If Ampere's law implies the Biot-Savart law, which implies Gauss's law for magnetism, does that mean...
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Studying electromagnetism, I came across the following fact:
- Maxwell's third equation (divergence of magnetic field is zero) can be derived from the Biot-Savart Law.
- The Biot-Savart Law can be derived from Maxwell-Ampère's Law
Hence, it seems that the four equations are redundant.
Unfortunately I've not found anything about this, so I ask: it is true?
EDIT:
For the sake of completeness, this is the proof of the 3rd equation (this is basically Griffith's proof):
begin{eqnarray}
vec{B}(r) &=& iiint_K frac{mu_0 i}{4 pi} frac{vec{j} times vec{r}}{r^3} dtau'\
&=& iiint_K frac{mu_0 i}{4 pi} vec{j} times nablaleft(-frac{1}{r}right) dtau'
end{eqnarray}
Applying divergence to both terms we obtain:
begin{eqnarray}
text{div} vec{B} &=& frac{mu_0 i}{4 pi} iiint_K text{div} left(vec{j} times nablaleft(-frac{1}{r}right)right) dtau'\
&=& frac{mu_0 i}{4 pi} iiint_K nabla times vec{j} cdot nablaleft(-frac{1}{r}right) - vec{j} cdot nabla times nablaleft(-frac{1}{r}right) dtau' \
&=& 0
end{eqnarray}
The last term is zero since the curl of a gradient is always zero and the divergence of $vec{j}$ is zero ($vec{j}$ depends on primed coordinates only).
And this is the derivation of the Biot-Savart Law from the Maxwell-Ampère Law: Is Biot-Savart law obtained empirically or can it be derived?
electromagnetism gauss-law maxwell-equations
asked Dec 21 '18 at 10:53
Mattia F.Mattia F.
1605
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add a comment |
$begingroup$
Studying electromagnetism, I came across the following fact:
- Maxwell's third equation (divergence of magnetic field is zero) can be derived from the Biot-Savart Law.
- The Biot-Savart Law can be derived from Maxwell-Ampère's Law
Hence, it seems that the four equations are redundant.
Unfortunately I've not found anything about this, so I ask: it is true?
EDIT:
For the sake of completeness, this is the proof of the 3rd equation (this is basically Griffith's proof):
begin{eqnarray}
vec{B}(r) &=& iiint_K frac{mu_0 i}{4 pi} frac{vec{j} times vec{r}}{r^3} dtau'\
&=& iiint_K frac{mu_0 i}{4 pi} vec{j} times nablaleft(-frac{1}{r}right) dtau'
end{eqnarray}
Applying divergence to both terms we obtain:
begin{eqnarray}
text{div} vec{B} &=& frac{mu_0 i}{4 pi} iiint_K text{div} left(vec{j} times nablaleft(-frac{1}{r}right)right) dtau'\
&=& frac{mu_0 i}{4 pi} iiint_K nabla times vec{j} cdot nablaleft(-frac{1}{r}right) - vec{j} cdot nabla times nablaleft(-frac{1}{r}right) dtau' \
&=& 0
end{eqnarray}
The last term is zero since the curl of a gradient is always zero and the divergence of $vec{j}$ is zero ($vec{j}$ depends on primed coordinates only).
And this is the derivation of the Biot-Savart Law from the Maxwell-Ampère Law: Is Biot-Savart law obtained empirically or can it be derived?
electromagnetism gauss-law maxwell-equations
asked Dec 21 '18 at 10:53
Mattia F.Mattia F.
1605
$endgroup$
1
$begingroup$
Please show the details of your proof.
$endgroup$
– my2cts
Dec 21 '18 at 10:59
$begingroup$
I've added the proof / references.
$endgroup$
– Mattia F.
Dec 21 '18 at 11:11
$begingroup$
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
$endgroup$
– Eric Duminil
Dec 21 '18 at 22:29
$begingroup$
That's not really what I meant. Maxwell's equations in a certain sense are the axioms of electromagnetism theory, so my question was whether one of the axioms could be derived from the other axioms (so it would actually be a theorem).
