Suppose a rocket at rest turns on its jet and is now moving with kinetic energy $$\frac12 mv^2$$ in the lab frame. Suppose the efficiency was maximum so that exactly $$\frac12mv^2$$ of chemical energy was just converted to this kinetic energy. Energy is conserved, obviously.

Now suppose I stay in the rocket's frame. In this frame, the rocket stays at rest but now the whole universe, say with mass $$M\gg m$$, is now moving towards me with energy $$\frac12 Mv^2$$. Energy was clearly not conserved in this frame. Presumably this is because it is an accelerating frame, and energy is only conserved in inertial frames. But my intuition says that as the acceleration approaches zero, the results should be the same as in an inertial frame. However, it is easy to check that this huge energy violation persists for arbitrarily small accelarations and at arbitrarily short times.

What is wrong with my thinking?

• you are trying to add pears with apples, you can't do that - conservation laws applies only in same reference frame Commented Feb 23, 2020 at 10:31
• The rocket frame is a single (albeit accelerating) frame. Commented Feb 23, 2020 at 12:20
• Why do you say that the rocket frame is accelerating? If I understand you correctly, you have considerd the case where the rocket has finished accelerating and has velocity $v$. Commented Feb 23, 2020 at 12:36
• @EricDavidKramer In rocket frame it has $0$ kinetic energy because it is at rest and only outside approaching objects has own kinetic energies. So you have misunderstood how and when apply conservation laws. Commented Feb 23, 2020 at 13:18
• @AgniusVasiliauskas I said it's at rest. You didn't answer my question, you just said it's wrong, which is what I also said. Commented Feb 24, 2020 at 11:36

What is wrong with my thinking?

There are a couple of points where you make key errors. The first (and least important) is here:

Suppose the efficiency was maximum so that exactly $$\frac{1}{2}mv^2$$ of chemical energy was just converted to this kinetic energy.

This is not possible for a rocket that begins at rest because it does not simultaneously conserve momentum. For a rocket at rest, at that moment, all of the power goes into the KE of the exhaust, not the rocket.

Even assuming no waste of any kind, the only time that 100% of the KE goes into the rocket is at the moment that the speed of the rocket equals the exhaust velocity. In general, for a constant $$\Delta PE$$ of the chemical potential energy in the fuel the $$\Delta KE$$ of the rocket depends strongly on the speed of the rocket.

In this frame, the rocket stays at rest but now the whole universe, say with mass 𝑀≫𝑚, is now moving towards me with energy $$\frac{1}{2}Mv^2$$. Energy was clearly not conserved in this frame. Presumable this is because it is an accelerating frame, and energy is only conserved in inertial frames. But my intuition says that as the acceleration approaches zero, the results should be the same as in an inertial frame. However, it is easy to check that this huge energy violation persists for arbitrarily small accelarations and at arbitrarily short times.

In the rocket frame the universe is subject to a fictitious gravitational acceleration $$g$$. So if the universe falls through this fictitious gravity field for a time $$t$$ then the velocity of the universe is $$v=gt$$. So we can write the KE of the universe as $$\frac{1}{2}M(gt)^2$$. If we expand that to first order in $$t$$ we get $$0+O(t^2)$$ and if we expand that to first order in $$g$$ we get $$0+O(g^2)$$. Therefore your intuition is correct.

Since you didn’t post your math it is impossible to know what led to the mistake, but at least in this case you are correct to intuit that as the time or the acceleration go to 0 so does the energy violation.

Be aware, if the acceleration is constant in time then it is possible to make a potential energy in this frame and recover conservation of energy. The universe falls through the gravitational field losing PE and gaining KE, thus conserving energy. That only works if the acceleration is constant.

• Nice answer. But the ratio of the total kinetic energy to the energy "spent" is still $M/m$ for abitrarily small times and accelerations. (There is probably an $\mathcal{O}(1)$ factor to account for my first mistake, but this ratio will still be huge.) Commented Feb 23, 2020 at 12:42
• Yes, the ratio is indeed huge. As the amount of KE spent goes to 0 the amount of KE violated goes to 0, but the ratio is indeed non-zero. I don’t known why you would expect the ratio to go to zero rather than the amount.
– Dale
Commented Feb 23, 2020 at 12:51
• It just means that the violation doesn't go away in the limit. Commented Feb 23, 2020 at 12:52
• I would say that the amount of violation does go to zero, just not the ratio. If your intuition is that the ratio should go to zero then your intuition is indeed wrong. It is the amount of violation that should and does go to zero. 1000000% of zero is zero even if it is still 1000000%
– Dale
Commented Feb 23, 2020 at 13:03

In an inertial frame, you should not see chemical energy being converted into kinetic energy. If you do, then you are also experiencing acceleration, and you will not be in an inertial frame.