The relativistic energy of a particle is given by the expression

\begin{equation} E^2 = m^2c^4 + p^2c^2 \end{equation}

The rest energy is $E_{0}=mc^2$ and the momentum is $p=mc$. In the rest frame, the kinetic energy is $T=E-mc^2$.

Ok, now in another frame of reference, we must include the Lorentz factor $\gamma$, where $\gamma=\frac{1}{\sqrt{1-v^2/c^2}}$.

In a different reference frame, momentum is $p=\gamma mc$ and the kinetic energy is $T=(\gamma-1)mc^2$.

Are these expressions correct? If so, I am confused. I have a question which asks me "The relativistic momentum is $p=mc$. What is the kinetic energy?".

Should I conclude this is $T=E-mc^2=mc$? That is, the kinetic energy is also $p=mc$? Or is the correct conclusion that $\gamma=1$ and therefore the kinetic energy is $T=0$?

  • 1
    $\begingroup$ The first equation is wrong, it should $E^2=p^2c^2+m^2c^4$. $\endgroup$
    – Kyle Kanos
    Sep 20, 2015 at 17:31
  • $\begingroup$ $$T=E−mc^2=mc$$ second part is not dimensionally correct. $\endgroup$ Sep 20, 2015 at 17:44
  • $\begingroup$ The first equation is wrong. The second equation is right. The third equation is wrong. The fourth equation is right. The fifth equation is right. The sixth equation is wrong. the seventh equation is right. The eighth equation is wrong. The first RHS of the ninth equation is right and the second RHS of the ninth equation is wrong. The tenth equation is wrong, and the 11th and the 12th... $\endgroup$
    – hft
    Sep 20, 2015 at 18:03
  • $\begingroup$ @hft Energy in a frame other than the rest frame is given by $E=\gamma mc^2$ and the kinetic energy is $T=(\gamma-1)mc^2$ The sixth equation is correct $\endgroup$ Sep 20, 2015 at 18:09

2 Answers 2


The expressions are not true in general. The first one should be $E^2 = m^2c^4 + p^2c^2$, and the momentum is in general $p = \gamma m v$. The rest energy is $E_0 = m c^2$ and it doesn't depend on the frame (by definition), and the kinetic energy is always $T=E-mc^2 = (\gamma - 1) mc^2$.

You are (understandably) confused because the question is not telling you that momentum is $mc$. You are being told that in a specific situation and in a specific frame, it just so happens that the momentum is equal to $mc$. You should be able to find the velocity from this, and then the kinetic energy.

Alright, since you're having trouble let's get our equations straight. First we define $\gamma$, which is a function of velocity $v$, as $1/\sqrt{1-v^2/c^2}$. The momentum $p$ of a particle with mass $m$ moving with velocity $v$ is given by $p = mv/\sqrt{1-v^2/c^2} = \gamma m v$. The expression $\gamma mv$ looks simpler but don't forget that $v$ is hidden inside $\gamma$.

There are two expressions for the energy. Obviously both are true and can be proved to be equal to each other; the only difference is whether you want the energy in terms of $p$ or $v$. So we have $E^2 = p^2c^2 + m^2c^4$ and $E = mc^2/\sqrt{1-v^2/c^2} = \gamma m c^2$. Kinetic energy $T$ is defined as $T = E-mc^2 = (\gamma-1)mc^2$. As always, this depends on $v$ through $\gamma$.

All these equations are true in any frame. The quantities themselves (such as $v$, $p$ or $E$) change when you change frames, but they change in such a way that all the equations remain correct.

Now, you have been told that in some particular frame, a particle is moving with $p=mc$. This will not be true in general, since the formula for $p$ is $mv/\sqrt{1-v^2/c^2}$; it just so happens that in our situation, $v$ is such that $p = mc$. This is an equation you can use to find $v$; knowing $v$, you can use the formula for kinetic energy $T$ (which, don't forget, depends on $v$) and find what you are being asked for.

  • $\begingroup$ "You should be able to find the velocity from this, and then the kinetic energy." That's true. However, this is a "self-check" multiple choice thing I am reading. All answers are in terms of m and c. $\endgroup$ Sep 20, 2015 at 17:59
  • $\begingroup$ @JianguoHisiang: That's great, because the kinetic energy will be expressed in terms of $m$ and $c$. $\endgroup$
    – Javier
    Sep 20, 2015 at 19:47
  • $\begingroup$ In this frame, $p=m$, and the kinetic energy is therefore $E=mc^2$. $\endgroup$ Sep 20, 2015 at 22:35
  • $\begingroup$ @JianguoHisiang: $p=m$ can't possibly be correct because the units are off. You know that in general, $p = mv/\sqrt{1-v^2/c^2}$. You also know that in this situation, $p = mc$. You can set those two equal to each other and find $v$. $\endgroup$
    – Javier
    Sep 20, 2015 at 23:48
  • $\begingroup$ I find $v=\frac{1}{\sqrt{2}}c$. As kinetic energy is $\frac{1}{2}mv^2$, I find the result is $\frac{1}{\sqrt{2}}m(\frac{1}{2}c^2)$ $\endgroup$ Sep 21, 2015 at 0:59

By definition, these equations are true in any frame.

Linear momentum is $p=\gamma mv$, energy is $E=\gamma mc^2$, rest energy is $E_{0}=mc^2$, kinetic energy is $T=(\gamma-1)mc^2$.

We've been given that in this frame $p=mc$. From this, we must conclude that $p=\gamma mv = mc$. Correct? $p=\gamma m v$ true in any frame.

This means $\gamma v = c$, i.e. $v/\sqrt{1-v^2/c^2} = c$

This is equivalent to $\sqrt{1-v^2/c^2}=v/c$ or $v=\frac{1}{\sqrt{2}}c$.

Now, using the expression for kinetic energy $T=(\gamma-1)mc^2$, we find $\gamma$. Substituting $v=\frac{1}{\sqrt{2}}c$ into $\gamma$, I find $\gamma=\frac{2}{\sqrt{3}}$, which is approximately 1.15.

Therefore, $T$ is around $(0.15)*mc^2$.


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