# Is the total energy conserved in a moving reference frame? [closed]

I have a ball attached to a spring and the spring is attached to a wall. There is no gravity for simplicity. In the rest RF the oscillating ball energy is conserved: T + U = const. In a moving RF it is not conserved. I would like to see the shortest answer to the question "Why?".

| <-- -->
|/\/\/\/\/\/\/O
|

=====> $V_{RF}-->$

Edit: For those who has doubt - I choose the moving RF velocity equal to the maximum ball velocity, for example. Then the ball with its maximum velocity in the rest RF looks as still and does not have any potential energy in the moving RF (the maximum velocity is attained at the equilibrium position where no force acts). So T = 0 and U = 0.

At the farthest right position the ball looks as moving with $-V_{RF}$ and having a potential energy $kx^2/2$, both energy terms are positive ($v^2 > 0, x^2 > 0$). So now the total energy of the ball is positive and thus is not conserved. It oscillates from zero to some maximum value. Why?

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## closed as not constructive by pho, Moshe, Matt Reece, Kostya, gigacyanFeb 8 '11 at 22:33

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Downvoters, give your short objections, please. –  Vladimir Kalitvianski Feb 8 '11 at 20:18
Minus one point for abusing the space that is reserved for questions for spreading primitive misconceptions about physics. Of course that energy is conserved in any - moving - reference frame. You may need to include the terms from the wall, too. But energy is conserved in any system with a time-translational invariance. –  Luboš Motl Feb 8 '11 at 20:19
Thanks, Lubosh, you are so kind as usual. But I do not want to include the wall. I ask another question. So it is -1 to you. –  Vladimir Kalitvianski Feb 8 '11 at 20:21
Maybe it is a good idea to rephrase the question and e.g. tag it as a [riddle]. Because it makes perfectly sense to consider such a RF and then address the apparent contradictions. One can learn a lot from studying such special situations. –  Gerard Feb 8 '11 at 22:41
I deleted the tail of the comment thread, since it was getting out of control. If there continue to be argumentative comments posted, I'll lock the question. –  David Z Feb 9 '11 at 0:15

We must consider the kinetic energy of the wall to see energy conservation. Taking the limit where the wall's mass goes to infinity does not alleviate this responsibility because although the wall's oscillations become infinitely small in that case, the oscillations in its energy do not.

To see this roughly, imagine that the wall is very massive and moving sinusoidally. In the center of mass frame, the maximum speed of the wall is $s$. Then the fluctuations in its kinetic energy are proportional to $M s^2$. Thus, in the rest frame, we can make the wall's energy fluctuations arbitrarily small by increasing $M$, because $s^2$ will decrease more rapidly.

Now go into a moving frame where the average speed of the wall is $v$. The fluctuations in its energy are now proportional to $M(v+s)^2 - Mv^2 \approx 2Mvs$ in the limit $v>>s$. For fixed $v$, we can make $s$ smaller and smaller by increasing $M$, but as we do so the fluctuations in the energy of the wall don't change size because $s \propto 1/M$.'

Next note that the fluctuations in the energy of the ball in this frame are proportional to $mvs'$, with $s'$ the speed of the ball in this frame. By conservation of momentum, $Ms = ms'$, so the fluctuations in energy of the ball and wall are about the same size in the moving reference frame. Thus, we must account for both of them.

I'll give a fuller answer in one dimension for simplicity. I'll first find the motion in the center of mass frame, then transform it and show that energy is conserved in the tranformed frame.

