I read this in my textbook: A charged particle or object is not affected by its own electric field.

Since I find this completely unintuitive and my mind is yelling "wrong! wrong! how could a particle even distinguish between its own field and the external fields?" I would really like to hear an explanation about why this is true, if it is.

(A corollary question that comes to my mind after thinking about this is... What about gravity and mass considering the analogous situation? And the other two forces?)

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    $\begingroup$ That statement is an approximation, and is just sweeping stuff under the rug that you don't have to worry about right now in your course/studies. $\endgroup$ – Chris Gerig Mar 8 '13 at 0:24
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    $\begingroup$ Maybe you are not aware of the Abraham-Lorentz Force. But don't spend too much time on it. A generalization of the AL force is the equally problematic Abraham-Lorentz-Dirac force. You can check it out in here. $\endgroup$ – PML Mar 8 '13 at 0:29
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    $\begingroup$ Btw, gravitational self force is an area of active study nowadays. This review is great but very hard to follow without appropriate background. $\endgroup$ – PML Mar 8 '13 at 0:33
  • $\begingroup$ @PML, good references! (+1), why not posting them as an answer? $\endgroup$ – Eduardo Guerras Valera Mar 8 '13 at 0:44
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    $\begingroup$ Just remember that a link to a reference doesn't make an answer by itself, but a good answer could definitely be based on it. $\endgroup$ – David Z Mar 8 '13 at 2:18

In Classic Electrodynamics it's well known that an accelerated charge will radiate energy and the radiated power is given by the Larmor Formula $$P=\frac{\mu_0q^2a^2}{6\pi c},$$ in SI units, where $c$ is the speed of light, $q$ is the charge of the particle and $\mu_0$ the magnetic constant.

Well, to incorporate the Larmor formula into Newtonian mechanics we claim that: if the particle is losing energy while accelerating then the effect can be expressed as if the particle feels the effect of a force opposing its movement. To account for this effect let's say that the 2nd Newton equation can be written as $$m \vec{\dot{v}}=\vec{F}_\text{ext}+\vec{F}_\text{rad},$$

where $\vec{F}_\text{ext}$ is an exterior force that accelerates the particle and $\vec{F}_\text{rad}$ is some force that would account for the energy (photons) radiated by the accelerated charge.

To continue we impose the following: $\vec{F}_\text{rad}$ has to be such that the work it does is equal to the energy lost by the particle accordingly to the Larmor formula;

We can the use this imposition to write that in an interval $t_2-t_1$, such that the system is in the same state at those instants, I mean the velocity and the acceleration of the particle is the same in $t_1$ and $t_2$ $^{\bf 1}$, the work done by $\vec{F}_\text{rad}$ is equal to the energy lost by the particle. Then: $$\int_{t_1}^{t_2}\vec{F}_\text{rad}\cdot \vec{v}~dt=-\int_{t_1}^{t_2}\frac{\mu_0q^2}{6\pi c}\dot{\vec{v}}\cdot \dot{\vec{v}}~dt,$$ where the relation $a^2=\dot{\vec{v}}\cdot \dot{\vec{v}}$ was used. Now, integrating by parts we find: $$\int_{t_1}^{t_2}\vec{F}_\text{rad}\cdot \vec{v}~dt=\frac{\mu_0q^2}{6\pi c}\int_{t_1}^{t_2}\ddot{\vec{v}}\cdot \vec{v}~dt-\frac{\mu_0q^2}{6\pi c}(\dot{\vec{v}}\cdot \vec{v})|_{t_1}^{t_2}.$$ Since the particle is in the same state in both the instants $t_1$ and $t_2$, the second term of the sum is zero and then we can write: $$\int_{t_1}^{t_2}\left[ \vec{F}_\text{rad}-\frac{\mu_0q^2}{6\pi c}\dot{\vec{a}} \right] \cdot \vec{v}~dt=0.$$ Since this must be true for all $\vec{v}$ you get the famous Abraham-Lorentz force.: $$\vec{F}_\text{rad}=\frac{\mu_0q^2}{6\pi c}\dot{\vec{a}}.$$

This equation has some serious problems when you try to use it. You can see those in this reference. Also in that reference Eric Poisson shows the special relativity generalization of the AL force called the Abraham-Lorentz-Dirac force.

As for the second part of the question: It's well known that any interaction that propagates at a finite velocity originates a self-force effect. In Classical mechanics only the electromagnetic interaction travels at finite velocity, $c$, so Newtonian gravity doesn't originate self-force effect since the interaction propagates at infinite velocity.

However, assuming that Lorentz invariance is correct (and at least we physicists believe so, and all the tests indicate towards that belief) nothing can propagate at infinite velocity and there is actually a maximum velocity allowed, $c$, so all interactions should originate a self-force effect (usually called radiation reaction).

PS: If anybody is still with me after this very long answer and you remember me writing that Newtonian gravity wouldn't originate radiation reaction, well in the modern way of looking at gravity (General Relativity) gravitational waves propagate at the speed of light so there must be a gravitational self-force effect... More on that you can check this reference in which Eric Poisson et al. deduce the first order correction to the movement of a particle in curved space-time.

$_{\bf 1}$ More on this you can check the book from griffiths - chapters 10 and 11.


This is the problem with classical electrodynamics. The self-energy of a charged point particle is infinite due to its interaction with its on potential.


$$W=\frac{q^2}{8\pi\epsilon_o}\int_0^{\infty}\frac{1 }{r^2}dr $$


  • $\begingroup$ Can someone explain why this got downvoted? Did user18764 say something wrong or was it just incomplete, or misleading? $\endgroup$ – user20250 Mar 8 '13 at 16:56
  • $\begingroup$ I think the point is that this self-energy is immeasurable, and is not really considered a problem in classical electrodynamics (it merely shifts the total energy by a constant, albeit infinite amount). $\endgroup$ – lionelbrits Dec 5 '13 at 15:47
  • $\begingroup$ The self-energy is actually related to the self-interaction, but this answer does not even mention that. $\endgroup$ – timur Sep 26 '17 at 5:04

The self force is the result of the radiated field created because of the force caused by another particle but it's not a force caused only by the particle to itself.

If we consider a particle with an extension and a charge density, there is no Coulomb force caused by a point of the particle over the rest of the point. The proof is easy: in the hydrogen atom, taking the first energy level the potential is caused only by the proton at the nucleous, if the electron would cause force over itself, the inner shells of the wave-function would change the potential in the outter shells.

Since the hydrogen atom energy levels have been evaluated empirically many time, this is a proof that a particles does not exert force over itself directly.

Best regards, Sergio


A correct treatment of the self-force problem, 100% rigorous and free of the paradoxes associated with the conventional textbook treatments is given here.

  • $\begingroup$ This is a link-only answer. The essence of the answer ought to be posted here. $\endgroup$ – sammy gerbil Feb 7 '20 at 2:58

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