I How a rigid body causes a reaction force?

AI Thread Summary
The discussion centers on the relationship between rigid bodies and reaction forces, particularly in the context of Newton's third law. It is argued that while rigid bodies theoretically do not deform, the action-reaction principle can be understood through molecular interactions and the conservation of momentum. Participants highlight that Newton's third law is fundamental and independent of specific mechanisms, yet it can be illustrated through scenarios like pushing objects in zero gravity. The conversation also touches on Noether's theorem, suggesting that conservation laws are linked to symmetries in physics, and clarifies that Newton's third law implies conservation of momentum, rather than the reverse. Overall, the dialogue emphasizes the complexities of understanding forces and motion in both classical and modern physics contexts.
Adesh
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The definition of rigid body says it cannot be deformed (theoretically). Now, Newton’s third law is caused (I mean the reaction force is caused ) due to the deformation of the body.

What I have learned is that every body is like a spring, when we push on it we compress it and hence feel a reaction force (equal to action force).

At molecular level, when our hands comes closer to a rigid body the molecules of our hand and the rigid body apply Coulombic force on each other and it’s perfectly understandable how action and reaction is there. But can we explain it without going to molecular level? I mean can we explain it just by using that concept of deformation.

Rigid body doesn’t deform on being applied with a force, then how it pushes back to pusher?
 
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Adesh said:
The definition of rigid body says it cannot be deformed (theoretically). Now, Newton’s third law is caused (I mean the reaction force is caused ) due to the deformation of the body
No no no. You get the action-reaction from a simple subatomic particle which does not deform.
 
anorlunda said:
No no no. You get the action-reaction from a simple subatomic particle which does not deform.
Means action and reaction can be explained only by that coulombic attraction/repulsion at molecular level?
 
Adesh said:
Means action and reaction can be explained only by that coulombic attraction/repulsion at molecular level?
No, it actually means that momentum is conserved. In this lecture, Professor Susskind does the math to show that the 3rd law is equivalent to conservation of momentum. No need to invoke gravity or Coulomb forces, it works for neutral particles too.
 
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anorlunda said:
No, it actually means that momentum is conserved. In this lecture, Professor Susskind does the math to show that the 3rd law is equivalent to conservation of momentum. No need to invoke gravity or Coulomb forces, it works for neutral particles too.
So, can you please explain me How reaction force is actually caused?
 
I don't know how much physics you have already. This is an I level thread.

Have you studied Noether's Theorem yet? If yes, then you can show that position translation symmetry leads to conservation of momentum.
 
anorlunda said:
I don't know how much physics you have already. This is an I level thread.

Have you studied Noether's Theorem yet? If yes, then you can show that position translation symmetry leads to conservation of momentum.
Wow! Today I learned a great thing from you, it’s the conservation of momentum that causes the Newton’s third law, not the deformation.

If we ignore the friction, if I push on a rigid body, it will start to move and since the initial momentum of the system (system is me and that rigid body , and I’m assuming everything was at rest initially) was zero therefore the final momentum also needs to be zero and hence I must move opposite to the rigid body and this motion must be caused by something and that something is the reaction force. Am I right and precise ?
 
Adesh said:
The definition of rigid body says it cannot be deformed (theoretically). Now, Newton’s third law is caused (I mean the reaction force is caused ) due to the deformation of the body.

What I have learned is that every body is like a spring, when we push on it we compress it and hence feel a reaction force (equal to action force).

At molecular level, when our hands comes closer to a rigid body the molecules of our hand and the rigid body apply Coulombic force on each other and it’s perfectly understandable how action and reaction is there. But can we explain it without going to molecular level? I mean can we explain it just by using that concept of deformation.

Rigid body doesn’t deform on being applied with a force, then how it pushes back to pusher?
First, Newton's third law is an incredible insight. When you push or pull something, you don't have a natural sense that it is pushing or pulling you (or at least I don't).

