# When we say a rigid body is a system of particles, what exactly are 'particles' here?

In Newtonian mechanics, a particle (in my knowledge) is a point-like mass with no shape and size, deformation, rotation and internal movements, which is an idealized model of an object which does have shape and size and can have deformation, rotation and internal movements.

Newton's Laws of Motion (in my knowledge) are originally given for particles but can be extended for extended bodies, like for a rigid body.

To derive Newton's Second Law for a rigid body we can consider (as my textbook says) it to be a collection of particles.

I do not understand that when we say a rigid body is a system of particles, what are we referring to 'particles' here? Atoms? a differential element of the material? or something else?

As an extension of this question, can we consider a deformable body as a collection of particles and develop the Newton's Second Law for it as well? If yes then what will be the 'particle' in the case of deformable body?

• Don’t mistake the model for the thing that is modeling
– wim
Jun 8 at 16:04
• Particles are Newton's least problem. Newtonian physics are utterly wrong on a very fundamental level (because there is no absolute time, interaction is not immediate, space is not Cartesian, take your pick). But that doesn't matter: If what you look at is not too small and not too large, the time frame is not too long and not too short, the masses are not too large etc. -- that is, as long as you are on roughly a human scale: It is a valid simplification whose results are good enough. Jun 8 at 20:30

TL;DR Never forget that we are talking about models here. We know that they cover only part of reality, but they are good enough and much easier to handle than "the full story".

A model need not correspond to each and every aspect of reality, only to those that we are interested in, that we need for our calculations. We want models that allow us to easily compute results that match reality to a degree that satisfies our needs.

So, if we talk about rigid bodies in Newtonian mechanics, we are interested in their movement and rotation as caused by external forces.

You mention the "wiggling of atoms" at the microscopic level (thermodynamics). You are absolutely correct. In an apple falling down from a tree, you don't have all atoms moving parallel, all with the same speed and direction, but instead each atom with its own chaotic movement.

We observed in many experiments that rigid macroscopic bodies react to forces the same way, regardless of their temperature (the amount of atom wiggling). So, a model of non-moving atoms in itself only represents absolute zero temperature, but we found out that it also matches our observations at higher temperatures. So, although we know that in reality we have a chaotic movement of particles, we can still use a model without that micro-level movement and get the correct result. This is an experiment-based justification for ignoring the "wiggling", but we can also do it mathematically.

If we take this chaotic micro-movement into account, we need to use statistics. And after some higher-mathematics calculations, we'll find out that, if we have enough atoms, the chaotic atom wiggling cancels itself out of all our formulas. The expected difference between the zero-temperature model and the wiggling-atoms model (as calculated based on a statistics model) quickly falls below any margin relevant for real life.

• Thanks Ralf. From what I understand, you say that I can consider even an atom, a particle, when talking about a system of particles like rigid body, apart from considering an infinitesimal element as a particle. Correct? I can model atoms as non-moving entities because their movement doesn't affect the motion of the rigid body. Jun 9 at 16:23
• Yes, you can sub-divide your rigid body into "particles" of any size you like, and treat the particles as point-masses. The smaller the "particle size", the better the approximation of the real-world results. When going down to atom-sized particles, you'll surely get a perfect match. Jun 10 at 8:42

In macroscopic mechanics (in other words, "usual" mechanics), a "particule" is a mesoscopic system.

The mesoscopic scale is an intermediary scale between microscopic and macroscopic such that:

• It's big enough to contain a large number of atoms/molecules, so their quantum nature mostly averages out and can be ignored.
• It's small enough to be point-like from our point of view.

From this comes the usual point of view of macroscopic physics, that matter is continuous. This is a good approximation, as most of physics until the 19th century showed, but it has its limits (when quantum effects start to get noticeable, usually at the atomic scale).

It's customary to use Avogadro's number to separate the different scales. If $$N$$ is the number of particles, the system is considered mesoscopic if:

$$1\ll N\ll\mathcal{N}_A$$

When you have something like "Let $$dV$$ be a very small volume..." in classic physics, it's usually a mesoscopic system.

