3
$\begingroup$

A non-inertial frame, such as a planet that does not have constant velocity in its orbit, would change the laws of physics in many ways. For example, if you stood on the surface of the planet as it accelerated and dropped something, it would seem as if a force pushed it in some direction as it fell.

But how would the people on the planet view the laws of physics? Also, how would the inhabitants react to a visit to Earth?

$\endgroup$
  • $\begingroup$ First of all for the OP, you would get a better answer in physics stack exchange, just be sure to say you are an amateur, if you are one, because they tend to be very formal. And for the @kingledion, he's not asking for the laws in another universe, and we have been in a non-inertial frame, earth, cars, everything that does not move in a straight line and does not have a constant velocity is in a non-inertial frame. In short, we know lots about this. He is not asking for a change in the laws, he is asking how they would change when you change your reference. $\endgroup$ – Ivan Lerner Dec 9 '16 at 20:31
  • 1
    $\begingroup$ I do think that Physics Stack Exchange would take this question, but I don't think it should be migrated. There's no reason to migrate a question if it's on-topic on the site it's originally asked on, and I think this sort of question is on-topic on Worldbuilding Stack Exchange. $\endgroup$ – HDE 226868 Dec 9 '16 at 20:36
  • $\begingroup$ I agree, just wanted to let the OP know. $\endgroup$ – Ivan Lerner Dec 9 '16 at 22:14
  • $\begingroup$ Planets don't have constant velocities in their orbits. You may want to check out Do transfer orbits toward the central star necessarily result in a higher velocity on arrival due to the star's gravity? (on Space Exploration), because it discusses this to some extent; Kepler's laws, and Newton's work based on them, are relevant. There's a reason why for example Wikipedia gives the mean orbital velocity, not merely the orbital velocity, for orbiting bodies. $\endgroup$ – a CVn Dec 9 '16 at 23:23
6
$\begingroup$

They could still tell the difference.

A non-inertial frame involves acceleration of some sort, i.e. non-uniform velocity. Such a frame can certainly have uniform speed, but the direction of motion of an object in the frame will be constantly changing. For instance, an observer on a merry-go-round is in a non-inertial frame, because they are moving in a circle.

We're in a non-inertial frame, because we're on Earth. There are two important reasons for this:

  1. Earth revolves around the Sun.
  2. Earth rotates on its axis.

Both of these mean that fictitious forces arise. The Coriolis force is perhaps the most famous one, and plays a role in ocean currents and other systems of circulating fluids.

Even though we're in a non-inertial frame, though, the effects are generally small. The acceleration of Earth around the Sun is relatively tiny; the same goes for the acceleration from its rotation. Therefore, our frame is approximately inertial, and we don't notice most of the fictitious forces.

However, we still know that we're in a non-inertial frame. We can create other non-inertial frames - such as merry-go-rounds and notice that there are fictitious forces there. Therefore, knowing that the Earth moves around the Sun, we can tell that we're in such a frame.

$\endgroup$
1
$\begingroup$

First, let's enumerate the laws:

1- F=ma (where F is the sum of all the forces acting on a body, and a is the resulting acceleration of the body itself)

2- If no force is acting on the body, it will move in a straight line without changing it's speed. (inertia)

3- For every force, there is an equal and opposite force on the other body.

Now let's see an example of an non-inertial frame: a car going in a circle. In this case, you know we feel a force push you to the outside of the curve, but why is that (from the inertial frame first)?

A change in direction, is also an acceleration and when the car accelerates to the left for example it makes a force on you to the same direction, but your body, if it was not under the action of a force, would go in a straight line (second law). So the car is pulling you to one side, but your whole body wants to go the other. This makes evident that an object which is not under the action of any force inside the car (if you throw a ball in the air for example) will go to the opposite side of the acceleration. That means the second law does not hold anymore, and have to be changed.

Now lets look at another situation in the car. The car pulls your body to one side, and you feel a force pushing you to the outside of the curve, just like in the first situation. But there is the door keeping you from being thrown out of the car, and to do that, in the frame of the car, the door must exert a force on you equal and opposite to the force you are exerting on it, since in the frame of the car you are perfectly still. That means the third law is still valid in a non-inertial frame.

To test the first law, you have to look at the equation. In the same situation of the first two points. We look at the mass, acceleration, and the forces, and if they behave in the same way they behave in the non-inertial frames, that means it is valid here as well. For our purposes here, an equation behaving the the same way it did before, means that even if you make changes in the mass, acceleration, or the forces present, the equality will still be true. In other words, the behavior of the variables (F, a, m) can change, but their relation to each other does not.

When we look at the mass, it is obvious that it does not change (if you do not consider relativity here, which I am completely ignoring), but what about the acceleration? If we want the laws in another frame, we have to think in that frames view, and the acceleration in this case would be the acceleration relative to the car, not the ground.

To organize a little this idea, look at this: in the cars frame, it is still, and so are the people sitting on it. If you throw a ball in the air, it will go to the opposite side to which the car is turning, so a force acts on it, and clearly an acceleration too, since it moves in a different direction from the direction you threw it.

From the grounds frame on the other hand, the car is not still, and when the ball looses contact with your hand, it begins to travel in a straight line (second law) so there is no force acting on it.

From this last two paragraphs, it is clear that there is a change in the acceleration and in the forces present in each frame, but as said earlier, if their relation is still the same, the law will be the same as well.

With this last considerations in mind, we look at the car again. A force that didn't exist in the ground appeared, but an acceleration that did not exist in the ground appeared as well. These two variables that exist in the equation changed, so it is possible that these changes balance to maintain the equality, and they indeed do, but showing that is out of the scope here. That means the first law is also valid on a non-inertial frame.

Conclusion

The first law does not change, but there are extra forces (called inertial forces) to consider now, that did not exist on the inertial frame.

The second law changes, and must state that if an object is not under the influence of any usual force (gravity, hand force, etc), it will move according to the inertial forces alone.

The third law does not change, but there is a little problem. In the case of the ball thrown in the air, where is the equal and opposing force? It does not exist, but this is actually not a problem, because there are no forces acting on it besides the inertial force, which in such a referential is considered to be another law, and not a real force. This is why inertial forces are not usually considered to be real forces. But for any other interaction that takes place in the car the third law is still valid.

$\endgroup$
  • 1
    $\begingroup$ It's not true that the second law fails to hold. It's still valid; it simply requires the introduction of a fictitious force term to balance things out, as I discussed in my answer. $\endgroup$ – HDE 226868 Dec 9 '16 at 22:06
  • $\begingroup$ Yeah, that's what I said in the conclusion, you have to change it a little. $\endgroup$ – Ivan Lerner Dec 10 '16 at 18:18

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.