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Today, a fellow student and I had a discussion and we can't figure out where we're making a mistake.

Suppose you have a car with $v_1$ = 10 m/s that's heading towards another car with $v_2$ = 0 m/s. The kinetic energy of the car is then $E_1 \propto v_1^2$ and $E_2 = 0$. Thus $E_{\text{sum}} = E_1 \propto v_1^2$. If I put myself in a moving reference frame, that starts in the middle of both cars with a velocity of $v_\text{ref} = \frac12 v_1$ = 5 m/s towards the still car, I perceive both cars moving with a velocity of $v'_1 = -v'_2$ = 5 m/s = $\frac12 v_1$. For the kinetic energy follows $E_{\text{sum}} = E_1 + E_2 \propto 2 \left(\frac12 v_1\right)^2 = \frac12 v_1^2$.

So it seems that in my reference frame, the collision is not as devastating, as it factually is. I know this can not be right, because energy is conserved and it should be conserved regardless of my reference frame, right? So, where is the hidden mistake?

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  • $\begingroup$ You're calculating relativistic energy not rest/invariant energy en.wikipedia.org/wiki/… $\endgroup$
    – Yogi DMT
    Oct 27 '16 at 19:56
  • $\begingroup$ Why should I calculate relativistcally for velocities v << c? $\endgroup$ Oct 27 '16 at 19:57
  • $\begingroup$ Please note that Phys.SE is a MathJax-enabled site; so use that to increase the readability of the post. For a quick look, please visit this meta Math.SE post. $\endgroup$
    – user36790
    Oct 27 '16 at 20:54
  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/1368/2451 , physics.stackexchange.com/q/230054/2451 and links therein. $\endgroup$
    – Qmechanic
    Oct 27 '16 at 22:50
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The total energy is not the relevant quantity. It's the change. How much of that kinetic energy is removed to do (harmful) work on the vehicles?

If we assume both cars have similar mass, then immediately after the collision, the crumpled wreckage is still moving with a speed approximately $\frac{1}{2}v_1$. Friction will eventually slow this to equal the velocity of the ground, but that's not where the damage comes from. (Imagine the collision is on an ice rink).

This means that a quarter of the moving car's energy is retained as kinetic energy, another quarter goes into raising the kinetic energy of the car at rest, and the other half of the initial $KE$ goes into deformation. This matches the energy loss you calculated in the frame where the center of mass was at rest.

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Conservation of energy does not imply that energy is invariant under a frame change. These are two different concepts. The kinetic energy depends on the reference frame so it is not invariant under frame transformation. Conservation means that in a given frame, the total quantity does not change. For example, for perfectly elastic collision, the KE before and after the collision is the same so the KE is conserved. But this conserved quantity will have difefrent values in different reference frames.

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  • $\begingroup$ I find the different values difficult to understand. In my example above, the crash seems to happen with half the kinetic energy. Yet, the effects are obviously the same. This seems contradicting. $\endgroup$ Oct 27 '16 at 20:03
  • $\begingroup$ This has nothing to do with the crash. The KE depends on the reference frame for any process. There is no contradiction in KE having different values in a different reference frame. $\endgroup$
    – nasu
    Oct 27 '16 at 20:07
  • $\begingroup$ Same reason why the 8m 20s old Sun that we see isn't the state the Sun is currently in, because the physics of the Sun doesn't care about what observers perceive. $\endgroup$
    – Yogi DMT
    Oct 27 '16 at 20:11
  • $\begingroup$ Maybe it will help to think that the effect of the crash depends not on the values of the KE before collision but on how much KE energy was dissipated during the collision. For perfectly elastic collision this is zero in any reference frame. For plastic collisions the KE energy converted to other forms is invariant even though the values of KE of the bodies are frame dependent. $\endgroup$
    – nasu
    Oct 27 '16 at 20:20
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While kinetic energy isn't a conserved quantity, we can look at how much kinetic energy is dissipated by the collision.

As an example, let's assume a completely inelastic collision, i.e., the cars stick together and have a final common velocity. And we must conserve momentum within the reference frame.

For the ground based observer, the initial kinetic energy is $\frac{1}{2}mv_1^2$ and the momentum is $mv_1$. After the collision, the cars are moving at velocity $V$, so $$2mV=mv_1$$ $$V=\frac{v_1}{2}.$$ That means the final kinetic energy is $$\frac{1}{2}\left(2m\right)V^2 = \frac{mv_1^2}{4},$$ for a kinetic energy loss of $\dfrac{mv_1^2}{4}$, most of which went into the destruction.