$endgroup$
– Mattia F.
Dec 23 '18 at 17:16
add a comment |
$begingroup$
Studying electromagnetism, I came across the following fact:
- Maxwell's third equation (divergence of magnetic field is zero) can be derived from the Biot-Savart Law.
- The Biot-Savart Law can be derived from Maxwell-Ampère's Law
Hence, it seems that the four equations are redundant.
Unfortunately I've not found anything about this, so I ask: it is true?
EDIT:
For the sake of completeness, this is the proof of the 3rd equation (this is basically Griffith's proof):
begin{eqnarray}
vec{B}(r) &=& iiint_K frac{mu_0 i}{4 pi} frac{vec{j} times vec{r}}{r^3} dtau'\
&=& iiint_K frac{mu_0 i}{4 pi} vec{j} times nablaleft(-frac{1}{r}right) dtau'
end{eqnarray}
Applying divergence to both terms we obtain:
begin{eqnarray}
text{div} vec{B} &=& frac{mu_0 i}{4 pi} iiint_K text{div} left(vec{j} times nablaleft(-frac{1}{r}right)right) dtau'\
&=& frac{mu_0 i}{4 pi} iiint_K nabla times vec{j} cdot nablaleft(-frac{1}{r}right) - vec{j} cdot nabla times nablaleft(-frac{1}{r}right) dtau' \
&=& 0
end{eqnarray}
The last term is zero since the curl of a gradient is always zero and the divergence of $vec{j}$ is zero ($vec{j}$ depends on primed coordinates only).
And this is the derivation of the Biot-Savart Law from the Maxwell-Ampère Law: Is Biot-Savart law obtained empirically or can it be derived?
electromagnetism gauss-law maxwell-equations
asked Dec 21 '18 at 10:53
Mattia F.Mattia F.
1605
$endgroup$
Studying electromagnetism, I came across the following fact:
- Maxwell's third equation (divergence of magnetic field is zero) can be derived from the Biot-Savart Law.
- The Biot-Savart Law can be derived from Maxwell-Ampère's Law
Hence, it seems that the four equations are redundant.
Unfortunately I've not found anything about this, so I ask: it is true?
EDIT:
For the sake of completeness, this is the proof of the 3rd equation (this is basically Griffith's proof):
begin{eqnarray}
vec{B}(r) &=& iiint_K frac{mu_0 i}{4 pi} frac{vec{j} times vec{r}}{r^3} dtau'\
&=& iiint_K frac{mu_0 i}{4 pi} vec{j} times nablaleft(-frac{1}{r}right) dtau'
end{eqnarray}
Applying divergence to both terms we obtain:
begin{eqnarray}
text{div} vec{B} &=& frac{mu_0 i}{4 pi} iiint_K text{div} left(vec{j} times nablaleft(-frac{1}{r}right)right) dtau'\
&=& frac{mu_0 i}{4 pi} iiint_K nabla times vec{j} cdot nablaleft(-frac{1}{r}right) - vec{j} cdot nabla times nablaleft(-frac{1}{r}right) dtau' \
&=& 0
end{eqnarray}
The last term is zero since the curl of a gradient is always zero and the divergence of $vec{j}$ is zero ($vec{j}$ depends on primed coordinates only).
And this is the derivation of the Biot-Savart Law from the Maxwell-Ampère Law: Is Biot-Savart law obtained empirically or can it be derived?
electromagnetism gauss-law maxwell-equations
electromagnetism gauss-law maxwell-equations
asked Dec 21 '18 at 10:53
Mattia F.Mattia F.
1605
asked Dec 21 '18 at 10:53
Mattia F.Mattia F.
1605
edited Dec 21 '18 at 16:06
Mattia F.
asked Dec 21 '18 at 10:53
Mattia F.Mattia F.
