Calculating the Invariant Interval for Two Events

AI Thread Summary
To find the time difference between two events given their coordinates, one must convert the spatial measurements into time using the speed of light (c). The time difference is calculated as t2 - t1, resulting in a value of (2/3)*10^-8 seconds. The discussion highlights confusion over the wording of the problem, but clarifies that it essentially involves unit conversion. After determining the time difference, the next steps involve finding a reference frame where this difference is negative and calculating the invariant interval using the provided formula. The conversation emphasizes the importance of understanding the context of the problem to avoid misinterpretation.
joriarty
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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.

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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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.

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?
 
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!
 
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