Calculating Spacetime Intervals for Simultaneous Events

AI Thread Summary
For events to be simultaneous, the invariant interval must be greater than zero, indicating a spacelike separation. The calculations show that the velocity derived from the Lorentz transformation is v = c/2, with the new spacetime coordinates for the second event being (-1, 2, 0, 0). However, there is confusion regarding the calculations, particularly in connecting steps and ensuring the correct application of the Lorentz transformation. The discussion highlights the need for clarity in deriving results and emphasizes that the invariant interval must be negative for events to occur at the same point in spacetime. Overall, the conversation revolves around the correct interpretation and application of spacetime intervals and velocities in relativistic contexts.
milkism
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Homework Statement
Analyse of two events with the use of invariant intervals
Relevant Equations
I= -c²t² + d²
Exercise:
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My solutions:
  1. 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 $$\beta = \frac{v}{c}$$.
  2. 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.
  3. No, for two events to occur at the same point (place) the invariant interval must be negative (timelike), which isn't.
 
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milkism said:
We get: $$\Delta (ct) = \beta (\Delta x) = \frac{0-1}{0-2} = \frac{v}{c} = v = \frac{c}{2}$$. Where $$\beta = \frac{v}{c}$$.
So you're claiming ##\Delta (ct) = \frac{c}{2}## so ##-1 = 1.5\times 10^8~\rm m/s##? Don't use an equal sign to connect steps.

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.
You may want to rethink this one.
 
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vela said:
So you're claiming ##\Delta (ct) = \frac{c}{2}## so ##-1 = 1.5\times 10^8~\rm m/s##? Don't use an equal sign to connect steps.You may want to rethink this one.
Are the spacetime coordinates wrong?
 
milkism said:
Are the spacetime coordinates wrong?
##v = \frac c 2## looks right for part (i), but your work seems very muddled to me. For example:

milkism said:
We get: $$\Delta (ct) = \beta (\Delta x) = \frac{0-1}{0-2} = \frac{v}{c} = v = \frac{c}{2}$$.
For part (ii) you need to work things out properly. You've effectively just guessed a (wrong) answer.
 
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PeroK said:
##v = \frac c 2## looks right for part (i), but your work seems very muddled to me. For example:For part (ii) you need to work things out properly. You've effectively just guessed a (wrong) answer.
Well, I'm lost.
 
You know the coordinates in ##S##. Try using your answer ##\beta = -1/2## and the Lorentz transformation and see what you get for the coordinates in the other frame.
 
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milkism said:
Well, I'm lost.
This problem, IMO, is simply asking you to solve an equation for ##v##. Namely:$$\gamma(1 - 2\frac v c)= -1$$
 
Haha, I thought we had to do something with invariant intervals for (ii), thanks! Is my solution for (iii) correct?
 
milkism said:
Is my solution for (iii) correct?
Yes. Note that you can prove it in this case by showing that no allowable ##v## would solve the relevant equation.
 
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PeroK said:
Yes. Note that you can prove it in this case by showing that no allowable ##v## would solve the relevant equation.
Because it would be bigger than the speed of light?
 
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milkism said:
Because it would be bigger than the speed of light?
I don't know. Try it!
 
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