# Energy is not frame independent

For a closed system we have energy conservation and this is deeply tied into the notion of the physics being unaltered under shifts in time. A salient point here is that we have fixed a frame in which to measure the energy. It is not true that the energy is the same measured in any inertial frame, what is true is that it is conserved.

A simple example

Let us consider a free particle of mass $$m$$ in one dimension. Let us pick some coordinates $$(x,t)$$. With respect to this frame we have that the kinetic energy

$$E = \frac{1}{2 m}p^{2}$$,

where the momentum $$p$$ is given by $$m v = m \dot{x}$$. The dot is the derivative with respect to time. (I am using the usuall abuses of notaion here, this should not confuse)

Now let us consider a Galilean transformation

$$x’ = x {-} ut$$
$$t’ =t$$.

Then we calculate how the momentum transforms

$$p = m \frac{d}{dt}\left( x’ – ut \right) = m \left( v’ +u \right)= p’ + mu$$.

Or $$p’ = p – mu$$.

This is exactly what you would expect. So on to the energy…

$$E = \frac{1}{2m}p^{2} = \frac{1}{2} \left(p’ + mu \right)^{2}$$
$$= E’ + \frac{1}{2}mu^{2} + up’ = E’ – \frac{1}{2}mu^{2}+ up$$

so

$$E’ = E + \frac{1}{2}mu^{2} -up$$.

Clearly $$E’ \neq E$$

.

Recap

Energy conservation should not be confused with energy being observer independent, it clearly is not.

# Marvin and Milo

Explore physics the exciting way, by trying out a simple and fun experiment. This month, Pouring light Don’t forget to check back next month for the next edition…

Do try this at home…
Marvin and Milo