Falling electric dipole contradicts equivalence principle?
$begingroup$
Consider an electric dipole consisting of charges $q$ and $-q$, both of mass $m$, separated by a distance $d$.
If the dipole is given an acceleration $a$ perpendicular to its moment the total electric force on it, due to each charge acting on the other, is given approximately by
$$F_e=frac{e^2a}{c^2d}$$
where we introduce $e^2 equiv q^2/4piepsilon_0$ for clarity. The exact expression is given in Electrostatic Levitation of a Dipole Eq(5).
Now suppose the dipole, initially oriented horizontally, is dropped in a vertical gravitational field of strength $g$.
Applying Newton's second law to the dipole as a whole we have: gravitational force (gravitational mass times field strength) plus electric force must equal the inertial mass times acceleration
$$2mg+F_e=2m a$$
Therefore the acceleration $a$ of the dipole is given by
$$a=glarge(1-frac{e^2}{2mc^2d}large)^{-1}$$
Thus the dipole is accelerating faster than gravity. An observer falling with the dipole will see it move away from him whereas in deep space the observer would not see the dipole move away.
Surely this contradicts the equivalence principle?
electromagnetism general-relativity classical-electrodynamics equivalence-principle
$endgroup$
add a comment |
$begingroup$
Consider an electric dipole consisting of charges $q$ and $-q$, both of mass $m$, separated by a distance $d$.
If the dipole is given an acceleration $a$ perpendicular to its moment the total electric force on it, due to each charge acting on the other, is given approximately by
$$F_e=frac{e^2a}{c^2d}$$
where we introduce $e^2 equiv q^2/4piepsilon_0$ for clarity. The exact expression is given in Electrostatic Levitation of a Dipole Eq(5).
Now suppose the dipole, initially oriented horizontally, is dropped in a vertical gravitational field of strength $g$.
Applying Newton's second law to the dipole as a whole we have: gravitational force (gravitational mass times field strength) plus electric force must equal the inertial mass times acceleration
$$2mg+F_e=2m a$$
Therefore the acceleration $a$ of the dipole is given by
$$a=glarge(1-frac{e^2}{2mc^2d}large)^{-1}$$
Thus the dipole is accelerating faster than gravity. An observer falling with the dipole will see it move away from him whereas in deep space the observer would not see the dipole move away.
Surely this contradicts the equivalence principle?
electromagnetism general-relativity classical-electrodynamics equivalence-principle
$endgroup$
$begingroup$
Related paper: arxiv.org/pdf/1311.5798.pdf
$endgroup$
– kkm
3 hours ago
$begingroup$
Possible duplicate of Does a charged particle accelerating in a gravitational field radiate?
$endgroup$
– Aaron Stevens
1 hour ago
add a comment |
$begingroup$
Consider an electric dipole consisting of charges $q$ and $-q$, both of mass $m$, separated by a distance $d$.
If the dipole is given an acceleration $a$ perpendicular to its moment the total electric force on it, due to each charge acting on the other, is given approximately by
$$F_e=frac{e^2a}{c^2d}$$
where we introduce $e^2 equiv q^2/4piepsilon_0$ for clarity. The exact expression is given in Electrostatic Levitation of a Dipole Eq(5).
Now suppose the dipole, initially oriented horizontally, is dropped in a vertical gravitational field of strength $g$.
Applying Newton's second law to the dipole as a whole we have: gravitational force (gravitational mass times field strength) plus electric force must equal the inertial mass times acceleration
$$2mg+F_e=2m a$$
Therefore the acceleration $a$ of the dipole is given by
$$a=glarge(1-frac{e^2}{2mc^2d}large)^{-1}$$
Thus the dipole is accelerating faster than gravity. An observer falling with the dipole will see it move away from him whereas in deep space the observer would not see the dipole move away.
Surely this contradicts the equivalence principle?
electromagnetism general-relativity classical-electrodynamics equivalence-principle
$endgroup$
Consider an electric dipole consisting of charges $q$ and $-q$, both of mass $m$, separated by a distance $d$.
If the dipole is given an acceleration $a$ perpendicular to its moment the total electric force on it, due to each charge acting on the other, is given approximately by
$$F_e=frac{e^2a}{c^2d}$$
where we introduce $e^2 equiv q^2/4piepsilon_0$ for clarity. The exact expression is given in Electrostatic Levitation of a Dipole Eq(5).
