In this paper the author discusses the Poincaré vector $\vec D$, and he wrote the Lorentz equation when magnetic charge is present.

Do you have any suggested reading for this and how he derived the equation?

$$\vec D = \vec r \times \dot{\vec r} - g \hat r $$

I also wanted to know what he means by g and how he got that?

The g is defined as magnetic strength with fixed magnetic monopole.

  • 2
    $\begingroup$ $g$ is like $q$, but for magnetism. $\vec{B}=g\hat{r}/r^2$, like $\vec{E}=q\hat{r}/r^2$. It is speculative “magnetic charge”. $\endgroup$ – G. Smith Aug 3 at 3:58
  • $\begingroup$ Do you have any good reading materials where I can learn how to derive the equation? $\endgroup$ – user193422 Aug 3 at 3:59
  • $\begingroup$ I believe the original paper is H. Poincaré, Remarques sur une expérience de M. Birkeland. C. R. Acad. Sci. Paris 123, 530-533 (1896). $\endgroup$ – G. Smith Aug 3 at 4:09
  • $\begingroup$ You should be able to show that its time derivative is zero, using the Lorentz force law. $\endgroup$ – G. Smith Aug 3 at 4:10
  • $\begingroup$ It may be derivable by computing the angular momentum in the electromagnetic field. Do you know about field energy, momentum, and angular momentum? $\endgroup$ – G. Smith Aug 3 at 4:14

I am not sure what you would consider an answer to your question, but I think one can just check that the vector $\vec{D}$ is indeed conserved. As usual, $\vec{r} = \vec{r}(t)$ describes the position of an electrically charged particle of charge $1$ and mass $1$ at any moment of time $t$. For brevity, denote $r = r(t) = \|\vec{r}(t)\| = \sqrt{\vec{r}(t) \cdot \vec{r}(t)\,}$, where $(\,\cdot \,)$ denotes the dot prodcut in three space.

Assume you are given a magnetic monopole, stationary at the origin of an inertial coordinate system, generating a magnetic field $$\vec{B}(\,\vec{r}\,) = g\,\frac{\vec{r}}{r^3}$$ where $g$ is a constant describing the strength of the magnetic field. Observe that its form is identical, possibly up to a sign, to Coulomb's electric field and Newton's gravitational field. The explanation for this formula can come from Maxwell's equations with magnetic monopoles. One can also explain it with Gauss' law, Helmholz decomposition and the intuitively reasonable requirement that the filed is rotationally symmetric and decreasing fast away from the source.