Let the ball have mass $m$ and the wall have mass $M$. The spring constant is $k$. Let the ball's position from equilibrium be $x$ and the wall's position be $X$. The Lagrangian is

$$\mathcal{L} = \frac{1}{2}\left(M\dot{X}^2 + m\dot{x}^2 - k(X-x)^2\right)$$

The equations of motion are

$$-k(X-x) = M\ddot{X}$$

$$k(X-x) = m\ddot{x}$$

Adding these gives the conservation of momentum

$$\frac{\textrm{d}}{\textrm{d}t}(M\dot{X} + m\dot{x}) = 0$$

In the center of mass frame this momentum is zero, so

$$\dot{x} = \frac{M\dot{X}}{m}$$

With the initial condition $x(0) = X(0) = 0$ we can integrate this to

$$X = -\frac{m}{M}x$$

The equation of motion for the small mass becomes

$$k(\frac{1}{M} + \frac{1}{m}) x = \ddot{x}$$

with the solution

$$x = A \sin(\omega t + \phi)$$ $$X = -\frac{m}{M}A\sin(\omega t + \phi)$$

with $\omega^2 = k(\frac{1}{M} + \frac{1}{m})$.

Now boost to a frame moving at speed $v$.

$$x = A \sin(\omega t + \phi) - vt$$ $$X = -\frac{m}{M}A\sin(\omega t + \phi) - vt$$

The potential energy is in the spring

$$U = \frac{1}{2}k(X-x)^2 = \frac{1}{2}k \left(A (1+\frac{m}{M})(\sin(\omega t + \phi) \right)^2$$

The kinetic energy is in the motion of the ball and wall

$$T = \frac{1}{2} \left(A \omega \cos(\omega t + \phi)\right)^2 \left(m + M(\frac{m}{M})^2\right)$$

Add these and applying the identity $\sin^2x+\cos^2x = 1$ gives

$$E = \frac{1}{2}k\left(A(1+\frac{m}{M})\right)^2$$

which is constant in time.

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@Vlad did you down-vote? The fact that it wasn't the language you wanted doesn't make it incorrect or deserve a down vote. –  Mark Eichenlaub Feb 8 '11 at 22:13
Yes, it was me. I said many times I do not want to consider the wall. –  Vladimir Kalitvianski Feb 8 '11 at 22:16
@Vlad It's there. It's reasonable to consider it. That was the point of my answer. Your behavior is annoying. –  Mark Eichenlaub Feb 8 '11 at 22:20
Sorry to hear that. –  Vladimir Kalitvianski Feb 8 '11 at 22:25
@Vlad That's just a general feature of mechanics. It's completely unenlightening in this scenario and as vacuous as explaining why something is accelerating by saying "because there is a force on it". –  Mark Eichenlaub Feb 8 '11 at 22:59

Suppose that the wall has some large but finite mass. Then the actual motion of the system involves oscillations of the wall as well as oscillations of the mass. If you work out the total energy of the entire system, you'll find that it's conserved.

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I know, Ted, I know. Where is your short answer to the post question? –  Vladimir Kalitvianski Feb 8 '11 at 21:16
If you're looking for an answer to the question "why is the energy of the ball not conserved?" then the answer is that there's no reason to expect the energy of just one part of a system to be conserved. Conservation of energy applies to the total energy. –  Ted Bunn Feb 8 '11 at 21:22
And I apply it to the total energy E = T + U of the ball. In the rest RF it is conserved, in a moving RF it is not. –  Vladimir Kalitvianski Feb 8 '11 at 21:31
The simple problem here is in sharp contrast to the "big science" (at first sight) in Vladimirs pages. I recommend to have a visit there. –  Georg Feb 8 '11 at 21:57
I don't understand the question and frankly doubt your seriousness in asking it. I'm done. –  Ted Bunn Feb 8 '11 at 22:48

Energy is not a Lorentz scalar (or Galilean scalar). In different reference frames, the values of energies wil also be different, but this does not mean that there is no energy conservation. The energy is still conserved in each reference frame.

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-1 for not answering my question. The energy is not only different but is not conserved. –  Vladimir Kalitvianski Feb 8 '11 at 21:06
@Vlad You're right that the answerer did not directly answer your question, but the name-calling is unnecessary. –  Mark Eichenlaub Feb 8 '11 at 21:58