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Facing a paradox: Earnshaw's theorem in one dimension



The 2019 Stack Overflow Developer Survey Results Are InDoes this example contradict Earnshaw's theorem in one dimension?Classify 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?










5












$begingroup$


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)?










share|cite|improve this question











$endgroup$
















    5












    $begingroup$


    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)?










    share|cite|improve this question











    $endgroup$














      5












      5








      5


      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)?










      share|cite|improve this question











      $endgroup$




      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















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Apr 4 at 16:44









      Aaron Stevens

      15k42453




      15k42453










      asked Apr 4 at 13:53









      SRSSRS

      6,781434125




      6,781434125




















          2 Answers
          2






          active

          oldest

          votes


















          10












          $begingroup$

          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











          $endgroup$












          • $begingroup$
            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
            Apr 4 at 14:57










          • $begingroup$
            @SRS Yes, that's true.
            $endgroup$
            – knzhou
            Apr 4 at 14:58










          • $begingroup$
            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
            Apr 4 at 15:37










          • $begingroup$
            @SRS The potential at a point charge is not defined (or you could say infinite)
            $endgroup$
            – Aaron Stevens
            Apr 4 at 16:33










          • $begingroup$
            I have to think more about it and I'll get back.
            $endgroup$
            – SRS
            Apr 4 at 16:37


















          4












          $begingroup$

          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









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            10












            $begingroup$

            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











            $endgroup$












            • $begingroup$
              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
              Apr 4 at 14:57










            • $begingroup$
              @SRS Yes, that's true.
              $endgroup$
              – knzhou
              Apr 4 at 14:58










            • $begingroup$
              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
              Apr 4 at 15:37










            • $begingroup$
              @SRS The potential at a point charge is not defined (or you could say infinite)
              $endgroup$
              – Aaron Stevens
              Apr 4 at 16:33










            • $begingroup$
              I have to think more about it and I'll get back.
              $endgroup$
              – SRS
              Apr 4 at 16:37















            10












            $begingroup$

            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











            $endgroup$












            • $begingroup$
              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
              Apr 4 at 14:57










            • $begingroup$
              @SRS Yes, that's true.
              $endgroup$
              – knzhou
              Apr 4 at 14:58










            • $begingroup$
              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
              Apr 4 at 15:37










            • $begingroup$
              @SRS The potential at a point charge is not defined (or you could say infinite)
              $endgroup$
              – Aaron Stevens
              Apr 4 at 16:33










            • $begingroup$
              I have to think more about it and I'll get back.
              $endgroup$
              – SRS
              Apr 4 at 16:37













            10












            10








            10





            $begingroup$

            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











            $endgroup$



            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














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 4 at 16:44









            Aaron Stevens

            15k42453




            15k42453










            answered Apr 4 at 13:58









            knzhouknzhou

            46.8k11126224




            46.8k11126224











            • $begingroup$
              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
              Apr 4 at 14:57










            • $begingroup$
              @SRS Yes, that's true.
              $endgroup$
              – knzhou
              Apr 4 at 14:58










            • $begingroup$
              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
              Apr 4 at 15:37










            • $begingroup$
              @SRS The potential at a point charge is not defined (or you could say infinite)
              $endgroup$
              – Aaron Stevens
              Apr 4 at 16:33










            • $begingroup$
              I have to think more about it and I'll get back.
              $endgroup$
              – SRS
              Apr 4 at 16:37
















            • $begingroup$
              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
              Apr 4 at 14:57










            • $begingroup$
              @SRS Yes, that's true.
              $endgroup$
              – knzhou
              Apr 4 at 14:58










            • $begingroup$
              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
              Apr 4 at 15:37










            • $begingroup$
              @SRS The potential at a point charge is not defined (or you could say infinite)
              $endgroup$
              – Aaron Stevens
              Apr 4 at 16:33










            • $begingroup$
              I have to think more about it and I'll get back.
              $endgroup$
              – SRS
              Apr 4 at 16:37















            $begingroup$
            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
            Apr 4 at 14:57




            $begingroup$
            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
            Apr 4 at 14:57












            $begingroup$
            @SRS Yes, that's true.
            $endgroup$
            – knzhou
            Apr 4 at 14:58




            $begingroup$
            @SRS Yes, that's true.
            $endgroup$
            – knzhou
            Apr 4 at 14:58












            $begingroup$
            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
            Apr 4 at 15:37




            $begingroup$
            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
            Apr 4 at 15:37












            $begingroup$
            @SRS The potential at a point charge is not defined (or you could say infinite)
            $endgroup$
            – Aaron Stevens
            Apr 4 at 16:33




            $begingroup$
            @SRS The potential at a point charge is not defined (or you could say infinite)
            $endgroup$
            – Aaron Stevens
            Apr 4 at 16:33












            $begingroup$
            I have to think more about it and I'll get back.
            $endgroup$
            – SRS
            Apr 4 at 16:37




            $begingroup$
            I have to think more about it and I'll get back.
            $endgroup$
            – SRS
            Apr 4 at 16:37











            4












            $begingroup$

            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









            $endgroup$

















              4












              $begingroup$

              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









              $endgroup$















                4












                4








                4





                $begingroup$

                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









                $endgroup$



                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












                share|cite|improve this answer



                share|cite|improve this answer










                answered Apr 4 at 13:59









                Aaron StevensAaron Stevens

                15k42453




                15k42453



























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