# 5.9 Relativistic energy  (Page 3/16)

 Page 3 / 16

## Solution for (a)

For part (a):

1. Identify the knowns: $v=0.990c;m=9.11\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\text{kg.}$
2. Identify the unknown: ${K}_{\text{rel}}.$
3. Express the answer as an equation: ${K}_{\text{rel}}=\left(\gamma -1\right)m{c}^{2}$ with $\gamma =\frac{1}{\sqrt{1-{u}^{2}\text{/}{c}^{2}}}.$
4. Do the calculation. First calculate $\gamma .$ Keep extra digits because this is an intermediate calculation:
$\begin{array}{cc}\hfill \gamma & =\frac{1}{\sqrt{1-\frac{{u}^{2}}{{c}^{2}}}}\hfill \\ & =\frac{1}{\sqrt{1-\frac{{\left(0.990c\right)}^{2}}{{c}^{2}}}}\hfill \\ & =7.0888.\hfill \end{array}$

Now use this value to calculate the kinetic energy:
$\begin{array}{cc}\hfill {K}_{\text{rel}}& =\left(\gamma -1\right)m{c}^{2}\hfill \\ & =\left(7.0888-1\right)\left(9.11\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\phantom{\rule{0.2em}{0ex}}\text{kg}\right)\left(3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}{\text{m/s}}^{2}\right)\hfill \\ & =4.9922\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}\phantom{\rule{0.2em}{0ex}}\text{J.}\hfill \end{array}$
5. Convert units:
$\begin{array}{cc}\hfill {K}_{\text{rel}}& =\left(4.9922\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}\phantom{\rule{0.2em}{0ex}}\text{J)}\left(\frac{1\phantom{\rule{0.2em}{0ex}}\text{MeV}}{1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}\phantom{\rule{0.2em}{0ex}}\text{J}}\right)\hfill \\ & =3.12\phantom{\rule{0.2em}{0ex}}\text{MeV.}\hfill \end{array}$

## Solution for (b)

For part (b):

1. List the knowns: $v=0.990c;\phantom{\rule{0.2em}{0ex}}m=9.11\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\text{kg.}$
2. List the unknown: ${K}_{\text{rel}}.$
3. Express the answer as an equation: ${K}_{\text{class}}=\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m{u}^{2}.$
4. Do the calculation:
$\begin{array}{cc}{K}_{\text{class}}& =\frac{1}{2}\phantom{\rule{0.2em}{0ex}}m{u}^{2}\hfill \\ & =\frac{1}{2}\left(9.11\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-31}\phantom{\rule{0.2em}{0ex}}\text{kg}\right){\left(0.990\right)}^{2}{\left(3.00\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{8}\phantom{\rule{0.2em}{0ex}}\text{m/s}\right)}^{2}\hfill \\ & =4.0179\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{J.}\hfill \end{array}$
5. Convert units:
$\begin{array}{cc}\hfill {K}_{\text{class}}& =4.0179\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-14}\phantom{\rule{0.2em}{0ex}}\text{J}\left(\frac{1\phantom{\rule{0.2em}{0ex}}\text{MeV}}{1.60\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-13}\phantom{\rule{0.2em}{0ex}}\text{J}}\right)\hfill \\ & =0.251\phantom{\rule{0.2em}{0ex}}\text{Mev.}\hfill \end{array}$

## Significance

As might be expected, because the velocity is 99.0% of the speed of light, the classical kinetic energy differs significantly from the correct relativistic value. Note also that the classical value is much smaller than the relativistic value. In fact, ${K}_{\text{rel}}\text{/}{K}_{\text{class}}=12.4$ in this case. This illustrates how difficult it is to get a mass moving close to the speed of light. Much more energy is needed than predicted classically. Ever-increasing amounts of energy are needed to get the velocity of a mass a little closer to that of light. An energy of 3 MeV is a very small amount for an electron, and it can be achieved with present-day particle accelerators. SLAC, for example, can accelerate electrons to over $50\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{9}\text{eV}=50,000\phantom{\rule{0.2em}{0ex}}\text{MeV.}$

Is there any point in getting v a little closer to c than 99.0% or 99.9%? The answer is yes. We learn a great deal by doing this. The energy that goes into a high-velocity mass can be converted into any other form, including into entirely new particles. In the Large Hadron Collider in [link] , charged particles are accelerated before entering the ring-like structure. There, two beams of particles are accelerated to their final speed of about 99.7% the speed of light in opposite directions, and made to collide, producing totally new species of particles. Most of what we know about the substructure of matter and the collection of exotic short-lived particles in nature has been learned this way. Patterns in the characteristics of these previously unknown particles hint at a basic substructure for all matter. These particles and some of their characteristics will be discussed in a later chapter on particle physics.

## Total relativistic energy

The expression for kinetic energy can be rearranged to:

$E=\frac{m{u}^{2}}{\sqrt{1-{u}^{2}\text{/}{c}^{2}}}=K+m{c}^{2}.$

Einstein argued in a separate article, also later published in 1905, that if the energy of a particle changes by $\Delta E,$ its mass changes by $\text{Δ}m=\text{Δ}E\text{/}{c}^{2}.$ Abundant experimental evidence since then confirms that $m{c}^{2}$ corresponds to the energy that the particle of mass m has when at rest. For example, when a neutral pion of mass m at rest decays into two photons, the photons have zero mass but are observed to have total energy corresponding to $m{c}^{2}$ for the pion. Similarly, when a particle of mass m decays into two or more particles with smaller total mass, the observed kinetic energy imparted to the products of the decay corresponds to the decrease in mass. Thus, E is the total relativistic energy of the particle, and $m{c}^{2}$ is its rest energy.

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