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Facing a paradox: Earnshaw's theorem in one dimension


Does 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?













3












$begingroup$


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











$endgroup$
















    3












    $begingroup$


    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











    $endgroup$














      3












      3








      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











      $endgroup$




      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




      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 6 hours ago









      Aaron Stevens

      14.2k42252




      14.2k42252










      asked 9 hours ago









      SRSSRS

      6,498433123




      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











          $endgroup$












          • $begingroup$
            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


















          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









          $endgroup$













            Your Answer





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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            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











            $endgroup$












            • $begingroup$
              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















            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











            $endgroup$












            • $begingroup$
              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













            8












            8








            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











            $endgroup$



            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 6 hours ago









            Aaron Stevens

            14.2k42252




            14.2k42252










            answered 8 hours ago









            knzhouknzhou

            46.1k11124222




            46.1k11124222











            • $begingroup$
              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
















            • $begingroup$
              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















            $begingroup$
            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




            $begingroup$
            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




            $begingroup$
            @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




            $begingroup$
            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




            $begingroup$
            @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




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











            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









            $endgroup$

















              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









              $endgroup$















                3












                3








                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









                $endgroup$



                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 8 hours ago









                Aaron StevensAaron Stevens

                14.2k42252




                14.2k42252



























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