According to Newton's third law, the forces in an action-reaction pair must have the same magnitude and opposite directions. But do they have to be the same kind of force (gravitational, electromagnetic, strong, weak)?

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    $\begingroup$ It is always worth noting that the "equal and opposite" form of Newton's third law needs to be generalized (by including the field momentum) with the introduction of electromagnetic dynamics. $\endgroup$ – dmckee --- ex-moderator kitten Dec 2 '19 at 0:31
  • $\begingroup$ see also physics.stackexchange.com/questions/96596 on the relation of Newton's third law and momentum conservation which in turn holds for any translationally invariant force. $\endgroup$ – Andre Holzner Dec 21 '19 at 20:19

We're talking about the "two sides" of the SAME force, so it must be yes, they are of the same type.

For example, if I attract you gravitationally, then you are also attracting me gravitationally. There is only one physical source of the force (in this case, gravity) and it is pulling us both equally and oppositely. Since there's only one thing, it has, and can only have, one type.

Or, to put it another way, according to Newton's third law: When I push against you, you are also pushing me back. It is the same physical thing doing both pushes, in this case electrical repulsion of our atoms.

Or like this: Imaging I support a book of 1N weight in my hand (against earth gravity). I can't support that book without supplying 1N of force to stop it falling. And that force will be provided by the electrical repulsion of the atoms in me vs the atoms in the weight. The book feels that 1N force upwards, and I feel it downwards on my hand. There's only one force doing that. There's ALSO a gravitational attraction pulling the weight towards the earth, balanced exactly by the gravitational attraction pulling the earth towards the weight. So there are (at least) two instances of the third law in play.

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    $\begingroup$ This argument would only work if forces were always perfectly symmetrical. But that's exactly what the OP is asking about :) You're begging the question. $\endgroup$ – Luaan Dec 2 '19 at 9:39
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    $\begingroup$ but they /are/ perfectly symmetrical, that's what Newton's law says. $\endgroup$ – Andrew Kay Dec 2 '19 at 11:04
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    $\begingroup$ I meant forces in a different sense. The magnitude of the forces acting on the object must be symmetrical, but that doesn't on its own mean that one of the forces cannot be electromagnetic, while the other is gravitational, for example. "It always is" is the correct answer (at least for the usual cases), but your explanation why doesn't feel very satisfying. Your edit makes this somewhat better, but it still doesn't quite show whether this is a fundamental symmetry in the universe, or if it just happens to be the case in your example. $\endgroup$ – Luaan Dec 2 '19 at 11:14
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    $\begingroup$ "How, even logically, could it be otherwise?" – What do you mean, how could it be otherwise? It could be otherwise like this: Imagine two particles A and B, such that A is exerting a gravitational (but not an electric) force on B, while B is exerting an equal-and-opposite electric (but not a gravitational) force on A. That's impossible given the laws of physics as we know them, but it's not impossible a priori. $\endgroup$ – Tanner Swett Dec 2 '19 at 12:11
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    $\begingroup$ @Luaan When I was trying to put together my own answer, I found that in all situations where the action and reaction were "mixed," that there was a representation of the same system with two separate force pairs, each consisting of an action/reaction of the same type. However, I was having fits trying to figure out a way to prove that must always be the case. $\endgroup$ – Cort Ammon Dec 3 '19 at 19:34

If you get down to the fundamental layer, what we perceive as Force on a macroscopic scale is really just exchange of field bosons. So when two particles interact via a boson exchange, it is the same force because it is mediated by the same boson.

The only way you could see asymmetric forces would be if the boson could oscillate into another form. For example, a proton could emit a photon (boson of electromagnetic force), intending to repel another proton electrically, but then the photon could oscillate into a graviton (gravitational force boson) and thus transmit a gravitational force to the second proton.

I've never heard of this happening...

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    $\begingroup$ This is the first satisfactory answer I've seen here, actually addressing the heart of the issue. Nice! $\endgroup$ – Apollys supports Monica Dec 3 '19 at 20:48

There is no such thing as "the force of A on B." Force is a thing that only exists between objects. That's all there is to it.

An important note is that force is not what physicists consider an observable. That is, you cannot directly measure a force. In classical physics, the only observables are distance, mass, and time. In this context, we infer the force on an object by measuring its mass (observable) and acceleration (length and time observables), and then applying Newton's 2nd law.

Danger: Extremely esoteric stuff below... An interesting philosophical detail here is that we then need to determine mass without using a force (such as gravity, spring, etc.). If we accept the equivalence of gravitational and inertial mass, then we can measure the mass of an object through a collision with another object of known mass.

