Invariant: \((\Delta s)^2 = (c\Delta t_0)^2 = (2L)^2\) — different observers measure different Δt and Δx, but they all compute the same value for (Δs)². It is the spacetime interval that is invariant, not the time itself.
Frame-dependent: \(c\Delta t\) and \(\Delta x\) individually.
Note: the data booklet writes the invariant as \((\Delta s)^2 = (c\Delta t)^2 - (\Delta x)^2\). Here \((\Delta s)^2 = (c\Delta t_0)^2 = (2L)^2\). So Δs = 2L = cΔt₀ is the spacetime interval — the invariant "length" of the event separation.
Slide 02 / 06
Theme E: Nuclear and quantum physics
E.2 – Quantum physics · Relativistic momentum
The problem — classical momentum is not conserved in all frames
•In frame S: two identical balls collide symmetrically. Classical momentum \(p=mv\) is conserved by symmetry.
•Switch to frame S′. Transform the speeds using the relativistic velocity addition formula and check whether \(p=mv\) is still conserved…
The speeds in S′ are found using the relativistic velocity addition formula (data booklet A.5):
\(\displaystyle u' = \frac{u - v}{1 - \dfrac{uv}{c^2}}\)
where v is the speed of S′ relative to S.
In frame S′, classical \(p = mv\) is not conserved, even though it was in S. We need a new definition that works in all frames.
•We require momentum conservation in all inertial frames — the same principle that requires the speed of light to be the same in all frames.
•This forces a new definition:
\(p = \gamma(v)\cdot m_0 \cdot v\)
relativistic momentum
where \(\displaystyle\gamma(v) = \frac{1}{\sqrt{1-v^2/c^2}}\) γ(v) means γ evaluated at that particle's own speed — in frame S, v = u; in frame S′, v = u′. Not multiplied.
At low speeds: γ → 1, so p → m₀v ✓ At v → c: γ → ∞, momentum grows without bound ✓
The γ factor and the Lorentz transformation come from the same spacetime geometry — built to match each other.
Slide 03 / 06
Theme E: Nuclear and quantum physics
E.2 – Quantum physics · Relativistic kinetic energy
Finding kinetic energy — same method as classical mechanics
•Kinetic energy = work done to accelerate a particle from rest:
\(\displaystyle E_k = \int F\;\mathrm{d}s \qquad\text{where}\quad F = \frac{dp}{dt},\quad p = \gamma m_0 v\)
The integration uses standard calculus techniques — ask your maths teacher. The result is what matters.
•The result of that integral is:
\(E_k = c\sqrt{m_0^2c^2 + p^2} - m_0c^2\)
kinetic energy
•When v = 0, p = 0 and Ek = 0 ✓ As v → c, the first term grows without bound ✓
The natural structure of the result
•The result has the form: (something) − (a constant)
E.2 – Quantum physics · Two special cases of \(E^2-(pc)^2=(m_0c^2)^2\)
Case 1 — massive particle at rest (p = 0)
•Set v = 0, so p = 0:
\(E^2 - 0 = (m_0c^2)^2 \quad\Rightarrow\quad E = m_0c^2\)
A stationary particle carries energy m₀c² just by having rest mass. This falls out of the triangle when the base collapses to zero — height and hypotenuse become the same line.
It's real energy: nuclear reactions convert rest energy into kinetic energy and radiation.
Case 2 — photon (m₀ = 0)
•Set m₀ = 0:
\(E^2 - (pc)^2 = 0 \quad\Rightarrow\quad E = pc \quad\Rightarrow\quad p = \dfrac{E}{c}\)
•Now use E = hf and c = fλ:
\(\displaystyle p = \frac{E}{c} = \frac{hf}{c} = \frac{h}{\lambda}\)
photon momentum
The result \(p = E/c\) follows directly from m₀ = 0 in the invariant.
Slide 06 / 06
Theme E: Nuclear and quantum physics
E.2 – Quantum physics · The de Broglie hypothesis
The de Broglie hypothesis
•The photon result \(\lambda = h/p\) connects a wave property (λ) to a particle property (p).
•de Broglie asked (1924): if waves have momentum, why shouldn't matter have a wavelength?
\(\lambda = \dfrac{h}{p}\)
de Broglie hypothesis
•"Any particle having a momentum p will have a wave associated with it having a wavelength h/p."
Written in terms of p — not mv — because p = γm₀v is the relativistically correct momentum. The equation works for slow particles and for photons alike.
In this course (IB DP): you will not be asked to use p = γm₀v. For the particles encountered, speeds are low enough that p ≈ m₀v, so \(\lambda = h/(mv)\) is acceptable. (For electrons accelerated through large voltages, speeds can reach a few percent of c — worth keeping in mind.)
•Confirmed by Davisson–Germer electron diffraction (1927).
The complete argument
1
Require momentum conservation in all frames → p = γm₀v
2
Work-energy integral with relativistic p → \(E_k = c\sqrt{m_0^2c^2+p^2}-m_0c^2\)
3
Define total energy \(E \equiv c\sqrt{m_0^2c^2+p^2}\)
4
Same Lorentz algebra as light clock → E² − (pc)² = (m₀c²)² is invariant
5
Set m₀ = 0 for photon → E = pc
6
E = hf and c = fλ → λ = h/p for photon
7
de Broglie: extend to all particles → λ = h/p universally
\(\lambda = \dfrac{h}{p} \approx \dfrac{h}{m_0 v}\quad\text{(for this course)}\)