# Work required to change an object's speed in different reference frames

Let $$\vec s$$, $$\vec v$$, $$\vec a$$ and $$\vec F$$ be displacement, velocity, acceleration, and force functions of time $$t$$. With respect to a absolute reference frame, such that $$\vec v = \vec 0$$ at $$t = 0$$, we can derive the kinetic energy of a particle of mass $$m$$ as below: \begin{align} E_k & = \int \vec F \cdot \vec {ds} \\ & = \int \vec F \cdot \vec v\; dt \\ & = \int m \vec a \cdot \vec v\; dt \\ & = m\int \vec a \cdot \vec v\; dt \\ \end{align} solving the integral: \begin{align} \int \vec a \cdot \vec v\; dt\; & = \vec v\;\cdot\int \vec a\;dt\; - \int\left(\vec v\,'\;\cdot \int \vec a\;dt\right)\; dt \\ & = \vec v\; \cdot \vec v\;- \int \vec a \cdot \vec v\; dt \\ \implies \int \vec a \cdot \vec v\; dt\;& = \frac {\vec v \cdot \vec v}{2} \\ \end{align} Thus: $$E_k = \frac {m\;(\vec v \cdot \vec v)}{2}$$ now let us say at some $$\Delta E_k$$ is the change in kinetic energy as velocity changes from $$\vec v_1$$ at $$t_1$$ to $$\vec v_2$$ at $$t_2$$ (that is $$\vec v(t_2) = \vec v_2$$). Integrating from $$t_1$$ to $$t_2$$ we get what we expect: \begin{align} \Delta E_k & = \int_{t_1}^{t_2} \vec F \cdot \vec {ds} \\ & = m \int_{t_1}^{t_2} \vec a \cdot \vec v\; dt \\ & = m \left[\frac {(\vec v_2 \cdot \vec v_2)}{2} - \frac {(\vec v_1 \cdot \vec v_1)}{2}\right] \\ \\ \Delta E_k & = \frac {m\; \left(\vec v_2 \cdot \vec v_2 - \vec v_1 \cdot \vec v_1\right)}{2} \\ \end{align} So far so good, now I try to get the same answer using a different line of reasoning. Let us shift to a inertial frame with velocity $$\vec v_1$$ with respect to the absolute reference frame above. Now $$\vec v_2$$ is equal to $$\vec v_2 - \vec v_1$$ in this new frame of reference and at $$t = t_1$$ velocity is $$\vec 0$$. So $$\Delta E_k$$ is: \begin{align} \Delta E_k & = \frac {m\;(\vec v_2 - \vec v_1) \cdot (\vec v_2 - \vec v_1)}{2} \\ & = \frac {m\;(\vec v_2 \cdot \vec v_2 + \vec v_1 \cdot \vec v_1 - 2\vec v_1 \vec v_2)}{2} \\ \end{align} this is obviously not equal to: $$\frac {m\; \left(\vec v_2 \cdot \vec v_2 - \vec v_1 \cdot \vec v_1\right)}{2}$$ where did I go wrong here? What am I missing? Here is a similar question, but there we have a plane of some mass $$M$$ to account for energy difference.

• What is an "absolute reference frame"? Nov 1, 2022 at 18:11
• @BobD basically a reference frame that is arbitrarily selected to represent some global inertial frame. link Nov 1, 2022 at 18:24
• According to the link "Even within the context of Newtonian mechanics, the modern view is that absolute space is unnecessary. Instead, the notion of inertial frame of reference has taken precedence," Why are you applying a concept which appears to be non mainstream physics? Nov 1, 2022 at 18:33
• I put that link to give an idea only, "absolute frame" in the context of this problem it is a sort of global intertial frame, no different from other frames other than the idea that we chose it to be global aka "absolute". This choice is arbitrary. Also I don't think this affects the question. Nov 1, 2022 at 18:48
• So it's just a preferred (for the purpose of observation) inertial reference? Nov 1, 2022 at 19:21

The simple answer is, that energy depends on the system you are in. In your second frame the work done to get a mass from 0 to v (which is seen from outside as v2) is less than the work you do on the first frame of reference. So you have to change your integral in the new system since you would not know about v1.
same with potential energy , E=mgh sometimes you use the door of your room as h=0 sometimes the tabletop or any other hight, so you do not even change your frame of reference.

When you change between two reference frames that are in relative uniform motion, the absolute value of the kinetic energy and the value of the work of a force may change (and thus the difference of the kinetic energy between two states of the system), but the theorem of the kinetic energy holds in both reference frames, as you may expect since classical physics is invariant to Galilean transformations.

Computations in your example. Maybe you forgot to change the coordinates in the work integral.

$$\displaystyle W = \int_{1}^{2} \mathbf{F} \cdot \mathbf{v} \, dt = \dfrac{1}{2} m |\mathbf{v}_2|^2 - \dfrac{1}{2} m |\mathbf{v}_1|^2 = \Delta K$$

The second observer measures velocities $$\tilde{\mathbf{v}} = \mathbf{v} - \mathbf{v}_0$$, and accelerations $$\tilde{\mathbf{a}} = \mathbf{a}$$

$$\displaystyle \tilde{W} = \int_{1}^{2} \mathbf{F} \cdot \tilde{\mathbf{v}} \, dt = \int_{1}^{2} m \tilde{\mathbf{a}} \cdot \tilde{\mathbf{v}} \, dt = \dfrac{1}{2} m |\tilde{\mathbf{v}}_2|^2 - \dfrac{1}{2} m |\tilde{\mathbf{v}}_1|^2 = \Delta \tilde{K}$$

The difference between the work evaluated by the two observers is

$$\displaystyle W - \tilde{W} = \int_{1}^{2} \mathbf{F} \cdot \mathbf{v}_0 \, dt = \int_{1}^{2} m \mathbf{a} \cdot \mathbf{v}_0 \, dt = m ( \mathbf{v}_2 - \mathbf{v}_1) \cdot \mathbf{v}_0 = m ( \tilde{\mathbf{v}}_2 - \tilde{\mathbf{v}}_1) \cdot \mathbf{v}_0 = m \Delta \mathbf{v} \cdot \mathbf{v}_0$$,

and the same holds for the difference of the kinetic energy difference, that reads

$$\Delta K - \Delta \tilde{K} = \dfrac{1}{2} m |\mathbf{v}_2|^2 - \dfrac{1}{2} m |\mathbf{v}_1|^2 - \dfrac{1}{2} m |\tilde{\mathbf{v}}_2|^2 + \dfrac{1}{2} m |\tilde{\mathbf{v}}_1|^2 =$$
$$\qquad \qquad \quad = \dfrac{1}{2}m(|\tilde{\mathbf{v}}_2|^2 + |{\mathbf{v}}_0|^2 + 2\tilde{\mathbf{v}_2} \cdot \mathbf{v}_0 - |\tilde{\mathbf{v}}_1|^2 - |{\mathbf{v}}_0|^2 - 2\tilde{\mathbf{v}_1} \cdot \mathbf{v}_0 - |\tilde{\mathbf{v}}_2|^2 + |\tilde{\mathbf{v}}_1|^2) =$$
$$\qquad \qquad \quad = m (\tilde{\mathbf{v}}_2 - \tilde{\mathbf{v}}_1) \cdot \mathbf{v}_0 = m \Delta \mathbf{v} \cdot \mathbf{v}_0$$.