Addressing comment below

The justification is the following: From a practical point of view it makes total sense to say "the force of A on B." For instance, an engineer might want to know the force a bushing puts on an axle. Or you might want to calculate the torque on a compass needle from an external magnetic field. It would be dumb to ask an engineer to calculate the mutual action between an axle and its bushing. However, the OP asked a deeper, more philosophical question that requires a subtler and more nuanced answer. This requires pulling away the shroud and pointing out that an N3 force pair does not actually consist of two forces. There is ONLY a mutual force. Thus the concept of the equal and opposite forces being of a different nature is nonexistent.

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    $\begingroup$ There is no such thing as "the force of A on B." Justification? If you take this perspective, Newton's second law is meaningless anyway. You're assuming the answer before you even start your response. $\endgroup$ – Apollys supports Monica Dec 3 '19 at 20:50
  • $\begingroup$ @ApollyssupportsMonica -- It's just a more mature and nuanced understanding of N3. In fact, it's closer to how Newton stated N3: "To every action there is always opposed an equal reaction: or the mutual actions of two bodies upon each other are always equal, and directed to contrary parts." Really, we don't need N3 at all. We can just say that momentum is conserved. The point is that the force or reaction only occurs as a mutual thing. So it is only ONE kind of thing. It can never be "two different kinds of force." $\endgroup$ – abalter Dec 4 '19 at 6:08
  • $\begingroup$ Why? You're just rephrasing/elaborating what you said above, but I still see no justification. (And yeah I meant Newton's third law in my previous comment, woops.) $\endgroup$ – Apollys supports Monica Dec 4 '19 at 21:07
  • $\begingroup$ @ApollyssupportsMonica please see my edit. And feel free to ask for further clarification if you want. $\endgroup$ – abalter Dec 4 '19 at 21:41

Yes, and it is one of the “tests” that one can use to check if two forces are a Newton third law pair.

The pairing always exists.

A body on a table has two forces on it.
The contact force on the body due to the table and the gravitational force on the body due to the Earth.

Although these two forces have the same magnitude and are opposite in direction they are not a Newton third law pair for the following reasons.
They act on the same body, Newton third law pairs must act on different bodies.
One is a contact force and the other is a gravitational force.
Take the table away the contact force disappears but the gravitational force acts on the body.

The Newton third law pair to the gravitational force on the body due to the Earth is the gravitational attraction on the Earth due to the body.

The idea that the Newton third law pairing always exists cannot be satisfied if the forces are of different types.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$ – tpg2114 Dec 2 '19 at 22:44

There is no such thing as "kind of force" in pure Newtonian physic. A force is defined only by its intensity and its direction.

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    $\begingroup$ And its intensity depends on properties such charge, or mass, or displacement, etc. If you change a physical property of the system and one of the forces changes, but not another, wouldn't you say they are different kinds? $\endgroup$ – electronpusher Dec 3 '19 at 10:29

For conservative system you get the force from this equation

$$F_i=-\frac{\partial U} {\partial q_i}$$

$$F_j=-\frac{\partial U} {\partial q_j}$$

where U is the potential energy and q is the generalized coordinate, thus action force is equal reaction force. Between two parts you have the same potential energy like spring gravitation ...



$$F_1=-\frac{\partial U} {\partial x_1}=-k\,(x_1-x_2)$$

$$F_2=-\frac{\partial U} {\partial x_2}=+k\,(x_1-x_2)$$




additional remarks:

every cut force (for example constraint or friction force or whats so ever) ,must obey Newton's third law ,if not your equation doesn't described correct the physical situation.

Let say you have two body , mass equal one , connected by a rigid rod , you get the constraint forces $F_1$ and $F_2$ by "cutting" the rod. the EOM's are:

$$\ddot{x}_1=+F_1$$ $$\ddot{x}_2=-F_2$$

the physic "tell" us that if the rod is rigid, $x_1-x_2$ must be zero, thus $F_2=-F_1$ this is Newton's third law action equal reaction