Now suppose the dipole, initially oriented horizontally, is dropped in a vertical gravitational field of strength $g$.
Applying Newton's second law to the dipole as a whole we have: gravitational force (gravitational mass times field strength) plus electric force must equal the inertial mass times acceleration
$$2mg+F_e=2m a$$
Therefore the acceleration $a$ of the dipole is given by
$$a=glarge(1-frac{e^2}{2mc^2d}large)^{-1}$$
Thus the dipole is accelerating faster than gravity. An observer falling with the dipole will see it move away from him whereas in deep space the observer would not see the dipole move away.
Surely this contradicts the equivalence principle?
electromagnetism general-relativity classical-electrodynamics equivalence-principle
electromagnetism general-relativity classical-electrodynamics equivalence-principle
edited 2 hours ago
John Eastmond
asked 4 hours ago
John EastmondJohn Eastmond
1,8891225
1,8891225
$begingroup$
Related paper: arxiv.org/pdf/1311.5798.pdf
$endgroup$
– kkm
3 hours ago
$begingroup$
Possible duplicate of Does a charged particle accelerating in a gravitational field radiate?
$endgroup$
– Aaron Stevens
1 hour ago
add a comment |
$begingroup$
Related paper: arxiv.org/pdf/1311.5798.pdf
$endgroup$
– kkm
3 hours ago
$begingroup$
Possible duplicate of Does a charged particle accelerating in a gravitational field radiate?
$endgroup$
– Aaron Stevens
1 hour ago
$begingroup$
Related paper: arxiv.org/pdf/1311.5798.pdf
$endgroup$
– kkm
3 hours ago
$begingroup$
Related paper: arxiv.org/pdf/1311.5798.pdf
$endgroup$
– kkm
3 hours ago
$begingroup$
Possible duplicate of Does a charged particle accelerating in a gravitational field radiate?
$endgroup$
– Aaron Stevens
1 hour ago
$begingroup$
Possible duplicate of Does a charged particle accelerating in a gravitational field radiate?
$endgroup$
– Aaron Stevens
1 hour ago
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The paper that claims this result was written in reply to An electric dipole in self-accelerated transverse motion, which claims that a dipole in zero gravitational field really can accelerate itself, indefinitely. So if you believe both of the results of these papers, the equivalence principle is satisfied; dipoles can have an extra, weird contribution to their acceleration in both a gravitational field and in a freely falling frame.
However, the analysis of point charges in electromagnetism, and especially their self-interaction, is full of subtleties. More recently, the paper Nonexistence of the self-accelerating dipole and related questions has claimed that both of these results are incorrect, but since the author of this paper is active on Physics.SE I'm going to refrain from attempting to summarize the paper, because I'll probably get it wrong!
From a more general perspective, we already know postulating perfectly pointlike charges leads to a ton of subtleties, even before bringing relativity into the mix. In the modern formulation of quantum field theory in curved spacetime, everything is manifestly covariant from the start, so the equivalence principle is satisfied by construction. Of course it's interesting to see how it can come about in a less fundamental theory like classical electromagnetism, but issues with that aren't going to bring all of relativity crashing down.
$endgroup$
1
$begingroup$
In addition to the points you raise, there are likely to be a couple of vexing issues: (1) that a boosted dipole isn't a dipole (related to Mansuripur's paradox); (2) there is no universally accepted way to define what is a radiation field in this kind of context.
$endgroup$
– Ben Crowell
3 hours ago
$begingroup$
The weird self-acceleration is merely a special case for extremely small separations of the two charges. John Eastmond's observation and question is valid even for dipoles that do not have such strange solutions.
$endgroup$
– Ján Lalinský
26 mins ago
add a comment |
$begingroup$
In your calculation you assume that gravitational mass $M_G$ of the system is $2m$ where $m$ is rest mass of a single particle, thus you assume it is independent of the mutual distance between the charged particles $d$. In other words, you do not take into account the force of gravity acting on the system due to the concentrated bound negative energy of EM field near the charged particles. However, since the system has lower inertial mass, it should also have lower gravitational mass.