The first thing to realize is that this is a law of nature. It's fundamental to the concept of force. And, it's essentially independent of the mechanism. The third law doesn't come with a defined mechanism that applies in all cases.

Like any law of nature, you can then look at specific physical scenarios and analyse how the law is enforced. A force generated by physically pushing something must, ultimately, have a mechanism at a molecular level. But, then, the electrostatic forces themselves may be explained by a mechanism (quantum mechanical or otherwise).

In short, the third law doesn't come with a explanation of how nature obeys the law in all cases.
 
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PeroK said:
In short, the third law doesn't come with a explanation of how nature obeys the law in all cases.
Yes, that’s a nice explanation. In Electromagnetism Newton’s 3rd law is no surprise but in macroscopic bodies it is a surprise (if we don’t assume conservation law to be more fundamental to Newton’s third law).
 
  • #10
Adesh said:
I must move opposite to the rigid body and this motion must be caused by something and that something is the reaction force. Am I right and precise ?
Yes.

Have you seen movies of people in zero gravity? For example people on the space station. In zero gravity it is easier to see that if you push an object, that you start moving the opposite direction, thus keeping the system (object and you) with constant momentum.

On Earth, friction often dominates making it hard to see.
 
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  • #11
anorlunda said:
Yes.

Have you seen movies of people in zero gravity? For example people on the space station. In zero gravity it is easier to see that if you push an object, that you start moving the opposite direction, thus keeping the system (object and you) with constant momentum.

On Earth, friction often dominates making it hard to see.
Yes, I have seen them. Thank you for teaching me that.
 
  • #12
anorlunda said:
In this lecture, Professor Susskind does the math to show that the 3rd law is equivalent to conservation of momentum
I intend to watch the video later, but I am a bit skeptical. If Newton's 3rd laws is equivalent to conservation of momentum then you could derive it starting from Newton's 2nd law (I think). It doesn't seems good because it would make Newton's 3rd law essentially redundant (because it would be a special case of the 2nd law).

An example: consider the scenario when you push something (like forniture) and it doesn't move. How can you "derive" Newton's 3rd law in this scenario using only the second law ? I can't find a way.
 
  • #13
dRic2 said:
If Newton's 3rd laws is equivalent to conservation of momentum then you could derive it starting from Newton's 2nd law (I think).

Starting with Newton's Third Law you can derive conservation of momentum. But they are not equivalent because you cannot derive Newton's Third from conservation of momentum.

And you can't derive it from Newton's 2nd Law.
 
  • #14
Adesh said:
What I have learned is that every body is like a spring, when we push on it we compress it and hence feel a reaction force (equal to action force).

This is true in Newtonian physics for the so-called contact forces like the normal force and the friction force. Both objects deform, so there is nothing to distinguish the so-called action from the reaction.

When a book rests on a table top, the book deforms the table top and the table top deforms the book. The book exerts a force on the table top and the table top exerts a force on the book. Either force could be called the action, either could be called the reaction.

In this sense, there really is no such thing as a truly rigid body. We very often model objects as rigid bodies because their deformation is negligibly small compared to other distances involved.

At molecular level, when our hands comes closer to a rigid body the molecules of our hand and the rigid body apply Coulombic force on each other and it’s perfectly understandable how action and reaction is there.

The molecules move closer together or further apart in response to those electromagnetic forces. That is the deformation.
 
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  • #15
Mister T said:
And you can't derive it from Newton's 2nd Law.
My bad. I misunderstood the statement made by @anorlunda
 
  • #16
dRic2 said:
Can't you derive conservation of momentum from the second law when F = 0?
Newton's First Law (not the 3rd law) is the special case of F=ma=0.
 
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  • #17
dRic2 said:
My bad. I misunderstood the statement made by @anorlunda

There's more to the 1st Law. It has, over the centuries, expanded it's limits of validity. At first it was a declaration that a state of uniform motion is equivalent to a state of rest. Since then it's been discovered that that idea is valid for electromagnetic phenomena as well as mechanical. The idea is now called the Principle of Relativity. Stated formally, it's the assertion that all inertial reference frames are equivalent. All that means, though, is that there is no way to distinguish between a state of rest and a state of uniform motion.
 
