Bertrand's and Earnshaw's theorems contradictionby Trifis Tags: bertrand, contradiction, earnshaw, theorems 

#1
May1212, 05:08 PM

P: 128

I think the title is selfexplanatory. The first theorem states that gravitational forces (1/r potentials in general) are able to produce stable orbits, whereas the second excludes stability! Can somebody help me to clear this out?




#2
May1212, 05:22 PM

P: 3,554





#3
May1212, 05:52 PM

P: 128

I need something more elaborate please. 



#4
May1312, 07:17 AM

P: 3,554

Bertrand's and Earnshaw's theorems contradiction
static = no movement
orbits = movement 



#5
May1312, 07:23 AM

P: 128

When an orbit has a stable point then the particle can as well stay at this point point forever without losing its dynamical stability.




#6
May1312, 08:21 AM

P: 3,554





#7
May1312, 09:13 AM

P: 102

Specifically, Earnshaw's Theorem states that in a static situation for pointlike particles, a 1/r potential does not have any maxima or minina (stable points) in an unoccupied region, since the sources themselves occupy space. When dynamics are added into the mix, there is an effective potential from the angular component which pushes away from the source and falls off as 1/r^{2}. For example, with gravity, the potential is a combination of angular repulsion ([itex]\frac{1}{2}\frac{mh^2}{r^2}[/itex]) and gravitational attraction ([itex]\frac{GMm}{r}[/itex]), which gives a total potential of [itex]U=\frac{1}{2}\frac{mh^2}{r^2}  \frac{GMm}{r}[/itex], and a minimum at [itex]r=\frac{h^2}{GM}[/itex].
(h is angular momentum per mass) 



#8
May2712, 04:06 PM

P: 128

Ok therefore it is the extra angular movement which provides the stability of the ORBIT and cannot be found in the static case.
On second thought it can be said that since Earnshaw applies only on 1/r forces (my oscillation argumantion was thereby false) there weren't any equilibrium states Keplerlike orbits first place to debate on in the first place ... 



#9
Dec2013, 02:53 AM

P: 108

so since Laplace says that there can be no local extrema then a charge at the center of a cube with charges at the 6 corners cannot be in electrostatic equilibrium since then U would be at a minimum? if the potential is like a saddle point for the center charge in a cube then in the xz plane it is at a max and yz it is at a minimum at the same time? (do you calculate the potential by superposition to find the saddle point?)
how do you know that the charge leaks out of every face of the cube? Griffiths doesn't say that much about this, is it better to read Purcell and Wave Electromagnetics at the same time? 


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