# Partitioning the kinetic energy into components in relativity

In classic physics, kinetic energy is defined as $$KE = \frac{1}{2}m v_x^2 + \frac{1}{2}m v_y^2 + \frac{1}{2}m v_z^2$$ So, by defining $KE_x = \frac{1}{2} m v_x^2$ , $KE_y = \frac{1}{2} m v_y^2$, $KE_z = \frac{1}{2} m v_z^2$, we can know the contribution of each components to the total kinetic energy.

However, in the case of relativistic kinetic energy, its definition is $$KE = \sqrt{(mc^2)^2+(p_x^2+p_y^2+p_z^2) c^2}-mc^2$$ Now, partitioning this into each components seems impossible.

Does this mean thinking of "x component" of kinetic energy becomes meaningless in relativistic theory?

I think it would still possible to determine relativistic components in each spacial direction. One would need to know the relative velocities $\beta = \frac{v}{c}$ w.r.t. light for each direction and multiply each component by the x,y, and z Gamma values. $\gamma = \frac{1}{\sqrt(1-\beta^2)}$

What you are looking for is a function of $f(v_i)$ such that $f(v_x) + f(v_y) + f(v_z) = KE$.

In the case of Newtonian physics, $f(v_i) = \frac{1}{2}mv_i^2$.

In the case of special relativity, it is impossible to find any such function.

Proof by contradiction: Suppose that such an $f$ did exist. Now imagine three objects of mass $m$, each with a different velocity:

Object $A$ is stationary. Its kinetic energy, denoted $KE_A$, is equal to $0$. This implies that $f(0)+f(0)+f(0)=0$, and therefore $f(0)=0$.

Object B is traveling in the $x$ direction with velocity $v_x=\frac{3c}{5}$. Therefore it has kinetic energy $KE_B = \frac{1}{4}mc^2$. Together with the result for object A, this implies that $f(\frac{3c}{5}) = \frac{1}{4}mc^2$.

Object C is traveling in both the $x$ direction and the $y$ direction. Its velocity components are $v_x=v_y=\frac{3c}{5}$. Its total velocity is $v = \sqrt{2}\frac{3c}{5}$. We have: $\gamma=\frac{1}{\sqrt{1-\frac{18}{25}}} = \frac{5}{\sqrt{7}}$. Therefore, $KE_C=\frac{5\sqrt{7}-7}{7}mc^2$. However, it must also be true that $KE_c = f(\frac{3c}{5}) + f(\frac{3c}{5}) = \frac{1}{2}mc^2$.