In this final lecture we will discuss one more relativistic correction - to energy. As we will see, while this correction is not quite as straightforward as the one for momentum, it has hugely important consequences, leading us to discover the equivalence between mass and energy.

Non-Relativistic Kinetic Energy

In classical, non-relativistic Newtonian mechanics, the kinetic energy of an object is defined as \[E_K = \frac{1}{2} mv^2,\] where \(m\) is the object’s mass and \(v\) is it’s velocity. As with momentum, this is the definition that we choose for kinetic energy precisely because it is this quantity which is conserved (providing we take into account all other types of energy in the system). Since \(v\) is frame-dependent, we expect \(E_K\) to also be frame dependent.

Relativistic Kinetic Energy

Based on the pattern we have seen so far, we might hope that relativistic energy might be as simple as \(E_K=(1/2) \gamma m_0v^2\). However, this is not the case. Instead, it is something very strange looking indeed. \[\boxed{E_K = (\gamma - 1) m_0c^2}.\] Firstly, we need not worry that there is no \(v\) in there, for this is hidden inside \(\gamma\).

Before we work out where this equation comes from, let’s check our requirement that this reduces to the familiar form for \(v \ll c\). It is not a simple as saying that \(\gamma = 1\) because then we have \(E_K = 0\) which is what we want for \(v=0\), but doesn’t tell us anything about small but non-zero values of \(v\).

To get there, we instead make use of the binomial expansion. For small values of \(x\): \[(1-x)^{-1/2} = \frac{1}{\sqrt{1-x}} \approx 1 + x/2.\]

If we expand out \(\gamma\) in the equation for relativistic kinetic energy, we have \[E_K = \bigg(\frac{1}{\sqrt{1-v^2/c^2}} -1\bigg) m_0c^2.\]

Then, using the binomial expansion, we obtain \[\begin{aligned} E_k &= (1 + v^2/2c^2- 1)m_0c^2 \\ E_k &=(1/2)m_0v^2, \end{aligned}\]

which is exactly the non-relativistic expression for kinetic energy. So once again, it turns out that our familiar definition is really an approximation which is good only for \(v \ll c\).

Derivation of Relativistic Kinetic Energy (Not Examinable)

To derive the equation for relativistic energy we need to recall that the change in kinetic energy of an object is given by the integral of the force applied over the distance it is applied through, i.e. \[E_K = \int F ds,\] where \(F\) is the force and \(s\) is the distance. Secondly, we recall that force is equal to the rate of change in momentum: \[F = \frac{dp}{dt}.\] Therefore, if we consider an object that has accelerated from rest to a velocity \(v\), the kinetic energy is given by: \[E_K = \int^{v=v}_{v=0} \frac{dp}{dt} ds.\] Now, \(ds/dt\) is simply velocity, \(v\), so \[E_K = \int^{v=v}_{v=0} v~dp,\] and relativistic momentum is given by \(p=\gamma m_0 u\), so that: \[E_K = \int^{v=v}_{v=0} vd\bigg(\frac{m_0v}{\sqrt{1-v^2/c^2}}\bigg) = \int^{v=v}_{v=0} v~d\bigg[(m_0v)\bigg(1-\frac{v^2}{c^2} \bigg)^{-1/2}\bigg],\] where we have expanded out \(\gamma\).

We want to get this integral in terms of \(dv\) so that we can integrate and apply the limits. To do this, we can use the chain rule to show that \[\begin{aligned} \frac{d}{dv} \bigg[(m_0v)\bigg(1-\frac{v^2}{c^2} \bigg)^{-1/2}\bigg] &= m_0\bigg(1-\frac{v^2}{c^2}\bigg)^{-1/2} + m_0\frac{v^2}{c^2}\bigg(1-\frac{v^2}{c^2}\bigg)^{-3/2}. \end{aligned}\] If we divide the first term top and bottom by \((1-v^2/c^2)\) to put everything over a common denominator we get: \[d\bigg[(m_0v)\bigg(1-\frac{v^2}{c^2} \bigg)^{-1/2}\bigg] = m_0\bigg(1-\frac{v^2}{c^2}\bigg)^{-3/2}~dv.\] Substitute this back into the equation for \(E_K\) and we get \[E_K = \int m_0 \bigg(1- \frac{v^2}{c^2}\bigg)^{-3/2}v~dv.\] This integral can be solved using the substitution \(w=\sqrt{1 / (1-v^2/c^2)}\), to give \[E_K = \frac{m_0c^2}{\sqrt{1-v^2/c^2}} - m_0c^2,\] and so \[\boxed{E_K = (\gamma - 1)m_0c^2}\] which is the expression for relativistic kinetic energy.