1605
asked Dec 21 '18 at 10:53
Mattia F.Mattia F.
1605
asked Dec 21 '18 at 10:53
Mattia F.Mattia F.
1605
1605
1
$begingroup$
Please show the details of your proof.
$endgroup$
– my2cts
Dec 21 '18 at 10:59
$begingroup$
I've added the proof / references.
$endgroup$
– Mattia F.
Dec 21 '18 at 11:11
$begingroup$
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
$endgroup$
– Eric Duminil
Dec 21 '18 at 22:29
$begingroup$
That's not really what I meant. Maxwell's equations in a certain sense are the axioms of electromagnetism theory, so my question was whether one of the axioms could be derived from the other axioms (so it would actually be a theorem).
$endgroup$
– Mattia F.
Dec 23 '18 at 17:16
add a comment |
1
$begingroup$
Please show the details of your proof.
$endgroup$
– my2cts
Dec 21 '18 at 10:59
$begingroup$
I've added the proof / references.
$endgroup$
– Mattia F.
Dec 21 '18 at 11:11
$begingroup$
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
$endgroup$
– Eric Duminil
Dec 21 '18 at 22:29
$begingroup$
That's not really what I meant. Maxwell's equations in a certain sense are the axioms of electromagnetism theory, so my question was whether one of the axioms could be derived from the other axioms (so it would actually be a theorem).
$endgroup$
– Mattia F.
Dec 23 '18 at 17:16
1
1
$begingroup$
Please show the details of your proof.
$endgroup$
– my2cts
Dec 21 '18 at 10:59
$begingroup$
Please show the details of your proof.
$endgroup$
– my2cts
Dec 21 '18 at 10:59
$begingroup$
I've added the proof / references.
$endgroup$
– Mattia F.
Dec 21 '18 at 11:11
$begingroup$
I've added the proof / references.
$endgroup$
– Mattia F.
Dec 21 '18 at 11:11
$begingroup$
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
$endgroup$
– Eric Duminil
Dec 21 '18 at 22:29
$begingroup$
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
$endgroup$
– Eric Duminil
Dec 21 '18 at 22:29
$begingroup$
That's not really what I meant. Maxwell's equations in a certain sense are the axioms of electromagnetism theory, so my question was whether one of the axioms could be derived from the other axioms (so it would actually be a theorem).
$endgroup$
– Mattia F.
Dec 23 '18 at 17:16
$begingroup$
That's not really what I meant. Maxwell's equations in a certain sense are the axioms of electromagnetism theory, so my question was whether one of the axioms could be derived from the other axioms (so it would actually be a theorem).
$endgroup$
– Mattia F.
Dec 23 '18 at 17:16
add a comment |
2 Answers
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Since there is no magnetic charge term in the Biot-Savart law, it is only correct if Gauss's law for magnetism ($nabla cdot mathbf{B} = 0$) is true and there are no magnetic monopoles. So it makes sense that Gauss's law can be derived from the Biot-Savart law.
However, the Biot-Savart law cannot be derived from the Maxwell-Ampère Law without implicitly assuming Gauss's law. In general, we know this both because of the lack of a magnetic charge term and because as Giorgio pointed out, the curl and divergence of a vector field are independent quantities.
The specific problem with the proof you cited is that it assumes that a continuous vector potential $mathbf{A}$ can be constructed such that $nabla times mathbf{A} = mathbf{B}$, which is not true if there are magnetic monopoles.
answered Dec 21 '18 at 12:31
ThorondorThorondor
1,089120
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add a comment |
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Curl and divergence of a vector field are independent quantities ( Helmholtz's theorem allows to reconstruct a vector field if both are known). So, it is impossible to deduce anything for each of these two quantities from the other.
answered Dec 21 '18 at 11:10
GiorgioPGiorgioP
2,413219
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$begingroup$
It's worth noting that the Biot-Savart Law is equivalent to Helmholtz's theorem for a divergence-free vector field. Similarly, Coulomb's Law is equivalent to Helmholtz's theorem for a curl-free vector field.
