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Facing a paradox: Earnshaw's theorem in one dimension

Revisiting Earnshaw's theorem from a different perspectiveClassify equilibrium points and find bifurcation points of a non-linear dynamic systemEarnshaw's theorem and springsEarnshaw's theorem for extended conducting bodiesPotential due to charge over infinite grounded plane conductor using the method of imagesRelation between electric field and dipole momentEarnshaw's theorm and Effective potentialDielectric liquid sucked up between two cylinders with a voltage differenceElectrostatics: Induced Boundary Dipole LayerWhy do we assume simply connected domains and continuously differentiable fields in electromagnetism theory?Does this example contradict Earnshaw's theorem in one dimension?

3

1

$begingroup$

Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$fracd^2Vdx^2=-fracrho(x)epsilon_0=-fracqdelta(x-x_0)epsilon_0.$$ If $q>0$, $V^primeprime(x_0)<0$, and if $q<0$, $V^primeprime(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?

electrostatics mathematical-physics potential classical-electrodynamics equilibrium

share|cite|improve this question

edited 5 hours ago

Aaron Stevens

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asked 8 hours ago

SRSSRS

6,498433123

$endgroup$

add a comment | 

3

1

$begingroup$

Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$fracd^2Vdx^2=-fracrho(x)epsilon_0=-fracqdelta(x-x_0)epsilon_0.$$ If $q>0$, $V^primeprime(x_0)<0$, and if $q<0$, $V^primeprime(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?

electrostatics mathematical-physics potential classical-electrodynamics equilibrium

share|cite|improve this question

edited 5 hours ago

Aaron Stevens

14.2k42252

asked 8 hours ago

SRSSRS

6,498433123

$endgroup$

add a comment | 

$begingroup$

Consider a one-dimensional situation on a straight line (say, $x$-axis). Let a charge of magnitude $q$ be located at $x=x_0$, the potential satisfies the Poisson's equation $$fracd^2Vdx^2=-fracrho(x)epsilon_0=-fracqdelta(x-x_0)epsilon_0.$$ If $q>0$, $V^primeprime(x_0)<0$, and if $q<0$, $V^primeprime(x_0)>0$. Therefore, it appears that the potential $V$ does have a minimum at $x=x_0$, for $q<0$. Does this imply that $x=x_0$ is a point of stable equilibrium? I must be missing something because this appears to violate Earnshaw's theorem (or it doesn't)?

electrostatics mathematical-physics potential classical-electrodynamics equilibrium

share|cite|improve this question

edited 5 hours ago

Aaron Stevens

14.2k42252

asked 8 hours ago

SRSSRS

6,498433123

$endgroup$

share|cite|improve this question

edited 5 hours ago

Aaron Stevens

14.2k42252

asked 8 hours ago

SRSSRS

6,498433123

share|cite|improve this question

edited 5 hours ago

Aaron Stevens

14.2k42252

asked 8 hours ago

SRSSRS

6,498433123

edited 5 hours ago

Aaron Stevens

14.2k42252

edited 5 hours ago

Aaron Stevens

14.2k42252

asked 8 hours ago

SRSSRS

6,498433123

asked 8 hours ago

SRSSRS

6,498433123

2 Answers
2

active

oldest

votes

8

$begingroup$

Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.

share|cite|improve this answer

edited 5 hours ago

Aaron Stevens

14.2k42252

answered 8 hours ago

knzhouknzhou

46.1k11124222

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
  $endgroup$
  – SRS
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SRS Yes, that's true.
  $endgroup$
  – knzhou
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
  $endgroup$
  – SRS
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SRS The potential at a point charge is not defined (or you could say infinite)
  $endgroup$
  – Aaron Stevens
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have to think more about it and I'll get back.
  $endgroup$
  – SRS
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

3

$begingroup$

So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$

This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.

share|cite|improve this answer

answered 8 hours ago

Aaron StevensAaron Stevens

14.2k42252

$endgroup$

add a comment | 

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8

$begingroup$

Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.

share|cite|improve this answer

edited 5 hours ago

Aaron Stevens

14.2k42252

answered 8 hours ago

knzhouknzhou

46.1k11124222

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
  $endgroup$
  – SRS
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SRS Yes, that's true.
  $endgroup$
  – knzhou
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
  $endgroup$
  – SRS
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SRS The potential at a point charge is not defined (or you could say infinite)
  $endgroup$
  – Aaron Stevens
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have to think more about it and I'll get back.
  $endgroup$
  – SRS
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

8

$begingroup$

Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.

share|cite|improve this answer

edited 5 hours ago

Aaron Stevens

14.2k42252

answered 8 hours ago

knzhouknzhou

46.1k11124222

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes. I meant Earnshaw's theorem. Thanks. Does it mean that there must be an Earnshaw's theorem for Newtonian gravitation? Because in a massless region, again one has $V^primeprime(x)=0$?
  $endgroup$
  – SRS
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SRS Yes, that's true.
  $endgroup$
  – knzhou
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I am not yet totally comfortable with this. If you have a charge at some point $x=x_0$, is it not correct to look at the behaviour of the potential at that point? @knzhou
  $endgroup$
  – SRS
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @SRS The potential at a point charge is not defined (or you could say infinite)
  $endgroup$
  – Aaron Stevens
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I have to think more about it and I'll get back.
  $endgroup$
  – SRS
  6 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

Your example does not contradict Earnshaw's theorem for electrostatics, because it rules out stable equilibrium in a region without charge, possibly containing fields made by charges outside that region. Here you're doing the exact opposite, looking at the only point in your situation with charge.

share|cite|improve this answer

edited 5 hours ago

Aaron Stevens

14.2k42252

answered 8 hours ago

knzhouknzhou

46.1k11124222

$endgroup$

share|cite|improve this answer

edited 5 hours ago

Aaron Stevens

14.2k42252

answered 8 hours ago

knzhouknzhou

46.1k11124222

edited 5 hours ago

Aaron Stevens

14.2k42252

edited 5 hours ago

Aaron Stevens

14.2k42252

answered 8 hours ago

knzhouknzhou

46.1k11124222

answered 8 hours ago

knzhouknzhou

46.1k11124222

3

$begingroup$

So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$

This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.

share|cite|improve this answer

answered 8 hours ago

Aaron StevensAaron Stevens

14.2k42252

$endgroup$

add a comment | 

3

$begingroup$

So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$

This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.

share|cite|improve this answer

answered 8 hours ago

Aaron StevensAaron Stevens

14.2k42252

$endgroup$

add a comment | 

$begingroup$

So technically $V''(x_0)$ doesn't have an actual value, since $delta(x-x_0)toinfty$ as $xto x_0$. However, if you understand the Dirac delta distribution to be a limit of a function whose peak "gets narrower" with its integral remaining constant, then this is fine and you could say there is a minimum at $x_0$ for $q<0$

This can be more easily understood by just thinking about the motion of a positive charge in this potential. It will move towards the negative charge, i.e. towards the minimum of the potential.

share|cite|improve this answer

answered 8 hours ago

Aaron StevensAaron Stevens

14.2k42252

$endgroup$

share|cite|improve this answer

answered 8 hours ago

Aaron StevensAaron Stevens

14.2k42252

answered 8 hours ago

Aaron StevensAaron Stevens

14.2k42252

answered 8 hours ago

Aaron StevensAaron Stevens

14.2k42252