I have had trouble reconciling the quadratic form of kinetic energy and reference frames traveling at different velocities for some times now. I'll give an example that confuses me for some illustration: A stationary rock (mass = 1kg) is dropped some height a gains potential energy = 4.5J In the reference frame of the rock it looses some amount of gravitational potential energy and gains kinetic energy equal to that (neglecting other lossy mechanisms)
\begin{equation} \Delta v = \sqrt{2\Delta PE / m} = 3ms^{-1} \end{equation} Now imagine the same scenario observed in a reference frame traveling at -4m/s \begin{equation} KE_{before} = 4^2/2 = 8 \end{equation} \begin{equation} KE_{after} \frac{4^2}{2} + 4.5= 8 + 4.5 = 12.25 \end{equation} \begin{equation} \Delta v = 1ms \end{equation} How is it possible that conserving energy results in different observed changes in speed?

I was under the impression the laws of physics were supposed to hold in any inertial frame, this appears to imply that there is a "right frame" which i know to be not true.

  • $\begingroup$ Did you not find my answer correct? $\endgroup$ – Chet Miller Jan 24 '16 at 2:11
  • $\begingroup$ I'm not sure i full understand your answer. The potential field in which the rock is moving doesn't change based on the reference frame so the change in potential energy is still the same. $\endgroup$ – Duke of Sam Jan 24 '16 at 12:42
  • $\begingroup$ No way. See the calculations I have edited into my Answer. $\endgroup$ – Chet Miller Jan 24 '16 at 12:54
  • $\begingroup$ Here is a followup question for you. If the rock is sitting motionless on a table in the stationary frame of reference, then, as reckoned from the decending frame of reference, it is rising at 4 m/s, and its potential energy is increasing. Yet its kinetic energy is remaining constant. How do you reconcile this? (I know the answer to this) $\endgroup$ – Chet Miller Jan 26 '16 at 13:55

From the moving frame of reference, the rock is moving upward with an initial velocity of 4 m/s, and, at the end of the same time interval, it is moving upward at 1 m/s. But now, as reckoned from the moving frame of reference, the change in potential energy is going to be different. This is because the datum for potential energy is moving downward, so, as reckoned from the moving frame, the potential energy increases (rather than decreases as in the case where the rock is falling). Potential energy is a frame-dependent quantity. In this problem, instead of the change in potential energy being -4.5 J after 0.3 sec., it is +7.5 J, as reckoned from the moving frame. The difference in potential energy here is 12.0 J, and is equal to the difference in elevation (4 m/s times 0.3 sec = 1.2 m) times g.

Here are some calculations to illustrate:


Taking the datum for PE for each of the reference frames the original elevation of the rock at time zero,

$$0+0=\frac{(-3)^2}{2}-4.5\tag{fixed frame of reference}$$ $$\frac{4^2}{2}+0=\frac{1^2}{2}+7.5\tag{moving reference frame}$$ Note that, when the moving reference frame moves downward by 1.2 meters during the 0.3 seconds, then, relative to an observer in this frame, the rock has moved upward 1.2 meters. So, according to this observer, the potential energy of the rock has increased by 12 J more than observed by the person in the fixed frame.


You are ignoring the change in the kinetic energy of the Earth as it and the rock accelerate towards their common centre of mass. If the rock speeds up in your moving frame then the Earth slows down in the moving frame and you have to consider both changes for the energy to balance. Obviously the change in the Earth's kinetic energy is tiny, but you need to include it to make the calculation work.

If you take the Earth as fixed then this implies there is an external force acting on the Earth that stops it accelerating towards the rock. You need to include this external force in your calculations of work done.

The calculation is not dissimilar to the one described in How am I able to stand up and walk down the aisle of a flying passenger jet?. In that case the energy discrepancy comes from ignoring the airplane moving in response to the force exerted by the walking passenger.

  • $\begingroup$ My original example was firing a bullet down the aisle of a passenger jet. Though I thought the problem could be made simpler with the example i gave. So the lesson to take from this is that energy is always conserved but the actual quantity attributed to everything will differ. This makes sense when I think about it in the context of other forms of energy such a potential energy. For some reason I didn't possess the same intuition for kinetic energy, for some reason being the most observable form of energy I assumed it would be immune to such things. $\endgroup$ – Duke of Sam Jan 23 '16 at 11:26

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