Relativity

Spacetime Diagrams

Start

Events, worldlines, and a geometric interpretation of space & time.

Lorentz Transformations

Derive the coordinate transformations that preserve the speed of light for every inertial observer.

Relativistic Momentum and Energy

Here

See how momentum and energy transform near the speed of light.

General Relativity

Finish

Connect accelerating frames to gravity through the equivalence principle, and move beyond flat spacetime.

Momentum and Energy

Spacetime diagrams show the geometry of special relativity, and the Lorentz transformations turn that geometry into equations. The next question is dynamical: once an object is moving close to the speed of light, how should we define its momentum and energy?

Classical Definitions

At everyday speeds, the familiar formulas

p = mv, \qquad K = \frac{1}{2}mv^2

work extremely well. Near the speed of light, they do not. These do not build in the fact that no material object can be pushed past $c$ , and they underestimate how rapidly the energy cost of speeding up grows as $|v|$ approaches $c$ .

Relativity fixes that by changing the momentum and energy formulas while keeping the conservation laws themselves intact.

Momentum

The cleanest way to write the relativistic formulas is to introduce the dimensionless speed

\beta = \frac{v}{c}

and the Lorentz factor

\gamma = \frac{1}{\sqrt{1-\beta^2}}.

With these, the expression for relativistic momentum is

p = \gamma mv

Why does momentum pick up a gamma factor?

In classical mechanics, momentum is mass times velocity measured per unit coordinate time:

p_{\text{classical}} = m\frac{dx}{dt}.

In relativity, different observers do not agree on the same coordinate-time interval $dt$ , so it is more natural to build the formula from the particle’s proper time $d\tau$ . Time dilation give the relationship between these two time intervals:

d\tau = \frac{dt}{\gamma} \qquad \Longrightarrow \qquad \frac{dt}{d\tau} = \gamma.

A clean relativistic generalization is then

p = m\frac{dx}{d\tau}.

Rewriting that in terms of the ordinary velocity $v = dx/dt$ :

p = m\frac{dx}{dt}\frac{dt}{d\tau} = mv\gamma.

This also passes the two checks we want. At low speed, $\gamma \approx 1$ , so

p \approx mv.

And as $|v| \to c$ , the factor $\gamma$ grows without bound, so (when applying a force) the momentum keeps increasing instead of allowing a massive particle to reach or exceed the speed of light.

Energy

The total relativistic energy is

E = \gamma mc^2.

When the particle is at rest, $\gamma = 1$ , so the energy does not vanish. Instead it reduces to the rest energy

E_0 = mc^2.

The kinetic part is whatever remains after subtracting that rest-energy baseline:

K = E - E_0 = (\gamma - 1)mc^2.

This makes an important distinction clear: rest energy is present even at $v = 0$ , while kinetic energy is the extra energy associated with motion.

How do we derive the relativistic energy formula?

In one dimension, the incremental work done on a particle is

dK = F\,dx.

Since $F = dp/dt$ and $v = dx/dt$ ,

dK = \frac{dp}{dt}\,dx = v\,dp.

Now substitute the relativistic momentum $p = \gamma mv$ :

dp = m\,d(\gamma v).

Differentiate $\gamma v$ with respect to $v$ :

\frac{d}{dv}(\gamma v) = \gamma + v\frac{d\gamma}{dv} = \gamma + \gamma^3\frac{v^2}{c^2} = \gamma^3.

Therefore

dp = m\gamma^3\,dv, \qquad dK = mv\gamma^3\,dv.

We can show that

\frac{d\gamma}{dv} = \gamma^3\frac{v}{c^2},

so the last expression becomes

dK = mc^2\,d\gamma.

Integrating from rest ( $\gamma = 1$ ) to speed $v$ gives

K = mc^2 \int_1^\gamma d\gamma = (\gamma - 1)mc^2.

We interpret this kinetic energy as some total energy $\gamma m c^2$ minus a baseline rest energy $mc^2$ .

Energy-momentum Relation

The momentum and energy formulas are tied together. Starting from

E = \gamma mc^2, \qquad p = \gamma mv,

we can eliminate $\gamma$ by computing

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

Factor out $\gamma^2 m^2 c^4$ :

E^2 - p^2c^2 = \gamma^2 m^2 c^4 \left(1 - \frac{v^2}{c^2}\right) = \gamma^2 m^2 c^4 (1 - \beta^2).

Since $\gamma^2 = 1/(1-\beta^2)$ , the final factor cancels:

E^2 - p^2c^2 = m^2 c^4.

This invariant relation is often written as

E^2 = p^2c^2 + m^2 c^4.

Not only is this true for massive particles, but it happens to work for massless particles as well. For example, the energy of a photon can be written as $E = pc$ . This immediately begs the question: how do we define the momentum carried by a photon? Classical electromagnetic waves carry energy and momentum, but the answer ultimately requires quantum mechanics if you want a massless particle theory of light.

Problem Solving

Compute Momentum and Energy from Beta

Problem

A particle moves with $\beta = 0.60$ . Find $\gamma$ , $p/(mc)$ , $E/(mc^2)$ , and $K/(mc^2)$ .

Given

$\beta = 0.60$

Use

\gamma = \frac{1}{\sqrt{1-\beta^2}}, \qquad \frac{p}{mc} = \gamma\beta, \qquad \frac{E}{mc^2} = \gamma, \qquad \frac{K}{mc^2} = \gamma - 1.

Solve

\gamma = \frac{1}{\sqrt{1-0.60^2}} = \frac{1}{\sqrt{0.64}} = 1.25.

Then

\frac{p}{mc} = (1.25)(0.60) = 0.75,

\frac{E}{mc^2} = 1.25, \qquad \frac{K}{mc^2} = 1.25 - 1 = 0.25.

Check

The total energy stays above $1$ because the rest-energy piece is always present, while the kinetic part is only the extra amount above that baseline.

Recover Speed from Total Energy

Problem

A particle has total energy $E = 2mc^2$ . Find its Lorentz factor $\gamma$ , its speed parameter $\beta$ , and its kinetic energy $K$ .

Given

$E = 2mc^2$

Use

Since

E = \gamma mc^2,

we have $\gamma = E/(mc^2)$ . Then

\beta = \sqrt{1 - \frac{1}{\gamma^2}}, \qquad K = (\gamma - 1)mc^2.

Solve

\gamma = 2.

\beta = \sqrt{1 - \frac{1}{2^2}} = \sqrt{\frac{3}{4}} \approx 0.866.

The kinetic energy is

K = (2 - 1)mc^2 = mc^2.

Check

Half of the total energy is the rest-energy baseline and the other half is kinetic energy. A speed of about $0.87c$ is consistent with a particle that is energetic but still below the speed of light.

Relativistic Dynamics Checkpoint

Question 1 of 3

A particle changes from $\beta = +0.60$ to $\beta = -0.60$. Which statement is correct?

Choose an answer to get instant feedback.