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.
Thanks for the heads up! 😀
1) Typo above: should be x’=x-ut
2) If one looks closely at your derivation one sees that energy is frame dependent but change in energy is not as the two, for a fixed mass m, are related by a constant additive factor. This, of course immediately implies the invariance of conservation of energy. Similarly, and in fact necessary to the observation regarding energy, one sees from that same work that change in momentum is invariant, implying the invariance of conservation of momentum. The invariance of change in momentum and energy is also a useful observation in the analysis of collisions — you cannot survive a car crash by simply viewing the collision in a different reference frame.
Just FYI — with the new LaTeX system, you can get display-style block equations with the same delimiters you’d use in LaTeX:
[ frac{-b pm sqrt{b^2 – 4ac}}{2a} ]
That’s using [ and the matching closing delimiter.
@DrRocket: thanks for the typo, now corrected and I hope that did not confuse anyone.
Your second point is absolutely correct. The thing that one might miss out naively is the “cross-term”, which is this additive factor.
@Xittenn: There seems to be some confusion of the forums, and elsewhere about the idea of conservation of something (Noether’s theorem) and the invariance of something. Energy and momentum are conserved in closed mechanical systems, but this does not mean that everyone will agree on the actual values.
I can handle that, thanks again. 🙂