Forest UCM Energy CentralForce
A central force is defined as a force depends only on separation distance
- [math]\displaystyle{ \vec{F} = f(\vec r) \hat r }[/math]
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Coulomb force and gravitation force.
Spherical Coordinates
Forest_UCM_NLM_Ch1_CoordSys#Spherical
Gradient in spherical coordinates
The differential change of [math]\displaystyle{ \vec r }[/math] in spherical coordinates occurs in three directions.
In the radial direction
- [math]\displaystyle{ dr \hat r }[/math]
In the polar angle direction
- [math]\displaystyle{ r d \theta \hat \theta }[/math]
In the azimuthal angle direction
- [math]\displaystyle{ r \sin \phi d \phi \hat \phi }[/math]
The differential force of the displacement vector in spherical coordinates is
- [math]\displaystyle{ d \vec r = dr \hat r + r d \theta \hat \theta + r \sin \phi d \phi \hat \phi }[/math]
The derivative may be represented as
- [math]\displaystyle{ df = f(x+dx) -f(x) = \frac{df}{dx} dx }[/math]
in three dimensional cartesian coordinates this may be written in terms of the gradient as
- [math]\displaystyle{ df = \frac{df}{dx} dx + \frac{df}{dy} dy + \frac{df}{dz} dz =\vec \nabla f \cdot d \vec r }[/math]
To determine the gradient in sperical coordinates on just compares the two equations
- [math]\displaystyle{ \frac{df}{dr} dr + \frac{df}{d \theta} d\theta + \frac{df}{d\phi} d\phi =\vec \nabla f \cdot d \vec r }[/math]
- [math]\displaystyle{ \vec \nabla f \cdot d \vec r = \vec \nabla f \cdot \left ( dr \hat r + r d \theta \hat \theta + r \sin \phi d \phi \hat \phi \right ) }[/math]
- [math]\displaystyle{ = \left ( \vec \nabla f \right )_r dr +\left ( \vec \nabla f \right )_{\theta} r d\theta +\left ( \vec \nabla f \right )_{\phi} r \sin \theta d\phi }[/math]
comparing terms of the above with
- [math]\displaystyle{ \left ( \vec \nabla f \right )_r dr =\frac{df}{dr} dr \Rightarrow \left ( \vec \nabla f \right )_r =\frac{df}{dr} }[/math]
- [math]\displaystyle{ \left ( \vec \nabla f \right )_{\theta} r d\theta =\frac{df}{d \theta} d\theta \Rightarrow \left ( \vec \nabla f \right )_{\theta} d\theta =\frac{1}{r} \frac{df}{d \theta} }[/math]
- [math]\displaystyle{ \left ( \vec \nabla f \right )_{\phi} r \sin \theta d\phi =\frac{df}{d\phi} d\phi \Rightarrow \left ( \vec \nabla f \right )_{\phi} d\phi = \frac{1}{ r \sin \theta } \frac{df}{d\phi} }[/math]
or
- [math]\displaystyle{ \vec \nabla = \hat r \frac{d}{dr} + \hat \theta \frac{1}{r} \frac{d}{d \theta} + \hat \phi \frac{1}{ r \sin \theta } \frac{d}{d\phi} }[/math]
The Central Force
Spherical symmetry
A central force needs to be spherically symmetric in order to be conservative.
- [math]\displaystyle{ \vec F = f(\vec r) \hat r }[/math]
If the force is conservative then a potential exists such that
- [math]\displaystyle{ \vec F = -\vec \nabla U }[/math]
- [math]\displaystyle{ = - \left ( \hat r \frac{d}{dr} + \hat \theta \frac{1}{r} \frac{d}{d \theta} + \hat \phi \frac{1}{ r \sin \theta } \frac{d}{d\phi} \right ) U }[/math]
in order to be only a function of [math]\displaystyle{ r }[/math] then
- [math]\displaystyle{ \hat \theta \frac{1}{r} \frac{d}{d \theta} U = 0 }[/math]
- [math]\displaystyle{ \hat \phi \frac{1}{ r \sin \theta } \frac{d}{d\phi} U = 0 }[/math]
If [math]\displaystyle{ \vec F }[/math] is a conservative central force then [math]\displaystyle{ \vec F }[/math] and [math]\displaystyle{ U }[/math] are spherically symmetric.
The converse is also true
- if the central force is spherically symmetric then it is a conservative force.