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Starting with the premises that spring constant $k$ and displacement, and therefore the elastic energy stored in a spring are frame invariant quantities, here's a thought experiment leading to the question.

A compressed spring containing Q joules of elastic energy rests on a frictionless horizontal surface with its axis parallel to the surface and one end fixed to the surface.  A block of material is placed in contact with the other end.  The spring is released, and a stationary observer correctly concludes that since Q joules of elastic energy have been transferred to the block, the block now has Q joules of KE.

A second observer, moving in the same direction as the block with velocity $v$ which happens to be equal to the final velocity of the block in the stationary frame, agrees that Q joules of energy were transferred to the block from the spring as kinetic energy, but observes that the result was that the block stopped and has zero KE.

That only works mathematically if the second observer calculated the initial KE as negative.   But KE is never actually negative, so it appears that the mathematics associated with frame dependence of KE does not describe reality when a frame invariant, non-KE form of energy is transferred to an object as KE.

How can this apparent contradiction between energy conservation and frame dependence of KE be explained?

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    $\begingroup$ Please define in how far kinetic energy and measured kinetic energy are different quantities. Also, if you talk about frame invariance, it would be nice to know the symmetry group you are talking about, just to be sure ... $\endgroup$ – Sanya Oct 29 '16 at 21:17
  • $\begingroup$ @Sanya You appear to be ready to answer the question based upon only the header. I say this because the difference between actual and measured KE are pretty clear in the body of the question. $\endgroup$ – D. Ennis Oct 29 '16 at 21:24
  • $\begingroup$ I read your whole post. There is not a concise definition in there that I can understand from it. $\endgroup$ – Sanya Oct 29 '16 at 21:26
  • $\begingroup$ Please highlight, maybe in bold, the exact part you think is a contradiction. Otherwise it's hard to see what your argument is. $\endgroup$ – knzhou Oct 29 '16 at 21:27
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    $\begingroup$ @D.Ennis : I am sorry, I still do not understand the distinction you are making between your question and the one I cited. You are asking if 'KE' is invariant while 'measured KE' is not. Like Sanya I do not understand the difference between these two terms. If you think there is one, please could you define what it is. ... In your example you seem to be asking "Where is the missing energy?" The related questions have already addressed this issue, as does MDC's answer. $\endgroup$ – sammy gerbil Oct 30 '16 at 19:16
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There are several conceptual errors.

  • You assume that the work done is the same in both frames. But work is relative as well. One way to see this is to note that $$W = \int \mathbf{F} \cdot \mathbf{v}\, dt.$$ Though your two observers agree on $\mathbf{F}$, they don't agree on $\mathbf{v}$, so they won't agree on $W$. In particular, in the second frame, the spring is doing negative work on the block.
  • You assume that the change in elastic potential energy is equal to the change in kinetic energy of the block. This is incorrect, because you have to account for the change in kinetic energy of whatever's on the other end of the spring.
  • One of the answers claims that the amount of energy stored in a spring is relative. This is incorrect, and you are correct.

If we get rid of these issues, the answer comes out right. Let's say the block has mass $m$, and is attached to a large object of mass $M \gg m$ on the other end of the spring. Consider your second observer. In their frame, the initial energy is $$\frac12 mv^2 + \frac12 Mv^2.$$ Next, we allow the spring to extend. The speed of the block changes from $v$ to $0$. By momentum conservation, the speed of the large object changes from $v$ to $v+(m/M)v$. The final energy is $$\frac12 M (v+(m/M)v)^2 \approx \frac12 Mv^2 + mv^2.$$ That is, the energy has increased by $mv^2/2$, as we expect. This is exactly the elastic potential energy initially stored in the spring, as you computed in the first observer's frame.

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  • $\begingroup$ Elegant and lucid. But I suggest that Q, the energy stored in the spring be added to the first equation. The moving observer sees that as part of the initial energy in the system. Then the last sentence could be more explicit about the fact that both the elastic energy and the block's kinetic energy went to zero, while the KE of the larger mass, M, increased by their sum. It's all in there now, but the reader has to dig it out. $\endgroup$ – D. Ennis Oct 31 '16 at 12:50
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I think Luca is right.