It is well known that systems with negative potential EM energy have inertial mass defect. In this case, the dipole is such a system, so it will have lower inertial mass than $2m$, thanks to its negative electrostatic potential energy $-frac{e^2}{d}$.
This "mass defect" effect comes from the forces of "acceleration electric fields" acting to speed up the charged particles. This you have taken into account by including force $F_{em/self} = frac{e^2}{c^2d}a$, which is the electromagnetic self-force acting on the dipole.
But defect in inertial mass should mean also defect in gravitational mass. Heuristically/naively, the gravitational mass to use in the formula $F_G = M_G g$ should correspond to total energy of the system via Einstein's formula
$$
E = M_Gc^2
$$
where $E$ is total energy of the system, including its internal potential energy. Using the Coulomb formula for potential energy,
$$
E = 2mc^2 - frac{e^2}{d}
$$
and so the gravitational mass of the dipole should be taken as
$$
M_{G} = 2m -frac{e^2}{c^2d}.
$$
Then, the Newtonian equation of motion turns out as follows. We have
$$
M_G g + F_{em/self} = 2ma;
$$
using the above expression for $M_G$ and $frac{e^2}{c^2d}a$ for $F_{em/self}$, we obtain
$$
left(2m - frac{e^2}{c^2d}right)g = 2ma - frac{e^2}{c^2d}a
$$
which always implies
$$
a = g,
$$
confirming that dipole will move in accordance with the equivalence principle.
$endgroup$
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
$begingroup$
The paper that claims this result was written in reply to An electric dipole in self-accelerated transverse motion, which claims that a dipole in zero gravitational field really can accelerate itself, indefinitely. So if you believe both of the results of these papers, the equivalence principle is satisfied; dipoles can have an extra, weird contribution to their acceleration in both a gravitational field and in a freely falling frame.
However, the analysis of point charges in electromagnetism, and especially their self-interaction, is full of subtleties. More recently, the paper Nonexistence of the self-accelerating dipole and related questions has claimed that both of these results are incorrect, but since the author of this paper is active on Physics.SE I'm going to refrain from attempting to summarize the paper, because I'll probably get it wrong!
From a more general perspective, we already know postulating perfectly pointlike charges leads to a ton of subtleties, even before bringing relativity into the mix. In the modern formulation of quantum field theory in curved spacetime, everything is manifestly covariant from the start, so the equivalence principle is satisfied by construction. Of course it's interesting to see how it can come about in a less fundamental theory like classical electromagnetism, but issues with that aren't going to bring all of relativity crashing down.
$endgroup$
1
$begingroup$
In addition to the points you raise, there are likely to be a couple of vexing issues: (1) that a boosted dipole isn't a dipole (related to Mansuripur's paradox); (2) there is no universally accepted way to define what is a radiation field in this kind of context.
$endgroup$
– Ben Crowell
3 hours ago
$begingroup$
The weird self-acceleration is merely a special case for extremely small separations of the two charges. John Eastmond's observation and question is valid even for dipoles that do not have such strange solutions.
$endgroup$
– Ján Lalinský
26 mins ago
add a comment |
$begingroup$
The paper that claims this result was written in reply to An electric dipole in self-accelerated transverse motion, which claims that a dipole in zero gravitational field really can accelerate itself, indefinitely. So if you believe both of the results of these papers, the equivalence principle is satisfied; dipoles can have an extra, weird contribution to their acceleration in both a gravitational field and in a freely falling frame.
However, the analysis of point charges in electromagnetism, and especially their self-interaction, is full of subtleties. More recently, the paper Nonexistence of the self-accelerating dipole and related questions has claimed that both of these results are incorrect, but since the author of this paper is active on Physics.SE I'm going to refrain from attempting to summarize the paper, because I'll probably get it wrong!
From a more general perspective, we already know postulating perfectly pointlike charges leads to a ton of subtleties, even before bringing relativity into the mix. In the modern formulation of quantum field theory in curved spacetime, everything is manifestly covariant from the start, so the equivalence principle is satisfied by construction. Of course it's interesting to see how it can come about in a less fundamental theory like classical electromagnetism, but issues with that aren't going to bring all of relativity crashing down.
