If Ampere's law implies the Biot-Savart law, which implies Gauss's law for magnetism, does that mean...












8














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?










share|cite|improve this question




















  • 1




    Please show the details of your proof.
    – my2cts
    Dec 21 '18 at 10:59










  • I've added the proof / references.
    – Mattia F.
    Dec 21 '18 at 11:11










  • With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
    – Eric Duminil
    Dec 21 '18 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 '18 at 17:16
















8














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?










share|cite|improve this question




















  • 1




    Please show the details of your proof.
    – my2cts
    Dec 21 '18 at 10:59










  • I've added the proof / references.
    – Mattia F.
    Dec 21 '18 at 11:11










  • With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
    – Eric Duminil
    Dec 21 '18 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 '18 at 17:16














8












8








8


3





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?










share|cite|improve this question















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






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edited Dec 21 '18 at 16:06

























asked Dec 21 '18 at 10:53









Mattia F.

1605




1605








  • 1




    Please show the details of your proof.
    – my2cts
    Dec 21 '18 at 10:59










  • I've added the proof / references.
    – Mattia F.
    Dec 21 '18 at 11:11










  • With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
    – Eric Duminil
    Dec 21 '18 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 '18 at 17:16














  • 1




    Please show the details of your proof.
    – my2cts
    Dec 21 '18 at 10:59










  • I've added the proof / references.
    – Mattia F.
    Dec 21 '18 at 11:11










  • With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
    – Eric Duminil
    Dec 21 '18 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 '18 at 17:16








1




1




Please show the details of your proof.
– my2cts
Dec 21 '18 at 10:59




Please show the details of your proof.
– my2cts
Dec 21 '18 at 10:59












I've added the proof / references.
– Mattia F.
Dec 21 '18 at 11:11




I've added the proof / references.
– Mattia F.
Dec 21 '18 at 11:11












With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
– Eric Duminil
Dec 21 '18 at 22:29




With your definition of "redundant", any math theorem is redundant. All you'd need would be axioms.
– Eric Duminil
Dec 21 '18 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 '18 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 '18 at 17:16










2 Answers
2






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oldest

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12














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.






share|cite|improve this answer































    6














    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.






    share|cite|improve this answer





















    • 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 '18 at 18:39











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    2 Answers
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    2 Answers
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    active

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    12














    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.






    share|cite|improve this answer




























      12














      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.






      share|cite|improve this answer


























        12












        12








        12






        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.






        share|cite|improve this answer














        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 21 '18 at 12:36

























        answered Dec 21 '18 at 12:31









        Thorondor

        1,089120




        1,089120























            6














            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.






            share|cite|improve this answer





















            • 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 '18 at 18:39
















            6














            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.






            share|cite|improve this answer





















            • 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 '18 at 18:39














            6












            6








            6






            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.






            share|cite|improve this answer












            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.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Dec 21 '18 at 11:10









            GiorgioP

            2,033218




            2,033218












            • 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 '18 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 '18 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 '18 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 '18 at 18:39


















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