If Ampere's law implies the Biot-Savart law, which implies Gauss's law for magnetism, does that mean...
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 at 10:53
Mattia F.
1605
New contributor
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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 at 10:53
Mattia F.
1605
New contributor
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Please show the details of your proof.
– my2cts
Dec 21 at 10:59
I've added the proof / references.
– Mattia F.
Dec 21 at 11:11
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
– Eric Duminil
Dec 21 at 22:29
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).
– Mattia F.
Dec 23 at 17:16
add a comment |
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 at 10:53
Mattia F.
1605
New contributor
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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 at 10:53
Mattia F.
1605
New contributor
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 21 at 10:53
Mattia F.
1605
New contributor
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Dec 21 at 16:06
asked Dec 21 at 10:53
Mattia F.
1605
New contributor
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Dec 21 at 10:53
Mattia F.
1605
asked Dec 21 at 10:53
Mattia F.
1605
1605
New contributor
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mattia F. is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
Please show the details of your proof.
– my2cts
Dec 21 at 10:59
I've added the proof / references.
– Mattia F.
Dec 21 at 11:11
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
– Eric Duminil
Dec 21 at 22:29
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).
– Mattia F.
Dec 23 at 17:16
add a comment |
1
Please show the details of your proof.
– my2cts
Dec 21 at 10:59
I've added the proof / references.
– Mattia F.
Dec 21 at 11:11
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
– Eric Duminil
Dec 21 at 22:29
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).
– Mattia F.
Dec 23 at 17:16
1
1
Please show the details of your proof.
– my2cts
Dec 21 at 10:59
Please show the details of your proof.
– my2cts
Dec 21 at 10:59
I've added the proof / references.
– Mattia F.
Dec 21 at 11:11
I've added the proof / references.
– Mattia F.
Dec 21 at 11:11
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
– Eric Duminil
Dec 21 at 22:29
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
– Eric Duminil
Dec 21 at 22:29
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).
– Mattia F.
Dec 23 at 17:16
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).
– Mattia F.
Dec 23 at 17:16
add a comment |
2 Answers
2
active
oldest
votes
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 at 12:31
Thorondor
1,07720
add a comment |
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 at 11:10
GiorgioP
1,795215
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.
– Michael Seifert
Dec 21 at 18:39
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "151"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Mattia F. is a new contributor. Be nice, and check out our Code of Conduct.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f449658%2fif-amperes-law-implies-the-biot-savart-law-which-implies-gausss-law-for-magne%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 at 12:31
Thorondor
1,07720
add a comment |
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 at 12:31
Thorondor
1,07720
add a comment |
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 at 12:31
Thorondor
1,07720
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 at 12:31
Thorondor
1,07720
edited Dec 21 at 12:36
answered Dec 21 at 12:31
Thorondor
1,07720
answered Dec 21 at 12:31
Thorondor
1,07720
answered Dec 21 at 12:31
Thorondor
1,07720
1,07720
add a comment |
add a comment |
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 at 11:10
GiorgioP
1,795215
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.
– Michael Seifert
Dec 21 at 18:39
add a comment |
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 at 11:10
GiorgioP
1,795215
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.
– Michael Seifert
Dec 21 at 18:39
add a comment |
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 at 11:10
GiorgioP
1,795215
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 at 11:10
GiorgioP
1,795215
answered Dec 21 at 11:10
GiorgioP
1,795215
answered Dec 21 at 11:10
GiorgioP
1,795215
answered Dec 21 at 11:10
GiorgioP
1,795215
1,795215
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.
– Michael Seifert
Dec 21 at 18:39
add a comment |
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.
– Michael Seifert
Dec 21 at 18:39
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.
– Michael Seifert
Dec 21 at 18:39
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.
– Michael Seifert
Dec 21 at 18:39
add a comment |
Mattia F. is a new contributor. Be nice, and check out our Code of Conduct.
Mattia F. is a new contributor. Be nice, and check out our Code of Conduct.
Mattia F. is a new contributor. Be nice, and check out our Code of Conduct.
Mattia F. is a new contributor. Be nice, and check out our Code of Conduct.
Thanks for contributing an answer to Physics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fphysics.stackexchange.com%2fquestions%2f449658%2fif-amperes-law-implies-the-biot-savart-law-which-implies-gausss-law-for-magne%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
Please show the details of your proof.
– my2cts
Dec 21 at 10:59
I've added the proof / references.
– Mattia F.
Dec 21 at 11:11
With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
– Eric Duminil
Dec 21 at 22:29
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).
– Mattia F.
Dec 23 at 17:16