$endgroup$
1
$begingroup$
In addition to the points you raise, there are likely to be a couple of vexing issues: (1) that a boosted dipole isn't a dipole (related to Mansuripur's paradox); (2) there is no universally accepted way to define what is a radiation field in this kind of context.
$endgroup$
– Ben Crowell
3 hours ago
$begingroup$
The weird self-acceleration is merely a special case for extremely small separations of the two charges. John Eastmond's observation and question is valid even for dipoles that do not have such strange solutions.
$endgroup$
– Ján Lalinský
26 mins ago
add a comment |
$begingroup$
The paper that claims this result was written in reply to An electric dipole in self-accelerated transverse motion, which claims that a dipole in zero gravitational field really can accelerate itself, indefinitely. So if you believe both of the results of these papers, the equivalence principle is satisfied; dipoles can have an extra, weird contribution to their acceleration in both a gravitational field and in a freely falling frame.
However, the analysis of point charges in electromagnetism, and especially their self-interaction, is full of subtleties. More recently, the paper Nonexistence of the self-accelerating dipole and related questions has claimed that both of these results are incorrect, but since the author of this paper is active on Physics.SE I'm going to refrain from attempting to summarize the paper, because I'll probably get it wrong!
From a more general perspective, we already know postulating perfectly pointlike charges leads to a ton of subtleties, even before bringing relativity into the mix. In the modern formulation of quantum field theory in curved spacetime, everything is manifestly covariant from the start, so the equivalence principle is satisfied by construction. Of course it's interesting to see how it can come about in a less fundamental theory like classical electromagnetism, but issues with that aren't going to bring all of relativity crashing down.
$endgroup$
The paper that claims this result was written in reply to An electric dipole in self-accelerated transverse motion, which claims that a dipole in zero gravitational field really can accelerate itself, indefinitely. So if you believe both of the results of these papers, the equivalence principle is satisfied; dipoles can have an extra, weird contribution to their acceleration in both a gravitational field and in a freely falling frame.
However, the analysis of point charges in electromagnetism, and especially their self-interaction, is full of subtleties. More recently, the paper Nonexistence of the self-accelerating dipole and related questions has claimed that both of these results are incorrect, but since the author of this paper is active on Physics.SE I'm going to refrain from attempting to summarize the paper, because I'll probably get it wrong!
From a more general perspective, we already know postulating perfectly pointlike charges leads to a ton of subtleties, even before bringing relativity into the mix. In the modern formulation of quantum field theory in curved spacetime, everything is manifestly covariant from the start, so the equivalence principle is satisfied by construction. Of course it's interesting to see how it can come about in a less fundamental theory like classical electromagnetism, but issues with that aren't going to bring all of relativity crashing down.
answered 3 hours ago
knzhouknzhou
43.2k11118206
43.2k11118206
1
$begingroup$
In addition to the points you raise, there are likely to be a couple of vexing issues: (1) that a boosted dipole isn't a dipole (related to Mansuripur's paradox); (2) there is no universally accepted way to define what is a radiation field in this kind of context.
$endgroup$
– Ben Crowell
3 hours ago
$begingroup$
The weird self-acceleration is merely a special case for extremely small separations of the two charges. John Eastmond's observation and question is valid even for dipoles that do not have such strange solutions.
$endgroup$
– Ján Lalinský
26 mins ago
add a comment |
1
$begingroup$
In addition to the points you raise, there are likely to be a couple of vexing issues: (1) that a boosted dipole isn't a dipole (related to Mansuripur's paradox); (2) there is no universally accepted way to define what is a radiation field in this kind of context.
$endgroup$
– Ben Crowell
3 hours ago
$begingroup$
The weird self-acceleration is merely a special case for extremely small separations of the two charges. John Eastmond's observation and question is valid even for dipoles that do not have such strange solutions.
$endgroup$
– Ján Lalinský
26 mins ago
1
1
$begingroup$
In addition to the points you raise, there are likely to be a couple of vexing issues: (1) that a boosted dipole isn't a dipole (related to Mansuripur's paradox); (2) there is no universally accepted way to define what is a radiation field in this kind of context.
$endgroup$
– Ben Crowell
3 hours ago
$begingroup$
In addition to the points you raise, there are likely to be a couple of vexing issues: (1) that a boosted dipole isn't a dipole (related to Mansuripur's paradox); (2) there is no universally accepted way to define what is a radiation field in this kind of context.
