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Revision Browser : relation between force and potential energy
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view 'relation between force and potential energy
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diff
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2009-03-06 18:11:50
- revision [
Version = 5
-->
Version 6
]
by
bci1
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% almost certainly you want these
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diff
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2009-03-06 18:10:31
- revision [
Version = 4
-->
Version 5
]
by
bci1
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derivation is true if and only if you impose (from the starting!) that the field's force $\mathbf{F}$ is irrotational, i.e. $\nabla\times\mathbf{F}=\marhbf{0}$, that is, $\nabla\times\mathbf{F}=\marhbf{0} \Leftrightarrow \marhbf{F}=-\nabla U$.
In another words, the field's force is conservative if and only if it is irrotational. So the conseravation of mechanical energy $dE/dt=d(T+U)/dt=0$ is a consequence of that theorem. Once you impose $\nabla\times\mathbf{F}=\marhbf{0}$, then you are proving the necessary condition for $\marhbf{F}=-\nabla U$. No problem about that. Another consequence about that theorem is that the ``work'' of the field's force is independent of the path described by the particle in its motion. That is, if $\Gamma_1$ and $\Gamma_2$ are two different paths, described by the particle, and joininig its initial and end position on the time interval $[t_1,t_2]$, then the line integrals
$\int_{\Gamma_1}\mathbf{F}\cdot d\mathbf{r}=
\int_{\Gamma_2}\mathbf{F}\cdot d\mathbf{r}$
must be equal and hence the work of the field's force, as the particle describes a closed path, must be zero, i.e.
$\oint\mathbf{F}\cdot d\mathbf{r}=0$.
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