Spacetime Interval
The spacetime interval measures the separation between two events in spacetime. $$ I = (ct)^2 - x^2 - y^2 - z^2 $$ It’s also known as the Minkowski interval.
It isn’t a “distance” in the usual Euclidean sense. Instead, it’s a quantity that remains unchanged when moving from one inertial reference frame to another. In other words, it’s the same for every inertial observer.
For this reason, it’s often called the spacetime invariant. It allows us to quantify the separation between two events A and B in spacetime.
The sign of (I) tells us the nature of the separation between the two events:
- If (I > 0), the separation is time-dominated (timelike)
The two events can be connected by an object moving at less than the speed of light (c). This means a causal relationship is possible: event A could influence event B. - If (I = 0), the separation is light-dominated (lightlike)
The two events can only be connected by something traveling exactly at the speed of light, such as a photon. - If (I < 0), the separation is space-dominated (spacelike)
The events cannot be connected, not even by light. In this case, no causal influence is possible because no signal can move fast enough to link them.
So, put simply, the spacetime interval tells us whether two events can influence each other or not.
It’s the relativistic analog of Euclidean distance, with one key difference: time and space enter the equation with opposite signs.
A practical example
Imagine a flash of light emitted from the origin (O).
For simplicity, let’s assume the flash travels along the x-axis, so the y and z coordinates remain unchanged.
Consider the following two events:
- Event A (the origin): \[ \begin{cases} t_A = 0 \\ x_A = 0 \\ y_A=0 \\ z_A=0 \end{cases} \] The position four-vector of event A is $$ x_A^\mu = (c \cdot t_A, x_A, y_A, z_A) = (c \cdot 0, 0, 0, 0) = (0, 0, 0, 0) $$
- Event B (arrival of the flash): \[ \begin{cases} t_B = 1 \\ \text{s} \\ x_B = c \cdot t = 3 \times 10^8 \; \text{m} \\ y_B = 0 \\ z_B = 0 \end{cases} \] The position four-vector of event B is $$ x_B^\mu = (ct_B, x_B, y_B, z_B) = (3 \times 10^8, \; 3 \times 10^8, \; 0, \; 0) $$
Now let’s calculate the spacetime interval:
$$ I = (ct_B - ct_A)^2 - (x_B - x_A)^2 - (y_B - y_A)^2 - (z_B - z_A)^2 $$
$$ I = (3 \times 10^8 - 0)^2 - (3 \times 10^8 - 0)^2 - (0-0)^2 - (0-0)^2 $$
$$ I = (3 \times 10^8)^2 - (3 \times 10^8)^2 - 0 - 0 $$
$$ I = 0 $$
Since $ I=0 $, the interval is lightlike, meaning event B can only be reached from A by a signal moving exactly at the speed of light.
In this case, the two events are separated by a zero spacetime interval. They lie on the light cone of the origin. Only light (or any other massless particle traveling at c) can connect them.
Example 2
Now let’s consider a signal leaving the origin (O) and arriving at a point in spacetime with \( t = 2 \,\text{s} \) and \( x = 3 \times 10^8 \,\text{m} \).
The events are:
- Event A (origin): \[ \begin{cases} t_A = 0 \\ x_A = 0 \\ y_A=0 \\ z_A=0 \end{cases} \] The position four-vector of event A is $$ x_A^\mu = (c \cdot t_A, x_A, y_A, z_A) = (c \cdot 0, 0, 0, 0) = (0, 0, 0, 0) $$
- Event B (arrival of the signal): \[ \begin{cases} t_B = 2 \,\text{s} \\ x_B = 3 \times 10^8 \,\text{m} \\ y_B=0 \\ z_B=0 \end{cases} \] The position four-vector of event B is $$ x_B^\mu = (ct_B, x_B, y_B, z_B) = (6 \times 10^8, \; 3 \times 10^8, \; 0, \; 0) $$
Now let’s compute the spacetime interval between the two events:
$$ I = (ct_B - ct_A)^2 - (x_B - x_A)^2 - (y_B - y_A)^2 - (z_B - z_A)^2 $$
$$ I = (6 \times 10^8 - 0 )^2 - (3 \times 10^8 - 0)^2 - (0-0)^2 - (0-0)^2 $$
$$ I = 36 \times 10^{16} - 9 \times 10^{16} $$
$$ I = 27 \times 10^{16} > 0 $$
Here the interval is $ I>0 $, which makes it timelike, because A and B can be connected by a physical object traveling at a speed less than \( c \). In other words, a massive particle or observer could move from A to B.
They lie within the light cone, and even a slower-than-light signal can connect them.
And so on.