  • #18
There are other expert threads that dig deeply in the nuances of Newton's Laws. This is an "I" level thread. Let's keep it that way.
 
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  • #19
Wait a moment! I understood from @anorlunda replies is that Newton’s Third Law is just a weaker form of Conservation of momentum. Although, in many books Conservation of Moemntum is derived from Newton’s Third Law but the reality is the other way round. Reaction force comes into play in order to ensure the momentum is conserved.

P.S. :- It’s now that I have received the notifs and by now @MisterT and @dRic2 have replied and reached a conclusion.
 
  • #20
dRic2 said:
I intend to watch the video later, but I am a bit skeptical. If Newton's 3rd laws is equivalent to conservation of momentum then you could derive it starting from Newton's 2nd law (I think). It doesn't seems good because it would make Newton's 3rd law essentially redundant (because it would be a special case of the 2nd law).

An example: consider the scenario when you push something (like forniture) and it doesn't move. How can you "derive" Newton's 3rd law in this scenario using only the second law ? I can't find a way.
You can reconstruct Newton's Laws from the fundamental symmetries of Galilei-Newton spacetime using Noether's theorem. This elucidate the issue very much and, even better, it also can be generalized to the special relativsistic case, where Newton's 3rd Law cannot hold, because there are no instantaneous actions at a distance, but momentum conservation of course holds, because Minkowski spacetime is invariant under spatial translations as is Galilei-Newton spacetime. This implies the convenience of field descriptions and locality of interactions in relativistic physics.
 
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  • #21
Adesh said:
... in many books Conservation of Moemntum is derived from Newton’s Third Law but the reality is the other way round...
Newton’s Third Law implies Conservation of Momentum, but not the other way round.
 
  • #22
OK I misspoke when I implied that the relationship between the 3rd law and conservation of momentum was bidirectional. My bad for oversimplifying.

But this is still an I level thread. Let's keep the level of replies and the vocabulary at the I level.

The original question had to to with the origin of the reactive forces.
 
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  • #23
anorlunda said:
OK I misspoke when I implied that the relationship between the 3rd law and conservation of momentum was bidirectional. My bad for oversimplifying.

But this is still an I level thread. Let's keep the level of replies and the vocabulary at the I level.

The original question had to to with the origin of the reactive forces.
So, can we start once again?
 
  • #24
A.T. said:
Newton’s Third Law implies Conservation of Momentum, but not the other way round.
Of course momentum conservation only implies Newton's 3rd law if you have instantaneous actions at a distance (see also my previous posting).
 
  • #25
Adesh said:
So, can we start once again?
On a fundamental level all forces are through interactions, and in Newtonian physics you tacitly assume actions at a distance, i.e., forces (often derivable from a gradient, i.e., being conservative forces) are only given by the positions of the particles. It's also almost always a good approximation to assume that there are only pair interactions, i.e., in a many-body system the interaction forces are a some over two-body forces.

The paradigmatic example is Newton's gravitational force. For two bodies you have
$$\vec{F}_{12}=-\frac{G}{|\vec{x}_1-\vec{x}_2|^3} (\vec{x}_1-\vec{x}_2), \quad \vec{F}_{21}=-\vec{F}_{12},$$
i.e., the force goes like the the inverse square of the distance and is directed along the line connecting the bodies, being always attractive.

If you have a closed system, from Newton's 2nd Law you immediately get momentum conservation:
$$m_1 \ddot{\vec{x}_1}=\vec{F}_{12}, \quad m_2 \ddot{\vec{x}}_2=\vec{F}_{21}=-\vec{F}_{12}.$$
Defining the total momentum of the system as
$$\vec{P}=m_1 \dot{\vec{x}}_1 + m_2 \dot{\vec{x}}_2,$$
you indeed find
$$\dot{\vec{P}}=m_1 \ddot{\vec{x}}_1 + m_2 \ddot{\vec{x}}_2 = \vec{F}_{12} + \vec{F}_{21}=0,$$
i.e., the total momentum is conserved.
 
