# What is the origin of this formula $\vec B = -\frac{1}{c^2}\vec v \times \vec E=-\frac{1}{ec^2}\vec v \times \nabla V(r)$?

For context, this is from the beginning of the proof for the Spin-orbit interaction term in atomic physics.

Classically, the magnetic field seen by a particle of charge $$e$$ moving with a velocity $$v$$ in an electric field $$\vec E$$ is given by $$\vec B = -\frac{1}{c^2}\vec v \times \vec E=-\frac{1}{ec^2}\vec v \times \nabla V(r)\tag{1}$$ where $$V(r)=\frac{-Ze^2}{4\pi \epsilon_0 r},$$ since the electron is moving in the electric field of the nucleus of charge $$Ze$$.

I know that Faraday's law is $$\nabla \times \vec E=-\frac{\partial \vec B}{\partial t}$$

and the Lorentz force is given by $$m\frac{dv}{dt}=e(\vec E + \vec v \times \vec B)$$

but eqn $$(1)$$ has neither of these forms so I can't figure out why eqn $$(1)$$ makes sense. For the RHS of equation $$(1)$$ I know that the Coulomb electrostatic force is given by $$F=-\nabla V(r)=\frac{Ze^2}{4\pi \epsilon_0 r^2}$$

But I am confused by the LHS as I have never seen it before.

Is anyone familiar with this equation $$\vec B = -\frac{1}{c^2}\vec v \times \vec E?$$

Or, put in another way is there a useful source of information (webpage etc.) that will explain where this equation originates?

I have already been given a thorough answer to this by a user which involves a relativistic treatment. Unfortunately, Lorentz transforms are way beyond my current scope of understanding.

Is there a 'simpler' way to intuit this formula $$\bbox[yellow,5px,border:2px solid blue]{\vec B = -\frac{1}{c^2}\vec v \times \vec E}$$ at the classical level?

I ask this because the quote given at the start of this post containing equation $$(1)$$ used the word "Classical" right at the beginning of the sentence, which implies there is a classical way to interpret $$(1)$$.

Thanks again.

• It comes from the Lorentz transformations of the electromagnetic fields.(See wikipedia: en.wikipedia.org/wiki/…)
– wcc
Mar 7, 2019 at 23:30

It comes from special relativity. You get it when you apply a Lorentz transform to a 4-potential and consider the Lorentz gauge. You can get an easy derivation from here: http://www.feynmanlectures.caltech.edu/II_26.html

• Thank you very much for your answer. Unfortunately, Lorentz transforms (and the Lorentz guage) are way beyond my current scope of understanding. Is there a 'simpler' way to intuit this formula $\vec B = -\frac{1}{c^2}\vec v \times \vec E$ at a classical level? Best regards. Mar 9, 2019 at 22:37
• There is no other derivation I know of, it seems to me that Lorentz transform is the only way to understand it... You can try and research a bit over special relativity. It's really not that difficult to grasp Mar 9, 2019 at 22:49
• Thanks for your reply, I have updated my question, so I will wait and see if anyone may have a classical way to think about this. If there is no other way then I guess it's time for me to start learning special relativity. Mar 9, 2019 at 22:54

Before the full relativistic understanding of electrodynamics was achieved, the equation $$\vec{B} = - \frac{1}{c^2} \vec{v} \times \vec{E}$$ was known from the Liénard-Wiechert potentials. The full derivation of these equations is far beyond the scope of a simple answer here; most upper-level undergraduate texts and graduate texts in electrodynamics devote a chapter to the subject. However, here's the outline of how derivation proceeds:

• Write the electric and magnetic fields in terms of a scalar potential $$V$$ and a vector potential $$\vec{A}$$: $$\vec{E} = - \nabla V - \partial \vec{A}/\partial t$$ and $$\vec{B} = \nabla \times \vec{A}$$.

• Write Gauss's Law and Ampère's Law in terms of the potentials $$V$$ and $$\vec{A}$$. Use gauge transformations to enforce the Lorenz gauge $$\partial V/\partial t + \nabla \cdot \vec{A} = 0$$. In this gauge, the equations become $$\Box^2 V = -\rho/\epsilon_0$$ and $$\Box^2 \vec{A} = - \mu_0 \vec{J}$$, where $$\Box^2$$ is the d'Alembertian operator.

• Solve these equations for $$V$$ and $$\vec{A}$$ given the charge distribution $$\rho = q \delta^{(3)}(\vec{r} - \vec{v} t)$$ and the current distribution $$\vec{J} = q \vec{v} \delta^{(3)}(\vec{r} - \vec{v} t)$$, assuming that $$|\vec{v}| \ll c$$. (These are the charge and current distributions corresponding to a point charge moving with constant velocity.)

• Take the appropriate derivatives of $$V$$ and $$\vec{A}$$ to obtain $$\vec{E}$$ and $$\vec{B}$$ for a point charge moving with constant velocity.

After all of this, you find that $$\vec{B} \approx - \frac{1}{c^2} \vec{v} \times \vec{E}$$; in other words, according to the electron, the proton is moving and so has a magnetic field in addition to its electric field.

Hopefully, the above description has convinced you that while you could find this result using pre-relativistic physics, you really don't want to. The description of the fields of moving charges in terms of special relativity is so much more elegant than the rather ugly PDE techniques required to derive these results in classical electrodynamics; it's not even close. I highly recommend that you read the chapter on "The Fields of Moving Charges" in Purcell & Morin's Electricity and Magnetism; it is beautifully written and utterly compelling.

• That's so close to the approach from special relativity! Before you get to solve the differential equations, it all looks so obviously like two lorentz invariant PDE's. It is so beautifull that Lorentz invariance comes up so naturally in the Lorentz gauge, it is like a hint from the universe that special relativity is the language of electrodynamics. Mar 10, 2019 at 14:24