Relativistic Momentum, Energy and E = mc² – Introduction to the Special Theory of Relativity (Senior 5 Physics) | Ganzaa

Introduction to the Special Theory of Relativity

Lessons

Senior 5 · Physics

Introduction to the Special Theory of Relativity

4 lessons

Lesson 4 of 4

Relativistic Momentum, Energy and E = mc²

25 min

The Sun is getting lighter

In Unit 7 you found that the Sun radiates about $3.83 \times 10^{26}$ W. Every second, that energy leaves for space as light. According to special relativity, energy has mass, so the Sun must lose mass at a rate of about $\frac{3.83 \times 10^{26}}{(3.00 \times 10^{8})^2} = 4.3 \times 10^{9}$ kg every second: more than four million tonnes. It has done this for 4.6 billion years and has still lost less than a tenth of a percent of its mass. This lesson shows where the famous equation $E = m c^{2}$ comes from, and why nothing with mass can be pushed to the speed of light.

Relativistic momentum

In Newtonian physics momentum is $p = m v$ , and it is conserved in collisions. Einstein showed that if momentum is to be conserved in every inertial frame, as the principle of relativity demands, it must be defined as

p = \gamma m_0 v = \frac{m_0 v}{\sqrt{1 - \frac{v^2}{c^2}}}

where $m_{0}$ is the rest mass, the mass measured in the object's own rest frame. At everyday speeds $\gamma \approx 1$ and $p = m_{0} v$ . As $v$ approaches $c$ , $\gamma$ and the momentum grow without limit. Many textbooks describe this by a relativistic mass $m = \gamma m_0$ , which increases with speed, so that $p = m v$ still holds; most physicists today prefer to keep "mass" for the rest mass and put the $\gamma$ in the formulas. Either way, the conclusion is the same: the faster an object moves, the harder it is to accelerate further, and an infinite force would be needed to reach $c$ . No object with mass can reach the speed of light.

Relativistic energy

The work done in accelerating a particle from rest gives its kinetic energy. With relativistic momentum the result is

E_k = (\gamma - 1)m_0c^2

This can be written $\gamma m_0c^2 = E_k + m_0c^2$ . Einstein interpreted each term:

E = \gamma m_0c^2 \quad \text{(total energy)}, \qquad E_0 = m_0c^2 \quad \text{(rest energy)}

A particle at rest still has energy $m_{0} c^{2}$ , simply because it has mass. Mass is a form of energy.

Check at low speeds. For $v \ll c$ , the binomial expansion gives $\gamma = \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}} \approx 1 + \frac{v^2}{2c^2}$ , so

E_k = (\gamma - 1)m_0c^2 \approx \frac{v^2}{2c^2} \cdot m_0c^2 = \frac{1}{2}m_0v^2

Newton's kinetic energy is the low-speed limit of Einstein's, as it must be.

Definition

The rest mass $m_{0}$ is the mass of an object measured in its own rest frame. Its rest energy is $E_{0} = m_{0} c^{2}$ ; its total energy is $E = \gamma m_0c^2$ ; its kinetic energy is $E_k = (\gamma - 1)m_0c^2$ . Mass–energy equivalence: any change in the energy of a system changes its mass by $\Delta m = \frac{\Delta E}{c^2}$ . Example: the rest energy of an electron is $(9.11 \times 10^{-31})(3.00 \times 10^{8})^2 = 8.20 \times 10^{-14}$ J $= 0.511$ MeV.

Worked example 1: an electron accelerated through 1.0 MV

An electron (rest energy 0.511 MeV) is accelerated from rest through a potential difference of 1.0 MV. Find (a) its kinetic energy, (b) $\gamma$ , (c) its speed. (d) What speed would classical physics predict?

Solution

(a) $E_{k} = e V = 1.0$ MeV.

(b) $(\gamma - 1)(0.511) = 1.0$ , so $\gamma = 1 + \frac{1.0}{0.511} = 2.96$ .

