Editing Kinetic energy (section)

===Frame of reference===

The speed, and thus the kinetic energy of a single object is frame-dependent (relative): it can take any non-negative value, by choosing a suitable [[inertial frame of reference]]. For example, a bullet passing an observer has kinetic energy in the reference frame of this observer. The same bullet is stationary to an observer moving with the same velocity as the bullet, and so has zero kinetic energy.<ref>{{cite book
|title=Introduction to the theory of relativity
|url=https://archive.org/details/introductiontoth0000sear_c2m9
|url-access=registration
|first1=Francis Weston
|last1=Sears
|first2=Robert W.
|last2=Brehme
|publisher=Addison-Wesley
|year=1968
|page=[https://archive.org/details/introductiontoth0000sear_c2m9/page/127 127]
}}, [https://books.google.com/books?id=cpzvAAAAMAAJ&q=%22in+its+own+rest+frame%22 Snippet view of page 127] {{Webarchive|url=https://web.archive.org/web/20200804041331/https://books.google.com/books?id=cpzvAAAAMAAJ&dq=%22in+its+own+rest+frame%22+%22kinetic+energy%22&q=%22in+its+own+rest+frame%22 |date=2020-08-04 }}</ref> By contrast, the total kinetic energy of a system of objects cannot be reduced to zero by a suitable choice of the inertial reference frame, unless all the objects have the same velocity. In any other case, the total kinetic energy has a non-zero minimum, as no inertial reference frame can be chosen in which all the objects are stationary. This minimum kinetic energy contributes to the system's [[invariant mass]], which is independent of the reference frame.

The total kinetic energy of a system depends on the [[inertial frame of reference]]: it is the sum of the total kinetic energy in a [[center of momentum frame]] and the kinetic energy the total mass would have if it were concentrated in the [[center of mass]].

This may be simply shown: let <math>\textstyle\mathbf{V}</math> be the relative velocity of the center of mass frame ''i'' in the frame ''k''. Since

<math display="block">
  v^2 = \left(v_i + V\right)^2
      = \left(\mathbf{v}_i + \mathbf{V}\right) \cdot \left(\mathbf{v}_i + \mathbf{V}\right)
      = \mathbf{v}_i \cdot \mathbf{v}_i + 2 \mathbf{v}_i \cdot \mathbf{V} + \mathbf{V} \cdot \mathbf{V}
      = v_i^2 + 2 \mathbf{v}_i \cdot \mathbf{V} + V^2,
</math>

Then,

<math display="block">  E_\text{k} = \int \frac{v^2}{2} dm = \int \frac{v_i^2}{2} dm + \mathbf{V} \cdot \int \mathbf{v}_i dm + \frac{V^2}{2} \int dm.</math>

However, let <math display="inline"> \int \frac{v_i^2}{2} dm = E_i </math> the kinetic energy in the center of mass frame, <math display="inline"> \int \mathbf{v}_i dm </math> would be simply the total momentum that is by definition zero in the center of mass frame, and let the total mass: <math display="inline"> \int dm = M </math>. Substituting, we get:<ref>[http://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1/node64.html Physics notes – Kinetic energy in the CM frame] {{webarchive|url=https://web.archive.org/web/20070611231147/http://www.phy.duke.edu/~rgb/Class/intro_physics_1/intro_physics_1/node64.html |date=2007-06-11 }}. [[Duke University|Duke]].edu. Accessed 2007-11-24.</ref>

<math display="block">E_\text{k} = E_i + \frac{M V^2}{2}.</math>

Thus the kinetic energy of a system is lowest to center of momentum reference frames, i.e., frames of reference in which the center of mass is stationary (either the [[center of mass frame]] or any other [[center of momentum frame]]). In any different frame of reference, there is additional kinetic energy corresponding to the total mass moving at the speed of the center of mass. The kinetic energy of the system in the [[center of momentum frame]] is a quantity that is invariant (all observers see it to be the same).