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Year 1 Sem 2/LC Classical Mechanics and Relativity 2/Lecture 12

From Ash's Lecture Notes

Lecture 12: Special Relativity IV: More Space-Time
Tue 03 Mar 2026 11:00

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Recap from last lecture:

  • Lorentz Transformations for space and time.
  • Lorentz Transformations for velocity/velocity addition.
  • A Lorentz Invariant: The space-time interval is invariant across frames of reference:
    $$\delta S^2 = c^2 \Delta t^2 - \Delta r^2$$

In this lecture:

  • Minkowski space and causality.
  • Relativistic energy and momentum.

Space-Time Invariance

Last lecture we showed that:

$$c^2 \Delta t^2 - \Delta r^2 = c^2 \Delta t^{\prime 2} - \Delta r^{\prime 2} = \quad (= 0 \text{ for light})$$

And we define the space-time interval $\Delta s$ as $\Delta S^2 = c^2 \Delta t^2 - r^2$. There is some relationship between space and time that keeps this quantity constant across frames. What does this quantity actually mean, however?

In a 3D set of axes, we have:

$$\Delta r^2 = \Delta x^2 + \Delta y^2 + \Delta z^2$$

And we have defined:

$$\Delta s^2 = c^2 \Delta t^2 - \Delta r^2 = c^2 \Delta t^2 - \left(\Delta x^2 + \Delta y^2 + \Delta z^2\right)$$

This looks almost like a 4D space (where one axis is time), but with some strange negative signs appearing? We don’t have the equipment yet to properly understand the negative signs, but can be in future courses.

Minkowski Space (Space-Time Diagrams)

In 2D, we consider one spatial coordinate and one time coordinate, where the $x$ axis represents our spatial coordinate and the vertical axis is given by $ct$.

Consider some kind of light-emitting event happening at the origin. This light travels at a fixed speed, so we end up with the path of light at 45 degrees on this space-time diagram:

These are given by the lines $c^2 t^2 = \pm x^2$ and are called “light cones”, as they form cones in higher dimensions. The positive velocity axis is called the “future cone”, the path taken by light emerging from event A. The negative velocity axis is called the “past cone”, and represents the path taken by light emerging at A.

Light cones in a 3D space (two spatial, one velocity)

Inside the future cone, events may be caused by the event happening at A. Things travelling at a velocity $<c$ can reach them.

The region inside the past cone gives regions in spacetime where events could have caused A.

Events outside the two cones cannot influence, nor can they be influenced by A. They cannot be causally connected in any way to the event at A.

We now have two events. A is defined as before, and a new event B separated from A by distance $\Delta \underline{r}$ and time $\Delta t$.

What is the condition that B lies within the light cone given by A? We require:

$$| \Delta \underline{r} | \leq c | \Delta t |$$

Or in invariant form:

$$\Delta S^2 = c^2 \Delta t^2 - \Delta r^2 \geq 0$$

We call $\Delta S$ the space-time interval, or invariant distance.

There are three possible cases:

  • $\Delta S^2 > 0$, “time-like intervals”. This is where $| \Delta \underline{r} | \leq c | \Delta t |$ and B is inside A-s light cone. $\Delta S \in \mathbb{R}$. Here, a signal can pass between A and B and the two events may be causally connected (where A causes B).
  • $\Delta S^2 = 0$. Here, $| \Delta \underline{r} | = c | \Delta t |$. This is where B sits on A’s light cone, and A/B may be causally connected. A signal can only travel between the two events if the signal is at velocity $c$. This is called a “light-like interval”
  • $\Delta S^2 < 0$, “space-like intervals”. This is where $| \Delta \underline{r} | > c | \Delta t |$. $\Delta S = \sqrt{c^2 \Delta t^2 - \Delta r^2} \in i\mathbb{R}$. No signal can pass between A and B, not even light. No causal connection is possible.

