0
$\begingroup$

Consider three objects: a rubber ball, a spaceship, and an isolated brick wall floating serenely in the vacuum of space.

From the perspective of an observer on the spaceship, the ball is moving toward the wall, while the wall and spaceship are stationary. The ball collides with the wall and bounces off. After that happens, the ball is now moving away from the wall (and, by a negligible amount, the wall is moving away from the ball) while the spaceship is still stationary. This makes sense.

However, as I understand it, from the ball's perspective, it started out stationary, while the wall was moving towards it, and the spaceship was moving in the same direction as the wall. Then an event occurred. Afterward, the ball is again stationary (from its perspective), but the wall and spaceship are both moving in roughly the opposite direction as before.

But the thing I don't understand is, from the ball's perspective, how can an event that only involved the ball and the wall have affected the movement of the spaceship?

$\endgroup$
1
$\begingroup$

From a classical perspective, we'd say that you're choosing a frame of reference in which the ball is always at rest. You can do so, but this will not be an inertial frame.

Because this is not an inertial frame, we don't have the luxury of assuming that object with zero forces on them remain unaccelerated in that frame.

$\endgroup$
  • $\begingroup$ One can also pick a noninertial frame and invent fictitious forces in order to make Newton's second law hold in that frame. That can be done in this example, but will not be very nice, because there is no predictability to these fictitious forces (as there would be in a rotating frame, for example). $\endgroup$ – user4552 Oct 14 '19 at 21:40
1
$\begingroup$

The equivalence of inertial frames means that you can choose any inertial frame to solve your physics problem.

It doesn't mean that an object's inertial rest frame represents "what it sees", or anything like that. The closest true statement is that it's sometimes easier to determine what an instrument like a camera or eye will record if you work in coordinates in which it's at rest. But even in that case it's often easier to work out the scene in other coordinates and then transform it using the aberration formula. (That's what I'd do in this case if you asked what a camera mounted to the ball would record.)

You don't consider yourself to be at rest, in general. If you're eating in a train's dining car, you'll likely use the car as a reference and think of yourself at rest. If you're looking out the window, you'll likely use the landscape as a reference and think of yourself as in motion. Both of those are fine. You can pick any reference body you want.

In your example of a ball bouncing off of a wall, you can use two inertial coordinate systems, one in which the ball is stationary before the bounce, and one in which it's stationary after, but all you're doing when you do that is making the problem harder to solve than it needs to be.

The rest of the universe doesn't react to the bouncing ball at all. Everything in the universe has different numeric properties (velocities, etc.) after the bounce than before, but only because you decided to use two different coordinate systems.

$\endgroup$
1
$\begingroup$

The event hasn't affected the movement of the spaceship. The spaceship's state of motion is entirely unchanged. You have just switched to a new frame of reference. Suppose the ball was an inflated one, and it had developed a leak. The thrust from the leaking air could have accelerated the ball so that it touched neither the wall nor the spaceship. From the ball's perspective it would look as if the spaceship and the wall were changing their state of motion, event though the event that triggered th apparent change, namely the air leak, hadn't involved either of them.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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