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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:

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

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

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    $\begingroup$ Do you denote the latter three coordinates with the dots because you ran out of letters, or do you actually mean that they are time derivatives of the first three? (If the latter, your particle would not be in 6 dimensions, but in three) $\endgroup$
    – ACuriousMind
    Aug 2, 2014 at 13:47
  • $\begingroup$ $\dot{x},\dot{y},\dot{z}$ are time derivatives of $x,y,z$. And you are right,the particle is in $3-$ dimension. Rather than saying $6-$ dimension, I should have said $6$ degrees of freedom. $\endgroup$
    – user105997
    Aug 2, 2014 at 16:14

2 Answers 2

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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.

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  • $\begingroup$ Here, the six dimensional space is something like phase space. The origin condition shows it. $\endgroup$
    – qfzklm
    Aug 2, 2014 at 16:06
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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.

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