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

  • #1
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0

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?
 

Answers and Replies

  • #2
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That fact ## E(\vec{x}(t) , \vec{\dot{x}}(t)) = E_0## is a consequence of ##E= K(\vec{\dot{x}})+ U(\vec{x})##
 
  • #3
3,740
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$$ f(x,y) =x^2 +y^2=R^2 $$
What kind of object is defined by f(x,y)=constant?
 
  • #4
8
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$$ f(x,y) =x^2 +y^2=R^2 $$
What kind of object is defined by f(x,y)=constant?
Wow, thanks, you're right.
 

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