In What Frame Is (t2 - t1)^2 Measured?

  • Thread starter Best of the Worst
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T = root((t2 - t1)^2 - (x2 - x1)^2) represents proper time as measured in any inertial frame, regardless of the frame in which (t2 - t1)^2 is measured. This is because T is an invariant, meaning it remains the same regardless of the frame of reference. In summary, T is proper time measured in someone's inertial frame, and (x2 - x1)^2 represents their movement through space. The equation is invariant and can be used to calculate proper time regardless of the frame of reference.
  • #1
Best of the Worst
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T = root((t2 - t1)^2 - (x2 - x1)^2)

As I understand it, T is proper time as measured in someone's intertial frame, and (x2 - x1)^2 is their movement through space... but in what frame is (t2 - t1)^2 measured?
 
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  • #2
t1,x1 and t2,x2 are both measured in somenone (anyone's) inertial frame.

Regardless of which inertial frame you chose to measure t1,x1,t2, and x2, the result T will be the same - it will be invariant.
 
  • #3
Awesome. Thank you kindly ;).
 
  • #4
Best of the Worst said:
T = root((t2 - t1)^2 - (x2 - x1)^2)

As I understand it, T is proper time as measured in someone's intertial frame, and (x2 - x1)^2 is their movement through space... but in what frame is (t2 - t1)^2 measured?
Please note that

[itex]\Delta[/itex]s2 = (t2 - t1)2 - (x2 - x1)2

may be either positive, negative or zero. [itex]\Delta[/itex]s2 may be either a proper time or a proper distance according to its sign.

Pete
 
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FAQ: In What Frame Is (t2 - t1)^2 Measured?

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