Do divergence and curl of Lorentz force have some physical meaning?

Question

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:

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

Epilogue

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

EDIT

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?

Answer 1

What is the meaning behind the rotor and divergence of an electro-magnetic force of Lorentz? I think, there are many considerations behind this problem, which could be a little confusing.

It could be much more easier, if it would be examined each member of each expression separately.

$\nabla \times {\vec{F}} = {q} \cdot (\nabla \times {\vec{E}}) + \nabla \times {(\vec{v} \times \vec{B})}$

The cross product shows, that in an electro-magnetic field, these electric and magnetic components of the fields, are perpendicular on the direction of movement, that may be parallel to the speed $\vec{v}$

The rotor of electric field is usually neglected. That is, an electric field has a symmetric appearance, which does not depend too much of a spatial rotation. Or if a dependency really exists, through the Ampere Law of the Magnetic Circuit, it is less stronger.

Instead, this electric field component behaviours as a long distance field, practically infinite. But at such big distances, the field intensity is too attenuated, and can be considered $0$. Or maybe the electric rotational component can be often neglected, due to this apparent property.

From this reason, the Lorentz force, it can be considered more, as a magnetic kind of force. But this term, could be replaced with a formula, that gets from the Faraday law of induction.

$\nabla \times {\vec{E}} = -\frac {d}{dt} {(\vec{B})}$

Note, that I could not use the symbols of partial derivation, $({\delta}/{\delta{t}})$, because in the general case, of a relative movement, the intensity of the induction $\vec{B}$ of the magnetic field, it has also a spatial variation, which is dependent of time.

In this way, all the terms can be expressed in function of an intensity of a magnetic induction, $\vec{B}$. This could have been also expressed in function of the intensity of the electric $\vec{E}$ field, too.

But at some shorter distances, the magnetic component of the force could be more stronger. This is why, the electric machines use more an electro-magnetic force, rather than the electric forces, and many of these machines, do have a magnetic core. (or the dielectric properties of the medium, aren't that so good)

$\nabla \times {\vec{F}} = -{q} \cdot \frac {d}{dt} {(\vec{B})} + \nabla \times {(\vec{v} \times \vec{B})}$

The second term can be also decomposed, and writen as a sum of rotors of v, respectively B:

$\nabla \times {(\vec{v} \times \vec{B})} = \vec{v} \cdot \nabla{\vec{B}} - \vec{B} \cdot \nabla{\vec{v}} + (\vec{B} \cdot \nabla)\vec{v} - (\vec{v} \cdot \nabla)\vec{B}$

First term in right member of equation, that divergence of induction B, can be considered $0$. According to Law of the Magnetic Stream, the divergence of this vector it is always null. This is because, the stream that is going out a surface, is equal to the incoming stream.

And this null result shows once again, it cannot exist any separation of some magnetic charges. (or in terms of the magnetism, there is no such macroscopic separation of the magnetic poles). This is meaning, that the magnetic poles, as usually, cannot exist independently as magnetic mono-poles. And other factors can be also considered $0$.

And the rotor of the speed v, can be also identified with an angular Ω velocity:

$\vec{Ω} = \nabla \times {\vec{v}}$

As you can see, several laws from the Maxwell's system of equations, can be also involved. There are several physical effects, that can't be considered separately, among one to each other. This is because, both the components, electric and magnetic, could participate together at the interaction.

About the expression of divergence of the Lorentz force, each term can be also seen partly.

$\nabla{\vec{F}} = {q} \cdot \nabla{\vec{E}} + \nabla{(\vec{v} \times \vec{B})}$

The spatial derivation of the electric field $\vec{E}$, this could be already expressed as a gradient. It is the expression of the electric field, depending of gradient of an electric potential V. It is quite the Laplace operator, or the Laplacian, being used in many differential equations.

$\vec{E} = -\nabla{V}$

$\nabla{\vec{E}} = -\nabla^2{V} = \delta{V}$

The second term, can be written as:

$\nabla{(\vec{v} \times \vec{B})} = (\nabla \times \vec{v}) \cdot {\vec{B}} - \vec{v} \cdot {(\nabla \times \vec{B})}$

The induction vector, can be also expressed as the rotor of a vector potential A: But a rotor of a rotor is considered it has a null result:

$\nabla \times (\nabla \times \vec{A}) = (\nabla \times \vec{A}) \cdot \vec{A} - \vec{A} \cdot (\nabla \times \vec{A}) = 0$

Where:

$\vec{B} = \nabla \times {\vec{A}}$

Integrating over all the space around a closed surface, the rotor of induction intensity, can be also reduced to a null divergence of the field according to the Theorem of Gauss-Ostrogradsky:

$\iint_S{(\nabla \times \vec{B}) \cdot d\vec{S}}=\iiint_{\mathscr{V}}{(\nabla{\vec{B})} \cdot d{\mathscr{V}}} = 0$

In theory, this integral may differ from a banal solution $0$, only by a constant, C. But the physical consequences are enough, to be considered, that some terms can be null.

So, anything could be reduced to the following equation:

$\nabla{\vec{F}} =-q \cdot {\nabla^2}{V} + \vec{Ω} \cdot (\nabla \times \vec{A})$

While an electric component of an electro-magnetic force, depends of the electric charge, the magnetic term cannot depend of a separation of magnetic charges.

There are also differences in the way, in which the electric potential is a scalar field, while a magnetic potential has a different behaviour, and describes a magnetic vector field. From this reason, the magnetic lines could be described as closed solenoidal field lines.

Sometimes, it's easier to express all the equations of field in a differential form. If it could be more difficult to solve that differential equations, then an integration could be applied. A mathematician would say, from the rotor and divergence, can be obtained the unknown fields.

In many problems of physics, the divergence, or a gradient is tested over some null values. When the divergence, or a gradient, are null, there are some maximum, or minimum solutions.

$\nabla{\vec{F}} = 0$

It can be applied this Lagrange condition, that can be tested for any possible solutions. Most of the times, there might be several spatial, or temporal limitation conditions, which can be applied at limit, like for example, could be the condition showing, that at an infinite distance, an intensity becomes $0$.

This could be quite a general method, which is being used, to quantify the values of the fields intensities.

This could lead also, to a relational link between the scalar potential $V$, and the magnetic potential vector $\vec{A}$, for a more proper substitution of this magnetic potential $\vec{A}$, which is unknown. Other convenient substitutions can be also made.

Sometimes in problems of electro-dynamics, it can be used a delayed potential, like this one:

$V_r = V - \vec{v} \cdot \vec{A}$

Or it is made a substitution of the electric field intensity $\vec{E}$, which at least formally, could verify the Maxwell's system of differential equations for the electro-magnetic field:

$\vec{E} = - \nabla {V} - \frac{d{}}{d{t}}(\vec{A})$