28.6 Relativistic Energy

Calculating Rest Energy: Rest Energy is Very Large

Calculate the rest energy of a 1.00-g mass.

Strategy

One gram is a small mass—less than half the mass of a penny. We can multiply this mass, in SI units, by the speed of light squared to find the equivalent rest energy.

Solution

1. Identify the knowns. $m=1\text{.00}\times {\text{10}}^{-3}\text{kg}$; $c=3\text{.}\text{00}\times {\text{10}}^8\text{m/s}$
2. Identify the unknown. $E_0$
3. Choose the appropriate equation. $E_0={\mathrm{mc}}^2$
4. Plug the knowns into the equation.
   $$\begin{array}{lll}{E}_0& =& {\mathrm{mc}}^2=(1.00\times {\text{10}}^{-3}\text{kg})(3.00\times {\text{10}}^8\text{m/s}{)}^2\\ & =& 9.00\times {\text{10}}^{\text{13}}\text{kg}\cdot {\text{m}}^2{\text{/s}}^2\end{array}$$

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5. Convert units.

   Noting that $1\text{kg}\cdot {\text{m}}^2{\text{/s}}^2=\mathrm{1\; J}$, we see the rest mass energy is

   $$E_0=9\text{.}\text{00}\times {\text{10}}^{\text{13}}\text{J}.$$

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Discussion

This is an enormous amount of energy for a 1.00-g mass. We do not notice this energy, because it is generally not available. Rest energy is large because the speed of light $c$ is a large number and $c^2$ is a very large number, so that ${\mathrm{mc}}^2$ is huge for any macroscopic mass. The $9\text{.}\text{00}\times {\text{10}}^{\text{13}}\text{J}$ rest mass energy for 1.00 g is about twice the energy released by the Hiroshima atomic bomb and about 10,000 times the kinetic energy of a large aircraft carrier. If a way can be found to convert rest mass energy into some other form (and all forms of energy can be converted into one another), then huge amounts of energy can be obtained from the destruction of mass.

Calculating Rest Mass: A Small Mass Increase due to Energy Input

A car battery is rated to be able to move 600 ampere-hours $\mathrm{(A·h)}$ of charge at 12.0 V. (a) Calculate the increase in rest mass of such a battery when it is taken from being fully depleted to being fully charged. (b) What percent increase is this, given the battery’s mass is 20.0 kg?

Strategy

In part (a), we first must find the energy stored in the battery, which equals what the battery can supply in the form of electrical potential energy. Since ${\text{PE}}_{\text{elec}}=\text{qV}$, we have to calculate the charge $q$ in $600\mathrm{A·h}$, which is the product of the current $I$ and the time $t$. We then multiply the result by 12.0 V. We can then calculate the battery’s increase in mass using $\mathrm{\Delta }E={\text{PE}}_{\text{elec}}=(\mathrm{\Delta }m)c^2$. Part (b) is a simple ratio converted to a percentage.

Solution for (a)

1. Identify the knowns. $I\cdot t=\text{600 A}\cdot \text{h}$; $V=\text{12}\text{.}0\text{V}$; $c=3\text{.}\text{00}\times {\text{10}}^8\text{m/s}$
2. Identify the unknown. $\mathrm{\Delta }m$
3. Choose the appropriate equation. ${\text{PE}}_{\text{elec}}=(\mathrm{\Delta }m)c^2$
4. Rearrange the equation to solve for the unknown. $\mathrm{\Delta }m=\frac{{\text{PE}}_{\text{elec}}}{c^2}$
5. Plug the knowns into the equation.
   $$\begin{array}{lll}\mathrm{\Delta }m& =& \frac{{\text{PE}}_{\text{elec}}}{c^2}\\ & =& \frac{\text{qV}}{c^2}\\ & =& \frac{(It)V}{c^2}\\ & =& \frac{(\text{600 A}\cdot \text{h})(\text{12.0 V})}{(3.00\times {\text{10}}^8{)}^2}\text{.}\end{array}$$

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   Write amperes A as coulombs per second (C/s), and convert hours to seconds.

   $$\begin{array}{lll}\mathrm{\Delta }m& =& \frac{(\text{600 C/s}\cdot \text{h}\left(\frac{\text{3600 s}}{\text{1 h}}\right)(\text{12.0 J/C})}{(3.00\times {\text{10}}^8\text{m/s}{)}^2}\\ & =& \frac{(2.16\times {\text{10}}^6\text{C})(\text{12.0 J/C})}{(3.00\times {\text{10}}^8\text{m/s}{)}^2}\end{array}$$

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   Using the conversion $1\text{kg}\cdot {\text{m}}^2{\text{/s}}^2=1\text{J}$, we can write the mass as

   $\mathrm{\Delta }m=2.88\times {\text{10}}^{-\text{10}}\text{kg}\text{.}$

Solution for (b)

1. Identify the knowns. $\mathrm{\Delta }m=2.88\times {\text{10}}^{-\text{10}}\text{kg}$; $m=\text{20.0 kg}$
2. Identify the unknown. % change
3. Choose the appropriate equation. $\text{\% increase}=\frac{\mathrm{\Delta }m}{m}\times \text{100\%}$
4. Plug the knowns into the equation.
   $$\begin{array}{lll}\text{\% increase}& =& \frac{\mathrm{\Delta }m}{m}\times \text{100\%}\\ & =& \frac{2.88\times {\text{10}}^{-\text{10}}\text{kg}}{\text{20.0 kg}}\times \text{100\%}\\ & =& 1.44\times {\text{10}}^{-9}\text{\%}\text{.}\end{array}$$

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Discussion

Both the actual increase in mass and the percent increase are very small, since energy is divided by $c^2$, a very large number. We would have to be able to measure the mass of the battery to a precision of a billionth of a percent, or 1 part in ${10}^{11}$, to notice this increase. It is no wonder that the mass variation is not readily observed. In fact, this change in mass is so small that we may question how you could verify it is real. The answer is found in nuclear processes in which the percentage of mass destroyed is large enough to be measured. The mass of the fuel of a nuclear reactor, for example, is measurably smaller when its energy has been used. In that case, stored energy has been released (converted mostly to heat and electricity) and the rest mass has decreased. This is also the case when you use the energy stored in a battery, except that the stored energy is much greater in nuclear processes, making the change in mass measurable in practice as well as in theory.

