1. I got quite confused reading a textbook about special relativity. The author defines the relativistic kinetic energy of a single particle as $$E_{kin}=\gamma m c^2.$$

  2. I looked it up on Wikipedia and the article tells me $$E_{kin}=\gamma m c^2-mc^2$$ which sounds more reasonable to me as I used to believe that $\gamma mc^2$ is the total energy and $mc^2$ the rest energy.

What is correct?

  • 7
    $\begingroup$ The author is wrong $\endgroup$
    – Paul
    Jan 25, 2017 at 13:53
  • 5
    $\begingroup$ Throw that book into the trash. $\endgroup$
    – garyp
    Jan 25, 2017 at 15:15
  • 2
    $\begingroup$ Which textbook? $\endgroup$
    – Qmechanic
    Mar 1, 2019 at 18:53

2 Answers 2


The second formula is correct. One can easily check it by performing a non-relativistic expansion

\begin{equation} \gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}} \simeq 1+\frac{v^2}{2c^2}, \end{equation}

which gives a familiar expression for the kinetic energy \begin{equation} E_{kin}=\gamma m c^2-mc^2=\frac{v^2}{2c^2} m c^2=\frac{m v^2}{2}. \end{equation}


Both expressions are valid, because definitions are not correct or wrong. Definitions are consensus that we take about elements in a formalism.

We can define anything as we want in a formalism, simply we have to respect basic logic rules, such as using always the same definition, once we chose one. Moreover, definitions must have physical meaning in theories of physical content.

So we have to ask ourselves, what is the physical meaning of kinetic energy? Usually, we define kinetic energy as the energy that changes due to work and both definitions of kinetic energy

$$E_{kin}^{(1)} \equiv mc^2 (\gamma -1)$$

$$E_{kin}^{(2)} \equiv mc^2 \gamma$$

are physically acceptable according to this criteria, because their time derivatives equal the rate of doing work

$$\frac{dE_{kin}^{(1)}}{dt} = \frac{dE_{kin}^{(2)}}{dt} = \mathbf{F}\mathbf{v}$$

which is expected because the difference between both definitions is a constant term, which vanish when differentiating.

In the momentum representation

$$E_{kin}^{(1)} \equiv \sqrt{m^2c^4 + \mathbf{p}^2c^2} - mc^2$$

$$E_{kin}^{(2)} \equiv \sqrt{m^2c^4 + \mathbf{p}^2c^2} $$

Most books use the first definition, but I have seen books using the definition $E_{kin}^{(2)}$. A reason for choosing this definition is given by Richard Talman in Geometric Mechanics; Toward a Unification of Classical Physics (Wiley 2007):

Here, we have used the symbol $E_{kin}$ which, since it includes the rest energy, differs by that much from being a generalization of the "kinetic energy" of Newtonian mechanics. Nevertheless it is convenient to have a symbol for the energy of a particle that accompanies its very existence and includes its energy of motion but does not include any "potential energy" due to its position in a field of force.

An example of article using the second definition of relativistic kinetic energy is Decay constants in the heavy quark limit in models à la Bakamjian and Thomas:

also for the kinetic energy $K(\{\mathbf{k}_i\})$, that can be taken to be of the non-relativistic $\frac{\mathbf{k}_i^2}{2m_i}$ or relativistic $\sqrt{\mathbf{k}_i^2 + m_i^2}$ forms.

  • 2
    $\begingroup$ I see your point. The thing that makes the "first" definition unpalatable is that it does not go to zero as the velocity goes to zero; it does not reduce to the non-relativistic form at low speed, but that doesn't matter. Any physical use of kinetic energy uses kinetic energy differences. We are familiar with PE needing an arbitrary zero, but so does KE. For example in non-relativistic mechanics we need to choose an arbitrary inertial frame in which we define velocity. At low speeds, the rest mass term cancels, so the "first" version works just as well as the "second". $\endgroup$
    – garyp
    Mar 1, 2019 at 18:45
  • 1
    $\begingroup$ Kinetic energy is not frame-invariant, so a proper definition is necessarily tied to a reference frame. because the word kinetic refers to motion, kinetic energy should be the energy associated with motion as measured in the specified reference frame and should, in that frame, be zero. Definitions matter because they allow us to communicate clearly. If someone is trying to define kinetic energy in a reference frame as $\gamma m c^2$, they are non-standard, at best. $\endgroup$
    – Bill N
    Mar 1, 2019 at 18:56
  • 1
    $\begingroup$ The physics is clearer if you do not associate potential energy with a particle, but with whatever it is that is actually holding the energy, such as an electromagnetic field. $\endgroup$ Mar 1, 2019 at 19:04
  • 1
    $\begingroup$ Who downvoted this answer? It's a good answer. $\endgroup$ Jul 11, 2021 at 10:50
  • 2
    $\begingroup$ "Definitions are consensus". Yes, they are. Defining kinetic energy such that it is non-zero in the rest frame is not the consensus view. $\endgroup$
    – ProfRob
    Jul 11, 2021 at 13:29

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