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 $\frac{1}{2}m(v+V)^2$. However, another observer on the frame calculates the energy as $\frac{1}{2}mv^2$. When each of these equations is plugged into the conservation of energy, they will result in 2 distinct results (I think).

  • $\begingroup$ @knzhou Wouldn't it be better if we tried to mark all those questions as duplicates of this one? Based on my quick search, this question is the oldest of the duplicates (and it has among the highest scores), so it would make the most sense to mark all the others as dupes of this one. $\endgroup$
    – user191954
    Sep 28 '18 at 5:15
  • $\begingroup$ @Chair If you think so, go ahead and vote! I didn’t pick this one because I don’t think the top answer here gets to the heart of the question — it just says “don’t think about where the extra energy comes from”. I also think that for very old questions which one came precisely first didn’t matter, the OPs are all long gone. $\endgroup$
    – knzhou
    Sep 28 '18 at 20:55

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 $K = \frac{1}{2}mv^2$, the ground observer will measure a kinetic energy of $\frac{1}{2}m(v + V)^2$, or

$$K + \sqrt{2Km}V + \frac{1}{2}mV^2$$

(in one dimension).

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,"

$$K = E - mc^2$$

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,

$$\begin{pmatrix}E/c \\ p\end{pmatrix} = \begin{pmatrix}\gamma_v mc \\ \gamma_v mv\end{pmatrix}$$

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,

$$\begin{pmatrix}E/c \\ p\end{pmatrix}_\text{ground} = \begin{pmatrix}\gamma & \gamma\beta \\ \gamma\beta & \gamma\end{pmatrix}\begin{pmatrix}E/c \\ p\end{pmatrix}_\text{train}$$

(where $\beta = V/c$ and $\gamma = 1/\sqrt{1 - \beta^2}$), so the energy as observed from the ground would be given by

$$E_\text{ground} = \gamma(E_\text{train} + \beta c p_\text{train})$$

The kinetic energy is obtained by subtracting $mc^2$ from the total energy, so you'd get

$$K_\text{ground} = \gamma(E_\text{train} + \beta c p_\text{train}) - mc^2$$

which works out to

$$K_\text{ground} = \gamma K_\text{train} + (\gamma - 1) mc^2 + \gamma\beta c p_\text{train}$$

where $K$ is the relativistic kinetic energy and $p$ is the relativistic momentum.

If you wanted it in terms of energy alone:

$$K_\text{ground} = \gamma K_\text{train} + (\gamma - 1) mc^2 + \gamma\beta\sqrt{K_\text{train}^2 + 2 mc^2 K_\text{train}}$$

You might start to notice a similarity to the non-relativistic expression above ($K + \sqrt{2Km}V + \frac{1}{2}mV^2$), and indeed, if you plug in some approximations that are valid at low speeds ($\gamma \approx 1$, $\gamma - 1 \approx V^2/c^2$, $K_\text{train} \approx \frac{1}{2}mv^2 \ll mc^2$), you will recover exactly that expression.


You should remain in one frame of reference when applying the law of conservation of energy. Then you should be fine.

  • $\begingroup$ Yes. Of course I must remain on the same frame for each case. $\endgroup$
    – xport
    Nov 28 '10 at 4:38

Let's approach it from a different angle. Recall that kinetic energy, K, is defined as the mass-energy difference between the dynamic mass (in motion), m, and it's rest mass, m0, so that K = (m-m0)c^2. So the question simplifies to "Does the mass measured in one frame of reference equal that measured in another frame of reference."

Let's look at an example: Frame 1 is the earth. Frame 2 is a space ship with velocity, v, relative to earth. The spaceship is initially at rest on earth and contains a 1 kg test mass. The spaceship accelerates from rest to velocity, v. Both observer's on earth and inside the spaceship then measure the mass of the test mass.

For the earth observer, the increase in the test mass follows Einstein's famous equation from Special Theory; m=m0/sqrt(1-v^2/c^2). The observer inside the spaceship measures

1) Inside the spaceship the observer knows he is accelerating away from earth. He can measure the acceleration, and can calculate he is traveling at velocity, v, relative to earth. Knowing this, he can calculate mass of the test mass using the ST relation. Same equation give the same result; the test mass increases by the same amount as that measured by earth observer above.

2) Trajectories of an accelerated particle are sometimes a function of its mass. For example within a cyclotron, the particle trajectory is a function of its mass. Furthermore, the trajectory is known to change as its mass increases with velocity. Since there can be only one trajectory regardless of the frame of reference, all frames of reference must logically conclude the test mass has the same mass as that measured on earth.

Both frames of reference measure precisely the same increase in the test mass. That is to say, mass is invariant between two inertial frames of reference. If mass changes in one frame of reference, it likewise changes in the other frame of reference. Consequently, Kinetic energy being a function of mass increase, is not relative, rather is conserved.

  • $\begingroup$ Beware! Relativistic mass is not a popular concept on this site. Sure, it can be handy, but it can also be rather misleading, so most modern physicists and physics teachers avoid it. $\endgroup$
    – PM 2Ring
    Apr 14 '18 at 22:26
  • $\begingroup$ I was unaware of the limitations and constraints on the site. I am a little baffled though. I frequently see general relativity tensors on this site which are definitely confusing. Just curious, but why do you believe relativistic mass is confusing? I always thought ST was one of the most intriguing subjects as undergrad physics students. My apologies. $\endgroup$
    – M. Pope
    Apr 16 '18 at 0:24
  • $\begingroup$ physics.stackexchange.com/questions/133376/… and the linked questions discuss the issues of relativistic mass. A simple example where it's misleading: thinking that we can turn a body into a black hole by giving it a high enough speed. But of course any body is traveling arbitrarily close to $c$ from some reference frame. $\endgroup$
    – PM 2Ring
    Apr 16 '18 at 1:11
  • $\begingroup$ There's no need to apologize, and there are many posts on this site that do reference relativistic mass. But many of the regulars avoid it, and prefer more modern treatments. OTOH, I don't know if answers using relativistic mass attract downvotes unless they reach incorrect conclusions. $\endgroup$
    – PM 2Ring
    Apr 16 '18 at 1:15
  • $\begingroup$ You mentioned a "more modern treatment" as a replacement for ST relativistic mass. I am curious about that, can you elaborate? And to your other point, why can't we turn an arbitrary mass into a black hole by imparting a high enough velocity? A black hole traveling at nearly the speed of light, sucking up everything in it's path, doesn't sound good but theoretically possible? $\endgroup$
    – M. Pope
    Apr 17 '18 at 15:06

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