If Ampere's law implies the Biot-Savart law, which implies Gauss's law for magnetism, does that mean Maxwell's equations are redundant?

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} d\tau'\\ &=& \iiint_K \frac{\mu_0 i}{4 \pi} \ \vec{j} \times \nabla\left(-\frac{1}{r}\right) d\tau' \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 \nabla\left(-\frac{1}{r}\right)\right) d\tau'\\ &=& \frac{\mu_0 i}{4 \pi} \iiint_K \nabla \times \vec{j} \cdot \nabla\left(-\frac{1}{r}\right) - \vec{j} \cdot \nabla \times \nabla\left(-\frac{1}{r}\right) d\tau' \\ &=& 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?

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

Thorondor · Accepted Answer · 2018-12-21T12:36:18.470

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.

score 6 · Answer 2 · answered Dec 21 '18 at 11:10

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.

answered Dec 21 '18 at 11:10

GiorgioP-DoomsdayClockIsAt-90

34,706

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

If Ampere's law implies the Biot-Savart law, which implies Gauss's law for magnetism, does that mean Maxwell's equations are redundant?

2 Answers2