$endgroup$
– Michael Seifert
Dec 21 '18 at 18:39
add a comment |
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2 Answers
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$begingroup$
Since there is no magnetic charge term in the Biot-Savart law, it is only correct if Gauss's law for magnetism ($nabla cdot mathbf{B} = 0$) is true and there are no magnetic monopoles. So it makes sense that Gauss's law can be derived from the Biot-Savart law.
However, the Biot-Savart law cannot be derived from the Maxwell-Ampère Law without implicitly assuming Gauss's law. In general, we know this both because of the lack of a magnetic charge term and because as Giorgio pointed out, the curl and divergence of a vector field are independent quantities.
The specific problem with the proof you cited is that it assumes that a continuous vector potential $mathbf{A}$ can be constructed such that $nabla times mathbf{A} = mathbf{B}$, which is not true if there are magnetic monopoles.
answered Dec 21 '18 at 12:31
ThorondorThorondor
1,089120
$endgroup$
add a comment |
$begingroup$
Since there is no magnetic charge term in the Biot-Savart law, it is only correct if Gauss's law for magnetism ($nabla cdot mathbf{B} = 0$) is true and there are no magnetic monopoles. So it makes sense that Gauss's law can be derived from the Biot-Savart law.
However, the Biot-Savart law cannot be derived from the Maxwell-Ampère Law without implicitly assuming Gauss's law. In general, we know this both because of the lack of a magnetic charge term and because as Giorgio pointed out, the curl and divergence of a vector field are independent quantities.
The specific problem with the proof you cited is that it assumes that a continuous vector potential $mathbf{A}$ can be constructed such that $nabla times mathbf{A} = mathbf{B}$, which is not true if there are magnetic monopoles.
answered Dec 21 '18 at 12:31
ThorondorThorondor
1,089120
$endgroup$
add a comment |
$begingroup$
Since there is no magnetic charge term in the Biot-Savart law, it is only correct if Gauss's law for magnetism ($nabla cdot mathbf{B} = 0$) is true and there are no magnetic monopoles. So it makes sense that Gauss's law can be derived from the Biot-Savart law.
However, the Biot-Savart law cannot be derived from the Maxwell-Ampère Law without implicitly assuming Gauss's law. In general, we know this both because of the lack of a magnetic charge term and because as Giorgio pointed out, the curl and divergence of a vector field are independent quantities.
The specific problem with the proof you cited is that it assumes that a continuous vector potential $mathbf{A}$ can be constructed such that $nabla times mathbf{A} = mathbf{B}$, which is not true if there are magnetic monopoles.
answered Dec 21 '18 at 12:31
ThorondorThorondor
1,089120
$endgroup$
Since there is no magnetic charge term in the Biot-Savart law, it is only correct if Gauss's law for magnetism ($nabla cdot mathbf{B} = 0$) is true and there are no magnetic monopoles. So it makes sense that Gauss's law can be derived from the Biot-Savart law.
However, the Biot-Savart law cannot be derived from the Maxwell-Ampère Law without implicitly assuming Gauss's law. In general, we know this both because of the lack of a magnetic charge term and because as Giorgio pointed out, the curl and divergence of a vector field are independent quantities.
The specific problem with the proof you cited is that it assumes that a continuous vector potential $mathbf{A}$ can be constructed such that $nabla times mathbf{A} = mathbf{B}$, which is not true if there are magnetic monopoles.
answered Dec 21 '18 at 12:31
ThorondorThorondor
1,089120
edited Dec 21 '18 at 12:36
answered Dec 21 '18 at 12:31
ThorondorThorondor
1,089120
answered Dec 21 '18 at 12:31
ThorondorThorondor
1,089120
answered Dec 21 '18 at 12:31
ThorondorThorondor
1,089120
1,089120
add a comment |
add a comment |
$begingroup$
Curl and divergence of a vector field are independent quantities ( Helmholtz's theorem allows to reconstruct a vector field if both are known). So, it is impossible to deduce anything for each of these two quantities from the other.
answered Dec 21 '18 at 11:10
GiorgioPGiorgioP
2,413219
$endgroup$
$begingroup$
It's worth noting that the Biot-Savart Law is equivalent to Helmholtz's theorem for a divergence-free vector field. Similarly, Coulomb's Law is equivalent to Helmholtz's theorem for a curl-free vector field.