(c) $\frac{v}{c} = \sqrt{1 - \frac{1}{\gamma^2}} = \sqrt{1 - \frac{1}{2.96^2}} = 0.94$ , so $v = 0.94c = 2.8 \times 10^{8}\,\mathrm{m\,s^{-1}}$ .

(d) $v = \sqrt{\frac{2E_k}{m}} = \sqrt{\frac{2(1.6 \times 10^{-13})}{9.11 \times 10^{-31}}} = 5.9 \times 10^{8}\,\mathrm{m\,s^{-1}}$ , about $2 c$ : impossible. The classical formula fails badly at these energies.

Energy and momentum together

Squaring $E = \gamma m_0c^2$ and $pc = \gamma m_0vc$ and subtracting:

E^2 - (pc)^2 = \gamma^2m_0^2c^4\left(1 - \frac{v^2}{c^2}\right) = m_0^2c^4

E^2 = (pc)^2 + \left(m_0c^2\right)^2

This energy–momentum relation holds in every frame. You used it in Unit 7 to derive the Compton shift. For a particle with no rest mass, such as a photon, it gives $E = p c$ , so $p = \frac{E}{c} = \frac{h}{\lambda}$ , and such a particle must always travel at exactly $c$ .

Mass and energy are equivalent

The rest-energy term means that whenever a system gains or loses energy $\Delta E$ , in any form, its mass changes by

\Delta m = \frac{\Delta E}{c^2}

Because $c^2 = 9.00 \times 10^{16}\,\mathrm{m^2\,s^{-2}}$ is so large, everyday energy changes give mass changes far too small to measure. In nuclear reactions, however, a measurable fraction of the mass (about 0.1% in fission, 0.7% in the fusion of hydrogen to helium) is converted into kinetic energy and radiation. This is the energy source of the Sun and of nuclear power stations, which you will study in Senior 6. A convenient unit: $1\,\mathrm{u}$ (atomic mass unit) is equivalent to 931.5 MeV.

Worked example 2: how much mass is that?

(a) 1.0 kg of water is heated from 20 °C to 100 °C (specific heat capacity $4200\,\mathrm{J\,kg^{-1}\,K^{-1}}$ ). By how much does its mass increase? (b) How much energy would be released if 1.0 g of mass were converted completely into energy?

Solution

(a) $\Delta E = mc_w\Delta\theta = (1.0)(4200)(80) = 3.4 \times 10^{5}$ J; $\Delta m = \frac{3.36 \times 10^{5}}{9.00 \times 10^{16}} = 3.7 \times 10^{-12}$ kg, far below what any balance can detect.

(b) $E = mc^2 = (1.0 \times 10^{-3})(9.00 \times 10^{16}) = 9.0 \times 10^{13}$ J, roughly the energy released by burning 2000 tonnes of petrol.

Worked example 3: protons in the Large Hadron Collider

Protons (rest energy 938 MeV) in the Large Hadron Collider at CERN reach a kinetic energy of 6.8 TeV ( $6.8 \times 10^{6}$ MeV). Find $\gamma$ and show how close their speed is to $c$ . For large $\gamma$ , $1 - \frac{v}{c} \approx \frac{1}{2\gamma^2}$ .

Solution

$\gamma = 1 + \frac{E_k}{m_0c^2} = 1 + \frac{6.8 \times 10^{6}}{938} = 7.25 \times 10^{3}$ .

$1 - \frac{v}{c} \approx \frac{1}{2(7250)^2} = 9.5 \times 10^{-9}$ , so $v = 0.999\,999\,990\,5c$ . Pouring in thousands of times the proton's rest energy brings it ever closer to $c$ , but never to it.

Warning

$E = m c^{2}$ is not "mass turns into energy only in bombs". It applies to every energy change: a compressed spring, a charged battery and a hot cup of tea are all slightly heavier. And use the right energy: kinetic energy is $(\gamma - 1)m_0c^2$ , not $\gamma m_0c^2$ and not $\frac{1}{2}m_0v^2$ at high speeds.

Tip

Work in MeV. For particles, use $m_{0} c^{2}$ in MeV (electron 0.511, proton 938) and $\gamma = 1 + \frac{E_k}{m_0c^2}$ . The numbers stay small and you avoid powers of ten.

