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?


5 Answers 5


This article says a change in kinetic energy (KE) remains constant in all inertial reference frames.


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 $v_s$

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?

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: $$ \Delta KE_2 = \Delta KE_1 + 0 + 0 = \Delta KE_1 $$

  • 1
    $\begingroup$ The linked article does specify a closed system. $\endgroup$
    – g s
    Mar 23, 2022 at 6:25
  • $\begingroup$ OK, I'll take your word for it. I did not read the entire article. $\endgroup$
    – hft
    Mar 23, 2022 at 17:30

Since the work-energy theorem is derived from the second Newton's law of motion, it is valid in any inertial reference frame. However, observers in two different inertial reference frames might not agree on work and kinetic energy values.

Example: A man starts pushing a cart of mass $m$ with constant force $F$ over a finite time period $\Delta t$, and the cart was initially at rest in the ground frame. What is the work and change in kinetic energy of the cart after $\Delta t$?

Ground reference frame

Since the cart is initially at rest as seen from the ground frame $v_0 = 0$, the velocity and displacement after $\Delta t$ are

$$ \begin{aligned} v_1 &= v_0 + a \Delta t = a \Delta t\\ \Delta s &= \frac{1}{2} a (\Delta t)^2 + v_0 \Delta t = \frac{1}{2} a (\Delta t)^2 \end{aligned} $$

where $a = F/m$ is the acceleration. The work and change in kinetic energy are

$$ \begin{aligned} W &= F \cdot \Delta s = ma \cdot \frac{1}{2} a (\Delta t)^2 = \frac{1}{2} m (a \Delta t)^2 \\ \Delta K &= \frac{1}{2} m v_1^2 - \frac{1}{2} m v_0^2 = \frac{1}{2} m v_1^2 = \frac{1}{2} m (a \Delta t)^2 \end{aligned} $$

Vehicle moving at constant velocity $V$ relative to the ground

In this (inertial) reference frame, the initial velocity is $v_0 = -V$, and the velocity and displacement after $\Delta t$ are

$$ \begin{aligned} v_1' &= v_0' + a \Delta t = -V + a \Delta t \\ \Delta s' &= \frac{1}{2} a (\Delta t)^2 + v_0 \Delta t = \frac{1}{2} a (\Delta t)^2 - V \Delta t \end{aligned} $$

where $a = F/m$ is acceleration which is equal to the one in ground reference frame. The work and change in kinetic energy are

$$ \begin{aligned} W' &= F \cdot \Delta s' = ma \cdot (\frac{1}{2} a (\Delta t)^2 - V \Delta t) = \frac{1}{2} m (a \Delta t)^2 - m V (a \Delta t) \\ \Delta K' &= \frac{1}{2} m v_1'^2 - \frac{1}{2} m v_0'^2 = \frac{1}{2} m (-V + a \Delta t)^2 - \frac{1}{2} m (-V)^2 = \frac{1}{2} m (a \Delta t)^2 - m V (a \Delta t) \end{aligned} $$


The work-energy theorem is valid in both reference frames, i.e. $\Delta K = W$ and $\Delta K' = W'$, but the observers in two different (inertial) reference frames do not agree on the absolute values of work and change in kinetic energy, i.e. $W \neq W'$ and $\Delta K \neq \Delta K'$.


Although the train is much more massive than the son, its mass is not infinite. When the son accelerates, the son does work on the train, pushing it (and all the passengers, including the father) ever so slightly backwards relative to an initially comoving inertial frame. That is to say, in the station frame, the train is slowed ever so slightly to pay for the son's gain in momentum.

We know the son's change in momentum is $m_s v_s$. Momentum is conserved, so the train's change in momentum must be equal and opposite, $-m_s v_s$.