• Thanks for the response. I see, so the particle is an infinitesimal volume, a mesoscopic system, when used in the context of rigid bodies. If we consider a deformable body, let's say, then will a particle still be an infinitesimal volume? Because now this infinitesimal volume undergoes deformation, can it be considered as a particle? Jun 7 at 15:17
• Yes. In most cases, a mesoscopic volume, being point-like, doesn't undergo deformation. In fact, in fluid mechanics, a "fluid particle" is a mesoscopic volume. The deformation comes from those particules moving relative to each others. Of course it's an approximation, but in macroscopic physics it's fine. Jun 7 at 15:20
• So, we can consider a differential element taken in a deformable body to be a particle (which undergoes no deformation, an approximation) when we are interested in only the motion of the particle. Even though sub-elements in this element will move at different velocities but these velocities will be very close to each other so that I can assume the entire element moves with the same velocity. If however, we were interested in studying the deformation of the element we wouldn't have considered it as a particle. That looks right? Jun 7 at 16:32
• This sounds more complicated than it really is, but yes. ;-) That's hidden behind the expession "averaged out" in my answer. The microscopic motions of the molecules are assumed to have no macroscopic effect (an assumption well-verified in practice) so they're ignored and only the "drift velocity" remains. Jun 7 at 16:35

The Newton mechanics of the "rigid body" model does not recognize any "natural" particles of a body (as e.g. atoms), nor any other internal body structure at all.

A finite-sized rigid body is assumed as being (at least, logically) divisible to infinitesimally small pieces that can be considered particles for the purposes of the Newton laws.

These pieces are assumed to carry the basic properties of the body (e.g. density) in a known manner (e.g. a constant density over the whole body).

Armed with this division, we can use a mathematical toolset known as integration in order to derive the whole body's behavior from the known behavior of its constituent particles.

• This. Newtonian physics are agnostic of the internal structure of matter. And for good reason: It doesn't, well, matter. Jun 8 at 20:26

We could consider any of those as valid options. The requirement is that classical physics describes the motion of these particles. Keep in mind that a rigid body can be described as a collection of particles with relative distances that do not change. After applying some constraints on the degrees of freedom of the whole collection you should end up with 3 displacements and 3 angles to describe the position and orientation of your rigid body.

• Thanks for the response. If we consider the atoms as particles (when analyzing rigid bodies) then apart from the motion that they get because the body (of which they are part of ) moves as a whole, they also wiggle due to the kinetic energy at the microscopic level. Wouldn't considering atoms as particles complicate our analysis? Jun 7 at 15:10
• See the other answer. We are in search of models, trying to solve problems. Jun 7 at 16:06

A particle is a small enough mass not to have any rotational properties, yet be able to carry momentum according to $$\boldsymbol{p}_i = m_i \boldsymbol{v}_i$$.

At the same time for a rigid body, all distances between all particles must remain constant, allowing only rotations between particles (meaning if you fix one particle all others will rotate about this one along a shared rotation axis).

Newtonian mechanics deals with a particle-like object/material point/point-like object/etc., which has negligeably small dimensions and whose internal structure can be ignored. Newton laws are formulates specifically for such objects - hence the insistence of the physics teachers on drawing all the force vectors coming from a single point.

Rigid bodies are then collections of such material points/particles, and the forces now can be applied to different particles within the same body (and the physics teacher will again insist on correctly drawing their application points.)

To summarize: here particle is a concept of Newtonian mechanics, unrelated to the elementary particles, which are the physical entities constituting the Universe, atoms of gas theory or any other true physical entities. Newtonian particle can be anything: an electron, an atom, a car, an elephant, a planet or a galaxy.

The rigid body can be said to be a collection or a system of particles because these particles are the fundamental units of a rigid body. Here the word 'particles' is referring to the atoms,molecules or the smallest possible unit that makes up that rigid body which is essentially atoms binded together. I can recommend you to read a great book on physics which will help you clarify your concepts which is 'Concepts of Physics' by the eminent Indian Phsysicist HC Verma. Good Luck !!!

• Considering atoms as particles will not complicate our analysis as you should know this fact when the body is in motion as a whole we can only tell the total kinetic energy of all the individual particles(i) KE = mi * vi(vector) dot vi(vector).We can never tell the kinetic energy of the individual particles.That's why no complications in our analysis. Jun 7 at 16:50
• Sorry, but this is incorrect. Of course, in reality macroscopic systems are made of atoms. But rigid body mechanics is a theory that existed long before quantum mechanics was discovered... If atoms were the "particules" of this mechanics, then quantum knowledge would be necessary. Macroscopic mechanics simply relies on a continuous model of matter, which works well (obviously), although it was its limitations. Jun 7 at 16:51
• The formula you're giving in your comment isn't valid for atoms. See quantum mechanics. It is, however, valid for the particles of rigid body mechanics, which are mesoscopic objects. Jun 7 at 16:53
• But this continuous model doesn't get complicated when sum of all the unit particles' Kinetic Energy is added it leads to the rigid body.And to tell you miss, the above texts which I have written are an excerpt from the famous textbook 'Concepts of Physics' by a famous Indian Physicist. Jun 7 at 17:07
• Please stop doing some advertisement for this physicist, it brings nothing to the discussion. Jun 7 at 17:10