Energy is frame-dependent, the infrastructure of energy is frame-independent

The concept of energy and the elaborate mathematical infrastructure that surrounds it are so fundamental to mechanics that they are generally introduced less than half-way through the first semester of the introductory course in a college classroom (true whether you are doing a force-first or momentum-first introduction).

But I'm pretty sure I'm not the only person to have said

"Wait a minute, how the #&%* did I miss that?!?"

when teaching that "elementary" infrastructure after using it for years. Not because I wasn't paying attention the first couple of times through (I had physics in high school before taking it in college), nor because "that" is deep or mysterious, but because I didn't have the context to appreciate the importance of "that" until much later in my education.

As far as I can tell there are several distinct things which qualify as "that" for various people, and today I want to talk about one of them.

Kinetic energy is frame dependent

Obvious if you think about it, right? I mean $T = \frac{1}{2} m v^2$ with $v$ depending on the frame.

But the Work-energy theorem
$$ W_\text{net} = \Delta T \;
$$ is fundamental to the whole infrastructure. How does that work (heh!) if kinetic energy depends on the observer's state of motion?

Work is frame dependent, too

Well, work depends on path length and *that* also depend on the observer's state of motion.
$$ \mathrm{d}W = \vec{F} \cdot \mathrm{d}\vec{s} \;,
$$ where the path $s$ can have a different shape and different length in different frames.

To illustrate how this works out let's consider a case in one dimension. Seen in frame $S$ a object of mass $m$ is subject to constant net force F starting from position $x_0$ with velocity $v_0$ so that it moves to position $x$ where it has velocity $v$. The choice to limit this to a constant acceleration case implies that
\begin{align}
x - x_0
&= \frac{v^2 - v_0^2}{2a} \nonumber\\
&= m\frac{v^2 - v_0^2}{2 F} \nonumber\\
F (x - x_0)& = \frac{1}{2} m (v^2 - v_0^2) \nonumber\\
W &= \Delta T
\end{align} as expected. or future reference, the elapsed time is $t = (v-v_0)/a = m(v-v_0)/F$.

Viewing the same event in frame $S'$ that moves with speed $-u$ with respect to frame $S$ and is initially co-located with $S$ we have
\begin{align}
\begin{aligned}
x_0' &= x_0 \\
x' &= x + u t \\
v_0' &= v_0 + u \\
v &= v + u \;,
\end{aligned}
\end{align} and of course $F' = F$; $m' = m$; $a' = a$; and $t' = t$.

We can perform the analysis in the primed frame two ways: we can use only primed symbols resulting in exactly the same work leading to
\begin{align}
W' = \Delta T' \;,
\end{align} or we can start with the primed quantities then switch to the un-primed quantities by substitution from (2). This give us
\begin{align}
x' - x_0' &= \frac{v'^2 - v_0'^2}{2 a} \nonumber\\
x - (x_0 + u t) &= \frac{1}{2a} \left[ (v + u)^2 - (v_o + u)^2 \right] \nonumber\\
\end{align} expanding the squares on the RHS and re-grouping leads to
\begin{align}
(x - x_0) + u t &= \frac{1}{2a} m\left[ v^2 - v_o^2 + 2u\left(v - v_0\right) \right] \;. \nonumber
\end{align}
Here we recognize the un-primed work and change in kinetic energy and also reduce the spare terms on the RHS to arrive at
\begin{align}
W + u t &= \Delta T + u t \;. \\
\end{align}

So changing frames changed the values of the work done and the difference in kinetic energy by the same amount, and the work-energy theorem continues to work even though the figures that appear in it can be very different.

Lessons

First of all, if you are using energy to work a problem you have to pick one frame of reference and stick with it because you can't meaningfully compare energies or works between frame of reference.

The second issue is a little more subtle. Most people who actually think about the classical energy infrastructure as taught in typical introductory books will eventually notice that the whole thing seems to rest on very arbitrary definitions. I mean, why should work be that formula, right? I intend to write a essay on why it isn't arbitrary at some point, but in the mean time it should be clear that not every conceivable foundation will have the property where the rules work as you change frames. So Galilean relativity puts strong limits on choice of infrastructure.