Kinetic Energy of a Particle Consider a particle of mass $m$ in $6$ dimension. Its coordinate w.r.t. origin $\left(0,0,0,0,0,0\right)$ is given as $\left(x,y,z,\dot{x},\dot{y},\dot{z}\right)$. If we denote $r = \sqrt{x^2+y^2+z^2}$, then which of the following two is the kinetic energy of this particle:


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*$T = \frac{1}{2}m\left(\dot{x}^2+\dot{y}^2+\dot{z}^2\right)~?$

*$T = \frac{1}{2}m\left(\frac{dr}{dt}\right)^2 = \frac{1}{2}m\frac{\left(x\dot{x}+y\dot{y}+z\dot{z}\right)^2}{r^2}~?$
 A: Hard for me to say what you're asking. If you have a particle of mass $M$ in three dimensions such that its positions is described by coordinates $\vec x(t)=(x(t),y(t),z(t))$, then the velocity vector $\vec v=\frac{d\vec x}{dt}=(\dot x(t),\dot y(t),\dot z(t))$
The kinetic energy is then defined as,
$$T=\frac{1}{2}M\vec v\cdot \vec v=\frac{1}{2}M\bigg(\dot x(t)^2+\dot y(t)^2+\dot z(t)^2\bigg)$$
If you're asking about a six dimensional space that has coordinates, $$\vec x(t)=(x_1(t),x_2(t),x_3(t),x_4(t),x_5(t),x_6(t))$$
$$\vec v = (\dot x_1(t),\dot x_2(t),\dot x_3(t),\dot x_4(t),\dot x_5(t),\dot x_6(t))$$
Then the kinetic energy is similar to the first one (assuming a 6-dim Euclidean metric)
$$T=\frac{1}{2}M\vec v\cdot \vec v=\frac{1}{2}M\bigg(\dot x_1(t)^2+\dot x_2(t)^2+\dot x_3(t)^2+\dot x_4(t)^2+\dot x_5(t)^2+\dot x_6(t)^2\bigg)$$
If the question is "why is that the form of the kinetic energy?" then you could say its a low velocity approximation of the energy of a particle. There are other neat motivations i've seen for that question as well.
Hope that helps.
A: The kinetic energy is $T = \frac{1}{2} m (\frac{d \vec{r}}{d t})^2$
$$\vec{r} = x \vec{i} + y \vec{j} + z \vec{k}$$
The first exprssion is right.
