How to find kinetic energy given relativistic linear momentum? 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 $T=0$? 
 A: 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.
A: 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 $(0.15)*mc^2$. 
