Facing a paradox: 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 $$frac{d^2V}{dx^2}=-frac{rho(x)}{epsilon_0}=-frac{qdelta(x-x_0)}{epsilon_0}.$$ 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







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$endgroup$

















    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 $$frac{d^2V}{dx^2}=-frac{rho(x)}{epsilon_0}=-frac{qdelta(x-x_0)}{epsilon_0}.$$ 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











    $endgroup$















      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 $$frac{d^2V}{dx^2}=-frac{rho(x)}{epsilon_0}=-frac{qdelta(x-x_0)}{epsilon_0}.$$ 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 $$frac{d^2V}{dx^2}=-frac{rho(x)}{epsilon_0}=-frac{qdelta(x-x_0)}{epsilon_0}.$$ 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 Apr 4 at 16:44









      Aaron Stevens

      15.6k42556




      15.6k42556










      asked Apr 4 at 13:53









      SRSSRS

      6,852435126




      6,852435126






















          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









          $endgroup$














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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

            15.6k42556




            15.6k42556










            answered Apr 4 at 13:58









            knzhouknzhou

            47.2k11128227




            47.2k11128227












            • $begingroup$
              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

                15.6k42556




                15.6k42556






























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