- #31
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This is a bit subtle, and I think there are already many answers to your question. So here are my 2cts in addition.
Usually in classical mechanics an external force is called "conservative", if it only depends on the position of the particle and if it has a scalar potential, i.e., if it is of the form
\vec{F}(\vec{x})=-\vec{\nabla} U(\vec{x}).
An interaction force between two particles is called conservative, if there exists a potential U(\vec{x}_1-\vec{x}_2) such that
\vec{F}_{12}=-\vec{\nabla}_1 U=-\vec{\nabla}_2 U=-F_{21}.
The specific dependence on the difference of the position vectors of the two particles is demanded by Newton's 3rd Law ("actio=reactio").
Other posters have already stated the conditions on the force/interaction field(s) that guarantee the existence of a potential and thus the criteria for a force/interaction being conservative.
Examples are the motion of a charged particle in an electrostatic field or the mutual gravitative attraction between bodies (in Newtonian approximation).
This is a pretty limited view on forces, however. E.g., there are nearly no relativistically covariant forces. An exception is the motion in an electrostatic field. Here, the equation of motion is governed by the more general Lorentz force, including the interaction with the magnetic field, but this force is not conservative in the above given sense, because the force depends on the velocity of the particle and not only on its position.
Nevertheless the conservation of energy holds for static (and only static!) electromagnetic fields, as can be easily proven by taking the line integral of the force along the particle's trajectory under the influence of the Lorentz force. It turns out that only the electric field does work on the particle while the magnetic field only changes the direction of the particle's trajectory. Indeed the power is
P=\vec{v} \cdot \vec{F}=\vec{v} \cdot q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=q \vec{v} \cdot \vec{E}.
The total energy of the particle is given by
\mathcal{E}=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}+q \Phi,
where \Phi is the potential of the electrostatic field,
\vec{E}=-\vec{\nabla} \Phi.
Usually in classical mechanics an external force is called "conservative", if it only depends on the position of the particle and if it has a scalar potential, i.e., if it is of the form
\vec{F}(\vec{x})=-\vec{\nabla} U(\vec{x}).
An interaction force between two particles is called conservative, if there exists a potential U(\vec{x}_1-\vec{x}_2) such that
\vec{F}_{12}=-\vec{\nabla}_1 U=-\vec{\nabla}_2 U=-F_{21}.
The specific dependence on the difference of the position vectors of the two particles is demanded by Newton's 3rd Law ("actio=reactio").
Other posters have already stated the conditions on the force/interaction field(s) that guarantee the existence of a potential and thus the criteria for a force/interaction being conservative.
Examples are the motion of a charged particle in an electrostatic field or the mutual gravitative attraction between bodies (in Newtonian approximation).
This is a pretty limited view on forces, however. E.g., there are nearly no relativistically covariant forces. An exception is the motion in an electrostatic field. Here, the equation of motion is governed by the more general Lorentz force, including the interaction with the magnetic field, but this force is not conservative in the above given sense, because the force depends on the velocity of the particle and not only on its position.
Nevertheless the conservation of energy holds for static (and only static!) electromagnetic fields, as can be easily proven by taking the line integral of the force along the particle's trajectory under the influence of the Lorentz force. It turns out that only the electric field does work on the particle while the magnetic field only changes the direction of the particle's trajectory. Indeed the power is
P=\vec{v} \cdot \vec{F}=\vec{v} \cdot q \left (\vec{E}+\frac{\vec{v}}{c} \times \vec{B} \right )=q \vec{v} \cdot \vec{E}.
The total energy of the particle is given by
\mathcal{E}=\frac{m c^2}{\sqrt{1-\vec{v}^2/c^2}}+q \Phi,
where \Phi is the potential of the electrostatic field,
\vec{E}=-\vec{\nabla} \Phi.