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We all have elaborative discussion in physics about classical mechanics as well as interaction of particles through forces and certain laws which all particles obey.

I want to ask,Does a particle exert a force on itself?

EDIT

Thanks for the respectful answers and comments.I edited this question in order to make it more elaborated.

I just want to convey that I assumed the particle to be a standard model of point mass in classical mechanics. As I don't know why there is a minimum requirement of two particles to interact with fundamental forces of nature,in the similar manner I wanted to ask does a particle exerts a force on itself?

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    $\begingroup$ Abraham–Lorentz force. $\endgroup$ – Keith McClary May 6 at 4:42
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    $\begingroup$ What's a particle? ;) $\endgroup$ – Guido May 6 at 13:41
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    $\begingroup$ Actually, if it weren't for the intense gravitational force pulling inward, an electron would be several feet in diameter. (This is, of course, pure bullsith, but can you prove that, with any practical demonstration? In fact, any forces that particles may exert on themselves are irrelevant, so long as they don't cause the particles to explode.) $\endgroup$ – Hot Licks May 6 at 17:03
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    $\begingroup$ By Newton, any force a particle exerts on itself will be cancelled out by an equal and opposite force it also exerts on itself. $\endgroup$ – OrangeDog May 7 at 10:25
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    $\begingroup$ Your title claims that a particle does not exert a force on itself. That is not an undisputed fact. $\endgroup$ – my2cts May 7 at 18:22
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This is one of those terribly simple questions which is also astonishingly insightful and surprisingly a big deal in physics. I'd like to commend you for the question!

The classical mechanics answer is "because we say it doesn't." One of the peculiarities about science is that it doesn't tell you the true answer, in the philosophical sense. Science provides you with models which have a historical track record of being very good at letting you predict future outcomes. Particles do not apply forces to themselves in classical mechanics because the classical models which were effective for predicting the state of systems did not have them apply forces.

Now one could provide a justification in classical mechanics. Newton's laws state that every action has an equal and opposite reaction. If I push on my table with 50N of force, it pushes back on me with 50N of force in the opposite direction. If you think about it, a particle which pushes on itself with some force is then pushed back by itself in the opposite direction with an equal force. This is like you pushing your hands together really hard. You apply a lot of force, but your hands don't move anywhere because you're just pushing on yourself. Every time you push, you push back.

Now it gets more interesting in quantum mechanics. Without getting into the details, in quantum mechanics, we find that particles do indeed interact with themselves. And they have to interact with their own interactions, and so on and so forth. So once we get down to more fundamental levels, we actually do see meaningful self-interactions of particles. We just don't see them in classical mechanics.

Why? Well, going back to the idea of science creating models of the universe, self-interactions are messy. QM has to do all sorts of clever integration and normalization tricks to make them sane. In classical mechanics, we didn't need self-interactions to properly model how systems evolve over time, so we didn't include any of that complexity. In QM, we found that the models without self-interaction simply weren't effective at predicting what we see. We were forced to bring in self-interaction terms to explain what we saw.

In fact, these self-interactions turn out to be a real bugger. You may have heard of "quantum gravity." One of the things quantum mechanics does not explain very well is gravity. Gravity on these scales is typically too small to measure directly, so we can only infer what it should do. On the other end of the spectrum, general relativity is substantially focused on modeling how gravity works on a universal scale (where objects are big enough that measuring gravitational effects is relatively easy). In general relativity, we see the concept of gravity as distortions in space time, creating all sorts of wonderful visual images of objects resting on rubber sheets, distorting the fabric it rests on.

Unfortunately, these distortions cause a huge problem for quantum mechanics. The normalization techniques they use to deal with all of those self-interaction terms don't work in the distorted spaces that general relativity predicts. The numbers balloon and explode off towards infinity. We predict infinite energy for all particles, and yet there's no reason to believe that is accurate. We simply cannot seem to combine the distortion of space time modeled by Einstein's relativity and the self-interactions of particles in quantum mechanics.

So you ask a very simple question. It's well phrased. In fact, it is so well phrased that I can conclude by saying the answer to your question is one of the great questions physics is searching for to this very day. Entire teams of scientists are trying to tease apart this question of self-interaction and they search for models of gravity which function correctly in the quantum realm!