Furthermore, it is assumed that there exists a time-independent electric field $\vec{E}(\,\vec{r}\,)$ generated from a rotationally-invariant electric potential $$V = V(\,\vec{r}\,) = V(r)$$ Then the electric field is equal to the minus gradient of the potential, i.e. $$\vec{E}(\,\vec{r}\,) = -\, \nabla\, V(r) = -\, v(r)\, \frac{\vec{r}}{r}$$ where $v(r) = \frac{dV}{dr}(r)$ The vector function $\vec{r}(t)$, describing the motion of the charged particle, satisfies the Lorenz law, which is equivalent to the system of second order ordinary differential equations: $$\frac{d^2\vec{r}}{dt^2} \,=\, g \, \frac{d\vec{r}}{dt} \times \frac{\vec{r}}{r^3} \, - \, v(\,r\,)\, \frac{\vec{r}}{r}$$ Just like in mechanics, you observe that the vectors $\vec{r}$ and $- \, v(\,r\,)\, \frac{\vec{r}}{r}$ are collinear, so their cross product is zero. Therefore, multiply both sides of the vector differential equation above by $\vec{r}$ from the left: $$\vec{r} \times \frac{d^2\vec{r}}{dt^2} \,=\, g \,\, \vec{r} \times \left(\frac{d\vec{r}}{dt} \times \frac{\vec{r}}{r^3}\right) \, - \, v(\,r\,)\, \vec{r} \times \frac{\vec{r}}{r}$$ and as are a result, the equation reduces to \begin{align*} \vec{r} \times \frac{d^2\vec{r}}{dt^2} \, &=\, g \,\, \vec{r} \times \left(\frac{d\vec{r}}{dt} \times \frac{\vec{r}}{r^3}\right) \\ &= \, g\,\, \frac{\big( \vec{r} \cdot \vec{r}\big)}{r^3}\, \frac{d\vec{r}}{dt} \, -\, g\,\, \left(\vec{r} \cdot \frac{d\vec{r}}{dt}\right) \frac{\vec{r}}{r^3}\\ &= \, g\,\, \frac{r^2}{r^3}\, \frac{d\vec{r}}{dt} \, -\, g\,\, \left(\vec{r} \cdot \frac{d\vec{r}}{dt}\right) \frac{\vec{r}}{r^3}\\ &= \, g\,\, \frac{1}{r}\, \frac{d\vec{r}}{dt} \, -\, g\,\, \frac{1}{r^3}\,\left(\vec{r} \cdot \frac{d\vec{r}}{dt}\right) \vec{r} \end{align*} One can check directly, that the left-hand side can be rewritten as $$\frac{d}{dt} \left(\vec{r} \times \frac{d\vec{r}}{dt}\right) = \frac{d\vec{r}}{dt} \times \frac{d\vec{r}}{dt} + \vec{r} \times \frac{d^2\vec{r}}{dt^2} = \vec{r} \times \frac{d^2\vec{r}}{dt^2}$$ and that the right-hand side can be rewritten as \begin{align*} \frac{d}{dt} \left(\,g\,\,\frac{\vec{r}}{r}\right) &= \frac{d}{dt} \left(\,g\, \,(\vec{r} \cdot \vec{r})^{-\frac{1}{2}} \, \vec{r}\,\right) = g\,\frac{d}{dt} \left( \,(\vec{r} \cdot \vec{r})^{-\frac{1}{2}} \, \right) \, \vec{r} \, + \, g\,\,\frac{1}{r} \, \frac{d\vec{r}}{dt}\\ &= g\, \left( \,-\,\frac{1}{2}\,(\vec{r} \cdot \vec{r})^{-\frac{1}{2}-1} \,\, 2\,\Big(\vec{r} \cdot \frac{d\vec{r}}{dt}\Big) \right) \, \vec{r} \, + \, g\,\,\frac{1}{r} \, \frac{d\vec{r}}{dt}\\ &= g\, \left( \,-\,(\vec{r} \cdot \vec{r})^{-\frac{3}{2}} \,\,\Big(\vec{r} \cdot \frac{d\vec{r}}{dt}\Big) \right) \, \vec{r} \, + \, g\,\,\frac{1}{r} \, \frac{d\vec{r}}{dt}\\ &= - \, g\, \frac{1}{r^3} \,\,\Big(\vec{r} \cdot \frac{d\vec{r}}{dt}\Big) \, \vec{r} \, + \, g\,\,\frac{1}{r} \, \frac{d\vec{r}}{dt}\\ &= \, g\,\,\frac{1}{r} \, \frac{d\vec{r}}{dt} \, - \, g\, \frac{1}{r^3} \,\,\Big(\vec{r} \cdot \frac{d\vec{r}}{dt}\Big) \, \vec{r} \end{align*} Consequently, we arrive at the identity $$\frac{d}{dt} \left(\vec{r} \times \frac{d\vec{r}}{dt}\right) = \vec{r} \times \frac{d^2\vec{r}}{dt^2} = \, g\,\,\frac{1}{r} \, \frac{d\vec{r}}{dt} \, - \, g\,\, \frac{1}{r^3} \,\,\Big(\vec{r} \cdot \frac{d\vec{r}}{dt}\Big) \, \vec{r} = \frac{d}{dt} \left(\,g\,\,\frac{\vec{r}}{r}\right)$$ or if we move the right-hand side to the left side of the identity we get $$\frac{d}{dt} \left(\vec{r} \times \frac{d\vec{r}}{dt}\right) \, - \, \frac{d}{dt} \left(\,g\,\,\frac{\vec{r}}{r}\right) = \vec{0}$$ Since the derivative is linear, we can rewrite the latter equation as follows $$\vec{0} = \frac{d}{dt} \left(\vec{r} \times \frac{d\vec{r}}{dt} \, - \, g\,\,\frac{\vec{r}}{r}\right)$$ which is possible if and only if for any solution $\vec{r} = \vec{r}(t)$ of the vector differential equation $$\frac{d^2\vec{r}}{dt^2} \,=\, g \, \frac{d\vec{r}}{dt} \times \frac{\vec{r}}{r^3} \, - \, v(\,r\,)\, \frac{\vec{r}}{r}$$ there exists a constant vector $\vec{D}$ such that $\vec{r}(t)$ must also satisfy the equation $$\vec{D} = \vec{r} \times \frac{d\vec{r}}{dt} \, - \, g\,\,\frac{\vec{r}}{r}$$


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