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  • #26
vanhees71 said:
On a fundamental level all forces are through interactions, and in Newtonian physics you tacitly assume actions at a distance, i.e., forces (often derivable from a gradient, i.e., being conservative forces) are only given by the positions of the particles. It's also almost always a good approximation to assume that there are only pair interactions, i.e., in a many-body system the interaction forces are a some over two-body forces.

The paradigmatic example is Newton's gravitational force. For two bodies you have
$$\vec{F}_{12}=-\frac{G}{|\vec{x}_1-\vec{x}_2|^3} (\vec{x}_1-\vec{x}_2), \quad \vec{F}_{21}=-\vec{F}_{12},$$
i.e., the force goes like the the inverse square of the distance and is directed along the line connecting the bodies, being always attractive.

If you have a closed system, from Newton's 2nd Law you immediately get momentum conservation:
$$m_1 \ddot{\vec{x}_1}=\vec{F}_{12}, \quad m_2 \ddot{\vec{x}}_2=\vec{F}_{21}=-\vec{F}_{12}.$$
Defining the total momentum of the system as
$$\vec{P}=m_1 \dot{\vec{x}}_1 + m_2 \dot{\vec{x}}_2,$$
you indeed find
$$\dot{\vec{P}}=m_1 \ddot{\vec{x}}_1 + m_2 \ddot{\vec{x}}_2 = \vec{F}_{12} + \vec{F}_{21}=0,$$
i.e., the total momentum is conserved.
When you wrote ##F_{21}=-F_{12}## did you use Newton’s Third Law?
 
  • #27
Yes: ##\vec{F}_{12}## is the interaction force on body 1 and ##\vec{F}_{21}## the one on body 2. According to Newton III, ##\vec{F}_{21}=-\vec{F}_{12}##.
 
  • #28
vanhees71 said:
Yes: ##\vec{F}_{12}## is the interaction force on body 1 and ##\vec{F}_{21}## the one on body 2. According to Newton III, ##\vec{F}_{21}=-\vec{F}_{12}##.
How the second body exerted force on the first body (if it cannot be deformed)?
 
  • #29
I'm not sure, what you are after here. Any body can be deformed. There is no strictly rigid body in nature, which follows alone from relativity, because the velocity of sound is ##<c## and not ##\infty##.

If you mean contact forces, it's not that easy, because you have to take into account quantum mechanics. In everyday life the interactions among bodies are solely electromagnetic. If two bodies are in contact with each other you have in good approximation electrostatic repulsion and also the Pauli effect due to the indistinguishability of the fermionic electrons.
 
  • #30
vanhees71 said:
I'm not sure, what you are after here. Any body can be deformed. There is no strictly rigid body in nature, which follows alone from relativity, because the velocity of sound is ##<c## and not ##\infty##.

If you mean contact forces, it's not that easy, because you have to take into account quantum mechanics. In everyday life the interactions among bodies are solely electromagnetic. If two bodies are in contact with each other you have in good approximation electrostatic repulsion and also the Pauli effect due to the indistinguishability of the fermionic electrons.
So, rigid body do deform and hence apply a reaction force?
 
  • #31
Adesh said:
So, rigid body do deform and hence apply a reaction force?
Objects are made of atoms and they do deform. Rigid Body is an approximation of reality. This is important to understand. A perfectly rigid body does not exist, so the question "how can a rigid body apply a reaction force if it can't be deformed?" is meaningless. In the approximation of Rigid Bodies, Objects are perfectly rigid and do not deform, yet they apply somehow reaction forces. You don't care how that happens. If you wish to investigate further you have to abandon the hypothesis of rigid bodies and enter in the realm of elasticity. Which is a total different subject.
 