$endgroup$
– Ben Crowell
3 hours ago
$begingroup$
The weird self-acceleration is merely a special case for extremely small separations of the two charges. John Eastmond's observation and question is valid even for dipoles that do not have such strange solutions.
$endgroup$
– Ján Lalinský
26 mins ago
$begingroup$
The weird self-acceleration is merely a special case for extremely small separations of the two charges. John Eastmond's observation and question is valid even for dipoles that do not have such strange solutions.
$endgroup$
– Ján Lalinský
26 mins ago
add a comment |
$begingroup$
In your calculation you assume that gravitational mass $M_G$ of the system is $2m$ where $m$ is rest mass of a single particle, thus you assume it is independent of the mutual distance between the charged particles $d$. In other words, you do not take into account the force of gravity acting on the system due to the concentrated bound negative energy of EM field near the charged particles. However, since the system has lower inertial mass, it should also have lower gravitational mass.
It is well known that systems with negative potential EM energy have inertial mass defect. In this case, the dipole is such a system, so it will have lower inertial mass than $2m$, thanks to its negative electrostatic potential energy $-frac{e^2}{d}$.
This "mass defect" effect comes from the forces of "acceleration electric fields" acting to speed up the charged particles. This you have taken into account by including force $F_{em/self} = frac{e^2}{c^2d}a$, which is the electromagnetic self-force acting on the dipole.
But defect in inertial mass should mean also defect in gravitational mass. Heuristically/naively, the gravitational mass to use in the formula $F_G = M_G g$ should correspond to total energy of the system via Einstein's formula
$$
E = M_Gc^2
$$
where $E$ is total energy of the system, including its internal potential energy. Using the Coulomb formula for potential energy,
$$
E = 2mc^2 - frac{e^2}{d}
$$
and so the gravitational mass of the dipole should be taken as
$$
M_{G} = 2m -frac{e^2}{c^2d}.
$$
Then, the Newtonian equation of motion turns out as follows. We have
$$
M_G g + F_{em/self} = 2ma;
$$
using the above expression for $M_G$ and $frac{e^2}{c^2d}a$ for $F_{em/self}$, we obtain
$$
left(2m - frac{e^2}{c^2d}right)g = 2ma - frac{e^2}{c^2d}a
$$
which always implies
$$
a = g,
$$
confirming that dipole will move in accordance with the equivalence principle.
$endgroup$
add a comment |
$begingroup$
In your calculation you assume that gravitational mass $M_G$ of the system is $2m$ where $m$ is rest mass of a single particle, thus you assume it is independent of the mutual distance between the charged particles $d$. In other words, you do not take into account the force of gravity acting on the system due to the concentrated bound negative energy of EM field near the charged particles. However, since the system has lower inertial mass, it should also have lower gravitational mass.
It is well known that systems with negative potential EM energy have inertial mass defect. In this case, the dipole is such a system, so it will have lower inertial mass than $2m$, thanks to its negative electrostatic potential energy $-frac{e^2}{d}$.
This "mass defect" effect comes from the forces of "acceleration electric fields" acting to speed up the charged particles. This you have taken into account by including force $F_{em/self} = frac{e^2}{c^2d}a$, which is the electromagnetic self-force acting on the dipole.
But defect in inertial mass should mean also defect in gravitational mass. Heuristically/naively, the gravitational mass to use in the formula $F_G = M_G g$ should correspond to total energy of the system via Einstein's formula
$$
E = M_Gc^2
$$
where $E$ is total energy of the system, including its internal potential energy. Using the Coulomb formula for potential energy,
$$
E = 2mc^2 - frac{e^2}{d}
$$
and so the gravitational mass of the dipole should be taken as
$$
M_{G} = 2m -frac{e^2}{c^2d}.
$$
Then, the Newtonian equation of motion turns out as follows. We have
$$
M_G g + F_{em/self} = 2ma;
$$
using the above expression for $M_G$ and $frac{e^2}{c^2d}a$ for $F_{em/self}$, we obtain
$$
left(2m - frac{e^2}{c^2d}right)g = 2ma - frac{e^2}{c^2d}a
$$
which always implies
$$
a = g,
$$
confirming that dipole will move in accordance with the equivalence principle.
