Does a particle exert force on itself?

Question

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 have edited this question in order to elaborate on it.

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?

Answer 1

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!

Answer 2

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).

Answer 3

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...

Answer 4

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.

Answer 5

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.

Answer 6

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.

Answer 7

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.

Answer 8

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.