I want to derive Gauss's law for magnetism,$$ \nabla \cdot \vec{B} = 0 \,.$$

The derivation in Griffiths Introduction to elecrodynamics uses$$ \nabla \cdot\vec{B} ~=~\frac{\mu_0}{4\pi} \int {\nabla \cdot \left(\vec{J} \times \frac{\vec{r}}{r^3} \right)} \,,$$but to use this, I would need $\vec{J} \times \frac{\vec{r}}{r^3}$ and $\nabla \cdot \left(\vec{J} \times \frac{\vec{r}}{r^3} \right)$ to be continuous.

Since $\vec{J}= \rho \vec{v}$, then if $\rho$ (charge density) has a continuous derivative, I'm done; but, I'm not sure why can I affirm this because clearly, I can construct charge distributions with jump discontinuities.

  • 1
    $\begingroup$ Can you prove this? Isn't this just saying that there are no magnetic monopoles, which is just something that is true about the universe (that we know so far) and cannot be proven? $\endgroup$ Commented May 28, 2018 at 22:06
  • $\begingroup$ Well, in fact i'm assuming biot-savart's law ( which is an experimental result) so I think it's a derivation instead that a proof. I think you can asume it to be true but a priori is not a obvious result and if the theory we are developing is correct it should be consistent with nature. Anyway that was just the context of my question, the real question is about why we can assume charge density to be continuous and have derivates $\endgroup$
    – Johanna
    Commented May 28, 2018 at 22:13
  • $\begingroup$ $\rm{div}(\vec{B})=0$ is an experimenal fact for all kinds of magnetic field configurations and time-dependencies. It is a separate, independent equation. $\endgroup$ Commented May 29, 2018 at 5:51
  • $\begingroup$ Indeed it may not be true. So you can't "prove it". $\endgroup$
    – ProfRob
    Commented May 29, 2018 at 8:35

1 Answer 1


I don't believe continuity is required. So long as the only source is some current $\vec{J}$ and not some monopole, you should be able to prove $\nabla \cdot\vec{B}=0$.

You'll notice in Griffiths when he presents the expression:

$$\nabla \cdot\vec{B} ~=~\frac{\mu_0}{4\pi} \int {\nabla \cdot \left(\vec{J} \times \frac{\vec{r}}{r^3} \right)} \,,$$

It's allowable to move the gradient operator inside the integral regardless of the current density because the bounds are fixed space (i.e. constants that are not functions of $\vec{r}$). This means we fall into the special case of the Leibniz integral rule where the gradient operator and the integral signs are interchangeable.

His argument that the RHS is zero comes from the following vector identities: $$\nabla \cdot(\vec{J} \times \frac{\vec{r}}{r^3})= \frac{\vec{r}}{r^3} \cdot (\nabla \times \vec{J}) - \vec{J} \cdot (\nabla \times \frac{\vec{r}}{r^3})$$ Because $\vec{J}$ is dependent on the source space $(x',y',z')$ and not the measurement space where we are taking the curl $(x,y,z)$, we can say the curl of the current density is zero. Now we need only to evaluate the second term, which we can expand using another product rule. $$\nabla \times \frac{\vec{r}}{r^3}= \frac{1}{r^3}(\nabla \times \vec{r})- \vec{r} \times (\nabla \frac{1}{r^3})$$ Both of these terms go to zero because the curl of $\vec{r}$ is zero, and because $\vec{r} \times \vec{r}$ is zero (I'm not sure exactly why Griffiths makes this last step, I would have believed that $\nabla \times \frac{\vec{r}}{r^3}=0$ from the same argument).

Regardless, the result does not depend on the continuity of the current density, just the space in which the problem is defined.

  • $\begingroup$ Thanks! It's really clear, but you don't need the continuity for changing the integral and the divergence (is at least leibnitz condition)? $\endgroup$
    – Johanna
    Commented May 29, 2018 at 23:44
  • $\begingroup$ So for the Leibniz condition, you can change the integral and derivative so long as the bounds are not functions of the variable the derivative operator acts on. I'll add a note to my answer: en.wikipedia.org/wiki/Leibniz_integral_rule $\endgroup$
    – kthaxt
    Commented May 30, 2018 at 3:01
  • $\begingroup$ For a counterexample where this is NOT allowed, look at another expression, such as Faraday's law of induction. If the body is moving, the bounds of the volume integral are not constants, but functions of time. So switching the integral sign and d/dt is not as simple in that case as switching the gradient operator in the above example. $\endgroup$
    – kthaxt
    Commented May 30, 2018 at 3:10
  • $\begingroup$ But anyway, in that case you need to ask for the continuity of the derivate of the function under the integral sign. On page 14 is the theorem with the conditions math.uconn.edu/~kconrad/blurbs/analysis/diffunderint.pdf $\endgroup$
    – Johanna
    Commented May 30, 2018 at 3:29
  • $\begingroup$ Oh I see. Hmm, I'm not certain, but maybe one answer is the same reason the curl of J is zero. $\vec{J}$ is dependent on the source space (x′,y′,z′) and not the measurement space where we are taking the curl (x,y,z). The integral is also over the source space. So perhaps there is no restraint on interchanging the two operators because the integral and the derivative operate on different variables? $\endgroup$
    – kthaxt
    Commented May 30, 2018 at 3:53

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