Facing a paradox: Earnshaw's theorem in one dimensionDoes 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?
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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?
$begingroup$
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
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
asked Apr 4 at 13:53
SRSSRS
6,746434125
$endgroup$
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
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
asked Apr 4 at 13:53
SRSSRS
6,746434125
$endgroup$
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
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
asked Apr 4 at 13:53
SRSSRS
6,746434125
$endgroup$
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
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
asked Apr 4 at 13:53
SRSSRS
6,746434125
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
asked Apr 4 at 13:53
SRSSRS
6,746434125
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
14.9k42453
asked Apr 4 at 13:53
SRSSRS
6,746434125
asked Apr 4 at 13:53
SRSSRS
6,746434125
asked Apr 4 at 13:53
SRSSRS
6,746434125
6,746434125
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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.
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
answered Apr 4 at 13:58
knzhouknzhou
46.7k11126224
$endgroup$
$begingroup$
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
|
show 1 more 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.
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
14.9k42453
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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.
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
answered Apr 4 at 13:58
knzhouknzhou
46.7k11126224
$endgroup$
$begingroup$
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
|
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.
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
answered Apr 4 at 13:58
knzhouknzhou
46.7k11126224
$endgroup$
$begingroup$
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
|
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.
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
answered Apr 4 at 13:58
knzhouknzhou
46.7k11126224
$endgroup$
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.
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
answered Apr 4 at 13:58
knzhouknzhou
46.7k11126224
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
edited Apr 4 at 16:44
Aaron Stevens
14.9k42453
14.9k42453
answered Apr 4 at 13:58
knzhouknzhou
46.7k11126224
answered Apr 4 at 13:58
knzhouknzhou
46.7k11126224
answered Apr 4 at 13:58
knzhouknzhou
46.7k11126224
46.7k11126224
$begingroup$
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
|
show 1 more comment
$begingroup$
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
|
show 1 more 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.
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
14.9k42453
$endgroup$
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.
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
14.9k42453
$endgroup$
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.
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
14.9k42453
$endgroup$
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.
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
14.9k42453
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
14.9k42453
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
14.9k42453
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
14.9k42453
14.9k42453
add a comment |
add a comment |
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