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Consider two electrons $e_1$, $e_2$, located at $P_{xyz}$ and $Q_{xyz}$, with velocities $u_{xyz}$ and $v_{xyz}$ respectively in the observer's inertial frame of reference, both with charge $e$.

I wish to determine the instantaneous electro-magnetic forces exerted by $e_1$ on $e_2$. I understand that I firstly need to calculate the Electric Field vector $\bf{E_{1Q}}$ and the Magnetic Field vector $\bf{B_{1Q}}$ which both occur at position Q due to the influence of electron $e_1$.

Secondly I need to determine (and then vector-sum) the forces exerted by those fields on the second electron $e_2$.

From this wikipedia article: Biot-Savart Law (section: "Point charge at constant velocity") I obtain expressions (derived by Oliver Heaviside, 1888, from Maxwell's Equations) for the Electric Field vector $E_{1Q}$ and the Magnetic Field vector $B_{1Q}$:-

$$ \bf{E_{1Q}} = \frac{e}{4\pi\epsilon_0} \frac{1-\frac{u^2}{c^2}}{\left(1-\frac{u^2}{c^2}\sin^2\theta_{ur}\right)^{3/2}}\frac{ \hat{r} }{\lvert r\rvert^2}$$

where $r$ is the displacement vector $r = Q_{xyz}-P_{xyz}$ ,$\hat{r}$ is the unit vector of $r$ and $\theta_{ur}$ is the angle between $e_1$'s velocity $u_{xyz}$ and $r$.

$$ \bf{B_{1Q}} = \frac{\textbf {u} \times \bf{E_{1Q}} }{c^2} $$

(where $\times$ is the vector product operator).

Next I can find the force $\mathbf{F_m}$ exerted by the magnetic field upon electron $e_2$ from the equation:-

$$\mathbf{F_m} = e \left( \bf{v} \times \bf{B_{1Q}} \right)$$

where the direction of $\mathbf{F_m}$ is obtained from Fleming's left-hand Motor rule.

However I am unclear as to how to calculate the force $\mathbf{F_E}$ exerted by the electric field upon target electron $e_2$. My textbook only gives the equation for the Coulomb force ($\mathbf{F_c}=e.\mathbf{E}$) exerted by an electric field on a static electron. I presume that the target electron velocity $\mathbf{v_{xyz}}$ needs to be taken into account somehow. Please can somebody tell me how?


As tomtom1-4 answered, I can use the Lorentz Force equation, which in SI units is:-

\begin{equation} \mathbf{F} = e ( \mathbf{E} + \mathbf{v} \times \mathbf{B}) \end{equation}

which is equivalent to:-

\begin{equation} \mathbf{F} = e ( \mathbf{E} + \mathbf{v} \times \frac{\textbf {u} \times\bf{E} }{c^2}) \end{equation}

\begin{equation} \mathbf{F} = e ( \mathbf{\hat{r}} \vert E\vert + \mathbf{v} \times \frac{\textbf {u} \times \mathbf{\hat{r}} \vert E\vert }{c^2}) \end{equation}

\begin{equation} \mathbf{F} = e \vert E\vert \left( \mathbf{\hat{r}} + \frac{1}{c^2} \mathbf{v} \times (\textbf {u} \times \mathbf{\hat{r}} ) \right). \end{equation}

Substituting for $\vert E \vert$ using the Heaviside expression for $E_{1Q}$ gives:-

\begin{equation} \mathbf{F} = \frac{ee}{4\pi\epsilon_0 r ^2} \left[\frac{1-\frac{u^2}{c^2}}{\left(1-\frac{u^2}{c^2}\sin^2\theta_{ur}\right)^{3/2}}\right] \left( \mathbf{\hat{r}} + \frac{\mathbf{v} \times (\textbf {u} \times \mathbf{\hat{r}} )}{c^2} \right). \end{equation}

For comparison here is the formula from Thomas Fritsch's (Darwin-based) answer:- $$\mathbf{F}=\frac{ee}{4\pi\epsilon_0r^2} \left(\hat{\mathbf{r}} + \frac{ \mathbf{u}(\hat{\mathbf{r}}\cdot\mathbf{v}) +\mathbf{v}(\hat{\mathbf{r}}\cdot\mathbf{u}) -\hat{\mathbf{r}}[\mathbf{u}_1\cdot\mathbf{v}+3(\mathbf{u}\cdot\hat{\mathbf{r}})(\hat{\mathbf{r}}\cdot\mathbf{v})] }{2 c^2} \right).$$

As per Ján Lalinský's comments, I'm beginning to see how, even in Galilean relativity, timing considerations make the situation I describe more complicated in terms of the time-variant E & B fields at point Q than they would be in the case of, for example, a Biot-Savart steady current element at point P.

