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Suppose there are three masses that are still relative to each other in space. They are positioned in an equilateral triangle. Let's accelerate one mass towards the other two with a force. The energy added to this system should be $F\cdot{ds}$. However, according to the particle that has been accelerated, the work done is double this amount assuming that the three particles are of the same mass. I don't think that I fully understand how does the conservation of energy really works.

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    $\begingroup$ energy is a frame dependent quantity $\endgroup$
    – Christoph
    Oct 24 '16 at 14:50
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    $\begingroup$ You have discovered the principle that energy is not invariant under change of reference frame. What is consistent throughout different frames of reference is the work-kinetic energy theorem. $\endgroup$ Oct 24 '16 at 14:51
  • $\begingroup$ It seems to me that even work done is different with both cases $\endgroup$
    – user86425
    Oct 24 '16 at 15:03
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Conservation of energy occurs within a given reference frame. If you change reference frames, you cannot use those rules.

A clear example of this occurs if you consider the energy of the system when considering the Earth and an airplane flying through the air. From the perspective of an observer on the ground, the airplane has kinetic energy of $\frac{1}{2}m_{plane}v^2$, and the earth has 0 kinetic energy. From the perspective of an observer on the plane, it is the plane that has 0 kinetic energy, and the earth has kinetic energy to the tune of $\frac{1}{2}m_{earth}v^2$. Needless to say, given that $m_{earth}\gt\gt m_{plane}$, the two observers will disagree greatly on the numeric value for the system's kinetic energy. However, if we consider changes in kinetic energy, both systems will find that energy is conserved (from their own perspective).

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  • $\begingroup$ but if work done is difference in energy, then why is this also different as seen in my example? $\endgroup$
    – user86425
    Oct 24 '16 at 14:58
  • $\begingroup$ In your example, the force is handwaved over. Force is also frame dependent. Consider skydivers. They feel no force of gravity when they jump. They feel "weightless," even though, to an outside observer, the force of gravity is accelerating them greatly. If you have two frames that are accelerating with respect to each other, they will disagree about forces. (the skydiver is accelerating towards the earth / the earth is accelerating towards the skydiver). However, if you pick one of the frames, the forces will be consistent within that frame. $\endgroup$
    – Cort Ammon
    Oct 24 '16 at 15:06
  • $\begingroup$ The famous version of this involves someone walking forward on an airplane. From the perspective of the airplane, the person is traveling slowly forward, so the work needed to walk forward is minimal. From the perspective of the ground, the person is traveling fast, so the same change in velocity appears require far more power than the human leg exerts. From the perspective of the person, they're pushing a plane backwards. $\endgroup$
    – Cort Ammon
    Oct 24 '16 at 15:08
  • $\begingroup$ However, if you focus on mass, momentum, and energy within a single frame, you find that, indeed, the numbers work out. Some of the frames have to account for odd effects (such as you walking forward causing the plane to slow down a hair, transferring momentum from it to the person, from the perspective of the ground observer). What matters is that the equations are always consistent within a reference frame, no matter which one you choose. It just happens that, in many of our problems, there is one clear "preferred" reference frame, so we get used to thinking that's the only way to think. $\endgroup$
    – Cort Ammon
    Oct 24 '16 at 15:09
  • $\begingroup$ but surely that is contradictory? How can work done be frame independent if we say that forces are relative? $\endgroup$
    – user86425
    Oct 24 '16 at 16:48
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Ok I think there are 2 distinct problems here. firstly it is that I cannot apply the same equations for energy in an accelerating coordinate system. It only works for inertial reference frames. Secondly it is that even under galilean transformations work done is not (and doesn't need to be) invariant which is what was addressed

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