Irodov 1.372: The Same Two Tenths of c Cost Three Times More

Problem Irodov 1.372

A particle of rest mass $m_0$ is sped up from $v_1 = 0.60c$ to $v_2 = 0.80c$ . Find the work $A$ that must be done on the particle to accomplish this, and compare it with the value $A_{\text{cl}} = \frac{1}{2}m_0(v_2^2-v_1^2)$ that the classical (Newtonian) expression for kinetic energy would predict for the same change in speed.

Given

$m_0$
$v_1 = 0.60c$
$v_2 = 0.80c$
$\beta_i = v_i/c$

Find

Find the work

A

done, and compare it with

\frac{1}{2}m_0(v_2^2-v_1^2)

Solution

Step 1Turn the work into a change of kinetic energy

The agent doing the work is the only thing acting on the particle, so the work-energy theorem in its relativistic form applies unchanged: the work equals the increase in kinetic energy. In relativity the kinetic energy is the total energy less the rest energy,

T = mc^2 - m_0c^2 = m_0c^2\left(\frac{1}{\sqrt{1-\beta^2}}-1\right) ,

so the work needed to go from $\beta_1$ to $\beta_2$ is

A = T_2 - T_1 = m_0c^2\left(\frac{1}{\sqrt{1-\beta_2^2}} -\frac{1}{\sqrt{1-\beta_1^2}}\right)

(1)

Step 2Evaluate the two Lorentz factors

Both speeds are chosen so the arithmetic closes exactly. At $\beta_1 = 0.60$ ,

\sqrt{1-\beta_1^2} = \sqrt{1-0.36} = 0.80, \qquad \frac{1}{\sqrt{1-\beta_1^2}} = \frac{5}{4} ,

and at $\beta_2 = 0.80$ ,

\sqrt{1-\beta_2^2} = \sqrt{1-0.64} = 0.60, \qquad \frac{1}{\sqrt{1-\beta_2^2}} = \frac{5}{3} .

So the particle’s kinetic energy rises from $T_1 = 0.25\,m_0c^2$ to $T_2 = 0.67\,m_0c^2$ .

Step 3Take the difference

Substituting the two factors into (1):

A = m_0c^2\left(\frac{5}{3}-\frac{5}{4}\right) = m_0c^2\,\frac{20-15}{12} = \frac{5}{12}\,m_0c^2 .

\begin{aligned} A &= m_0c^2\left(\frac{1}{\sqrt{1-\beta_2^2}} -\frac{1}{\sqrt{1-\beta_1^2}}\right) \\ &= 0.42\,m_0c^2 \end{aligned}

(2)

Step 4Compare with what the classical formula would have charged

The classical work-energy theorem gives the difference of $\frac{1}{2}m_0v^2$ at the two speeds:

A_{\text{cl}} = \frac{1}{2}m_0\left(v_2^2-v_1^2\right) = \frac{1}{2}m_0c^2\left(0.64-0.36\right) = 0.14\,m_0c^2 .

The true cost is three times the classical estimate.

Kinetic energy against speed: the gold curve relativistic, γ - 1; the royal one classical, β^2/2. Each answer is its own curve's rise between the marked abscissas 0.60 and 0.80, gold 0.42 m_0c^2 against royal 0.14 m_0c^2. — Kinetic energy against speed: the gold curve relativistic, $\gamma - 1$ ; the royal one classical, $\beta^2/2$ . Each answer is its own curve’s rise between the marked abscissas $0.60$ and $0.80$ , gold $0.42\,m_0c^2$ against royal $0.14\,m_0c^2$ .

Check

The two formulas must agree when the speeds are small, and they do. Expanding the Lorentz factor,

\frac{1}{\sqrt{1-\beta^2}} = 1 + \frac{1}{2}\beta^2 + \frac{3}{8}\beta^4 + \dots ,

turns (1) into $A \approx \frac{1}{2}m_0c^2\left(\beta_2^2-\beta_1^2\right)$ , which is exactly $A_{\text{cl}}$ . At the speeds actually asked for, the neglected $\beta^4$ term alone contributes $\frac{3}{8}\left(0.4096-0.1296\right) = 0.105$ in units of $m_0c^2$ , three quarters of the whole classical answer, so the expansion has no business being truncated here and the factor of three is no surprise.

Remark

Where the increment is applied matters far more than how big it is. The same $0.20c$ of speed bought at the bottom of the scale, from rest to $0.20c$ , costs only $m_0c^2\left(1/\sqrt{0.96}-1\right) = 0.02\,m_0c^2$ , about twenty times less than the identical increment bought between $0.60c$ and $0.80c$ . Classically the two would differ by a factor of seven, and only because $v^2$ is not linear.