Mass and energy (A-level only)

AQA·Physics·3.12.3.5

Mass–energy equivalence

Mass and energy are equivalent: a mass corresponds to an amount of energy. The relationship is

E=mc2E = mc^2

Here, $$E$$ is energy, $$m$$ is the corresponding mass, and $$c$$ is the speed of light in a vacuum, approximately $$3.00 \times 10^8\ \mathrm{m,s^{-1}}$$. Because $$c^2$$ is very large, even a small mass corresponds to a large amount of energy.

Rest energy

A body has energy associated with its mass even when it is stationary. This is its rest energy, $$E_0$$. For a body with rest mass $$m_0$$,

E0=m0c2E_0 = m_0c^2

Rest energy is not energy the body has because it is moving. It is the energy associated with its rest mass.

Changes in mass and energy

A change in a system’s energy corresponds to a change in its mass. For the system as a whole at rest,

ΔE=Δmc2\Delta E = \Delta m c^2

Here, $$\Delta E$$ is the change in energy and $$\Delta m$$ is the corresponding change in mass. If the system loses energy, its mass decreases; if it gains energy, its mass increases. The large value of $$c^2$$ means that an energy change can correspond to a very small mass change.

Relativistic mass and kinetic energy at speed

A moving body has energy associated with its motion as well as energy associated with its rest mass. At speeds close to the speed of light, its total energy rises much more steeply with speed than the familiar low-speed kinetic energy formula predicts.

For a body with rest mass $$m_0$$ moving at speed $$v$$, its total relativistic energy $$E$$ is:

E=m0c21−v2c2=γm0c2,γ=11−v2c2E=\frac{m_0c^2}{\sqrt{1-\frac{v^2}{c^2}}}=\gamma m_0c^2,\qquad \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}

Here, $$c$$ is the speed of light in vacuum, and $$\gamma$$ is the Lorentz factor. At rest, $$v=0$$ and $$\gamma=1$$, so the body’s rest energy is $$m_0c^2$$. As speed increases, $$\gamma$$ increases and so does the total energy.

In the relativistic-mass convention, the mass associated with the moving body is written as $$m=\gamma m_0$$, making its total energy $$E=mc^2$$. The rest mass $$m_0$$ itself does not change with speed.

Kinetic energy at speed

Kinetic energy is the extra energy the body has because it is moving: total energy minus rest energy.

Ek=E−m0c2=(γ−1)m0c2E_k=E-m_0c^2=(\gamma-1)m_0c^2

At rest, $$\gamma=1$$, so $$E_k=0$$. At speeds much smaller than $$c$$, relativistic mass is approximately $$m_0$$ and the kinetic energy is approximately $$E_k=\tfrac12m_0v^2$$.

Relativistic mass and kinetic energy graphs

On a graph against speed, relativistic mass starts at $$m_0$$, while kinetic energy starts at zero. Both curves rise increasingly steeply as $$v$$ approaches $$c$$. The denominator in the Lorentz factor approaches zero, so both relativistic mass and kinetic energy tend to infinity.

A body with nonzero rest mass would therefore require an unlimited amount of energy to reach $$c$$. No finite supply of energy can accelerate it to the speed of light.

Bertozzi’s experiment

Bertozzi’s experiment tested how the speed of an electron changes as its kinetic energy increases. Pulses of electrons were accelerated to different kinetic energies. For each energy, the time taken by a pulse to travel a known distance was measured, giving its speed from $$v=d/t$$.

The electrons also struck a target and heated it. The heating provided a check on the energy delivered by the electrons.

Bertozzi’s experiment: Classical prediction and result

The classical expression relates kinetic energy $$E_k$$ to mass $$m$$ and speed $$v$$:

Ek=12mv2E_k = \frac{1}{2}mv^2

For an electron of fixed mass, this expression predicts that speed will continue to increase without limit as kinetic energy increases. It therefore predicts that an electron given enough energy could exceed the speed of light.

The measurements showed a different result. As kinetic energy increased, electron speed increased but approached the speed of light, $$c$$, rather than increasing without limit.

Bertozzi’s experiment: Relativistic relationship

The relativistic expression for an electron’s kinetic energy is:

Ek=mc2(11−v2/c2−1)E_k = mc^2\left(\frac{1}{\sqrt{1-v^2/c^2}}-1\right)

Here $$m$$ is the electron’s rest mass. As $$v$$ approaches $$c$$, increasingly large amounts of kinetic energy are needed to increase the speed further. The electron’s speed approaches $$c$$ but does not reach it. Bertozzi’s measurements provided direct evidence for this relativistic relationship rather than the classical prediction.

Mass and energy (A-level only) | AQA A-level Physics (7408) Revision Notes | KnowledgeSet