Calculating the Invariant Interval for Two Events

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.
  • #1
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Homework Statement



Consider two events
[tex]ct_{1}\; =\; 3\; m,\; x_{1}\; =\; 2\; m,\; ct_{2}\; =\; 5\; m,\; x_{2}\; =\; 6m[/tex]

  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 [tex]\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}[/tex] in both frames.

Homework Equations



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.
 
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  • #2
joriarty said:

Homework Statement



Consider two events
[tex]ct_{1}\; =\; 3\; m,\; x_{1}\; =\; 2\; m,\; ct_{2}\; =\; 5\; m,\; x_{2}\; =\; 6m[/tex]

  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 [tex]\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}[/tex] in both frames.

Homework Equations



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?
 
  • #3
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 :smile: 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.

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