Causality and Time Order

It is possible for observers in two different frames to disagree on the order in which events have taken place. If we have two events with $t_1, t_2$ in $\Sigma$ and $t_1^\prime$, $t_2^\prime$ in $\Sigma^\prime$, there are possible cases where:

  • $t_2 > t_1$.
  • $t_1^\prime > t_2^\prime$.

This may seem uncomfortable - does this cause a breakdown in causality?

Suppose we have a frame $\Sigma$ with two events, $(x_1, t_1)$, $(x_2, t_2)$, such that $\Delta t = t_2 - t_1 > 0$.

Frame $\Sigma^\prime$ moves with speed $v$ relative to $\Sigma$ and in $\Sigma^\prime$ the same two events have coordinates $(x_1^\prime, t_1^\prime)$ and $(x_2^\prime, t_2^\prime)$. For the observers to disagree on causality, we have $\Delta t^\prime t_2^\prime - t_1^\prime < 0$. We begin by applying the Lorentz Transformation to $\Delta t^\prime$:

$$\Delta t^\prime = \gamma \left(t_2 - \frac{v}{c^2} x_2\right) - \gamma \left(t_1 - \frac{v}{c^2} x_1\right) < 0$$
$$\implies \gamma(t_2 - t_1) + \frac{\gamma v}{c^2} \left(-x_2 + x_1\right) <0$$
$$\Delta t - \frac{v}{c^2} \Delta x < 0$$

I.e. $\Delta t < \frac{v}{c^2} \Delta x$ or $c \Delta t < \frac{v}{c} \Delta x$. We know that $v/c < 1$, hence $c \Delta t < \Delta x$. This means that:

$$\Delta S = \sqrt{c^2 \Delta t^2 - \Delta x^2}$$

Which is the square root of a negative number, hence this gives a space-like interval. This means that while we can define two points in space-time where two observers will disagree about their order, the space-time invariant is imaginary hence these events cannot be causally linked. This means that observers disagreeing on event order does not break causality.

Simultaneity

Consider a relativistic ladder passing through a stationary garage. The ladder’s proper length is longer than the proper length of the garage, so the ladder could not fit inside the garage if both were stationary.

However, viewed from the rest frame of the garage, the ladder is length contracted. From this frame, the ladder can momentarily (before smashing into the back door) fit entirely in the garage and it would even be possible to briefly close the front door.

From the ladder’s rest frame however, the garage is length contracted and not the ladder. Therefore, from this frame, there’s no possible way to fit the ladder into the garage.

How can it simultaneously be possible and impossible depending on the rest frame being considered?

Problem

Let the ladder have proper length $l_0^L = 20$m and the garage have proper length $l_0^G = 10$m. The ladder moves at $0.9$c wrt the garage.

$$\gamma = \frac{1}{\sqrt{1 - \beta^2}} = \frac{1}{\sqrt{1-0.9^2}} \approx 2.29$$

In the garage’s rest frame, the ladder length is $L^L = \frac{20}{2.29} = 8.7$m. So, from the garage’s POV, the ladder fits!

We now consider the length of the garage from the ladder’s rest frame: $L^G = \frac{10}{2.29} = 4.4$m, so from the ladder’s POV, the ladder cannot fit! This creates a paradox.

Solution

We can resolve this problem because the concept of simultaneity is frame dependant. There are two things that happened here:

  • A. The front of the ladder reached the back garage door (the first figure)
  • B. The back of the ladder reaches the front garage door.

We’ve been assuming that these events occur at exactly the same time, simultaneously, but this isn’t the case.

The time taken between these events has to obey a Lorentz transformation. If they are the same in some frame (i.e. the garage, for the ladder to fit perfectly inside), then $\Delta t_{AB} = 0$. In the rest frame of the ladder:

$$\Delta t_{AB}^\prime = \gamma \left(\Delta t_{AB} - \frac{v}{c^2} \Delta x_{AB}\right) = \frac{\gamma v}{c^2} \Delta x_{AB} \not = 0$$

Hence, if the garage perceives the events as being simultaneous, they cannot be observed as being simultaneous in the ladder’s frame.

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