# 28.6 Relativistic energy  (Page 7/12)

 Page 7 / 12

A photon decays into an electron-positron pair. What is the kinetic energy of the electron if its speed is $0.992c$ ?

$\begin{array}{lll}{\text{KE}}_{\text{rel}}& =& \left(\gamma -1\right){\mathrm{mc}}^{2}=\left(\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}-1\right){\mathrm{mc}}^{2}\\ & =& \left(\frac{1}{\sqrt{1-\frac{\left(\text{0.992}c{\right)}^{2}}{{c}^{2}}}}-1\right)\left(\text{9.11}×{\text{10}}^{-\text{31}}\phantom{\rule{0.25em}{0ex}}\text{kg}\right)\left(\text{3.00}×{\text{10}}^{8}\phantom{\rule{0.25em}{0ex}}\text{m/s}{\right)}^{2}=\text{5.67}×{\text{10}}^{-\text{13}}\phantom{\rule{0.25em}{0ex}}\text{J}\end{array}$

## Test prep for ap courses

A mass of 50 g is completely converted into energy. What is the energy that will be obtained when such a conversion takes place?

Show that relativistic kinetic energy becomes the same as classical kinetic energy when $v=c$ .

Relativistic kinetic energy is given as ${\text{KE}}_{\text{rel}}=\left(\gamma -1\right)m{c}^{2}$

where $\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$

Classical kinetic energy is given as ${\text{KE}}_{\text{class}}=\frac{1}{2}m{v}^{2}$

At low velocities $v=0$ , a binomial expansion and subsequent approximation of $\gamma$ gives:

$\gamma =1+\frac{1{v}^{2}}{2{c}^{2}}$ or $\gamma -1=\frac{1{v}^{2}}{2{c}^{2}}$

Substituting $\gamma -1$ in the expression for ${\text{KE}}_{\text{rel}}$ gives

${\text{KE}}_{\text{rel}}=\left[\frac{1{v}^{2}}{2{c}^{2}}\right]m{c}^{2}=\frac{1}{2}m{v}^{2}={\text{KE}}_{\text{class}}$

Hence, relativistic kinetic energy becomes classical kinetic energy when $v\ll c$ .

The relativistic energy of a particle in terms of momentum is given by:

1. $E=\sqrt{{p}^{2}{c}^{2}+{m}_{0}^{2}{c}^{4}}$
2. $E=\sqrt{{p}^{2}{c}^{2}+{m}_{0}^{4}{c}^{4}}$
3. $E=\sqrt{{p}^{2}{c}^{2}+{m}_{0}^{2}{c}^{2}}$
4. $E=\sqrt{{p}^{2}{c}^{4}+{m}_{0}^{2}{c}^{2}}$

## Section summary

• Relativistic energy is conserved as long as we define it to include the possibility of mass changing to energy.
• Total Energy is defined as: $E={\mathrm{\gamma mc}}^{2}$ , where $\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$ .
• Rest energy is ${E}_{0}={\mathrm{mc}}^{2}$ , meaning that mass is a form of energy. If energy is stored in an object, its mass increases. Mass can be destroyed to release energy.
• We do not ordinarily notice the increase or decrease in mass of an object because the change in mass is so small for a large increase in energy.
• The relativistic work-energy theorem is ${W}_{\text{net}}=E-{E}_{0}=\gamma {\mathrm{mc}}^{2}-{\mathrm{mc}}^{2}=\left(\gamma -1\right){\mathrm{mc}}^{2}$ .
• Relativistically, ${W}_{\text{net}}={\text{KE}}_{\text{rel}}$ , where ${\text{KE}}_{\text{rel}}$ is the relativistic kinetic energy.
• Relativistic kinetic energy is ${\text{KE}}_{\text{rel}}=\left(\gamma -1\right){\mathrm{mc}}^{2}$ , where $\gamma =\frac{1}{\sqrt{1-\frac{{v}^{2}}{{c}^{2}}}}$ . At low velocities, relativistic kinetic energy reduces to classical kinetic energy.
• No object with mass can attain the speed of light because an infinite amount of work and an infinite amount of energy input is required to accelerate a mass to the speed of light.
• The equation ${E}^{2}=\left(\mathrm{pc}{\right)}^{2}+\left({\mathrm{mc}}^{2}{\right)}^{2}$ relates the relativistic total energy $E$ and the relativistic momentum $p$ . At extremely high velocities, the rest energy ${\mathrm{mc}}^{2}$ becomes negligible, and $E=\mathrm{pc}$ .

## Conceptual questions

How are the classical laws of conservation of energy and conservation of mass modified by modern relativity?

What happens to the mass of water in a pot when it cools, assuming no molecules escape or are added? Is this observable in practice? Explain.

Consider a thought experiment. You place an expanded balloon of air on weighing scales outside in the early morning. The balloon stays on the scales and you are able to measure changes in its mass. Does the mass of the balloon change as the day progresses? Discuss the difficulties in carrying out this experiment.

The mass of the fuel in a nuclear reactor decreases by an observable amount as it puts out energy. Is the same true for the coal and oxygen combined in a conventional power plant? If so, is this observable in practice for the coal and oxygen? Explain.

We know that the velocity of an object with mass has an upper limit of $c$ . Is there an upper limit on its momentum? Its energy? Explain.

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