Time ago I started thinking about this: if we take the well known Lorentz Force expression, namely

$$\mathbf{F} = q\left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right)$$

and we operate $\nabla\cdot \mathbf{F}$ and $\nabla\times\mathbf{F}$, what do we obtain? I performed the calculations, but I have to say that

1) I don't know if they are exact (at least until where I stopped)

2) I ignore their physical meanings

The question is

Do they have some interpretation?

Calculation of $\nabla\cdot\mathbf{F}$

$$ \begin{align*} \nabla\cdot\mathbf{F} & = \nabla\cdot\left(q\left(\mathbf{E} + \mathbf{v}\times\mathbf{B}\right)\right) \\\\ & = q\nabla\cdot\mathbf{E} + q\nabla\cdot (\mathbf{v}\times\mathbf{B}) \end{align*} $$

Using Maxwell equations, and considering $\nabla\times\mathbf{v} = \mathbf{\Omega}$ as the vorticity (that should be the definition of the curl of a velocity) we gain

$$\boxed{\nabla\cdot\mathbf{F} = \frac{q\rho}{\epsilon_0} + q\mathbf{\Omega}\cdot\mathbf{B} - q\mu_0 \mathbf{v}\cdot \left(\mathbf{J} + \epsilon_0\frac{\partial \mathbf{E}}{\partial t}\right)}$$

Calculation of $\nabla\times\mathbf{F}$

$$ \begin{align*} \nabla\times\mathbf{F} & = q\left[\nabla\times\mathbf{E} + \nabla\times(\mathbf{v}\times\mathbf{B}\right] \\\\ & = q\nabla\times\mathbf{E} + q\left[\mathbf{v}(\nabla\cdot\mathbf{B}) - \mathbf{B}(\nabla\cdot \mathbf{v}) + (\mathbf{B}\cdot\nabla)\mathbf{v} - (\mathbf{v}\cdot\nabla)\mathbf{B}\right] \end{align*} $$

Again, using Maxwell equations, and defining (if it has any sense..) $\mathbf{a} = \nabla\cdot \mathbf{v}$, we get:

$$\boxed{\nabla\times\mathbf{F} = -q\frac{\partial\mathbf{B}}{\partial t} + q\left[-\mathbf{B}\mathbf{a} + (\mathbf{B}\cdot\nabla)\mathbf{v} - (\mathbf{v}\cdot\nabla)\mathbf{B}\right]}$$


So, if all of that is correct, what now?


Thanks to a page linked in the comment I understood that (thanks Rob Jeffires):

From the solenoidal law $\nabla \cdot {\bf B}=0$ always, and $\nabla \cdot {\bf v} = \partial/\partial t(\nabla \cdot {\bf r})=0$. Furthermore, $({\bf B}\cdot \nabla){\bf v} = ({\bf B}\cdot \frac{\partial}{\partial t} \nabla){\bf r} = 0$, so $$ \nabla \times {\bf F} = - q\left[\frac{\partial {\bf B}}{\partial t} + \frac{\partial {\bf B}}{\partial x} \frac{\partial x}{\partial t} + \frac{\partial {\bf B}}{\partial y} \frac{\partial y}{\partial t} + \frac{\partial {\bf B}}{\partial z} \frac{\partial z}{\partial t}\right] $$ $$\nabla \times {\bf F} = - q\frac{d {\bf B}}{d t}$$ and the force is only conservative in the case of stationary magnetic (and hence electric) fields.

Now again:

  • Can we solve this equation to obtain a different form for $\mathbf{F}$?

  • What about the divergence?

  • 1
    $\begingroup$ I think you are missing a dot product symbol in the first boxed equation between the $\mathbf{v}$ and ( )'s. $\endgroup$ – honeste_vivere Jan 27 '16 at 15:02
  • 1
    $\begingroup$ All these expressions make perfect sense inside some integrals, and using the Stokes theorem. Then you can relate the work -- being $\int F\cdot dr=\iint\nabla\times F\cdot dS$ in terms of the voltage drop $\int E\cdot dr=\int\nabla\varphi\cdot dr=\varphi_{2}-\varphi_{1}$ for instance. For the magnetic field, it may be more transparent to use the gauge potential. In integrals there is no difference of course. $\endgroup$ – FraSchelle Feb 3 '16 at 8:00
  • 1
    $\begingroup$ The physical meaning of $\nabla\times F$ and $\nabla\cdot F$ directly, as far as I can see, are the usual ones : the longitudinal and transverse component of the force field. But be warn that the transverse and longitudinal components of the force is not related directly to the transverse and longitudinal component of the gauge fields, because of the cross product $v\times B$ in the Lorentz force. Using the Maxwell equations (as you tried) it must be possible to express the longitudinal and transverse component of the force using only one component of the field. $\endgroup$ – FraSchelle Feb 3 '16 at 8:04
  • $\begingroup$ As you have perhaps seen, the curl of the Lorentz force is only zero for static fields. So your edit is wrong. physics.stackexchange.com/questions/118498/… $\endgroup$ – Rob Jeffries Nov 2 '16 at 7:27
  • $\begingroup$ @RobJeffries Uhh Thank you so much for that page! Going to edit the edit lol. What about the divergence instead? $\endgroup$ – Les Adieux Nov 2 '16 at 7:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.