Year 1 Sem 2/LC Classical Mechanics and Relativity 2/Lecture 14

Lecture 14: Special Relativity V: Relativistic Mechanics
Tue 10 Mar 2026 11:00

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Recap of last lecture:

Minkowski Space (space-time)
Time-like and space-like intervals.
Ensuring that this new model does not violate causality.
Simultaneity.

Relativistic Momentum

We generally use:

$$\pmb{p} = m \pmb{v}$$

But this is just an approximation for $v \ll c$. This would:

Allow $v>c$ for an object with sufficient momentum.
Not allow massless objects like photons to have any momentum, but we know this is true.
Is not obviously conserved.

Instead, we (proof non-examinable) use:

$$p = \gamma m v$$

This is the “true” expression for momentum, but for small speeds it collapses nicely to $p = mv$ as $\gamma \to 1$ so we don’t notice in standard mechanics.

This doesn’t resolve the issue with massless photons yet, but does fix the first point!

Relativistic Kinetic and Total Energy

By definition, we have:

$$F = \frac{dp}{dt}$$

Now, momentum has an extra factor of $\gamma$. Suppose we have a massive object which is initially stationary. We apply a force to that object, in order to give it a kinetic energy. This KE is denoted $K_A$ at the point where it reaches speed $v_A$ under the effect of the force. We assume that all work done is entirely transferred to kinetic energy stores.

$$K_A = \int_{0}^{A} F \, dx = \int_{0}^{A} \frac{dp}{dt} \, dx$$

And we can use the chain rule to say:

$$dx = \frac{dx}{dt} dt \implies K_A = \int_{0}^{A} \frac{dp}{dt} \frac{dx}{dt} \, dt = \int_{0}^{A} \frac{dp}{dt} v \, dt \tag{1}$$

We can also say that:

$$\frac{d}{dt} \left(pv\right) = p \frac{dv}{dt} + v \frac{dp}{dt}$$

So substituting this into eqn. 1 we have:

$$K_A = \int_{0}^{A} \left[\frac{d}{dt} (pv) - p \frac{dv}{dt}\right] \, dt$$

$$K_A = \int_{0}^{A} \frac{d}{dt} \left(pv\right) \, dt - \int_{0}^{A} p \frac{dv}{dt} \, dt$$

$$K_A = \left[pv\right]_0^A - \int_{0}^{A} p \, dv$$

$$K_A = \gamma m V_a \cdot V_a - \int_{0}^{v_a} \gamma m v \, dv$$

$$K_A = \gamma mv^2 - m \int_{0}^{v_A} v \left(1 - \frac{v^2}{c^2}\right)^{-1/2} \, dv$$

Note that:

$$\frac{d}{dv} \left(\left[1 - \frac{v^2}{c^2}\right]^{+1/2}\right) = \frac{1}{2} \left(1 - \frac{v^2}{c^2}\right)^{-1/2} \left(- \frac{2v}{c^2}\right) = \frac{-v}{c^2} \left(1 - \frac{v^2}{c^2}\right)^{-1/2}$$

Hence:

$$K_A = \gamma m v_A^2 = m \left[-c^2 \left(1 - \frac{v^2}{c^2}\right)^{+1/2}\right]_0^{v_A}$$

$$K_A = \gamma mv_A^2 + m c^2 \left(\left[1 - \frac{v_A^2}{c^2}\right]^{+1/2} - 1\right)$$

$$K_A = \gamma mv_A^2 + mc^2 \left(\frac{1}{\gamma} - 1\right)$$

And multiplying through by gamma to allow factorisation:

$$K_A = \gamma mv_A^2 + \gamma mc^2 \left(1 - \frac{V_A^2}{c^2}\right) - mc^2$$

$$K_A = \gamma mv_A^2 + \gamma mc^2 - \gamma mv_A^2 - mc^2s$$

$$\boxed{K_A = mc^2 \left(\gamma - 1\right)}$$

This gives the kinetic energy of an object with speed $v_A$ with $\gamma$ as standard. This yields the low energy limit $K = mv^2 / 2$ as $V_A / c \to 0$ as expected. Einstein suggested the breakdown of this into two terms:

$$(\gamma - 1) mc^2 = \gamma \underbrace{mc^2}_\text{total energy} - \underbrace{mc^2}_\text{mass energy}$$

For almost all purposes, and all purposes in this course, we will refer to the total energy and not the overall total energy:

$$E_\text{tot} = E \gamma m c^2$$

This is not Lorentz invariant, but is conserved in decays/collisions etc.

Forces

By definition we have:

$$\vec{F} = \frac{d \vec{p}}{dt} = \frac{d}{dt} \left(\gamma m v\right)$$

This can be performed to show that:

$$F = \gamma^3 ma$$

As $\gamma \to \infty$ as $v \to c$, it gets harder and harder to accelerate a particle as the particle becomes ultra-relativistic.

Lorentz Transformations for Energy and Momentum

There are equivalent transformations for energy and momentum. It turns out than we can use exactly the same Lorentz transformations if we make the following substitutions:

$$\left[ct, x, y, z\right] \to \left[\frac{E}{c}, p_x, p_y, p_z\right]$$

This “grouping three coordinates with another quantity” is broadly called a 4-vector.

Where we previously had:

$$x = \gamma \left(x^\prime + vt^\prime\right) \quad , \quad t = \gamma \left(t^\prime + \frac{v}{c^2}x^\prime\right)$$

We now substitute $x \to p_x$, $ct \to E/c \implies t \to E/c^2$:

$$p_x = \gamma \left(p_x^\prime + v \frac{E^\prime}{c^2}\right)$$

$$\frac{E}{c^2} = \gamma \left(\frac{E^\prime}{c^2} + \frac{v}{c^2} p_x^\prime\right) \implies E = \gamma \left(E^\prime + v p_x\right)$$

And to transform backwards we use the same logic of replacing all the primed variables with unprimed variables and reverse the sign of the speed. We rarely use this in practice. Instead, we want to find an energy-momentum quantity which is Lorentz Invariant.

Energy-Momentum Invariant

Recall that:

$p = \gamma mv$
Instead of kinetic energy we use total energy $E = E_\text{tot} = \gamma mc^2 = \text{KE} - \text{mass energy}$

We want to continue using $(t, x) \to \left(\frac{E}{c^2}, p_x\right)$. Using the space-time invariant, we have:

$$c^2 t^2 - x^2 - y^2 - z^2$$

Is the energy-momentum invariant simply the space-time invariant but with the energy/momentum transformations?

$$E^2 - (p_x^2 - p_y^2 - p_z^2)c^2$$

This is just a postulate, but does turn out to be a Lorentz invariant. To check its true, we can substitute:

$$E^2 - p^2 c^2 = \gamma m^2 c^4 - \gamma^2 m^2 v^2 c^2$$

$$= \gamma^2 m^2 c^2 \left(c^2 - v^2\right) = \frac{m^2 c^2 \left(c^2 - v^2\right)}{1 - \frac{v^2}{c^2}} = \frac{m^2 c^2 \left(c^2 - v^2\right)}{1/c^2 (c^2 - v^2)}$$

Hence:

$$E_\text{tot} - p^2 c^2 = m^2 c^4$$

As the mass is a constant between frames, as is the speed of light, this is invariant between different frames. This gives us the invariant:

$$\boxed{E^2 = m^2 c^4 + \vec{p}^2 c^2}$$

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