Comparing Kinetic Energy: Relativistic Energy Versus Classical Kinetic Energy

An electron has a velocity $v=0\text{.}\text{990}c$. (a) Calculate the kinetic energy in MeV of the electron. (b) Compare this with the classical value for kinetic energy at this velocity. (The mass of an electron is $9\text{.}\text{11}\times {\text{10}}^{-\text{31}}\text{kg}$.)

Strategy

The expression for relativistic kinetic energy is always correct, but for (a) it must be used since the velocity is highly relativistic (close to $c$). First, we will calculate the relativistic factor $\gamma$, and then use it to determine the relativistic kinetic energy. For (b), we will calculate the classical kinetic energy (which would be close to the relativistic value if $v$ were less than a few percent of $c$) and see that it is not the same.

Solution for (a)

1. Identify the knowns. $v=0\text{.}\text{990}c$; $m=9\text{.}\text{11}\times {\text{10}}^{-\text{31}}\text{kg}$
2. Identify the unknown. ${\text{KE}}_{\text{rel}}$
3. Choose the appropriate equation. ${\text{KE}}_{\text{rel}}=\left(\gamma -1\right){\mathrm{mc}}^2$
4. Plug the knowns into the equation.

   First calculate $\gamma$. We will carry extra digits because this is an intermediate calculation.

   $$\begin{array}{lll}\gamma & =& \frac{1}{\sqrt{1-\frac{v^2}{c^2}}}\\ & =& \frac{1}{\sqrt{1-\frac{(0\text{.}\text{990}\text{c}{)}^2}{c^2}}}\\ & =& \frac{1}{\sqrt{1-(0\text{.}\text{990}{)}^2}}\\ & =& 7.0888\end{array}$$

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   Next, we use this value to calculate the kinetic energy.

   $$\begin{array}{lll}{\text{KE}}_{\text{rel}}& =& (\gamma -1){\mathrm{mc}}^2\\ & =& (\text{7.0888}-1)(9.11\times {\text{10}}^{–\text{31}}\text{kg})(3.00\times {\text{10}}^8\text{m/s}{)}^2\\ & =& 4.99\times {\text{10}}^{\text{–13}}\text{J}\end{array}$$

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5. Convert units.
   $$\begin{array}{lll}{\text{KE}}_{\text{rel}}& =& (4.99\times {\text{10}}^{\text{–13}}\text{J})\left(\frac{\text{1 MeV}}{1.60\times {\text{10}}^{–\text{13}}\text{J}}\right)\\ & =& \text{3.12 MeV}\end{array}$$

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Solution for (b)

1. List the knowns. $v=0\text{.}\text{990}c$; $m=9\text{.}\text{11}\times {\text{10}}^{-\text{31}}\text{kg}$
2. List the unknown. ${\text{KE}}_{\text{class}}$
3. Choose the appropriate equation. ${\text{KE}}_{\text{class}}=\frac{1}{2}{\mathrm{mv}}^2$
4. Plug the knowns into the equation.
   $$\begin{array}{lll}{\text{KE}}_{\text{class}}& =& \frac{1}{2}{\mathrm{mv}}^2\\ & =& \frac{1}{2}(9.00\times {\text{10}}^{–\text{31}}\text{kg})(0.990{)}^2(3.00\times {\text{10}}^8\text{m/s}{)}^2\\ & =& 4.02\times {\text{10}}^{–\text{14}}\text{J}\end{array}$$

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5. Convert units.
   $$\begin{array}{lll}{\text{KE}}_{\text{class}}& =& 4.02\times {\text{10}}^{–\text{14}}\text{J}\left(\frac{\text{1 MeV}}{1.60\times {\text{10}}^{–\text{13}}\text{J}}\right)\\ & =& \text{0.251 MeV}\end{array}$$

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Discussion

As might be expected, since the velocity is 99.0% of the speed of light, the classical kinetic energy is significantly off from the correct relativistic value. Note also that the classical value is much smaller than the relativistic value. In fact, ${\text{KE}}_{\text{rel}}{\text{/KE}}_{\text{class}}=\text{12}\text{.}4$ here. This is some indication of how difficult it is to get a mass moving close to the speed of light. Much more energy is required than predicted classically. Some people interpret this extra energy as going into increasing the mass of the system, but, as discussed in Relativistic Momentum, this cannot be verified unambiguously. What is certain is that 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 $\text{50}\times {\text{10}}^9\text{eV}=\text{50,000 MeV}$.

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 to any other form, including into entirely new masses. (See Figure 28.23.) 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. Particles are accelerated to extremely relativistic energies and made to collide with other particles, producing totally new species of particles. 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 covered in Particle Physics.

Figure 28.23 The Fermi National Accelerator Laboratory, near Batavia, Illinois, was a subatomic particle collider that accelerated protons and antiprotons to attain energies up to 1 Tev (a trillion electronvolts). The circular ponds near the rings were built to dissipate waste heat. This accelerator was shut down in September 2011. (credit: Fermilab, Reidar Hahn)