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marin [14]
3 years ago
10

Spacetime interval: What is the interval between two events if in some given inertial reference frame the events are separated b

y: (a) 7.5 x 10 m and 3s? (b) 5x10 m and 0.58? (c) 5x 10"m and 58?
Physics
1 answer:
shepuryov [24]3 years ago
5 0

Answer:

  • a. \Delta s ^2 = 8.0888 \ 10^{17} m^2
  • b. \Delta s ^2 = 3.0234 \ 10^{16} m^2
  • c. \Delta s ^2 = 3.0234 \ 10^{20} m^2

Explanation:

The spacetime interval \Delta s^2 is given by

\Delta s ^2 = \Delta (c t) ^ 2 - \Delta \vec{x}^2

please, be aware this is the definition for the signature ( + - - - ), for the signature (- + + + ) the spacetime interval is given by:

\Delta s ^2 = - \Delta (c t) ^ 2 + \Delta \vec{x}^2.

Lets work with the signature ( + - - - ), and, if needed in the other signature, we can multiply our interval by -1.

<h3>a.</h3>

\Delta \vec{x}^2 = (7.5 \ 10 \ m)^2

\Delta \vec{x}^2 = 5,625 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 3 \ s)^2

\Delta (c t) ^ 2 = (899,377,374 \ m)^2

\Delta (c t) ^ 2 = 8.0888 \ 10^{17} m^2

so

\Delta s ^2 = 8.0888 \ 10^{17} m^2 - 5,625 m^2

\Delta s ^2 = 8.0888 \ 10^{17} m^2

<h3>b.</h3>

\Delta \vec{x}^2 = (5 \ 10 \ m)^2

\Delta \vec{x}^2 = 2,500 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 0.58 \ s)^2

\Delta (c t) ^ 2 = (173,879,625.6 \ m)^2

\Delta (c t) ^ 2 = 3.0234 \ 10^{16} m^2

so

\Delta s ^2 = 3.0234 \ 10^{16} m^2 - 2,500 m^2

\Delta s ^2 = 3.0234 \ 10^{16} m^2

<h3>c.</h3>

\Delta \vec{x}^2 = (5 \ 10 \ m)^2

\Delta \vec{x}^2 = 2,500 m^2

\Delta (c t) ^ 2 = (299,792,458 \frac{m}{s} \ 58 \ s)^2

\Delta (c t) ^ 2 = (1.73879 \ 10^{10} \ m)^2

\Delta (c t) ^ 2 = 3.0234 \ 10^{20} m^2

so

\Delta s ^2 = 3.0234 \ 10^{20} m^2 - 2,500 m^2

\Delta s ^2 = 3.0234 \ 10^{20} m^2

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