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I Curl of the Magnetic Field of an Infinite Wire

  1. May 5, 2017 #1
    I'm familiar with the relationship [tex]\nabla\cdot\frac{\hat{r}}{r^2}=4\pi\delta(r)[/tex] in classical electromagnetism, where [itex]\hat{r}[/itex] is the separation unit vector, that is, the field vector minus the source vector. This is result can be motivated by applying the divergence theorem to a single point charge.

    I stumbled across a similar relationship when playing with Stokes' theorem and the magnetic field of an infinite straight wire with a steady current [itex]I[/itex]: it is [tex]\nabla\times\frac{\hat{\phi}}{r}=2\pi\delta^2(r) \, \hat{z}[/tex]

    The magnetic field of an infinite straight wire is [tex]B=\frac{\mu_0I}{2\pi r} \, \hat{\phi}[/tex] Stokes' theorem states [tex]\iint_S(\nabla\times B)\cdot d a=\oint B\cdot d \ell[/tex] The right side of the equation is [itex]\mu_0I[/itex]. The left side of the equation, however, is zero since [tex]\nabla\times\frac{\hat{\phi}}{r}=0[/tex] using the standard definition of the gradient in cylindrical coordinates. Using the relationship I introduced in the second paragraph seems to fix this problem. This relationship is also able to satisfy Stokes' theorem for a finite straight wire.

    So far, I have only been able to find one tangential reference to this relationship in Ben Niehoff's response to the question "Is B with curl 0 possible?" If this formula is familiar to anyone, or if anyone can reference literature that specifically addresses this formula, I would greatly appreciate it.
  2. jcsd
  3. May 7, 2017 #2


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    The problem is that you cannot use cylinder coordinates to study the ##\delta##-distribution like behavior along the ##z## axis, because there the coordinates are singular, because the Jacobian between Cartesian and cylinder coordinates is ##\det (\partial(x,y,z)/\partial(\rho,\varphi,z)=r##, which is ##0## for ##\rho=0##, i.e., along the ##z## axis.

    Also your first formula is a bit problematic. Correct is
    $$\vec{\nabla} \cdot \frac{\vec{r}}{r^3}=-\Delta \frac{1}{r}=4 \pi \delta^{(3)}(\vec{x}),$$
    but you cannot equal this simply to ##4 \pi \delta(r)## (which simply doesn't make sense as a equation of distributions; it's even dimensionally wrong).

    To answer your question, you can however use this correct equation (which is derived in electrostatics when calculating the Green's function of the negative Laplace operator). In magnetostatics you have (in Heaviside-Lorentz units)
    $$\vec{\nabla} \times \vec{B}=\frac{1}{c} \vec{j}, \quad \vec{\nabla} \cdot \vec{B}=0.$$
    The latter equation implies that there is a vector potential
    $$\vec{B}=\vec{\nabla} \times \vec{A},$$
    which is only defined up to a gradient field (for given ##\vec{B}##). Thus we can impose one constraint condition ("gauge-fixing condition"). Here the most convenient choice is
    $$\vec{\nabla} \cdot \vec{A}=0.$$
    Then you have (in Cartesian coordinates!)
    $$\vec{\nabla} \times \vec{B}=\vec{\nabla} \times (\vec{\nabla} \times \vec{A})=\vec{\nabla} (\vec{\nabla} \cdot \vec{A})-\Delta \vec{A}=-\Delta \vec{A}=\frac{1}{c} \vec{j}.$$
    Now for your case of an infinitely long infinitesimally thin wire along the ##z## axis you have
    $$\vec{j}=\vec{e}_z I \delta(x) \delta(y).$$
    Using the Green's function of the negative Laplace operator leads to
    $$\vec{A}(\vec{r})=\int_{\mathbb{R}^3} \frac{\vec{j}(\vec{r}')}{4 \pi c |\vec{r}'-\vec{r}|}=\vec{e}_z \frac{I}{4 \pi c} \int_{-\infty}^{\infty} \mathrm{d} z' \frac{1}{\sqrt{(z'-z)^2+\rho^2}}=\vec{e}_z \frac{I}{4 \pi c} \int_{-\infty}^{\infty} \mathrm{d} z' \frac{1}{\sqrt{z^{\prime2}+\rho^2}}.$$
    This diverges of course, but we can add an arbitrary constant to the potential. So we can calculate instead
    $$\vec{A}_{\text{reg}}(\vec{r})=\vec{e}_z \frac{I}{4 \pi c} \int_{-\infty}^{\infty} \mathrm{d} z' \left (\frac{1}{\sqrt{z^{\prime2}+\rho^2}} - \frac{1}{\sqrt{z^{\prime2}+\rho_0^2}} \right )=-\vec{e}_z \frac{I}{2 \pi} \ln \left (\frac{\rho}{\rho_0} \right).$$
    In the last step I used Mathematica and, of course, I have assumed ##\rho, \quad \rho_0>0## for ##\rho=0## the integral doesn't exist at all.

    Now evaluating the magnetic field, indeed gives your result
    $$\vec{B}=\vec{\nabla} \times \vec{B}=\frac{I}{2 \pi \rho} \hat{\phi}, \quad \rho \neq 0.$$
    Our derivation shows that indeed
    $$\vec{\nabla} \times \vec{B}=I \delta(x) \delta(y) \vec{e}_z,$$
    $$\vec{\nabla} \times \frac{\hat{\phi}}{2 \pi \rho}=\delta(x) \delta(y) \vec{e}_z.$$
    It's clear that you can't write the right-hand side in terms of cylinder coordinates, because ##\delta(x) \delta(y)## has its support at ##x=y=0##, i.e., ##\rho=0## along the ##z## axis, where these coordinates are singular and their Jacobian with Cartesian coordinates vanishes.
  4. May 7, 2017 #3
    Thank you!

    I didn't know you could write the current density for an infinite wire. Very neat.
    Did you mean [tex]\vec{B}=\vec{\nabla} \times \vec{A}=\frac{I}{2 \pi \rho} \hat{\phi}, \quad \rho \neq 0?[/tex]
    I don't have a very good grasp on the theory behind the Dirac delta since you typically only learn how to use them undergraduate E&M or Quantum course when they're useful. Very handy objects, though.
  5. May 8, 2017 #4


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    Yes, I should have put a larger blank after the comma.
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