Does the inertial frame of reference idea work for constant velocity upward?

Answer 1

Inertial reference frameFirst of all, it's better to specify that an experiment done by two different obrervers in two different reference frames gives the same result only if the two reference frames are "inertial" (that is, one neither rotating nor accelerating relatively to the other, and viceversa). This is due to the relativity principle, stating that "All physical laws are the same in every inertial reference frame."

So the correct question should be "Does the 'inertial frame of reference' idea work for constant velocity upward?".

You are right when you say that, in this case, for scientists moving upward, getting far away from Earth, gravitation force gets weaker and weaker, and so the experiment they do gives different result from the same experiment done on the ground. The apparent paradox is solved if we analize the relativity principle better: it just says that the laws are the same, but does not tell anything about conditions. The laboratory on the ground has, as a condition, a constant filed (gravitation field, also called "acceleration of gravity", g = - GmE /rE2 ≈ 9.8 m/s2), whilst the laboratory moving upward has a different condition, that is, a decreasing field (g(t) = - GmE /(rE + vt)2): the laws are the same, but the conditions are not.

When we say that an experiment done in an inertial reference frame gives the same result of the experiment done in another inertial reference frame, we leave unsaid that there must be the same conditions (when you say "the same experiment", you mean "the same experiment and the same conditions"). In fact, as you noticed, if we restore a constant field in the laboratory moving upward, instead of a decreasing one, the conditions are the same, and the experiment gives the same result, preserving the validity of relativity principle and the idea of inertial reference frame.

The way you proposed to restore a constant field is not completely correct: a constant acceleration would be the perfect way only if the laboratory was very far from Earth (far from any massive body, actually): if it doesn't feel any gravitation force, then a constant acceleration of 9.8 m/s2 (that is about the value of g at the ground) would be equivalent to the situation on the Earth (and this perfect equivalence is the idea the general relativity is based on). But in our case, the laboratory moving upward is not free from the gravitation field generated by the Earth, and so the way to compensate and restore a constand field is more complex. I would say that, if we want to find what kind of motion of the laboratory would restore a constant field of the value - GmE /rE2 ≈ 9.8 m/s2, we should reason as follows.

Considering that the laboratory is moving along the radius direction (with origin in the center of mass of the Earth), we can use scalars instead of vectors. Given an unknown motion h(t), the acceleration is a(t) = d2h/dt2; knowing that the gravitation field is g(t) = - GmE /(rE + h(t))2, and since we want total acceleration to be equal to - GmE /rE2, it's

- GmE /rE2 = g(t) + a(t) →

→ - GmE /rE2 = - GmE /(rE + h(t))2 + d2h/dt2.

This should be the differential equation whose solution h = h(t) represents the equation of motion the laboratory must have in order to feel a constant field of the same value of the one felt by the laboratory on the ground.

I hope this explanation is clear. The important concept is that the idea of inertial reference frame is valid everywhere, if you consider the same conditions for the experiment (in other words, it simly states that the constant velocity a system may have does not affect the laws of physics; so much so that it has no sense to ask the velocity of a system without specifying the (inertial) reference frame relatively the whom you want to know the velocity).

Answer 2

First of all, it's better to specify that an experiment done by two different obrervers in two different reference frames gives the same result only if the two reference frames are "inertial" (that is, one neither rotating nor accelerating relatively to the other, and viceversa). This is due to the relativity principle, stating that "All physical laws are the same in every inertial reference frame."

So the correct question should be "Does the 'inertial frame of reference' idea work for constant velocity upward?".

You are right when you say that, in this case, for scientists moving upward, getting far away from Earth, gravitation force gets weaker and weaker, and so the experiment they do gives different result from the same experiment done on the ground. The apparent paradox is solved if we analize the relativity principle better: it just says that the laws are the same, but does not tell anything about conditions. The laboratory on the ground has, as a condition, a constant filed (gravitation field, also called "acceleration of gravity", g = - GmE /rE2 ≈ 9.8 m/s2), whilst the laboratory moving upward has a different condition, that is, a decreasing field (g(t) = - GmE /(rE + vt)2): the laws are the same, but the conditions are not.

When we say that an experiment done in an inertial reference frame gives the same result of the experiment done in another inertial reference frame, we leave unsaid that there must be the same conditions (when you say "the same experiment", you mean "the same experiment and the same conditions"). In fact, as you noticed, if we restore a constant field in the laboratory moving upward, instead of a decreasing one, the conditions are the same, and the experiment gives the same result, preserving the validity of relativity principle and the idea of inertial reference frame.

The way you proposed to restore a constant field is not completely correct: a constant acceleration would be the perfect way only if the laboratory was very far from Earth (far from any massive body, actually): if it doesn't feel any gravitation force, then a constant acceleration of 9.8 m/s2 (that is about the value of g at the ground) would be equivalent to the situation on the Earth (and this perfect equivalence is the idea the general relativity is based on). But in our case, the laboratory moving upward is not free from the gravitation field generated by the Earth, and so the way to compensate and restore a constand field is more complex. I would say that, if we want to find what kind of motion of the laboratory would restore a constant field of the value - GmE /rE2 ≈ 9.8 m/s2, we should reason as follows.

Considering that the laboratory is moving along the radius direction (with origin in the center of mass of the Earth), we can use scalars instead of vectors. Given an unknown motion h(t), the acceleration is a(t) = d2h/dt2; knowing that the gravitation field is g(t) = - GmE /(rE + h(t))2, and since we want total acceleration to be equal to - GmE /rE2, it's

- GmE /rE2 = g(t) + a(t) →

→ - GmE /rE2 = - GmE /(rE + h(t))2 + d2h/dt2.

This should be the differential equation whose solution h = h(t) represents the equation of motion the laboratory must have in order to feel a constant field of the same value of the one felt by the laboratory on the ground.

I hope this explanation is clear. The important concept is that the idea of inertial reference frame is valid everywhere, if you consider the same conditions for the experiment (in other words, it simly states that the constant velocity a system may have does not affect the laws of physics; so much so that it has no sense to ask the velocity of a system without specifying the (inertial) reference frame relatively the whom you want to know the velocity).

Answer 3

Yes, the concept of an inertial frame of reference still applies to an object moving at a constant velocity upward. As long as there are no external forces acting on the object, it will continue moving in a straight line with a constant velocity in that frame of reference.