# How does Kinetic Energy transform from one frame of reference to another?

Let's say I observe a charge q floating in the middle of space. I set up another charge Q a distance d away from charge q was causes a force F to act on charge q. As charge q moves away from charge Q, I adjust the position of Q so that it always remains a distance d away from q. I do this until charge q has moved a distance d. Now since the distance between q and Q is being kept constant, the force on charge q will also be constant. This constant force F over a distance d means that charge q will have had an amount of work W done on it equal to F$$\times$$d. This work manifests itself as a higher kinetic energy E:

$$E=F\times d = \frac{kqQ}{d}$$

(k is some constant designed to make the units work out.)

Which means that I can work out q's speed v:

$$v=\sqrt \frac{2E}{m} = \sqrt \frac{2Fd}{m} = \sqrt \frac{2kqQ}{md}$$

This is all straightforward.

Now let's say that some other observer witnesses me performing this experiment. She is moving along the same axis as the experiment at a speed v$$_{RF}$$ (RF for Reference Frame). She is moving away from the experiment and according to her, charge q which previously had a speed v$$_{RF}$$, is further boosted along by my experiment to a speed v'. I now want to ask the question, how fast is speed v'?

I could do this 2 ways:

1: The observer initially sees charge q going at speed v$$_{RF}$$ which means she calculates its Kinetic Energy to be:

$$E_{RF} = \frac{mv_{RF}^2}{2}$$

Once I've done my experiment and transferred energy $$E=\frac{kqQ}{d}$$ to charge q, the observer sees this additional amount E added to the amount $$E_{RF}$$, or E' = $$E_{RF}$$ + E. Therefore, according to the observer, charge q travels at speed:

$$v'=\sqrt \frac{2E'}{m} = \sqrt \frac{2(E_{RF}+E)}{m}$$

2: Using the velocity addition rule to get the speed that the charge appears to be moving at relative to the observer, we just add speed v that I observe charge q to be going at to the speed of the observer v$$_{RF}$$.

$$v' = v + v_{RF}$$

We have already worked out that the speed v is equal to:

$$v=\sqrt \frac{2E}{m}$$

We can also see out that v$$_{RF}$$ is equal to:

$$v_{RF}=\sqrt \frac{2E_{RF}}{m}$$

Therefore, the speed that the observer measures charge q to be going at is:

$$v' = \sqrt \frac{2E}{m} + \sqrt \frac{2E_{RF}}{m}$$

So which is it?

$$v'= \sqrt \frac{2(E_{RF}+E)}{m}$$

Or

$$v' = \sqrt \frac{2E}{m} + \sqrt \frac{2E_{RF}}{m}$$

Putting it another way, what adds together, the speeds or the kinetic energies?

On the one hand the speed transformation rule is found in every elementary physics textbook. On the other hand, it's hard to argue that E shouldn't be added to E$$_{RF}$$, given that we worked out that $$E = \frac{kqQ}{d}$$, and none of these factors depend on frame of reference. Yet, obviously they cannot both be right. So which one is it?

Kinetic energy is not invariant under Galilean-transformations$$^*$$. Therefore, you can't take the energy added in the first scenario and add it to the initial energy in the second scenario. If you want to work with kinetic energy in the second scenario, you must use velocities in that frame. Therefore, the velocity addition is correct. The energy addition is not.
$$^*$$ This can easily be seen in a simple example. If I see a ball moving by me, I observe it to have kinetic energy. If someone is moving with the ball, they observe it to have no kinetic energy.