$endgroup$
add a comment |
$begingroup$
In your calculation you assume that gravitational mass $M_G$ of the system is $2m$ where $m$ is rest mass of a single particle, thus you assume it is independent of the mutual distance between the charged particles $d$. In other words, you do not take into account the force of gravity acting on the system due to the concentrated bound negative energy of EM field near the charged particles. However, since the system has lower inertial mass, it should also have lower gravitational mass.
It is well known that systems with negative potential EM energy have inertial mass defect. In this case, the dipole is such a system, so it will have lower inertial mass than $2m$, thanks to its negative electrostatic potential energy $-frac{e^2}{d}$.
This "mass defect" effect comes from the forces of "acceleration electric fields" acting to speed up the charged particles. This you have taken into account by including force $F_{em/self} = frac{e^2}{c^2d}a$, which is the electromagnetic self-force acting on the dipole.
But defect in inertial mass should mean also defect in gravitational mass. Heuristically/naively, the gravitational mass to use in the formula $F_G = M_G g$ should correspond to total energy of the system via Einstein's formula
$$
E = M_Gc^2
$$
where $E$ is total energy of the system, including its internal potential energy. Using the Coulomb formula for potential energy,
$$
E = 2mc^2 - frac{e^2}{d}
$$
and so the gravitational mass of the dipole should be taken as
$$
M_{G} = 2m -frac{e^2}{c^2d}.
$$
Then, the Newtonian equation of motion turns out as follows. We have
$$
M_G g + F_{em/self} = 2ma;
$$
using the above expression for $M_G$ and $frac{e^2}{c^2d}a$ for $F_{em/self}$, we obtain
$$
left(2m - frac{e^2}{c^2d}right)g = 2ma - frac{e^2}{c^2d}a
$$
which always implies
$$
a = g,
$$
confirming that dipole will move in accordance with the equivalence principle.
$endgroup$
In your calculation you assume that gravitational mass $M_G$ of the system is $2m$ where $m$ is rest mass of a single particle, thus you assume it is independent of the mutual distance between the charged particles $d$. In other words, you do not take into account the force of gravity acting on the system due to the concentrated bound negative energy of EM field near the charged particles. However, since the system has lower inertial mass, it should also have lower gravitational mass.
It is well known that systems with negative potential EM energy have inertial mass defect. In this case, the dipole is such a system, so it will have lower inertial mass than $2m$, thanks to its negative electrostatic potential energy $-frac{e^2}{d}$.
This "mass defect" effect comes from the forces of "acceleration electric fields" acting to speed up the charged particles. This you have taken into account by including force $F_{em/self} = frac{e^2}{c^2d}a$, which is the electromagnetic self-force acting on the dipole.
But defect in inertial mass should mean also defect in gravitational mass. Heuristically/naively, the gravitational mass to use in the formula $F_G = M_G g$ should correspond to total energy of the system via Einstein's formula
$$
E = M_Gc^2
$$
where $E$ is total energy of the system, including its internal potential energy. Using the Coulomb formula for potential energy,
$$
E = 2mc^2 - frac{e^2}{d}
$$
and so the gravitational mass of the dipole should be taken as
$$
M_{G} = 2m -frac{e^2}{c^2d}.
$$
Then, the Newtonian equation of motion turns out as follows. We have
$$
M_G g + F_{em/self} = 2ma;
$$
using the above expression for $M_G$ and $frac{e^2}{c^2d}a$ for $F_{em/self}$, we obtain
$$
left(2m - frac{e^2}{c^2d}right)g = 2ma - frac{e^2}{c^2d}a
$$
which always implies
$$
a = g,
$$
confirming that dipole will move in accordance with the equivalence principle.
edited 33 mins ago
answered 49 mins ago
Ján LalinskýJán Lalinský
14.5k1334
14.5k1334
add a comment |
add a comment |
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$begingroup$
Related paper: arxiv.org/pdf/1311.5798.pdf
$endgroup$
– kkm
3 hours ago
$begingroup$
Possible duplicate of Does a charged particle accelerating in a gravitational field radiate?
$endgroup$
– Aaron Stevens
1 hour ago