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.