Total Energy

If we expand out our newly discovered expression for kinetic energy \[E_K = \gamma mc^2 - mc^2,\] we notice that it consists of one quantity subtracted from another. We might wonder what these two terms represent physically. As we will show below, it turns out that this equation really means: \[\textrm{kinetic energy = total energy - rest energy}\] and so, \[\boxed{E_{total} = \gamma mc^2}\]

Rest Energy and Mass-Energy Equivalence

In the rest frame of the object, where \(v=0\), then \(\gamma = 1\) and so \[\boxed{E=mc^2}\] This shows that a non-moving object, i.e. an object viewed from its rest frame, still has some energy. We call this the rest energy.

We see that this rest energy depends only on the mass, \(m\), and a constant, \(c\). Since mass and energy are linked by a fixed constant, this demonstrates an equivalence between the two. In particle accelerators we can see this in action, as kinetic energy is converted to rest energy, and hence rest mass. So if we smash two particles together with a certain total rest mass, \(m_0\), it doesn’t follow that the rest mass of a new particle created must be equal to \(m_0\).

Energy and Mass Units

The SI Unit for energy is the Joule, J. However, the kinds of objects we most often observe moving at speeds which are an appreciable fraction of \(c\) (and hence where the effects of special relativity become noticeable) tend to be particles (for example in a particle accelerator). In this case, the Joule is not a sensible unit to use. Instead we use electron-volts (eV). \[1~\mathrm{eV} = 1.6 \times 10^{-19}~\mathrm{J}\] (This comes from the energy gained by an electron moving across a voltage of 1 V, but that isn’t important here.)

Since we know that rest mass is equivalent to rest energy, linked by the constant \(c^2\), \(E_0=m_0c^2\), we can also specify the rest mass in eV type units.
A particle with a rest energy of \(1~\mathrm{eV}\) would have a rest mass of \(1~\mathrm{eV/c^2}\).
In practice we will actually find ourselves working with larger units of energy and mass, in which case we will talk about \(\mathrm{MeV}\) and \(\mathrm{MeV/c^2}\), or \(\mathrm{GeV}\) and \(\mathrm{GeV/c^2}\)
We can also use these kinds of units for momentum. Recall that \(p=\gamma mv\). If we specify mass in \(\rm MeV/c^2\) and velocity as a fraction of \(c\), then we can specify our momentum in units of \(\rm MeV/c\).
Using these kinds of units makes dealing with mass/energy/momentum problems in particle physics and relativity much simpler (since the \(c\)s also cancel out) and you should be very comfortable with them going into the exam.

Example Energy and Momentum Calculations

Demonstration that Massless Particles Travel at the Speed of Light

It is sometimes useful to write energy in terms of momentum. In classical mechanics, since \(E_K=(1/2)mv^2\) and \(p=mv\), we have \[E = \frac{p^2}{2m}\] The relativistic equivalent is: \[E^2 = p^2c^2 + m_0^2c^4\] We can see this by recognising that \(p=\gamma m_0v\) and so \(p^2 = \gamma^2m_0^2v^2.\) So we can write \[c^2p^2 = \gamma^2m_0^2c^4\frac{u^2}{c^2}.\] From \(\gamma^2 = 1/(1-u^2/c^2)\), a bit of algebra shows us that \[c^2p^2 = \gamma^2 m_0^2c^4 - m_0^2c^4.\] Since \(E=\gamma m_0c^2\), the first term is simply \(E^2\), giving us \[\boxed{E^2 = p^2c^2 + m_0^2c^4}\]

This is an important equation which relates the total energy, rest energy and momentum of a particle (without explicitly including the velocity). When the energy is much greater than the rest energy, it becomes \(E\approx pc\).

Continuing, from \(E=\gamma m_0 c^2\) and p = \(\gamma m_0v\) \[p = \frac{Ev}{c^2}\] Now, if we consider a massless particle, so that \(m_0=0\), then \(E^2=p^2c^2\) and we have the exact equation \[E=pc.\] Substituting in \(p=Ev/c^2\) gives us \[E = \frac{Evc}{c^2}\] which immediately tells us that \[v = c.\] This shows that massless particles must have a velocity equal to \(c\), the speed of light!

The more observant might notice that, if \(m_0 = 0\), then we have a division by zero when we calculated \(p=Ev/c^2\) (and if \(v=c\) then \(\gamma\) is undefined!). So this ‘derivation’, as we have presented it, is not entirely convincing. However, it turns out that \(p=Ev/c^2\) is true when \(m_0=0\) and \(v=c\), in which case the only possible reconciliation with \(E=pc\) is that \(v=c\). More on this will have to wait until you learn about the energy-momentum 4-vector!