<s>[INST] 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?
 [/INST] 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 $KE_c = f(\frac{3c}{5}) + f(\frac{3c}{5}) = \frac{1}{2}mc^2$.
Since these two answers are different, we have a contradiction.
Q.E.D.
</s>