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    $\begingroup$ This is a decent popularization, but I think it's doing a common unsatisfying thing with quantum gravity. The numbers "balloon and explode off towards infinity" in just about all quantum field theories; gravity is not special in this sense at all. The problems with quantum gravity are more subtle, and are covered elsewhere on this site. $\endgroup$ – knzhou May 5 at 21:08
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    $\begingroup$ @knzhou My understanding was that the explosions off to infinity could be dealt with via renormalization, but the curvature of space from gravity distorted things such that the math of renormalization no longer worked. Obviously comments aren't the place for correcting QM misconceptions, but is that far from the truth? $\endgroup$ – Cort Ammon May 6 at 14:24
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    $\begingroup$ Just a note: a classical charged particle exerts a force on itself, a classical gravitating mass exerts a force on itself. It is only that 1) if the forces are contained within a finite isolated body, its center of mass does not exert a force on itself (but a body and/or a particle is rarely isolated), and 2) in the Newtonian limit the gravitational self-force vanishes. It is tempting to make this about the classical vs. quantum realm, but it is more that the self-forces are negligible for the situations treated in a 101 classical mechanics course. $\endgroup$ – Void May 6 at 15:11
  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – ACuriousMind May 7 at 19:40
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    $\begingroup$ Well, self-interactions aren't really interactions of a particle with itself. It is an interaction of more than one particles of the same kind. Correct me if I am wrong. $\endgroup$ – Dvij Mankad May 21 at 14:43
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This question is never addressed by teachers, altough students start asking it more and more every year (surprisingly). Here are two possible arguments.

  1. A particle is meant to have 0 volume. Maybe you're used to exert a force on yourself, but you are an extended body. Particles are points in space. I find it quite hard to exert a force on the same point. Your stating that the sender is the same as the receiver. It's like saying that one point is gaining momentum from itself! Because forces are a gain in momentum, after all. So how can we expect that some point increases its momentum alone? That violates the conservation of momentum principle.

  2. A visual example (because this question usually arises in Electromagnetism with Coulomb's law):

    $$\vec{F}=K \frac{Qq}{r^2} \hat{r}$$

If $r=0$, the force is not defined, what's more, the vector $\hat{r}$ doesn't even exist. How could such force "know" where to point to? A point is spherically symmetric. What "arrow" (vector) would the force follow? If all directions are equivalent...

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    $\begingroup$ An accelerated charge does exert a force onto itself in general. That's called radiation reaction force, or Abraham-Lorentz force. $\endgroup$ – Ruslan May 5 at 15:02
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    $\begingroup$ A charged particle at rest outside an uncharged black hole, or outside an uncharged straight cosmic string, also exerts an electrostatic force on itself. Whenever there is no symmetry to rule it out, you can expect that a self-force does exist! $\endgroup$ – G. Smith May 5 at 16:54
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    $\begingroup$ The two points in this answer make a spherical cow assumption, by saying a particle is a point. $\endgroup$ – Denis de Bernardy May 5 at 18:36
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    $\begingroup$ The Standard Model of particle physics assumes that all elementary particles are point particles. Any other assumption is speculative. The Standard Model works well, whereas cows are obviously not spherical. $\endgroup$ – G. Smith May 5 at 20:59
  • $\begingroup$ @G.Smith Still, models of non-point electron were abundant in early XX c, although they seem to almost always had some errors in mathematical calculations. Rohrlich gives an interesting account of them in his "Classical Charged Particles" (and also claims to provide a resolution to self-interaction problem in classical ED). $\endgroup$ – Joker_vD May 6 at 10:40
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Well a point particle is just an idealization that has spherical symmetry, and we can imagine that in reality we have some finite volume associated with the "point", in which the total charge is distributed. The argument, at least in electromagnetism, is that the spherical symmetry of the charge together with its own spherically symmetric field will lead to a cancellation when computing the total force of the field on the charge distribution.

So we relax the idealization of a point particle and think of it as a little ball with radius $a$ and some uniform charge distribution: $\rho= \rho_{o}$ for $r<{a}$, and $\rho=0$ otherwise.

