You have successfully discovered that the kinetic energy depends on the reference frame.

That is actually true. What is amazing, however, is that while the value of the energy is frame DEpendent, once you've chosen a reference frame, the law of conservation of energy itself is **NOT** reference frame-dependent -- every reference frame will observe a constant energy, even if the exact number they measure is different. So, when you balance your conservation of energy equation in the two frames, you'll find different numbers for the total energy, but you will also see that the energy before and after an elastic collision will be that same number.

So, let's derive the conservation of energy in two reference frames. I'm going to model an elastic collision between two particles. In the first reference frame, I am going to assume that the second particle is stationary, and we have:

$$\begin{align}
\frac{1}{2}m_{1}v_{i}^{2} + \frac{1}{2}m_{2}0^{2} &= \frac{1}{2}m_{1}v_{1}^{2} + \frac{1}{2}m_{2}v_{2}^{2}\\
m_{1}v_{i}^2 &= m_{1}v_{1}^{2} + m_{2}v_{2}^{2}
\end{align}$$

to save myself time and energy, I'm going to call $\frac{m_{2}}{m_{1}} = R$, and we have:

$$v_{i}^{2} = v_{1}^{2} + Rv_{2}^{2}$$

Now, what happens if we shift to a different reference frame, moving to the right with speed v? This is essentially the same thing as subtracting $v$ from all of these terms. We thus have:

$$\begin{align}
(v_{i}-v)^{2} + R(-v)^{2} &= (v_{1}-v)^{2} + R(v_{2}-v)^{2}\\
v_{i}^{2} -2v_{i}v + v^{2} + Rv^{2} &= v_{1}^{2} - 2 vv_{1} + v^{2} + Rv_{2}^{2}-2Rv_{2}v + Rv^{2}\\
v_{i}^{2} -2v_{i}v &= v_{1}^{2}- 2vv_{1} + Rv_{2}^{2}-2Rv_{2}v\\
v_{i}^{2} &= v_{1}^{2} + Rv_{2}^{2} + 2v(v_{i} - v_{1} - R v_{2})
\end{align}$$

So, what gives? It looks like the first equation, except we have this extra $2v(v_{i} - v_{1} - R v_{2})$ term? Well, remember that momentum has to be conserved too. In our first frame, we have the conservation of momentum equation (remember that the second particle has initial velocity zero:

$$\begin{align}
m_{1}v_{i} + m_{2}(0) &= m_{1}v_{1} + m_{2}v_{2}\\
v_{i} &= v_{1} + Rv_{2}\\
v_{i} - v_{1} - Rv_{2} &=0
\end{align}$$

And there you go! If momentum is conserved in our first frame, then apparently energy is conserved in all frames!

You do not need any of those formulae. Visualize an ideal fluid element (no viscosity) in its rest frame. If it is incompressible, its volume does not change and all work done on it by pressure forces in any time interval equals change in its kinetic energy. If the flow is also adiabatic, no heat is transferred from one element to another.

Since no work and no heat is being transferred to the element in the frame where the element is at rest, the change in its internal energy is zero. If internal energy is supposed to be function of temperature only, this means that also temperature does not change.

## Best Answer

I can't see where this has any utility at all. The point of having any equation, differential or algebraic, is to put constraints on a system. We then solve these equations to obtain an unknown subject to the constraint. However in this case, we already know the answer and the equation gives us no new constraints and hence no new info information.