# If the velocity is a relative quantity, will it make inconsistent equations when applying it to the

If the velocity is a relative quantity, will it make inconsistent equations when applying it to the conservation of energy equations?
For example:
In the train moving at $V$ relative to ground, there is an object moving at $v$ relative to the frame in the same direction the frame moves. Observer on the ground calculates the object kinetic energy as $\frac{1}{2}m\left(v+V{\right)}^{2}$. However, another observer on the frame calculates the energy as $\frac{1}{2}m{v}^{2}$.
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owerswogsnz
Yes, kinetic energy is a relative quantity. As you might guess, this means that when you're using energy conservation, you have to stay within a single frame of reference; all that energy conservation tells you is that the amount of energy as measured in any one frame stays the same over time. You can't meaningfully compare the amount of energy measured in frame A (e.g. the ground) to the amount of energy measured in frame B (e.g. the train).
However, you can convert an amount of kinetic energy measured in one frame to another frame, if you know their relative velocity. If you're working at low speeds, the easy (approximate) way to do this is to just calculate the relative velocity, as you did. So if the train observer measures a kinetic energy $K=\frac{1}{2}m{v}^{2}$, the ground observer will measure a kinetic energy of $\frac{1}{2}m\left(v+V{\right)}^{2}$, or
$K+\sqrt{2Km}V+\frac{1}{2}m{V}^{2}$
If you get up to higher speeds, or you want an exact expression, you'll have to use the relativistic definition of energy. In special relativity, the kinetic energy is given by the difference between the total energy and the "rest energy,"
$K=E-m{c}^{2}$
One way to figure out the transformation rule is to use the fact that the total energy is part of a four-vector, along with the relativistic momentum,
$\left(\begin{array}{c}E/c\\ p\end{array}\right)=\left(\begin{array}{c}{\gamma }_{v}mc\\ {\gamma }_{v}mv\end{array}\right)$
where ${\gamma }_{v}=1/\sqrt{1-{v}^{2}/{c}^{2}}$. This four-vector transforms under the Lorentz transformation as you shift from one reference frame to another,
${\left(\begin{array}{c}E/c\\ p\end{array}\right)}_{\text{ground}}=\left(\begin{array}{cc}\gamma & \gamma \beta \\ \gamma \beta & \gamma \end{array}\right){\left(\begin{array}{c}E/c\\ p\end{array}\right)}_{\text{train}}$
(where $\beta =V/c$ and $\gamma =1/\sqrt{1-{\beta }^{2}}$), so the energy as observed from the ground would be given by
${E}_{\text{ground}}=\gamma \left({E}_{\text{train}}+\beta c{p}_{\text{train}}\right)$
The kinetic energy is obtained by subtracting $m{c}^{2}$ from the total energy, so you'd get
${K}_{\text{ground}}=\gamma \left({E}_{\text{train}}+\beta c{p}_{\text{train}}\right)-m{c}^{2}$
which works out to
${K}_{\text{ground}}=\gamma {K}_{\text{train}}+\left(\gamma -1\right)m{c}^{2}+\gamma \beta c{p}_{\text{train}}$
where $K$ is the relativistic kinetic energy and $p$ is the relativistic momentum.
If you wanted it in terms of energy alone:
${K}_{\text{ground}}=\gamma {K}_{\text{train}}+\left(\gamma -1\right)m{c}^{2}+\gamma \beta \sqrt{{K}_{\text{train}}^{2}+2m{c}^{2}{K}_{\text{train}}}$
You might start to notice a similarity to the non-relativistic expression above ($K+\sqrt{2Km}V+\frac{1}{2}m{V}^{2}$), and indeed, if you plug in some approximations that are valid at low speeds ($\gamma \approx 1$, $\gamma -1\approx {V}^{2}/{c}^{2}$, ${K}_{\text{train}}\approx \frac{1}{2}m{v}^{2}\ll m{c}^{2}$), you will recover exactly that expression.