Now let's consider the center of mass reference frame:$$\text{Kinetic energy = }\frac{1}{2}m\left(\frac{v_1}{2}\right)^2+\frac{1}{2}m\left(\frac{-v_1}{2}\right)^2 = \frac{mv_1^2}{4}$$ $$\text{Momentum = } 0$$

When we conserve momentum after the collision, the final velocity of the cars, now stuck together, is zero! The final kinetic energy in the center of mass is zero, so the loss of energy to the collision is $\dfrac{mv_1^2}{4}$, exactly the same as the loss in the first reference frame.

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Bear with me, because after the numerical examples I'm going to try to get to the heart of your question.

If $car_1$ with mass M were traveling at $10m/s$ because its motor had burned some fuel to accelerate it to that velocity, it would have $50Mm^2/s^2$ (or 25|M| J) of KE.  If it collided with $car_2$ of mass M which was resting at $0m/s$, and you were a stationary observer, you would see the two cars move off together after the collision with a combined mass of 2M at $5m/s$, according to the rules of conservation of momentum, with 25|M| J of KE, and you would conclude that the other 25|M| J of KE had been transferred to thermal and sound energy in the collision.

If you were running toward $car_2$ at 5m/s and the same collision occurred, you would see each of the two cars as moving at $5m/s$, and calculate that they each had 12.5|M| J of KE, and you would perceive them as coming to a halt when they collided.  Again you would say that 25|M| J of KE had been transferred to thermal and sound energy.   The crash was the same.

If you were running toward $car_1$ at $10m/s$ you would see $car_1$ moving at $20m/s$ and calculate that it had 200|M| J of KE, and would see $car_2$ moving at $10m/s$ with 50|M| J of KE.   After the crash they would appear to be moving at $15m/s$, and with their combined mass of 2M would appear to have 225|M| J of KE.   So yet again you would say that 25|M| J of KE had been transferred to thermal and sound energy.   The crash was the same.

These are examples of what is meant by kinetic energy being reference frame dependent. The calculated pre-collision KE for each car can vary greatly with the chosen frame of reference, yet they predict the same result in objective reality.   All observers would see the same final effects on the cars.   So KE is dependent on the frame of reference in the context of mathematical models, for the purposes of those mathematical models.   And there are examples of such modeling which are more sophisticated than this one.   Yet they all work out like this one to be true to reality because they must.   And what is this reality?

It is that there is only a certain amount of energy in the universe, and if some of it is transferred to an object in the form of kinetic energy, then that's where it stays until it gets transferred out of the object to something else.   In all of the scenarios above, the kinetic energy was actually in $car_1$, and its amount was 50|M| J.   That is because 50|m| J of the energy which had been stored as chemical energy in a certain amount of fuel and oxygen was transferred to $car_1$.   That fact does not depend on the frame of reference.   Energy transfer is objective.   However, once the car is moving, observers in different reference frames will measure different velocities and calculate different values of KE.   But the frame of reference models worked, because total energy is conserved even if we don't know or care which car really had how much kinetic energy. So you were correct when you said, ". . . energy is conserved and it should be conserved regardless of my reference frame, right?"   Yes, right.

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    $\begingroup$ You seem to confuse KE with total energy. And the statement that KE is not frame dependent is misleading, as it is definitely not true. The fact that is frame dependent does not contradict conservation of KE or conservation of total energy. Rather than help eliminate the confusion of the person asking teh question you contribute to confusion, $\endgroup$
    – nasu
    Oct 28 '16 at 14:15
  • $\begingroup$ @nasu That's a pretty hefty pronouncement. I don't recognize your authority to make it. Please restrict your criticism to ideas. I don't believe that you understood my answer, but apparently the OP did. And by the way, kinetic energy is not a conserved quantity. Only total energy is. Do we agree on that? $\endgroup$
    – D. Ennis
    Oct 28 '16 at 16:13
  • $\begingroup$ I did not comment on anything but your ideas expressed in the answer. And the comment is not based on my "authority" but on basic classical mechanics. KE may be conserved in some processes, like elastic collisions. But this has nothing to do with frame invariance as these are two distinct concepts. $\endgroup$
    – nasu
    Oct 28 '16 at 19:09
  • $\begingroup$ Well this gave me the perspective I needed. So thank you :-) $\endgroup$ Oct 28 '16 at 22:05
  • $\begingroup$ @nasu Is the amount of water in a graduated cylinder frame dependent? No. Is the amount of energy stored in an alkaline cell frame dependent? No. Is the amount of energy stored in a compressed spring frame dependent? No. Is the amount of energy stored in the motion of an object frame dependent? No. Is the measurement of the energy stored in the motion of an object frame dependent? Yes. Do the different values in different frames predict different outcomes? No. Agree? $\endgroup$
    – D. Ennis
    Oct 29 '16 at 5:15

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