  • 1
    $\begingroup$ Related : Magnetic field due to a single moving charge. See equations (01a) and (01b) in my answer. $\endgroup$
    – Frobenius
    Oct 2, 2020 at 18:37
  • $\begingroup$ @Frobenius, thanks I'm having a look at your answer there. $\endgroup$
    – steveOw
    Oct 2, 2020 at 20:09
  • 1
    $\begingroup$ @Froebenius. Thanks very much (inserted a basic 2D picture). $\endgroup$
    – steveOw
    Oct 3, 2020 at 23:00
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    $\begingroup$ There are approximate expressions that are usable in cases where radiation/retardation is negligible such as the Darwin Lagrangian - implied fields of moving charges. However, these do not and can't give the force exactly, because of special relativity which requires that the force can't be a function of instantaneous values of position, velocities, accelerations only (there is no concept of simultaneity for two electrons). $\endgroup$ Oct 4, 2020 at 0:50
  • 1
    $\begingroup$ Finite speed of interaction means the real force differs from those simple approximations by a part that can be related to radiation or force retardation. In other words, the forces depend on more things than just positions, velocities and accelerations of the particles. There is a so-called Feynman-Heaviside formula which takes this problem into account in case the fields are purely retarded, see feynmanlectures.caltech.edu/II_21.html $\endgroup$ Oct 4, 2020 at 0:52

2 Answers 2


No, what you have should be correct. The force on a charge $e$ from an electromagnetic field is given by the Lorenz force \begin{equation} \mathbf{F} = e ( \mathbf{E} + \frac{\mathbf{v}}{c} \times \mathbf{B}) . \end{equation} You have seperated this into the parts of the electric and magnetic field. But the formula \begin{equation} \mathbf{F}_\mathbf{E} = e \mathbf{E} \end{equation} still holds. And it is independent of the velocity of the second electron.

  • $\begingroup$ Thanks, I'd forgotten about the Lorentz force equation. I note your equation for $F$ is in cgs units. (I need to check my units are consistent!) $\endgroup$
    – steveOw
    Oct 2, 2020 at 20:51

It seems you are looking for something like the Darwin interaction.

The simple Newton-Coulomb Lagrangian of two charged particles (in SI-units) $$L=\frac{1}{2}m_1v_1^2+\frac{1}{2}m_2v_2^2-\frac{q_1q_2}{4\pi\epsilon_0r}$$ can be improved by

  • handling both particles relativistically,
  • describing one particle reacting to the electric and magnetic field generated by the other particle,
  • and then Taylor-expanding up to terms of order $\frac{1}{c^2}$

Then you get the Darwin Lagrangian (in SI-units), given by $$\begin{align} L&=\frac{1}{2}m_1v_1^2+\frac{1}{8c^2}m_1v_1^4 \\ &+\frac{1}{2}m_2v_2^2+\frac{1}{8c^2}m_2v_2^4 \\ &-\frac{q_1q_2}{4\pi\epsilon_0r} \\ &+\frac{q_1q_2}{4\pi\epsilon_0r}\frac{1}{2c^2} [\mathbf{v}_1\cdot\mathbf{v}_2+(\mathbf{v}_1\cdot\hat{\mathbf{r}})(\hat{\mathbf{r}}\cdot\mathbf{v}_2)] \end{align} \tag{1}$$

You see:

  • It still describes the interaction as an instantaneous interaction, depending only on the current positions and velocities of the two particles. By doing so it neglects any retardation effects due to the electromagnetic field propagating with the finite speed of light.
  • The third line of (1) is the well-known Coulomb interaction.
  • The fourth line is a magnetic interaction (the so-called Darwin interaction).

From this Lagrangian (1) you can find the equations of motions $$\begin{align} \frac{d\mathbf{p}_1}{dt}&=\frac{q_1q_2}{4\pi\epsilon_0r^2}\hat{\mathbf{r}} \\ &+ \frac{q_1q_2}{4\pi\epsilon_0r^2}\frac{1}{2 c^2}\{ \mathbf{v}_1(\hat{\mathbf{r}}\cdot\mathbf{v}_2) +\mathbf{v}_2(\hat{\mathbf{r}}\cdot\mathbf{v}_1) -\hat{\mathbf{r}}[\mathbf{v}_1\cdot\mathbf{v}_2+3(\mathbf{v}_1\cdot\hat{\mathbf{r}})(\hat{\mathbf{r}}\cdot\mathbf{v}_2)] \} \\ \frac{d\mathbf{p}_2}{dt}&=\text{similar as above with $\hat{\mathbf{r}}$ replaced by $-\hat{\mathbf{r}}$} \end{align}$$

The first term on the right side is the well-known Coulomb force.
The second term on the right side is a magnetic force (proportional to the velocities of the two particles, and with a quite complicated directional dependency).

  • $\begingroup$ Thanks I need to ponder this. (Note to self this uses cgs units). I'm not clear what 𝑟̂ 𝑟̂ means? $\endgroup$
    – steveOw
    Oct 2, 2020 at 22:19
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    $\begingroup$ @steveOw The $\hat{\mathbf{r}}\hat{\mathbf{r}}$ used in the Wikipedia article seems to be meant as the tensor product. I have resolved this tensor product to make it easier. I also added the $4\pi\epsilon_0$ factors to make the equations SI-conform, instead of the CGS equations from Wikipedia. $\endgroup$ Oct 2, 2020 at 22:23
  • $\begingroup$ Thanks that is very helpful to me! $\endgroup$
    – steveOw
    Oct 2, 2020 at 22:28
  • $\begingroup$ Am I right in thinking that the Feynman-Heaviside formula (referred to in comments to my question by Ján Lalinský) is a more accurate formula because it does take into account "the retardation effects due to the electromagnetic field propagating with the finite speed of light" which you mentioned in your answer. $\endgroup$
    – steveOw
    Oct 4, 2020 at 22:28

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