# Energy and motion from $R^6 \rightarrow R$

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## Main Question or Discussion Point

We know that energy is a function of space and velocity and it’s constant (in ideal case) though time.
So $E(\vec{x}(t) , \vec{\dot{x}}(t)) = E_0$

where $\vec{x} , \vec{\dot{x}} \in \mathbb{R}^3$.
So my function is $E : \mathbb{R}^6 \rightarrow \mathbb{R}$.

Then there is my question: my teacher said that the motion will not be in a 6-dimensional space but on a 5-dimensional *surface*.
I don’t understand that statement (and maybe I understood it wrongly).
Anybody can help me?

Related Classical Physics News on Phys.org
That fact $E(\vec{x}(t) , \vec{\dot{x}}(t)) = E_0$ is a consequence of $E= K(\vec{\dot{x}})+ U(\vec{x})$

$$f(x,y) =x^2 +y^2=R^2$$
What kind of object is defined by f(x,y)=constant?

$$f(x,y) =x^2 +y^2=R^2$$
What kind of object is defined by f(x,y)=constant?
Wow, thanks, you're right.