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?
 A: 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.
A: 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. $\gamma = \frac{1}{\sqrt(1-\beta^2)}$
A: Although Ricky Tensor is totally right, I like to understand special relativity visually through hyperbolas.  Consider this space-time diagram:

The slope of the green lines represents the speed of light.  The purple arrow represents your motion over 1 second, as perceived by yourself: you don't move.  The blue arrow represents that same motion perceived by someone you're moving past at the origin - in this case, you can see that your speed as seen by that observer is about $\frac35$ of light speed.
The law of special relativity says your displacement will reach the same red hyperbola (whose asymptote is the speed of light) regardless of your speed relative to the observer.  Conversely, any displacement that ends at that hyperbola must have taken 1 second for the person moving.  That's what they call the "proper time" of the motion.
At the same time, by changing the axis labels, you could think of these arrows as representing your "spacetime velocity": you move by the purple arrow per second of proper time as seen by yourself, or you move by the blue arrow per second of proper time as seen by that other observer.  That's what they call the 4-velocity, because it has a time component in addition to the spatial ones.  In this picture for example, the time component seen by yourself is 1 second of perceived time per 1 second of proper time, while that seen by the observer is about 1.2 seconds of observer time per 1 second of proper time.  However, we multiply the time component by the speed of light so that it has units of velocity like the others.  So the time component here is actually just $c$ for yourself, or $1.2c$ for the observer.
Without relativity, we'd expect the blue arrow to end at the same time value as the purple one, because we'd think the time of your motion is the same for everybody.  But as you can see, in relativity it's longer.
Likewise, the spatial component of your 4-velocity is bigger than your regular velocity, because the distance is stretched out, but the proper time is still 1 second.
Now to get relativistic energy, you actually just multiply the 4-velocity by $mc$, and look at the time component.  For the purple 4-velocity, remember the time component is $c$, so that turns into $mc^2$.  But for the blue one, it's slightly bigger - in fact, when you do the math, it turns out to be bigger by $\frac12mv^2$, to first order.  And really, that's why we call that time component "energy": because for low speeds, it has the same formula as Newtonian kinetic energy ... plus this curious constant offset of $mc^2$, which Einstein, by association, realized must be energy bound up in the mass itself!
But that Newtonian term is only the first-order approximation; for high speeds, you have to add in additional terms.  And that's because the red curve is a hyperbola, NOT a parabola.  It just looks like a parabola close to the center (low speeds).
And to answer your question, you can see what happens when you add more spatial dimensions.  With a second spatial dimension, the hyperbola becomes a hyperboloid of revolution about the time axis.  Again, near the center it looks like a parabaloid whose value is $mc^2 + \frac12mv^2$, and since $v^2 = v_x^2 + v_y^2$, you can separate the component contributions.  That's actually because the cross-sections of a paraboloid are identical parabolas with different offsets.  But the same is not true of a hyperboloid, so the general (high speed) formula is not separable like that.