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  • #32
vanhees71 said:
Of course momentum conservation only implies Newton's 3rd law if you have instantaneous actions at a distance (see also my previous posting).
But in Newtonian Mechanics, can't you obtain conservation of momentum from the first and the second law ?
 
  • #33
How so? You have to make an assumption about the forces to get momentum conservation. The 1st and 2nd law don't tell anything about the forces!
 
  • #34
vanhees71 said:
Of course momentum conservation only implies Newton's 3rd law if you have instantaneous actions at a distance (see also my previous posting).
Even for instantaneous interactions momentum conservation doesn't imply Newton's 3rd law, as soon you have 3 or more interacting bodies.
 
  • #35
Can you give an example? I don't see this so easily.
 
  • #36
vanhees71 said:
How so? You have to make an assumption about the forces to get momentum conservation. The 1st and 2nd law don't tell anything about the forces!
I naively think that if you have a system of particles Newton's 2nd law says ##\sum_i \dot p_i = F_{tot}## so, assuming the system to be isolated, you get that the total momentum is conserved. In a scattering event this implies ##(m_1v_1 + m_2v_2)_1 = (m_2v_2 + m_2v_2)_2##. Do I have some terrible misconception ?
 
  • #37
But how do you get that ##\vec{F}_{\text{tot}}=0## if you don't assume this through the 3rd Law?
 
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  • #38
ah I see. My life was a lie up until now.
 
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  • #39
A.T. said:
Even for instantaneous interactions momentum conservation doesn't imply Newton's 3rd law, as soon you have 3 or more interacting bodies.
vanhees71 said:
Can you give an example? I don't see this so easily.
This satisfies momentum conservation, but violates Newton's 3rd law:
F12 = 1N, F21 = -2N,
F13 = 2N, F31 = -2N,
F23 = 2N, F32 = -1N
 
  • #40
Adesh said:
hence apply a reaction force?
It is not clear what you mean by "reaction force".

It sounds like you are thinking of Newton's third law as some kind of manifestation of Hooke's law -- that a force exerted on an object causes a deflection, that a deflection causes a stress and that the stress results in a "reaction" force. But Newton's third law does not have that sort of mechanical explanation. It is far more basic.

It is a statement that one object cannot influence another without itself being influenced. This does not happen as a result of a deflection. It happens during a deflection. It happens (for forces at a distance) even without a deflection.

The term "reaction" is unfortunate. It suggests that the force of one object on the other is a cause and the force of the other object on the one is an effect. The truth (at least in the Newtonian model) is that both forces occur together. Neither causes the other.
 
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  • #41
Adesh said:
So, rigid body do deform and hence apply a reaction force?
If I run fast towards a concrete wall and collide with it, I am sure the wall will exert noticeable and painful reactive force on my body without suffering noticeable deformation itself.

When we put sustancial compression force on a bearing steel ball, it does not deform much; actually, it creates noticeable dents on flat surfaces of hardened steel that compress it.
The ball may deform some, but not much even when compressed with 20 tons of force.
At point of violent rupture, the multitude of sparks and speed of flying debris suggest that a huge internal stress had been created at molecular level.

 
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  • #42
Adesh said:
So, rigid body do deform and hence apply a reaction force?
I meanwhile lost the focus here, so it the problem that you want
to understand Newtons Laws? Or is is the basic conception of a rigid body?
Sorry, but I would love your update, what is the problem?
 
  • #43
Omega0 said:
I meanwhile lost the focus here, so it the problem that you want
to understand Newtons Laws? Or is is the basic conception of a rigid body?
Sorry, but I would love your update, what is the problem?
My problem is how a Newton’s 3rd Law is caused? I mean if I apply a force on some block then how it exerts force on me? For Gravitational force and Electromagnetic forces “action-reaction” is quite obvious but on macroscopic level how can we give an explanation?

By the way, I have received so many great replies.
 
