This is not about special relativity, so assume speeds are much less than $c$.
This article says a change in kinetic energy (KE) remains constant in all inertial reference frames.
So the kinetic energy depends upon the measurement frame of reference.
But whatever inertial (non-accelerating) reference frame you use,
changes of kinetic energy will be unaffected by this choice.
I understand how a change in potential energy (PE) ($mg\Delta h$) is constant no matter what the reference $0$ is.
But I don't get how a change in KE doesn't depend on the frame of reference. I'll provide a contradicting example.
Consider a father+son sitting in a moving train with velocity $v_t$. After some time, the son gets up and starts running with velocity $v_s$
The father's frame:
change in KE of son = $\frac{1}{2}mv_s^2$
Station frame:
change in KE of son = $\frac{1}{2}m(v_t+v_s)^2-\frac{1}{2}mv_t^2 =\frac{1}{2}mv_s^2 + \color{red}{mv_tv_s} $
These two are clearly not same. What is wrong in my thinking?
Best Answer
Nothing is wrong with your thinking. These two values are different.
The article is either wrong, or contains some caveat about a closed system (according to a commenter the article does contain such a caveat).
To see the difference with a closed system, consider your same example, but don't allow external forces (the father and son are the only interacting masses in the system).
In the closed-system example assume further that the father and son have the same mass. In a closed system the only way the son could start running is due to a force applied the father. There will be an equal and opposite force on the father due to the son. This leads to conservation of momentum. The son will move with velocity $v_s$ and the father with velocity $-v_s$ (recall, we assumed they have equal mass, and here further assume this example is one dimensional).
So we have:
Frame One:
Change in KE = $\frac{1}{2}mv_s^2 + \frac{1}{2}m(-v_s)^2 - (0 + 0) = m v_s^2$
Frame Two:
Change in KE = $\frac{1}{2}m(v_t+v_s)^2 + \frac{1}{2}m(v_t-v_s)^2 - (\frac{1}{2}mv_t^2 + \frac{1}{2}mv_t^2) = m {v_s}^2$
Although the above example is very specific, the result holds with much more generality (for any closed system).
In a closed system, the changes in momentum of each mass can only occur due to forces from other masses. By Newton's third law there will always be pairs of equal and opposite forces that make momentum changes, therefore the sum over all the forces is zero, and therefore the total momentum $\mathbf P$ is conserved. To see this more explicitly, consider the total momentum in Frame 1: $$ \mathbf P = \sum_i m_i \mathbf{v}_i\;, $$ where the sum is over all the masses in the system.
This quantity ($\mathbf P$) is conserved, in a closed system, because: $$ \frac{d\mathbf P}{dt} = \sum_i m_i \mathbf{a}_i = \sum_i \mathbf{F}_i = 0\;. $$ Note that if there were any external forces, the total momentum of the system would not be conserved, but there are no external forces since this is a closed system.
To see in more detail why the final equality holds in the equation above, recall that $\mathbf{F}_i = \sum_j \mathbf{F}_{ij}\;,$ where $\mathbf{F}_{ij}$ means "the force on mass i due to mass j." Newton's 3rd Law states that $\mathbf{F}_{ij} = -\mathbf{F}_{ji}$. Therefore: $$ \sum_i \mathbf{F}_i = \sum_{i,j} \mathbf{F}_{ij} = \sum_{i,j} \mathbf{F}_{ji} = -\sum_{i,j}\mathbf{F}_{ij} = -\sum_i \mathbf{F}_i\;, $$ where the second equality is due to the fact that we can rename the dummy variable $i$ to $j$ and $j$ to $i$, and the third equality holds due to Newton's 3rd Law. The above equation states that $\sum_i \mathbf{F}_i$ is equal to the negative of itself, which means it is zero.
The total Kinetic energy in Frame 1 is $$ KE_1 = \sum_i \frac{1}{2} m_i v_i^2 $$
The total Kinetic Energy in Frame 2 is $$ KE_2 = \sum_i \frac{1}{2} m_i (\mathbf{v}_i + \mathbf{V})^2 = KE_1 + \mathbf{P}\cdot \mathbf{V} + \frac{1}{2}MV^2\;, $$ where $\mathbf V$ is the (constant) relative velocity between the inertial frames and where $M = \sum_i m_i$.
Because $\mathbf{P}$, $M$, and $\mathbf{V}$ are constant, we thus have: $$ \Delta KE_2 = \Delta KE_1 + 0 + 0 = \Delta KE_1 $$