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Suppose, I'm on top of a hill, and I drop a ball from some height. As there are no external forces acting on the Earth-ball system, I can safely claim that the total energy is conserved. I know that, as the ball falls from a certain height, it (the system) loses potential energy and gains kinetic energy.

Depending on where I took my reference frame level, the value of the potential energy at all the different points would change. However, the change of potential energy is independent of my reference level. Similarly, the velocity doesn't depend on where I'm standing. So I can say $K_f+U_f=K_i+U_i$. More generally, I can say $\Delta K+\Delta U=0$.

At any point along the fall, the total energy is the kinetic energy at that point added to the potential energy to that point. This creates a small problem. Even though the change in total energy is $0$, the value of the total energy depends on where I took my reference level. In this case, it is the value of the potential energy that depends on my choice of reference.

Another case is, suppose I, the observer, jump with the ball, such that I'm moving with some speed. In this frame the ball moves at a different speed as compared to the first one. Since the work done by gravity remains the same, I can argue that the potential energy of the system remains the same (depending on our reference level).

The problem now is that the ball has some kinetic energy different from the kinetic energy in the first example. So, even though the total change in energy is $0$, the value of the total energy is not the same anymore.

Another example, consider a car moving in the $x$ direction. The total energy is equal to the kinetic energy and that is $E$ suppose. With respect to some observer in the car, the kinetic energy and by extension, the total energy is $0$. So, as we move from one frame to the other, the total energy seems to have changed.

What does energy conservation mean exactly?

Does it mean total mechanical energy remains exactly the same in all inertial frames. For example if it is $E_1$ in one frame, it must be $E_1$ in all other frames?

Or does it mean that if the total energy is conserved in one inertial frame, i.e. the change in total energy is $0$ in one frame, then the change in total energy must also be $0$ in all other inertial frames?

Thus the value of the total energy at a point, seems to depend upon, the speed of the observer's inertial frame (kinetic energy depends on this), and the reference level where the observer assumes potential is $0$ (the potential energy at a point depend on this).

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    $\begingroup$ The end of your post asks about inertial frames, but one of the examples is falling with the ball, which is not an inertial reference frame in classical physics. Did you mean to include a different inertial reference frame at moves relative to the first one? $\endgroup$ Dec 7, 2021 at 2:31
  • $\begingroup$ @BioPhysicist yes, that is exactly what I meant. I'm so sorry I completely forgot about the fact that the ball is accelerating downward, and so my frame must be non-inertial. I'll edit that right away. $\endgroup$
    – RayPalmer
    Dec 7, 2021 at 2:49
  • $\begingroup$ consider the definition of potential energy, it's an integral from one point to the other point. so it doesn't matter where you put the coordinate. same as $\Delta U$. $\endgroup$
    – leave2014
    Dec 7, 2021 at 3:01
  • $\begingroup$ @leave2014 but I can add an arbitrary constant to the potential energy, which would change it's value. For example if I take the ground as reference, then potential energy at height $h$ is $mgh$. If I take the height at reference, then the potential energy there is $0$. However the difference doesn't change. The integral describes this difference. I suppose it would be useless to talk about the potential energy at a point. It would be more useful to talk about $\Delta U$ instead. $\endgroup$
    – RayPalmer
    Dec 7, 2021 at 3:08
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    $\begingroup$ @RayPalmer $mgh$ is a simplified formula for using, it's a result of a definite integral between $h=h$ and $h=0$ by putting the coordinate on the ground where the gravity force cannot do more work say "$U=0$". $\endgroup$
    – leave2014
    Dec 7, 2021 at 3:21

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What does energy conservation mean exactly?

It means $\Delta E = 0$, or if you know calculus $\frac{d}{dt}E=0$. The total energy is constant, it does not change as time goes on.

Does it mean total mechanical energy remains exactly the same in all inertial frames. For example if it is 𝐸1 in one frame, it must be 𝐸1 in all other frames?

What you are describing is called frame invariance. Energy is conserved, but it is not invariant. Different inertial frames will disagree on the value of energy, but they will all agree that it is constant over time.

Or does it mean that if the total energy is conserved in one inertial frame, i.e. the change in total energy is 0 in one frame, then the change in total energy must also be 0 in all other inertial frames?

Yes.

Most of your examples are based on this misunderstanding of energy and are simply resolved by recognizing that your analysis is correct, but does not contradict the conservation of energy. However, there is one small error here:

suppose I, the observer, jump with the ball, such that I'm moving with some speed. In this frame the ball moves at a different speed as compared to the first one. Since the work done by gravity remains the same, I can argue that the potential energy of the system remains the same (depending on our reference level).

The work done by gravity on the ball does not remain the same in different reference frames. This problem does demonstrate energy conservation, but neither energy nor work is invariant. To show energy conservation requires using the conservation of momentum also, and careful accounting for the change in the KE of the earth.

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