  • #44
A.T. said:
This satisfies momentum conservation, but violates Newton's 3rd law:
F12 = 1N, F21 = -2N,
F13 = 2N, F31 = -2N,
F23 = 2N, F32 = -1N
Well, but does it fulfill all the requirements of a proper dynamical law within Galilei-Newton spacetime? I doubt it from using Noether's theorem, which leads to
$$V=\frac{1}{2} \sum_{i \neq j} V_2(|\vec{x}_i-\vec{x}_j|),$$
as long as only two-body forces are in the game.
 
  • #45
Adesh said:
For Gravitational force and Electromagnetic forces “action-reaction” is quite obvious but on macroscopic level how can we give an explanation?
In the Newtonian model, forces, masses, displacements, velocities and accelerations all combine linearly. If you can accept Newton's third law at the microscopic level, it is a mathematical fact that Newton's third law at the macroscopic level follows.

The force of one macroscopic object on another is the sum of the forces of all the microscopic components of the one object on all the microscopic components of the other.
 
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  • #46
vanhees71 said:
Well, but does it fulfill all the requirements of a proper dynamical law within Galilei-Newton spacetime?
The question was if momentum conservation alone implies Newton's 3rd law. With additional requirements you can obviously deduce Newton's 3rd law.
 
  • #47
That's an interesting question indeed. I.e., you ask for the most general two-body force law if you consider only translation invariance. Of course, using the argument of the Noether theorem you also need to assume that you have a conservative interaction-force law. Then I think also Newton III follows, because the two-body potential is ##V_2(\vec{x}_1,\vec{x}_2)##, where ##\vec{x}_1## and ##\vec{x}_2## are the position vectors of the two particles. Of course you can uniquely rewrite it as a function ##V_2'(\vec{R},\vec{r})## with ##\vec{R}=\vec{r}_1+\vec{r}_2## and ##\vec{r}=\vec{x}_1-\vec{x}_2##. Now to make the Hamiltonian invariant under the translation by an arbitrary vector ##\vec{a}##, ##\vec{x}_1 \rightarrow \vec{x}_1`+\vec{a}## and ##\vec{x}_2 \rightarrow \vec{x}_2+\vec{a}##. This implies that ##V_2'(\vec{R},\vec{r})=\tilde{V}_2(\vec{r})## and thus
$$\vec{F}_{12}=-\vec{\nabla}_1 \tilde{V}_2(\vec{r})=+\vec{\nabla}_2 \tilde{V}_2(\vec{r})=-\vec{F}_{21}.$$
 
  • #48
jbriggs444 said:
In the Newtonian model, forces, masses, displacements, velocities and accelerations all combine linearly. If you can accept Newton's third law at the microscopic level, it is a mathematical fact that Newton's third law at the macroscopic level follows.

The force of one macroscopic object on another is the sum of the forces of all the microscopic components of the one object on all the microscopic components of the other.
Yeah, I know that the contact force is actually Electromagnetic force between the constituent molecules.

But I think at the time of Newton this fact was quite unknown. So, is there any explanation why Newton’s Third Law occurs/exists purely on the basis of macroscopic level?
 
  • #49
A.T. said:
The question was if momentum conservation alone implies Newton's 3rd law. With additional requirements you can obviously deduce Newton's 3rd law.
vanhees71 said:
That's an interesting question indeed. I.e., you ask for the most general two-body force law...
Two-body law is an additional restriction, beyond momentum conservation. You could have a three-body force law that satisfies momentum conservation, but not Newton's 3rd law.
 
  • #50
Have a look at this:

https://www.physicsforums.com/threa...y-when-carefully-analysed.979739/post-6263544

and the quoted manuscript about "many-body forces". Also there translation invariance leads to momentum conservation and an extended kind of Newton's 3rd postulate. Of course you are right, all that relies on Hamilton's principle and Noether's theorem and you may find forces that violate the 3rd law and momentum conservation. The only question is whether they describe anything in nature (as far as Newtonian mechanics can be used as an approximation).
 
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