The Heaviside-Feynman formula (see Feynman Lectures vol I Ch.28, vol II Ch. 21) gives the electric and magnetic fields measured at an observation point $P$ due to an arbitrarily moving charge $q$

$$ \mathbf{E} = -\frac{q}{4 \pi \epsilon_0} \left\{ \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{ret} + \frac{[r]_{ret}}{c} \frac{\partial}{\partial t} \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{ret} + \frac{1}{c^2} \frac{\partial^2}{\partial t^2} [\mathbf{\hat r}]_{ret} \right\} $$ $$ \mathbf{B} = -\frac{1}{c} [\mathbf{\hat r}]_{ret} \times \mathbf{E} $$

where $[\mathbf{\hat r}]_{ret}$ and $[r]_{ret}$ is the unit vector and distance from the observation point $P$ at time $t$ to the retarded position of charge $q$ at time $t - [r]_{ret}/c$ (hence the minus signs).

This formula is remarkable in that it is completely relational. It does not refer to any external reference frame. The fields at point $P$ only depend on the vector $\mathbf{r}$ from point $P$ to the retarded position of charge $q$ and its first and second order rates of change with respect to local time $t$.

Now one can imagine two ways in which the vector $\mathbf{r}$ from point $P$ to charge $q$ can change. One could move charge $q$ and keep point $P$ fixed or one could move point $P$ and keep charge $q$ fixed. The above formula is valid for the former situation giving the fields at a fixed point $P$ due to a moving charge $q$. This is the conventional retarded solution of Maxwell's equations.

But what about the latter situation in which the observation point $P$ moves and the charge $q$ is fixed. The relational nature of the formula implies to me that it should still apply in this situation. Perhaps this is the situation in which the advanced Heaviside-Feynman formula is valid given by

$$ \mathbf{E} = -\frac{q}{4 \pi \epsilon_0} \left\{ \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{adv} - \frac{[r]_{adv}}{c} \frac{\partial}{\partial t} \left[ \frac{\mathbf{\hat r}}{r^2} \right]_{adv} + \frac{1}{c^2} \frac{\partial^2}{\partial t^2} [\mathbf{\hat r}]_{adv} \right\} $$ $$ \mathbf{B} = -\frac{1}{c} [\mathbf{\hat r}]_{adv} \times \mathbf{E} $$

where $[\mathbf{\hat r}]_{adv}$ and $[r]_{adv}$ is the unit vector and distance from the observation point $P$ at time $t$ to the advanced position of charge $q$ at time $t + [r]_{adv}/c$. The advanced Heaviside-Feynman formula is the time-reverse of the conventional retarded formula.

This interpretation of the advanced formula, if valid, implies that an accelerated observer inside a fixed charged insulating spherical shell would measure an electric field whose strength is proportional to the acceleration. This implies that an accelerated charge feels a kind of electromagnetic inertial force and thus has an electromagnetic inertia due to the presence of the charged spherical shell.

For example imagine an electron with charge $-e$ inside such a fixed charged insulating spherical shell at potential $+V$ volts. Using the above advanced Heaviside-Feynman formula one can calculate that this electromagnetic inertia $m_{em}$ is given by $$m_{em} = \frac{2}{3} \frac{eV}{c^2}$$ For a shell charged to a high voltage $V=1000000$ volts this electromagnetic inertia would be of similar order to the electron's native mass and should therefore be easily observable. It would probably be important that the spherical shell is a charged insulator rather than a conductor because it is assumed that the charges inside the shell remain fixed.

Finally, there is a close analogy between Maxwell's equations and Einstein's field equations in the limit of weak gravitational fields. There is a clear gravitational analogue of the advanced Heaviside-Feynman formula. Thus one would expect that a mass accelerated inside a fixed spherical shell of mass should experience a gravitational inertial force in a manner analogous to the above electrical example (just substitute mass $m$, gravitational potential $\phi$ for charge $e$, electrical potential $V$ in above expression).

Perhaps this is the origin of inertia as hypothesised in Mach's Principle?

  • $\begingroup$ Where did you find the name "Heaviside-Feynman formula" for this formula? $\endgroup$
    – a06e
    Mar 23, 2013 at 16:51
  • $\begingroup$ In Classical Electrodynamics by J.D.Jackson page 247-248 $\endgroup$ Mar 23, 2013 at 23:40

4 Answers 4


You can only calculate electric fields or magnetic fields after fixing a reference frame, so no, you can't move P around in that formula. It is assumed in that formula that you are working in a specific frame. The formula is invariant with respect to translating both $\mathbf{r}$ and $P$ by the same displacement, but not with respect to boosting them by the same velocity. If you boost them, the fields transform. See http://en.wikipedia.org/wiki/Classical_electromagnetism_and_special_relativity


In the first place, those expressions are only approximated; they only give the external classical field in absence of gravity, and without considering radiation reaction effects and other nuisances.

In the second place, the expression is not relational. The correct functional dependence is $\mathbf{E}=\mathbf{E}(x,t)$ and $\mathbf{B}=\mathbf{B}(x,t)$ where $(x,t)$ is the observation point as described from a given frame. Precisely the partial time derivatives of $\mathbf{r}$ acts only on the position of the source ($x$ is held constant), thus generating the well-known source velocity terms.

Third, there are not implications for such ill-defined terms as "electromagnetic inertia".


I would like to try to clarify my argument.

The scalar potential $\phi$ and vector potential $A$ at a distance $r$ from charge $q$ are given approximately by

$$\phi = \frac{q}{r}$$

$$\mathbf{A} = \frac{q\mathbf v}{r}$$

where the constants have been suppressed.

The corresponding electric and magnetic fields are given by

$$\mathbf{E} = -\nabla \phi - \frac{\partial \mathbf A}{\partial t}$$

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

The gradient and curl terms fall off as $1/r^2$ so that at large distances from the charge $q$ we only have an electric field $E$ given approximately by

$$\mathbf{E} = - \frac{\partial \mathbf A}{\partial t}$$


$$\mathbf{A} = \frac{q\mathbf v}{r}$$

Thus at large distances an accelerating charge produces an electric field that is proportional to its acceleration.

But velocity is relative. Instead of the charge moving with instantaneous velocity $v$ with respect to a fixed observer the charge could be fixed and the observer could be moving with instantaneous velocity $-v$.

If the observer changed his velocity with time then in his comoving frame he would experience a changing $A$-field.

My proposal is that this "apparently" changing $A$-field should still have physical consequences. i.e. there should be an induced electric field in the observer's accelerating frame.

I realise that this is non-standard as Maxwell's equations are usually only applied in an inertial frame. My hypothesis is that the retarded solutions apply to an inertial frame whereas the previously unused advanced solutions apply to accelerated frames as in the example above.


Can I add a few ideas that might be relevant to your interesting question:

  1. Suppose the acceleration inside the charged sphere happens due to gravity. Then, since by equivalence principle the frame of the accelerating particle is inertial, Maxwell's equations should apply and the charged (in this frame) accelerating sphere should produce an ''inertial-type'' force like you suggest. Right? If this is not so, I would like to know why not.

  2. Same type of reasoning holds with ''gravitational charges'', at least approximately, and Sciama for example has investigated the possibility of explaining inertial mass this way: e.g. On the origin of Inertia.

  3. In Weber's electrodynamics (which is completely relational) predicts an ''inertial term'' for accelerating particles. (Also in rotating frames it predicts a ''Coriolis term''). See e.g. Andre Assis' Lecture in Youtube


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