Year 1 Sem 2/LC Classical Mechanics and Relativity 2/Lecture 17

Lecture 17: Special Relativity VII: Kinematics Examples, Doppler and Natural Units
Tue 24 Mar 2026 11:00

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Example I

As viewed in a lab frame, a cosmic ray proton of energy $E_p$ hits a stationary proton. What is the minimum threshold energy of $E_p$ that allows the reaction:

$$p + p \to p + p + \pi$$

To take place, given $m_p = 938 \mathrm{MeV / c^2}$ and $m_\pi = 104 \mathrm{MeV / c^2}$?

Let the cosmic ray proton be proton one with $E_p, p_p$ and the stationary proton be proton 2 with $m_p$.

We can consider energy and momentum in the lab frame, initially (before the collision):

$$\left(\Sigma E\right)_\text{initial}^\text{lab} = E_p + M_p c^2 \qquad \text{and} \qquad \left(\Sigma p\right)_\text{initial}^\text{lab} = p_p \tag{1}$$

For a threshold energy, the produced particles have no relative motion (i.e. at rest in the centre of mass frame). This means that all available relativistic energy goes into particle production with no excess.

Hence in the centre of mass frame finally, we have two stationary protons and one stationary pion. This is the lowest energy state we can have in the centre of mass frame, with all these particles still existing.

$$\left(\Sigma E\right)^\text{CMS}_\text{final} = (2 m_p + m_\pi) c^2 \qquad \text{and} \qquad \left(\Sigma p\right)^\text{CMS}_\text{final} = 0 \tag{2}$$

We can then apply the energy momentum invariant:

$$s = \left[\left(\Sigma E\right)^2 - \left(\Sigma p\right)^2\right]_\text{final}^\text{lab} = \left[\left(\Sigma E\right)^2 - \left(\Sigma p\right)^2\right]_\text{CMS}^\text{final} \tag{3}$$

Inserting 1, 2 into 3:

$$\left(E_p + m_p c^2\right)^2 - p_p^2c^2 = \left(2 m_p + m_\pi\right)^2 c^4$$

$$\implies E_p^2 + m_p^2 c^4 + m_p c^2 E_p - p_p^2 c^2 = \left(2 m_p + m_\pi\right)^2 c^4$$

For a single particle, $E_p^2 - p_p^2 c^2 = m_p^2 c^4$, hence:

$$2 m_p^2 c^4 + 2 m_p c^2 E_p = \left(2 m_p + m_\pi\right)^2 c^4$$

$$E_p = \frac{\left(2 m_p + m_\pi\right)^2 c^4 - 2 m_p^2 c^4}{2 m_p c^2}$$

And we have known masses $m_p, m_\pi$ hence (noting that the masses are in units energy per speed of light squared, so the $c$s cancel out):

$$E_p = \frac{\left(2 \times 938 + 104\right)^2-2\times 938^2}{2 \times 938} = 1228 \mathrm{MeV}$$

Example II

In an LHC experiment, a $B_0$ meson ($b \bar{d}$) with mass $5 \mathrm{GeV / c^2}$ is produced in a lab frame momentum of $50 \mathrm{GeV / c^2}$. The meson has proper decay time $t_0 = 2 \times 10^{-12}$s. How far does the experiment observe the meson fly before decay?

Let the range of the $B_0$ in the lab frame be:

$$d_\text{lab} = v_\text{lab} t_\text{lab}$$

There is however a time dilation between the lab frame and the rest frame, hence:

$$t_\text{lab} = \gamma t_0$$

$$\implies d_\text{lab} = \gamma v_\text{lab} t_0$$

To find $\gamma v_\text{lab}$ note that $p_\text{lab} = \gamma m v_\text{lab} \implies \gamma v_\text{lab} = \frac{p_\text{lab}}{m}$. Hence:

$$d_\text{lab} = \frac{p_\text{lab} t_0}{m}$$

Relativistic Longitudinal Doppler Effect

We want to build an analogue for the sonic doppler effect for light. This is especially important in astronomy, i.e. tracking the movement of cosmic spectral lines to determine speeds of distant sources.

We will only considerer the longitudinal effect, i.e. the movement of bodies directly towards or away from an observer in their line of sight. We will not (yet) consider transverse effects in this course.

This comes from two components:

The standard doppler effect (as for sonic waves).
Relativistic time dilation.

The derivation isn’t examinable, but the result for a receding source (redshift) is:

$$f = f_0 \sqrt{\frac{c - v}{c + v}}$$

And for an approaching source:

$$f = f_0 \sqrt{\frac{c + v}{c -v}}$$

Problem Solving

The energy-momentum invariant is extremely useful. A lot of problems are solved by equating $E^2 - p^2c^2$ before and after a reaction or a collision.
If a problem requires finding $\gamma$ or $v \gamma$ it’s often useful to use $p = \gamma mv$ or $E = \gamma m v^2$ so solve for them instead of manually faffing with $u$ and $c$.
Given a rest frame lifetime $t_0$, the range is given by $v \gamma t_0$ or $p t_0 / m$.

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