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

[Best of the Worst]

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?

 

[pervect]

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.

 

[Best of the Worst]

Awesome. Thank you kindly ;).

 

[pmb_phy]

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

$\Delta$s² = (t₂ - t₁)² - (x₂ - x₁)²

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

Pete