- #1

milkism

- 118

- 15

- Homework Statement
- Analyse of two events with the use of invariant intervals

- Relevant Equations
- I= -c²t² + d²

Exercise:

My solutions:

My solutions:

- For events to be simultaneous, the invariant interval must be bigger than zero (spacelike). I got $$I = -c^2 \Delta t^2 + \Delta x^2 + \Delta y^2 + \Delta z^2 = -(0-1)^2 + (0-2)^2 + (0-0)^2 + (0-0)^2 = -1 + 4 = 3 >0$$. Which is indeed greater than zero, to find the velocity, I will use the first Lorentz-transformation formula with four vectors $$\overline{x}^0 = \gamma \left( x^0 - \beta x^1 \right) = \Delta (c\overline{t}) = \gamma (\Delta (ct) - \beta (\Delta x))$$, we want $$\Delta \overline{t} = 0$$. We get: $$\Delta (ct) = \beta (\Delta x) = \frac{0-1}{0-2} = \frac{v}{c} = v = \frac{c}{2}$$. Where $$\beta = \frac{v}{c}$$.
- The new spacetime coordinates for second event will be (-1,2,0,0), the invariant interval doesn't change, meaning it's still spacelike, so the velocity will be the same but opposite sign.
- No, for two events to occur at the same point (place) the invariant interval must be negative (timelike), which isn't.