  • $\begingroup$ You are calculating the total force on each particle, and with this U they just happen to balance. Which is not what the 3rd law is about, I believe. What if e.g. U = mg(x1 + x2) $\endgroup$ – Andrew Kay Dec 2 '19 at 11:09
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    $\begingroup$ Writing the force as the derivative of some random potential energy will not necessarily represent a force. Example forces like friction which are non conservative you will not get the force to be a gradient of potential energy $\endgroup$ – SK Dash Dec 2 '19 at 11:56
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    $\begingroup$ The maths is lovely, but what's the answer? Yes or no? $\endgroup$ – Lightness Races in Orbit Dec 2 '19 at 16:09
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    $\begingroup$ That should probably be in the answer. Along with a justification for how your equations prove it, if that is your intent. $\endgroup$ – Apollys supports Monica Dec 3 '19 at 1:30
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    $\begingroup$ This would be a better answer if (a) it clearly stated the answer, (b) 2/3 of the material was cut because it's unnecessary, and (c) it was at the level of the OP rather than showing off with generalized coordinates, etc. $\endgroup$ – user4552 Dec 3 '19 at 15:06

Whether or not the forces must have the same flavour is a choice of definition.

Two particles (A and B) exert a force on one another Newton tells us that the magnitudes of those two forces will be equal but their directions will be opposite. The question is, are the two forces of the same "flavour" (Electromagnetic, gravitational).

Suppose that their WERE two particles that did infact have an interaction where the a force on one from some source (say Electric) was matched by an equal and opposite force on the other from a different flavour (gravity). If this were ever experimentally observed one could draw one of two conclusions:

  • The reaction force has a different flavour to the action force in this specific case
  • The two forces we thought were fundamentally different actually have the same flavour.

Which one you pick is just words and definitions, their is no physics different between them. I think people would more likely take the second definition.

The same logic can be turned on its head. Imagine that someone who learned physics in a weird way tells you about two different flavours of force they learned about: one is "The negative E force" that pulls negative charges towards positive ones, and the other is a completely different force of a different "flavour" called "the Positive E force" that pulls positive charges towards negative ones.

This person agrees with you about what happens when a positively charged particle interacts with a negatively charged one (they attract each other). The only difference is the language they use (how they cut up the "flavours").


Newton's third law says nothing about the type of forces involved, so in itself it does not require the forces to be of the same type; however, they must be because the four fundamental forces act independently of each other. For example, two masses will attract each other gravitationally, each having an equal but opposite impact upon the other. If they are both positively charged they will also repel each other electrically, and the repulsion will affect each of them equally.


Sorry, but I will go against "mainstream view" here. Yes, origin of forces can be different in a Newton third law force pair. It also depends how we will define a "force". For example, if we will define object weight in operational terms,- weight is the force it exerts on it's support. So, consider this situation of body lying on table:

enter image description here

Then weight of body acts on table surface, and this induces reactive normal force of table which acts on body bottom surface. Thus, these forces acts on different bodies and is a perfect candidates for Newton third law force pair. However their nature is completely different. Weight has gravitational origin and reactive normal force has inter-molecular Van der Waals force origin of table-body surface contact. So they roots are completely different.

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    $\begingroup$ The weight of the body is basically the force of gravity, which acts on the body. The normal force you have indicated also acts on the body. So, these 2 forces do not form an action-reaction pair. $\endgroup$ – Thomas Dec 3 '19 at 8:36
  • $\begingroup$ Have you read my post fully ? I said, that it depends how you choose metrics. I chose weight as "the force it exerts on its support". So, No in such definition weight acts on table surface not on body. Van der Waals force also originates at table-body contact surface. Van der Waals force doesn't care about body COM. It only affects contact points. However, later this normal force is "transferred" to body COM by it's lattice. We just draw these forces starting at COM, because it is comfortable to do so in free body diagram $\endgroup$ – Agnius Vasiliauskas Dec 3 '19 at 8:50
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    $\begingroup$ The contact force the body exerts on the table and the contact force the table exerts on the body indeed form a Newton's third law pair. But I don't think that's what you were getting at. $\endgroup$ – Thomas Dec 3 '19 at 8:53
  • $\begingroup$ But I do. No need to say The contact force the body exerts on the table, because it is simply same body weight expressed in operational terms. I.e. just say "weight" and you will get my picture. I agree that gravitational force is a different thing, which acts on body COM, but I'm not talking about it here. $\endgroup$ – Agnius Vasiliauskas Dec 3 '19 at 8:59
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    $\begingroup$ This is wrong. For example, if I accelerate the table upward, then N and W will be unequal. This is not a violation of the 3rd law because those aren't the forces involved in the third law. Another way to see that this is wrong is that the 3rd law involves A acting on B and B acting on A. In your pair of N and W, you have 3 objects involved. $\endgroup$ – user4552 Dec 3 '19 at 15:04

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