18.2.1 Einstein’s Equation (E=mc2)

From the special theory of relativity, Einstein realized that apart from the kinetic energy due to its motion, a particle also has energy of mc² even when it is resting (see Appendix A if you’re interested). This energy came to be known as the rest energy.

From here, Einstein argues that the mass of a system m (when it is at rest) is a manifestation of the total energy of the system. This includes the KE of any moving constituents (e.g. atoms) in the system, the energies of any photons trapped in the system, and any form of potential energy whether positive and negative. In short, if a system has (rest) mass m, then the system has total (rest) energy

$\displaystyle E=m{{c}^{2}}$

This idea came to be known as the mass-energy equivalence principle.

Let’s think of some implications of this principle:

When a torchlight is switched on, it is emitting photons and thus losing energy continuously. Shouldn’t its mass be decreasing continuously then?
Hot water has higher internal energy than cold water (mainly due to the higher KE of the vibrating H2O molecules). Shouldn’t hot water weigh more than cold water then?
Shouldn’t the mass of an ice cube be lower than the mass of the water that formed it since the H2O molecules in the ice cube has lower PE?
To ionize a hydrogen atom, energy must be supplied. Does this not mean that the mass of a proton and an electron is higher than the mass of a hydrogen atom?

Based on Einstein’s equation, the change in mass can be calculated by $\displaystyle \Delta m=\frac{{\Delta E}}{{{{c}^{2}}}}$ . So let’s crunch the numbers for our four examples.

As you can see, the predicted (percentage) change in mass is too small to be measured for ordinary energy transactions. So where can we possibly find any evidence to support the mass-energy equivalence principle? The answer lies in the atomic nucleus.

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Concept Test

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	System	Energy gained/lost	Change in mass $\displaystyle \Delta m=\frac{{\Delta E}}{{{{c}^{2}}}}$
1	Shining a 1 W torchlight for 1 min	$\displaystyle 1\times 60=60\text{ J}$	$\displaystyle \frac{{60}}{{{{{(3.00\times {{{10}}^{8}})}}^{2}}}}=6.67\times {{10}^{{-16}}}\text{ kg}$
2	Heating up 1 kg of water by 1 °C	4200 J	$\displaystyle \frac{{4200}}{{{{{(3.00\times {{{10}}^{8}})}}^{2}}}}=4.67\times {{10}^{{-14}}}\text{ kg}$
3	Freezing 1 kg of water	33600 J	$\displaystyle \frac{{33600}}{{{{{(3.00\times {{{10}}^{8}})}}^{2}}}}=3.73\times {{10}^{{-13}}}\text{ kg}$
4	Ionizing 1 mole of hydrogen atoms	$\displaystyle {{N}_{A}}\times 13.6\text{ eV}$	$\displaystyle =\frac{{(6.02\times {{{10}}^{{23}}})(13.6\times 1.60\times {{{10}}^{{-19}}})}}{{{{{(3.00\times {{{10}}^{8}})}}^{2}}}}=1.46\times {{10}^{{-11}}}\text{ kg}$

xmPhysics

18.2.1 Einstein’s Equation (E=mc2)

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