1
$\begingroup$

Strange question here.

If you would, assume a game of marbles on a planetary scale, one player throws a marble very close to the speed of light travelling along a direction x, another throws a marble perfectly perpendicular to the flight path of the first, in a direction y. The two marbles collide in such a way that the speed of the first marble in the direction x doesn't change. On collission, the second marble comes to a complete halt.

Here's my question:

In the reference frame of the very-close-to-light-speed marble, the second marble would be travelling quite a bit slower than a neutral observer would see it. So how fast would that marble now be travelling in the y direction? Would it go as fast in the y direction as a neutral observer would see it? Or as fast as the light-speed marble would see it?

Very weird question, but I'm curious.

Thanks!

Improve this question
$\endgroup$
2
  • $\begingroup$ Does the "planetary scale" refers to a size of each of the two marbles (and, consequently, of all of their particulate components) up-scaled to the size of the planet most typical for us (the earth) and its particulate components, or does it refer to the game being played with ordinary marbles in a circle whose width would equal the diameter of such a planet? $\endgroup$
    – Edouard
    Commented Mar 15, 2022 at 15:55
  • $\begingroup$ Please clarify your specific problem or provide additional details to highlight exactly what you need. As it's currently written, it's hard to tell exactly what you're asking. $\endgroup$
    – Community Bot
    Commented Mar 15, 2022 at 15:59

1 Answer 1

Reset to default
2
$\begingroup$

This question would be very messy, but certainly doable.

First, all of the initial masses and velocities are known, and the final masses and velocities are unknown. So that is six unknowns: two final masses, two final x components of velocity, and two final y components of velocity. The final velocity of the second object is zero, so that determines two, and the final x component of the first objects velocity is equal to the known initial value, so that is a third. So that leaves three unknowns.

We can write conservation of energy to get one equation. Then we can write conservation of momentum which gives us two more equations, one for the x and one for the y components. So in the end we get three equations.

Solve the three equations in three unknowns and then you have everything known in the initial frame. Once you have that, then you can simply transform to any other frame of interest using the Lorentz transform. I leave the messy details as an exercise for the interested reader.

So how fast would that marble now be travelling in the y direction? Would it go as fast in the y direction as a neutral observer would see it? Or as fast as the light-speed marble would see it?

In the neutral observer’s frame it goes as fast as the neutral observer sees it and in the near-light-speed marble’s frame it goes as fast as the near-light-speed marble sees it. Velocity is not “absolute” so different frames disagree about its value. All frames agree that energy and momentum were conserved, but not their values.

Improve this answer
$\endgroup$
2
  • $\begingroup$ In spite of having read about "inertial frames of reference" dozens of times, I (the initial reviewer) had not been able to pick the practical problems out of the OP's question, and, especially, I'd neglected a practical application of that Lorentz transformation whose social details (Einstein's contact with Lorentz, during those spells when politics were quiet enough to allow free travel within Europe) I'd also seen mentioned time and time again....So, I'm truly impressed with Dale's answer, and delighted with the size of his PSE reputation in relation to my own. $\endgroup$
    – Edouard
    Commented Mar 20, 2022 at 10:08
  • 1
    $\begingroup$ One really has to wonder how much learning we're missing during the current autocratic boor's inhumane and horrific fireworks display. $\endgroup$
    – Edouard
    Commented Mar 20, 2022 at 10:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.