We first consider the $r<a$ region and draw a nice little Gaussian sphere of radius $r$ inside of the ball. We have: $$\int_{} \vec{E}\cdot{d\vec{A}} =\dfrac{Q_{enc}}{\epsilon_{0}}$$ $$4\pi r^{2}E(r) = \frac{1}{\epsilon_{0}}\frac{4}{3}\pi r^{3}\rho_{0} \qquad , \qquad r<a$$

Now we say that the total charge in this ball is $q=\frac{4}{3}\pi r^{3}\rho_{0}$, then we can take the previous line and do $$4\pi r^{2}E(r) = \frac{1}{\epsilon_{0}}\frac{4}{3}\pi a^{3}*\frac{r^{3}}{a^3}\rho_{0}=\frac{q}{\epsilon_0}\frac{r^{3}}{a^{3}}\rho_0$$

or

$$\vec{E}(r)=\frac{q}{4\pi\epsilon_{0}}\frac{r}{a^{3}}\hat{r} \qquad,\qquad r<a$$

Outside the ball, we have the usual: $$\vec{E}(r)=\frac{q}{4\pi\epsilon_{0}}\frac{1}{r^{2}}\hat{r} \qquad,\qquad r>a$$

So we see that even if the ball has a finite volume, it still looks like a point generating a spherically symmetric field if we're looking from the outside. This justifies our treatment of a point charge as a spherical distribution of charge instead (the point limit is just when $a$ goes to $0$).

Now we've established that the field that this finite-sized ball generates is also spherically symmetric, with the origin taken to be the origin of the ball. Since we now have a spherically symmetric charge distribution, centered at the origin of a spherically symmetric field, then the force that charge distribution feels from its own field is now

$$\vec{F}=\int \vec{E} \, dq =\int_{sphere}\vec{E} \rho dV = \int_{sphere} E(r)\hat{r}\rho dV$$

which will cancel due to spherical symmetry. I think this argument works in most cases where we have a spherically symmetric interaction (Coulomb, gravitational, etc).

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  • $\begingroup$ "A point particle is just an idealization that has spherical symmetry" and the fact that it has spherical symmetry both seem to make a spherical cow assumption. $\endgroup$ – Denis de Bernardy May 5 at 19:03
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    $\begingroup$ If the sphere is in uniform motion (no acceleration), then there's a cylindrical symetry around the velocity vector. Since the electromagnetic field distribution is dipolar in this case, there's still no force exerted on the sphere by itself. But if the sphere is accelerated, there are an instantaneous velocity and acceleration vectors. These vectors destroy the spherical or cylindrical symetry, which implies that there may be an electromagnetic force. This is the origin of the radiation reaction self-force on the particle. $\endgroup$ – Cham May 5 at 19:56
  • $\begingroup$ "we can imagine that in reality we have some finite volume associated with the "point" - we have no reason to do so, though... $\endgroup$ – AnoE May 8 at 11:12
  • $\begingroup$ @AnoE the equations above demonstrate that they are equivalent as far as the electric fields they generate, which is really the only physical quantity that we have to work with that can describe the system. this tells us that these models are equivalent from an electrostatic standpoint. now, we have no reason to assume that the fundamental charges are really 0 dimensional, right? in either case, were assuming an approximate model which makes a mathematical analysis possible. whether we assume 0D or finite D, the answer will not change $\endgroup$ – Wai-Ga D Ho May 8 at 15:28
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What even is a particle in classical mechanics?

Particles do exist in the real world, but their discovery pretty much made the invention of quantum mechanics necessary.

So to answer this question, you have to set up some straw man of a "classical mechanics particle" and then destroy that. For instance, we may pretend that atoms have the exact same properties as the bulk material, they're just for inexplicable reasons indivisible.

At this point, we cannot say any more whether particles do or do not exert forces on themselves. The particle might exert a gravitational force on itself, compressing it every so slightly. We could not detect this force, because it would always be there and it would linearly add up with other forces. Instead, this force would show up as part of the physical properties of the material, in particular its density. And in classical mechanics, those properties are mostly treated as constants of nature.

