# Calculating the Invariant Interval for Two Events

• joriarty
In summary, the conversation is about finding the time difference between two events given their coordinates, finding a reference frame where the time difference is the negative of the original frame, and calculating the invariant interval in both frames. The first question is a conversion from distance to time, using the speed of light as c. The time difference is found to be (2/3)*10^-8 s. The other two problems can be solved using the given information.

## Homework Statement

Consider two events
$$ct_{1}\; =\; 3\; m,\; x_{1}\; =\; 2\; m,\; ct_{2}\; =\; 5\; m,\; x_{2}\; =\; 6m$$

1. What is the time difference between the two events?
2. Find a reference frame for which the time difference is the negative of the time difference in the original frame.
3. Calculate the invariant interval $$\left( \Delta s \right)^{2}=+c^{2}\left( t_{2}-t_{1} \right)^{2}-\left( x_{2}-x_{1} \right)^{2}-\left( y_{2}-y_{1} \right)^{2}-\left( z_{2}-z_{1} \right)^{2}$$ in both frames.

## The Attempt at a Solution

I honestly don't have any idea what the first question is actually asking. How can I calculate a time difference when I am only given these distance figures with no velocities? It seems like an extremely poorly worded question.

Last edited:
joriarty said:

## Homework Statement

Consider two events
$$ct_{1}\; =\; 3\; m,\; x_{1}\; =\; 2\; m,\; ct_{2}\; =\; 5\; m,\; x_{2}\; =\; 6m$$

1. What is the time difference between the two events?
2. Find a reference frame for which the time difference is the negative of the time difference in the original frame.
3. Calculate the invariant interval $$\left( \Delta s \right)^{2}=+c^{2}\left( t_{2}-t_{1} \right)^{2}-\left( x_{2}-x_{1} \right)^{2}-\left( y_{2}-y_{1} \right)^{2}-\left( z_{2}-z_{1} \right)^{2}$$ in both frames.

## The Attempt at a Solution

I honestly don't have any idea what the first question is actually asking. How can I calculate a time difference when I am only given these distance figures with no velocities? It seems like an extremely poorly worded question.
The first question is just a unit conversion. What does c stand for?

Well c = 3*108 ms-1. So t1 = 1*10-8 s and t2 = (5/3)*10-8 s.

I guess assuming these are both observed from a laboratory frame of reference then the time difference is just t2 - t1 = (2/3)*10-8 s?

It was a simple question, but the obscure way in which it was asked threw me off. Typical me, struggling with the simple stuff but I can much more easily understand the complicated things Thank you for your help. Hopefully I can do the other two problems myself!

## What is the Invariant Interval?

The Invariant Interval is a measurement in the theory of special relativity that describes the distance between two events in a four-dimensional space-time. It is a fundamental concept that remains constant for all observers, regardless of their relative motion.

## How is the Invariant Interval calculated?

The Invariant Interval is calculated using the Minkowski metric, which takes into account both time and space in a four-dimensional space-time. It is represented by the equation s² = c²Δt² - Δx² - Δy² - Δz², where s is the Invariant Interval, c is the speed of light, and Δt, Δx, Δy, and Δz are the differences in time and space between the two events.

## What does a positive or negative Invariant Interval indicate?

A positive Invariant Interval indicates that the two events are separated by both time and space, while a negative Invariant Interval indicates that the two events are separated by time only. A zero Invariant Interval indicates that the two events are at the same position in space-time.

## Why is the Invariant Interval important?

The Invariant Interval is important because it is a fundamental concept in the theory of special relativity that helps us understand the relationships between time and space in a four-dimensional space-time. It also plays a crucial role in the calculation of physical quantities, such as momentum and energy.

## Can the Invariant Interval change for different observers?

No, the Invariant Interval remains constant for all observers, regardless of their relative motion. This is a fundamental principle of special relativity, which states that the laws of physics should be the same for all observers in uniform motion.