Facing a paradox: 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 $$frac{d^2V}{dx^2}=-frac{rho(x)}{epsilon_0}=-frac{qdelta(x-x_0)}{epsilon_0}.$$ 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
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
asked Apr 4 at 13:53
SRSSRS
6,852435126
$endgroup$
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 $$frac{d^2V}{dx^2}=-frac{rho(x)}{epsilon_0}=-frac{qdelta(x-x_0)}{epsilon_0}.$$ 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
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
asked Apr 4 at 13:53
SRSSRS
6,852435126
$endgroup$
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 $$frac{d^2V}{dx^2}=-frac{rho(x)}{epsilon_0}=-frac{qdelta(x-x_0)}{epsilon_0}.$$ 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
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
asked Apr 4 at 13:53
SRSSRS
6,852435126
$endgroup$
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
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
asked Apr 4 at 13:53
SRSSRS
6,852435126
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
asked Apr 4 at 13:53
SRSSRS
6,852435126
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
15.6k42556
asked Apr 4 at 13:53
SRSSRS
6,852435126
asked Apr 4 at 13:53
SRSSRS
6,852435126
asked Apr 4 at 13:53
SRSSRS
6,852435126
6,852435126
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
15.6k42556
answered Apr 4 at 13:58
knzhouknzhou
47.2k11128227
$endgroup$
$begingroup$
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
15.6k42556
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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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
15.6k42556
answered Apr 4 at 13:58
knzhouknzhou
47.2k11128227
$endgroup$
$begingroup$
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
15.6k42556
answered Apr 4 at 13:58
knzhouknzhou
47.2k11128227
$endgroup$
$begingroup$
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
15.6k42556
answered Apr 4 at 13:58
knzhouknzhou
47.2k11128227
$endgroup$
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
15.6k42556
answered Apr 4 at 13:58
knzhouknzhou
47.2k11128227
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
edited Apr 4 at 16:44
Aaron Stevens
15.6k42556
15.6k42556
answered Apr 4 at 13:58
knzhouknzhou
47.2k11128227
answered Apr 4 at 13:58
knzhouknzhou
47.2k11128227
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
|
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
15.6k42556
$endgroup$
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
15.6k42556
$endgroup$
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
15.6k42556
$endgroup$
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
15.6k42556
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
15.6k42556
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
15.6k42556
answered Apr 4 at 13:59
Aaron StevensAaron Stevens
15.6k42556
15.6k42556
add a comment |
add a comment |
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