$endgroup$
– Michael Seifert
Dec 21 '18 at 18:39
add a comment |
$begingroup$
Curl and divergence of a vector field are independent quantities ( Helmholtz's theorem allows to reconstruct a vector field if both are known). So, it is impossible to deduce anything for each of these two quantities from the other.
answered Dec 21 '18 at 11:10
GiorgioPGiorgioP
2,413219
$endgroup$
$begingroup$
It's worth noting that the Biot-Savart Law is equivalent to Helmholtz's theorem for a divergence-free vector field. Similarly, Coulomb's Law is equivalent to Helmholtz's theorem for a curl-free vector field.
$endgroup$
– Michael Seifert
Dec 21 '18 at 18:39
add a comment |
$begingroup$
Curl and divergence of a vector field are independent quantities ( Helmholtz's theorem allows to reconstruct a vector field if both are known). So, it is impossible to deduce anything for each of these two quantities from the other.
answered Dec 21 '18 at 11:10
GiorgioPGiorgioP
2,413219
$endgroup$
Curl and divergence of a vector field are independent quantities ( Helmholtz's theorem allows to reconstruct a vector field if both are known). So, it is impossible to deduce anything for each of these two quantities from the other.
answered Dec 21 '18 at 11:10
GiorgioPGiorgioP
2,413219
answered Dec 21 '18 at 11:10
GiorgioPGiorgioP
2,413219
answered Dec 21 '18 at 11:10
GiorgioPGiorgioP
2,413219
answered Dec 21 '18 at 11:10
GiorgioPGiorgioP
2,413219
2,413219
$begingroup$
It's worth noting that the Biot-Savart Law is equivalent to Helmholtz's theorem for a divergence-free vector field. Similarly, Coulomb's Law is equivalent to Helmholtz's theorem for a curl-free vector field.
$endgroup$
– Michael Seifert
Dec 21 '18 at 18:39
add a comment |
$begingroup$
It's worth noting that the Biot-Savart Law is equivalent to Helmholtz's theorem for a divergence-free vector field. Similarly, Coulomb's Law is equivalent to Helmholtz's theorem for a curl-free vector field.
$endgroup$
– Michael Seifert
Dec 21 '18 at 18:39
$begingroup$
It's worth noting that the Biot-Savart Law is equivalent to Helmholtz's theorem for a divergence-free vector field. Similarly, Coulomb's Law is equivalent to Helmholtz's theorem for a curl-free vector field.
$endgroup$
– Michael Seifert
Dec 21 '18 at 18:39
$begingroup$
It's worth noting that the Biot-Savart Law is equivalent to Helmholtz's theorem for a divergence-free vector field. Similarly, Coulomb's Law is equivalent to Helmholtz's theorem for a curl-free vector field.
$endgroup$
– Michael Seifert
Dec 21 '18 at 18:39
add a comment |
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$begingroup$
Please show the details of your proof.
$endgroup$
– my2cts
Dec 21 '18 at 10:59
$begingroup$
I've added the proof / references.
$endgroup$
– Mattia F.
Dec 21 '18 at 11:11
$begingroup$
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
$endgroup$
– Eric Duminil
Dec 21 '18 at 22:29
$begingroup$
That's not really what I meant. Maxwell's equations in a certain sense are the axioms of electromagnetism theory, so my question was whether one of the axioms could be derived from the other axioms (so it would actually be a theorem).
$endgroup$
– Mattia F.
Dec 23 '18 at 17:16