Would this be correct for kinetic energy in special relativity? In galilean relativity $$p=mv$$ and $$KE=\frac{1}{2}mv^2$$
If I understand it in special relativity the equation for momentum is $$p=\frac{mv}{\sqrt{1-\frac{v^2}{c^2}}}$$  In galilean relativity there is the equation $$\frac{p}{m}=v$$ while in special relativity the equation is $$\frac{p}{m}=\frac{v}{\sqrt{1-\frac{v^2}{c^2}}}$$ In galilean relativity $$KE=\frac{1}{2}m\frac{{p}^2}{{m}^2}$$ and if this equation holds in special relativity it produces the equation $$KE=\frac{1}{2}m\frac{v^2}{1-\frac{v^2}{c^2}}$$  Is this a correct way to express kinetic energy in special relativity?
 A: It is useful to understand how the expression for kinetic energy is arrived at in non relativistic mechanics and apply that for the relativistic case. Let's say we have a body at stand still and apply a force $F = \frac{d(mv)}{dt}$ to get it moving. The energy gained by the body is then
$
E_k = \int_0^t Fdx = \int_0^t \frac{d(mv)}{dt} dx = \int_0^tmv dv = \frac{1}{2} m v^2
$
Doing the same in relativistic mechanics you need to take into account the body gets heavier (or more "inertial") by a factor $\gamma = (1-\frac{v^2}{c^2})^{-1/2}$ with $m$ now its rest mass and the integral works out differently. Calculating the work done we will find:
$
E_k =  \int_0^t Fdx =\int_0^t \frac{d(\gamma mv)}{dt} dx = \int_0^tv d(m\gamma v)
$
This integral is slightly more involved but will result in $E_k = (\gamma -1 ) mc^2$. Physically a lot more work has to be done to accelerate the body as it gets heavier.
Note that in the classical case you can write $E_k = \int_0^t \frac{p}{m} dp = \frac{p^2}{2m}$ but not anymore in relativistic mechanics as you can see from the integral expression in the above which becomes
$\int_0^tv d(m\gamma v) = \int_0^t\frac{p}{\gamma m} dp$.
So your expression for $E_k$ in terms of momentum is incorrect.
A: Kinetic energy in Special Relativity is $KE=(\gamma -1)mc^2$, where $\gamma $ is the Lorentz factor.  This is derived from the relativistic equation for total energy, $E^2 = (pc)^2 + ((mc)^2)^2$; the negative term removes the rest energy from the total energy.
The Galilean kinetic energy is the limit of this expression for small velocities.
