In Newtonian mechanics, the kinetic Energy is a form of energy given by $\frac{1}{2}mv^2$.
Since Newtonian Mechanics is a limiting case of the more general General Relativity (when the curvature of Space-Time is insignificant), can we also assume a similar Energy exists for a velocity of a moving body in General Relativity? How do we define it, for various frames of reference w.r.t. the object in question? Wouldn't the energy content of the object be different for each frame?
The relativistic energy equation is $E=\gamma mc^2$ where $\gamma$ is the Lorentz factor $\gamma=\frac{1}{\sqrt{1−\frac{v^2}{c^2}}}$. We set $\beta=\frac{v}{c}$ giving $γ=\frac{1}{\sqrt{1−\beta^2}}$. Doing a Maclaurin expansion of the Lorentz factor around $\beta$ gives you get $\gamma \approx 1+\frac{1}{2}\beta^2+\frac{3}{8}\beta^4$. When we combine the expansion back into the energy equation we get:
$E\approx mc^2+\frac{1}{2}mc^2\frac{v^2}{c^2}+\frac{3}{8}mc^2\frac{v^4}{c^4}+...$. So when velocity is $0$ this becomes the equation $E=mc^2$. This is the mass energy conversion equation that everyone knows and loves. It indicates how much energy you have if you convert mass into energy. As velocity increases the other terms are important. The first term simplified to $\frac{1}{2}mv^2$ is the kinetic energy term giving the connection to Newtonian physics. This is also the term for things like temperature as temperature is the average random molecular kinetic energy in a system. The third term $\frac{3}{8}m\frac{v^4}{c^2}$ is the first relativistic correction in special relativity. Note that velocity must be very high ($\approx 0.75c$) before this term comes into noticeable effect.
So yes the energy is dependent on your frame of reference.
