# How does constant thrust avoid quadratic kinetic energy accumulation?

I apologize in advance, but I'm going to make the problem look worse first so that I can explain it:

Imagine that instead of you measuring the velocity from the station where the rocket is launched - let's call it station A - you measure instead from station B, which is moving away from station A at some high speed. And imagine that station A launches two identical rockets, one toward you and one away from you.

Since we're operating under the well-tested current theory of special relativity, we understand that we can either say "B is moving away from A" or with equal correctness "A is moving away from B" since we can declare either one as "stationary" for purposes of measuring velocities with respect to our chosen coordinate system. So let's start over and say that Station A is moving at speed away from Station B and launches two rockets.

Of course, before they launch from A, both rockets will have a certain kinetic energy based on their motion, and that energy will be equal for the two identical rockets. When we launch them, however, that begins to change: one of them is starting to move away from us even faster, while the other begins moving away from us slower. After some time, in fact, one will be moving away from us at twice the speed it originally had - that is, four times the kinetic energy - while at the same time the other is moving away from us at zero relative velocity: no kinetic energy whatsoever! Now we're really mixed up, because both rockets did the exact same thing but one ended up with a very high kinetic energy and one with no energy whatsoever.

The issue I hope to highlight with that example is that kinetic energy is frame-dependent and not an intrinsic energy for an object. That's why you can safely ride along in a car moving at 100km/h along a road but not be safely struck by a car moving at 100km/h along a road; if you travel with the car, it has zero kinetic energy in your frame, but if you're standing on the road it has a very high kinetic energy in your frame. Kinetic energy is relative.

On another note, I said before that the two rockets are doing "the same thing," but in any given single frame of reference that's not quite true. Obviously, in my example above, from the frame of station B one rocket is losing kinetic energy and the other is gaining kinetic energy, for example. But that's only considering the rockets themselves: those are being propelled by launching photons the other direction, and those photons are expelled with a certain energy themselves. If you were to measure the energy of the photons themselves, (attaching a small mirror to the back of the rocket headed toward you so that you can see the "exhaust") you would notice that over time the exhaust of the rocket moving away from you seems less energetic and the exhaust of the rocket moving toward you seems more energetic. That is to say, photons launched toward you are less energetic than photons launched away from you... which was the same conclusion we came to with the rockets themselves launching from Station A, and again is a consequence of the relativity of kinetic energy. But this time we can more directly say that one is "red-shifted" and the other "blue-shifted."

EDIT: It may also be simpler to imagine the solution if you use a normal reaction drive with massive exhaust. The exhaust is low mass but leaves at high speed; the rocket is higher mass and thus gains less velocity from the reaction. The kinetic energy increases by the same amount in opposite directions for each body of mass (taking all the exhaust as a "body"), otherwise we're getting energy for free from somewhere. Also, since massive exhaust has subliminal expulsion speed, you can have a case where the rocket is accelerating away from you but the exhaust is also moving away from you, which isn't possible with photon drives. But in all cases, photons or not, the total energy of the system is conserved: measured from an inertial coordinate system, the rocket "gains" the same amount of energy that the exhaust "loses," and the measured exchange rate does increase quadratically.

## TL;DR:

As the rocket goes faster, the emitted photons carry less energy in the rest frame. This means more energy is available for the rocket. Conservation of energy and momentum are shown to give exactly the result we expect, even in the classical low-speed limit.

To figure out what is going on, we need to consider what happens to both momentum and energy of rocket and photons.

Without loss of generality we can assume the photons emitted are blue in the frame of reference of the rocket - it makes the description easier.

When the rocket is moving slowly, the photons still look blue in the frame of reference of a stationary observer: they appear to carry all the energy. As the rocket accelerates, the photons will be Doppler shifted, and become "redder" - they carry less energy away after the interaction with the rocket. So for the same number of photons per unit time, the rocket feels the same thrust, but (in the stationary frame) it results in greater increase in energy as the photons have less energy.

Mathematically, if we emit $N$ photons of wavelength $\lambda_0$ in a short time interval, then for a rocket of mass $m$ with velocity $v$ the total change in momentum is

$$m\Delta v = N\frac{h}{\lambda}$$

The energy increase of the rocket is

\begin{align} \Delta E &= \frac12 m (v_1^2 - v_0^2) \\ &= \frac12 m (v_1+v_0)(v_1-v_0) \\ &= v\Delta p\\ &= \frac{Nvh}{\lambda_0}\end{align}

The energy of the (Doppler shifted) photons is (in the non-relativistic approximation)

$$\lambda_1 = (1+\frac{v}{c})\lambda_0$$

And the energy is

\begin{align} E &= N\frac{hc}{\lambda_1}\\ &= N\frac{hc}{\left(1+\frac{v}{c}\right)\lambda_0}\\ &\approx N\frac{hc}{\lambda_0}\left(1-\frac{v}{c}\right)\end{align}

As you can see, the sum of this energy plus the energy gained by the rocket is constant - as the rocket goes faster, the photons carry away $N\frac{hv}{\lambda_0}$ less energy - and that is precisely the energy gained by the rocket.

If the rocket reaches relativistic velocities, the calculation is more complex but the conclusion is the same. A proper calculation would also not ignore the mass loss of the rocket due to the energy emitted, but again that doesn't detract from the fundamental principle at play.