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    $\begingroup$ Hello Sir, I thought a particle to be just a tiny point mass! $\endgroup$ – Unique May 7 at 9:46
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This exact question is considered at the end of Jackson's (somewhat infamous) Classical Electrodynamics. I think it would be appropriate to simply quote the relevant passage:

In the preceding chapters the problems of electrodynamics have been divided into two classes: one in which the sources of charge and current are specified and the resulting electromagnetic fields are calculated, and the other in which the external electromagnetic fields are specified and the motions of charged particles or currents are calculated...

It is evident that this manner of handling problems in electrodynamics can be of only approximate validity. The motion of charged particles in external force fields necessarily involves the emission of radiation whenever the charges are accelerated. The emitted radiation carries off energy, momentum, and angular momentum and so must influence the subsequent motion of the charged particles. Consequently the motion of the sources of radiation is determined, in part, by the manner of emission of the radiation. A correct treatment must include the reaction of the radiation on the motion of the sources.

Why is it that we have taken so long in our discussion of electrodynamics to face this fact? Why is it that many answers calculated in an apparently erroneous way agree so well with experiment? A partial answer to the first question lies in the second. There are very many problems in electrodynamics that can be put with negligible error into one of the two categories described in the first paragraph. Hence it is worthwhile discussing them without the added and unnecessary complication of including reaction effects. The remaining answer to the first question is that a completely satisfactory classical treatment of the reactive effects of radiation does not exist. The difficulties presented by this problem touch one of the most fundamental aspects of physics, the nature of an elementary particle. Although partial solutions, workable within limited areas, can be given, the basic problem remains unsolved.

There are ways to try to handle these self-interactions in the classical context which he discusses in this chapter, i.e. the Abraham-Lorentz force, but it is not fully satisfactory.

However, a naive answer to the question is that really particles are excitations of fields, classical mechanics is simply a certain limit of quantum field theory, and therefore these self-interactions should be considered within that context. This is also not entirely satisfactory, as in quantum field theory it is assumed that the fields interact with themselves, and this interaction is treated only perturbatively. Ultimately there is no universally-accepted, non-perturbative description of what these interactions really are, though string theorists might disagree with me there.

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This answer may a bit technical but the clearest argument that there is always self interaction, that is, a force of a particle on itself comes from lagrangian formalism. If we calculate the EM potential of a charge then the source of the potential, the charge, is given by $q=dL/dV$. This means that $L$ must contain a self interaction term $qV$, which leads to a self force. This is true in classical and in quantum electrodynamics. If this term were absent the charge would have no field at all!

In classical ED the self force is ignored, because attempts to describe have so far been problematic. In QED it gives rise to infinities. Renormalisation techniques in QED are successfully used to tame the infinities and extract physically meaningful, even very accurate effects so called radiation effects originating from the self interaction.

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  • $\begingroup$ A point particle charge $q$ does not have to obey equation such as $q = \partial L/\partial V$, because what is $V$ at the point of point particle? External potential? Then there is no connection between $q,V$. Total potential? Then there is connection, but $V$ is infinite at the very point you would like to apply that equation and the Lagrangian cannot depend on $V$ at that point. $\endgroup$ – Ján Lalinský May 6 at 19:14
  • $\begingroup$ @JanLalinsky Isn't that exactly the point of this question? Also, I repeat, without self interaction term the point charge has no field so it does obey such an equation. $\endgroup$ – my2cts May 6 at 20:23
  • $\begingroup$ My point is that your argument is wrong, in fact the Lagrangian does not have to contain a self-interaction term in order a charged particle to produce a field. There is a family of consistent non-quantum-theoretical theories that demonstrate this - action at a distance electrodynamics, by Tetrode, Fokker, Frenkel, Feynman and Wheeler etc. $\endgroup$ – Ján Lalinský May 6 at 21:45
  • $\begingroup$ @JanLalinsky Standard lagrangians contain self interaction or else charges would it produce fields. Calling my post "wrong" overstates your position. Although interesting, these theories are not mainstream physics. What is their status anyway? See en.m.wikipedia.org/wiki/Wheeler%E2%80%93Feynman_absorber_theory $\endgroup$ – my2cts May 6 at 21:58
  • $\begingroup$ Those theories are deficient in that they do not capture some phenomena involving charges such as pair creation/destruction. But they are an example that there is no necessity to self-interaction to have a consistent theory of interacting particles that is also consistent with macroscopic EM theory. $\endgroup$ – Ján Lalinský May 7 at 10:50
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The difficulties presented by this problem touch one of the most fundamental aspects of physics, the nature of the elementary particle. Although partial solutions, workable within limited areas, can be given, the basic problem remains unsolved. One might hope that the transition from classical to quantum-mechanical treatments would remove the difficulties. While there is still hope that this may eventually occur, the present quantum-mechanical discussions are beset with even more elaborate troubles than the classical ones. It is one of the triumphs of comparatively recent years (~ 1948–1950) that the concepts of Lorentz covariance and gauge invariance were exploited sufficiently cleverly to circumvent these difficulties in quantum electrodynamics and so allow the calculation of very small radiative effects to extremely high precision, in full agreement with experiment. From a fundamental point of view, however, the difficulties remain.