The son's mass is much less than the train, so when we solve the system of equations for the exchange of momentum between the son and the train (which we can model as an elastic collision) we find that the train's acceleration is small enough that we can treat $v_t$ as approximately the same value before and after the acceleration as long as we remember that the acceleration is a real nonzero value. Specifically, we are making that the assumption that

$|(v_t + \Delta v_t)^2 - v_t^2| \gg |\Delta v_t|$

We now have all we need to find the work done by the son on the train as measured by an observer on the station. Assuming constant forces and using a minus sign because the force is exerted opposite the direction of the son's change in velocity:

$F = -m_s\Delta v_s / \Delta t = (m_s v_s)/\Delta t$

$\Delta s = v_t \Delta t$

$W = F \Delta s = -m_s v_s\Delta s/\Delta t = -m_s v_s v_t$

The extra $m_s v_s v_t$ term in the son's kinetic energy as measured measured by the observer on the station is the amount of work the son did on the train, that is, the amount by which the train's kinetic energy was reduced during the son's acceleration.

In the father's frame, then, the train's kinetic energy started at 0 and ended at 0, and the son's kinetic energy increased by $0.5 m_s v_s^2$, so the total kinetic energy of the system increased by $0.5 m_s v_s^2$ during the interaction.

In the station frame, the train's kinetic energy started at some value $0.5 m_t v_t ^2$ and ended at $0.5 m_t (v_t + \Delta v_t)^2 = 0.5 m_t v_t ^2 - m_s v_s v_t$, so it changed by $-m_s v_s v_t$. The son's kinetic energy increased by $0.5 m_s v_s^2 + m_s v_s v_t$. So the total kinetic energy of the system increased by $0.5 m_s v_s^2$ during the interaction.


As hft said, the rule applies to a closed system. Here is a way to see why such a system has an invariant change in kinetic energy. First, the kinetic energy $K$ in any frame is related to the kinetic energy $K_{\mathrm{in\ cm}}$ in the center-of-mass frame by $$K = K_{\mathrm{in\ cm}} + \tfrac{1}{2}M|\mathbf{v}_{\mathrm{of\ cm}}|^2 = K_{\mathrm{in\ cm}} + \frac{|\mathbf{P}|^2}{2M},$$ where $M$ is the total mass, $\mathbf{v}_{\mathrm{of\ cm}}$ is the velocity of the center of mass, and $\mathbf{P} = M\mathbf{v}_{\mathrm{of\ cm}}$ is the total momentum. Since total mass and total momentum are both conserved for a closed system ($\Delta M = 0$ and $\Delta\mathbf{P} = \mathbf{0}$), it follows that $$\Delta K = \Delta K_{\mathrm{in\ cm}}.$$ This says the change in kinetic energy is the same in every inertial frame (equal to that in the center-of-mass frame).

  • $\begingroup$ I think you are missing a $\mathbf P\cdot \mathbf{v}_{of cm}$ term on the RHS (and middle) of your first equations line. Both are constant and so both drop out of the change. $\endgroup$
    – hft
    Mar 23, 2022 at 17:35
  • $\begingroup$ $K = K_{in cm} + \mathbf P \cdot \mathbf v_{of cm} + \frac{1}{2}Mv^2_{ofcm}$ $\endgroup$
    – hft
    Mar 23, 2022 at 17:37
  • $\begingroup$ @hft I disagree -- e.g., for a single particle, $K_{\mathrm{in\ cm}} = 0$ and your formula would give an incorrect (3x too large) kinetic energy. Did you refer to the article I linked? $\endgroup$
    – nanoman
    Mar 24, 2022 at 0:37
  • $\begingroup$ The P.v term just comes from expanding the quadradic velocity term. You can see my expanded answer for why. In your example of a single particle in CM frame, P = zero, so the formula does not "give an incorrect kinetic energy." $\endgroup$
    – hft
    Mar 24, 2022 at 1:26
  • $\begingroup$ Anyways, I see that your answer applies only to boosts relative to the CM frame, for which P is always zero anyways (as specified in the article you linked). My (expanded) answer applies to any two frames, for which you can not drop the P.v term (I use P to mean the total momentum in Frame 1, which is not necessarily the CM frame). Both answers are correct, but generally the P term can not be dropped. $\endgroup$
    – hft
    Mar 24, 2022 at 2:03

The change in kinetic energy varies from frame to frame.

In your example, consider an external force working on the son. If the son doesn't move in his father's frame, the work done is 0, so $\Delta K$ is 0. But in the station's frame, that force resulted in some displacement ($v_t*\Delta t$).

So as work done varies from frame to frame, by the work-energy theorem, the kinetic energy also varies.


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