factor to 2nd order: $\gamma \approx 1 + \beta^2/2$ . What is $\gamma$ ?

Your answer

3 of 6

For $v = 0.2c$ , the classical KE is $\frac{1}{2}mv^2$ . What is this in units of $mc^2$ ?

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4 of 6

For $v = 0.3c$ , the first relativistic correction to KE is $3mv^4/(8c^2) = 3mc^2\beta^4/8$ . In units of $mc^2$ , what is this correction?

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5 of 6

Using the 4-term binomial expansion $(1+x)^{-1} \approx 1 - x + x^2 - x^3$ with $x = 0.3$ , approximate $1/1.3$ .

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6 of 6

At $v = 0.6c$ , the exact $\gamma = 1/\sqrt{1-0.36} = 1.25$ . The leading binomial approximation gives $\gamma \approx 1 + \beta^2/2 = 1.18$ . What is the fractional error $|\gamma_{\rm approx} - \gamma_{\rm exact}|/\gamma_{\rm exact}$ ?

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Key Takeaways

The binomial series $(1+x)^\alpha = 1 + \alpha x + \frac{\alpha(\alpha-1)}{2}x^2 + \ldots$ works for any real $\alpha$ and $|x| < 1$ .
The key skill: identify $x$ and $\alpha$ . For $\gamma = (1-\beta^2)^{-1/2}$ , use $\alpha = -1/2$ and $x = -\beta^2$ .
Relativistic KE expands as $K = \frac{1}{2}mv^2 + \frac{3mv^4}{8c^2} + \ldots$ . Classical mechanics is the leading term.
The Newtonian limit of any relativistic formula is its Taylor expansion in $v/c$ to leading order.
The binomial approximation is valid when $|x| \ll 1$ ; at $v = 0.1c$ the Lorentz factor error is about 0.4%.