We have to remember that, in the moving frame, the compressed spring and block (before the spring is released ) will also be observed to be moving, and so they will already possess both kinetic and potential energy.

This original kinetic energy can always be subtracted from the total kinetic energy measured after the spring is released.

The resulting amount of kinetic energy should be identical to what was measured by the stationary observer.

$(KE_{spring + block})_{before}+(PE_{spring + block})_{before}=(KE_{spring})_{after}+(KE_{block})_{after}$

So, although the block may be seen to be moving either slower or faster in the moving frame than what it would be in the stationary frame, it is in fact the very motion of the frame which should correct for the "missing kinetic energy".

This same argument (with some basic modifications) can be shown to be true even if relativistic effects are significant.

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I think you are making a false point, based on an ambiguity on the definition of Hooke's law: $$ \vec f = - k \vec x $$

means that you take $\vec x\ $ as the shift from the rest position $\vec x_0\ $ of the spring. In a system in which the spring is at rest, this shift is the same as the $\vec s\ $ in the work formula $$W =\int \vec f\cdot \vec s = \frac12 k x^2 $$

Now, when you compute work done in another reference system, you cannot use this formula. In your example, if an observer moves towards the block (let's say from left to right), he will see it moving slower, so he will measure his KE at midpoint as less than $ \frac12 k x^2 $, but the block, in his reference system, has moved towards the left some time after the spring was released (its $v_0 < 0 $ in that reference frame), so the work made by the spring in that reference system is negative until the velocity of the block becomes positive, that is until the speed of the block in the static reference system exceeds the speed of the moving reference system. So, it's no surprise that its KE is inferior: in the moving reference frame, the spring made less work on the block. What remains true is that $\Delta K = L$: to see this, just take the case in which $\ v_R = \frac12 v_{MAX}$. Then $\ v_0' = - v_{MAX}' \ $ and $\Delta K' = 0$, but as you can see from the integral, $L' = 0$ too. (I'm using the apostrophe for the moving reference frame)

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  • $\begingroup$ At best you are showing that the moving observer would calculate a different amount of energy released, based upon observation of the spring's motion, than would the stationary observer. However, they didn't do that. They agreed a priori that Q joules were stored and released. I could have used an iron nail shot from a solenoid instead of the block and spring. $\endgroup$ – D. Ennis Oct 29 '16 at 23:10
  • $\begingroup$ D.Ennis you are using the formula $U = \frac12 k x^2$. This is correct only in the reference frame where the spring is stationary. $\endgroup$ – Luca Oct 29 '16 at 23:15
  • $\begingroup$ Maybe our comments crossed in the mail. $\endgroup$ – D. Ennis Oct 29 '16 at 23:20
  • $\begingroup$ I don't understand what you mean. My opinion about your question is that it is based on the false premise that in different reference frames the energy stored is the same: you can use the formula $U= \frac12 k x^2$ only in the static frame. D.ennis, potential energy and kynetic energy are not invariant, their variations are. $\endgroup$ – Luca Oct 29 '16 at 23:27
  • $\begingroup$ I will try once more. Both observers agree that k is k, x² is x², and therefore that the energy stored in the spring is ½kx², which is Q joules. If you would like, imagine that they compressed the spring together in the stationary frame, and then one of them stepped into the moving frame. $\endgroup$ – D. Ennis Oct 29 '16 at 23:42
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according to the frame moving with v velocity both block and spring were moving with v velocity in opposite direction when spring force acts on the block it acts in direction opposite to the direction of relative velocity of block with respect frame moving with velocity v.Now in the your frame of reference i.e. the frame moving with velocity v the block is moving in direction opposite to the conservative spring force acting on it and when it finally stops its K.E has been converted into P.E..Now as far a your given energy Q is concerned w.r.t stationary frame it is 1/2kx^2 as block is moving in direction of spring force but w.r.t. frame moving with velocity v it -Q as now work done by spring on block is -1/2kx^2.To be more precise in stationary frame initial potential energy is Q which is changing into kinetic energy of block and in frame moving with velocity v initial potential energy of spring is Q and kinetic energy of block is 1/2mv^2,this kinetic energy is changing into potential energy of spring as work done by spring force i.e. conservative force is negative in this frame.

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protected by Qmechanic Nov 1 '16 at 14:01

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