John David Jackson, Classical Electrodynamics.

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Interesting question. The majority of the present answers seems to limit the possibility of self-interaction to the case of charges, referring in a direct or indirect way to the radiation reaction force. References to self-interaction in QFT, although interesting, seem to go beyond the limits of the original question, which is explicitly in the realm of classical mechanics and also implicitly, taking into account that the concept of force is pivotal in classical mechanics, but not in QM.

Without any claim to write the ultimate answer, I would like to add a few thoughts from a more general perspective, entirely based on classical mechanics.

  1. radiation reaction, or similar mechanisms, are not truly self interaction forces. They can be seen as interaction of a particle with itself mediated by the interaction with a different system which allows a feedback mechanism. Such a feedback cannot be instantaneous, but this is not a problem: retarded potentials (and therefore retarded forces) are almost obvious in the case of electromagnetic (EM) interaction. But also without EM fields, retarded self interaction may be mediated by the presence of a continuum fluid. However, the key point is that in all those cases, the self interaction is an effect of the existence of a second physical system. Integrating out such second system, results in an effective self-interaction.

  2. A real self interaction should correspond to a force depending only on the state variables (position and velocity) and characteristic properties of just one particle. This excludes typical one-body interactions. For example, even though a viscous force $-\gamma {\bf v}$ apparently depends only on the velocity of one particle, we know that the meaning of that velocity is the relative velocity of the particle with respect to the surrounding fluid. Moreover the friction coefficient $\gamma$ depends on quantities characterizing the surrounding fluid.

  3. We arrive to the key point: a real self-interaction would imply a force acting on one isolated particle. However, the presence of such self-interaction would undermine at the basis the whole Newtonian mechanics, because it would imply that an isolated particle would not move in a straight line with constant speed. Or, said in a different way, we would not have the possibility of defining inertial systems.

Therefore, my partial conclusion is that a real self-interaction is excluded by the principles of Newtonian mechanics. On the experimental side, such non-Newtonian behavior has never been observed, at the best of my knowledge.

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  • $\begingroup$ It is not obvious why isolated point particle should move in a straight line with constant speed, or why failure of one single particle to do that would preclude our ability to define inertial systems. For example we could “dequantize” Dirac equation in such a way that there is zitterbewegung of point particles as a pure classical effect. This would probably qualify as self-interaction via state variables of single point particle (without external systems). $\endgroup$ – A.V.S. Jun 7 at 16:28
  • $\begingroup$ @A.V.S Dirac equation and zitterbewegung are not classical mechanics stuff. Maybe it could not be obvious why isolated point particle should move in a straight line with constant speed, but it is one of the modern formulation of the first principle of dynamics. If an isolated particle could self-accelerate, please, explain how would you define an inertial system. $\endgroup$ – GiorgioP Jun 7 at 20:37
  • $\begingroup$ That is why I said “dequantize” as in “build classical mechanical model of a concept usually discussed in QM context”. See e.g. here for internally self consistent models of self-accelerating point particles. If we do include self-acceleration then inertial system could be defined through postulating observers that do not self-accelerate. And it is conflating assumptions (sometimes implicit) and necessary requirements from mathematical consistency that I object to. $\endgroup$ – A.V.S. Jun 7 at 22:06

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