The physics behind Einstein's most famous equation, by Lindsay DeMarchi and Fabio Pacucci (TED-Ed). Watch for why mass counts as energy, and how the Sun turns a little of its mass into the light that reaches us.

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Practice

Write down the expressions for the momentum, total energy, rest energy and kinetic energy of a particle of rest mass $m_{0}$ moving at speed $v$ .
Show that $(\gamma - 1)m_0c^2 \approx \frac{1}{2}m_0v^2$ when $v \ll c$ .
Find the speed of an electron whose kinetic energy equals its rest energy.
A proton moves at $0.80 c$ . Find its momentum and its kinetic energy in MeV ( $m_{0} c^{2} = 938$ MeV).
The Sun radiates $3.83 \times 10^{26}$ W. How much mass does it lose in one year ( $3.16 \times 10^{7}$ s)?
Use $E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}$ to find the momentum, in MeV/c, of an electron with total energy 2.0 MeV.
Explain, using the relativistic expression for kinetic energy, why no amount of work can accelerate an electron to the speed of light.
Ishimwe says: "Relativity proves Newton's mechanics is wrong, so we should stop teaching it." Evaluate his statement.

Show answersHide answers

$p = \gamma m_0v$ ; $E = \gamma m_0c^2$ ; $E_{0} = m_{0} c^{2}$ ; $E_k = (\gamma - 1)m_0c^2$ .
$\gamma = \left(1 - \frac{v^2}{c^2}\right)^{-\frac{1}{2}} \approx 1 + \frac{v^2}{2c^2}$ by the binomial expansion, so $(\gamma - 1)m_0c^2 \approx \frac{v^2}{2c^2}m_0c^2 = \frac{1}{2}m_0v^2$ .
$\gamma - 1 = 1$ , $\gamma = 2$ : $\frac{v}{c} = \sqrt{1 - \frac{1}{4}} = 0.87$ .
$\gamma = \frac{1}{0.60} = 1.67$ . $pc = \gamma m_0c^2\frac{v}{c} = (1.67)(938)(0.80) = 1.25 \times 10^{3}$ MeV, so $p = 1250$ MeV/c ( $6.7 \times 10^{-19}\,\mathrm{kg\,m\,s^{-1}}$ ). $E_{k} = (0.667) (938) = 625$ MeV.
$\Delta m = \frac{(3.83 \times 10^{26})(3.16 \times 10^{7})}{9.00 \times 10^{16}} = 1.3 \times 10^{17}$ kg.
$(p c)^{2} = {2.0}^{2} - {0.511}^{2} = 3.74$ , $p c = 1.93$ MeV, so $p = 1.93$ MeV/c.
$E_k = (\gamma - 1)m_0c^2$ and $\gamma \to \infty$ as $v \to c$ . Reaching $c$ would need infinite kinetic energy, so infinite work; any finite work gives a speed below $c$ .
Not justified. Newton's mechanics is the low-speed limit of relativity: at everyday speeds the two agree to better than one part in a billion, and Newton's equations are far simpler. Engineers design bridges, cars and even most spacecraft with Newton. Relativity is needed when speeds approach $c$ or extreme precision (GPS) is required. Both should be taught, with their range of validity.

Key points

Relativistic momentum $p = \gamma m_0v$ ; nothing with mass reaches $c$ .
Total energy $E = \gamma m_0c^2$ , rest energy $m_{0} c^{2}$ , kinetic energy $(\gamma - 1)m_0c^2 \to \frac{1}{2}m_0v^2$ at low speed.
$E^{2} = (p c)^{2} + (m_{0} c^{2})^{2}$ ; for photons $E = p c$ .
Mass–energy equivalence: $\Delta E = \Delta m\,c^2$ , the energy source of the Sun and of nuclear reactions.

1The Sun is getting lighter

2Relativistic momentum

3Relativistic energy

4Energy and momentum together

5Mass and energy are equivalent

The Sun is getting lighter

Relativistic momentum

Relativistic energy

Energy and momentum together

Mass and energy are equivalent