Forces and numbers

With coronavirus forcing people's confinement around the world, many of us are reminded of the time when Isaac Newton flew away from the bubonic plague to seek refuge in a farm belonging to his family and during the two years of isolation that followed, he produced genial work. I don't have his genius nor the time he disposed of, since unfortunately I keep teleworking and I am also dispersed with other studies, but I will at least share some thoughts on Newton's laws of motion that have been haunting me these days. No discovery at all, of course, but at least a way of presenting things that I deem pedagogical and is not usually mentioned.

The gist of the message is that the strength of a force is not something that is given prior to the interaction between two objects, but depends on how much of the respective masses gets involved. One could simplify this as a question of the "number of soldiers being called to service". A question that requires two steps: 

• only if the acting force is gravity, then the number is 100%, because gravitational force gives particles no option to take part or not in the war; things are transparent to gravity, so the force affects all its targets equally, at least in local phenomena where the field is uniform; 
• in other cases, like in contact forces, the force triggers on the constituents of a body a chain reaction, so how many particles are involved depends on their readiness to cooperate in the interaction, which in turn has to do with the cohesion of the material (whether it is solid or fluid);
• in particular, in an interaction between a medium and an object falling due to gravity, the mass that will cooperate in the interaction will be 100% of the mass of the less massive object (when the support material is a solid) or the mass of the displaced fluid (when the support medium is a fluid).

Things are transparent to gravity and hence do not resist it

Gravity is a force that acts at distance. In that, it is like electrostatic force. In fact, the formulas for both are very similar: G (Mm)/r^2 and k(Qq)/r^2. And in both cases the way to act at a distance is not mysterious: both a charge Q and a mass M create attraction "fields" over their surrounding spaces (well, in the case of Q that is for opposite charges; for like charges, it is a "repulsion" field). A massive or an electrically charged object send around messages like this:  <<I am here with my mass "M" or charge "Q", so if you are a mass "m" or a charge "q", respectively, please be accelerated..." The first time that the field is created or if it changes, the message will take some time to reach its targets (if the Sun disappeared, the effect on the orbit of the Earth would only be perceived after some minutes), but once the field is there, it accelerates whatever falls under its net. The question for us is: at what rate? Let us look at the factors that compose the above-mentioned force formulas:

- Intentionally we leave aside, for the time being, the constants "G" and "k".

- We choose to reason on the basis of a uniform field: we consider a given distance "r" and then we assume that any additional distances till one target or the other are negligible, so "r^2" is a fixed value. 

- The next step is dividing by the mass of the object which is under the influence of the field. This results from Newton's second law of motion: Force = mass times acceleration (F = ma), so to guess acceleration you must divide F by m.  In the case of electrostatic force, this means that don't you know in advance how much acceleration you will get, unless you are also given the mass of the object in question. Instead, things are easier in the case of a gravitational field, because "by chance" you also had "m" at the numerator of the force formula, so when you divide by "m" as well, both terms cancel out. This means in practice that, in order to obtain the acceleration caused by, say, the Earth on any object located in its gravitational field, you don't need to know the mass of the concerned object: you just need M (the mass of the source), G (which is a constant) and r (which we deem fixed in a uniform field). The "passive" mass is hence irrelevant.

This is the equivalence principle, which was first enunciated by Galileo. All objects are accelerated towards a given gravitational source, say the Earth, at the same rate, no matter their respective masses. Of course the crash between the Earth and a neutron star (an incredibly massive object) will happen in much less time than the crash between the Earth and a feather, but that is only because the Earth itself will be much more accelerated by the star than by the feather. But both star and feather will pass by a "neutral" reference point (an inertial one, i.e. one that is not accelerated itself) at the same time, because as I said the two things are accelerated equally. Certainly, in a real-life case, at the surface of the Earth, we have air resistance, but that is a complication we are not considering now, as we are analyzing a "pure" gravitational attraction. So, without an atmosphere, if you drop one feather one day and repeat the experiment a week later with a steel ball and you record the meeting times, would they be the same? Well, strictly speaking, no, because as I said in terms of meeting times, it does count how much the object attracts the Earth; so, in theory, there should be a tiny difference in favor of the steel ball (less meeting time), although this advantage would be ridiculously small and utterly negligible. However, if we released feather and ball together, they would be jointly attracting the Earth at the same time, as if they were the same object, so even such an unperceivable difference would not arise.

So, by looking at the sheer mathematical equation, we see a crucial difference between gravitational and electrostatic force: how much m1 or m2 will get accelerated, depends (in a uniform field) only on one variable factor, i.e. M; instead, in order to guess how much q1 or q2 will get accelerated by Q, you need to know much more: not only Q, but also if we talk about q1 or q2 and also if q1, for instance, has mass m or m' or whatever. 

What is the physical difference that this math is reflecting?  

Sometimes (even in text-books) you hear that this is so because there is a funny coincidence: the Earth attracts more the steel ball than the feather, but the steel ball would "resist" more, so there is a trade-off between the two things and thus ultimately the accelerations experienced by ball and feather would be identical. I dislike this explanation, because a variable "resistance" implies a variable "action": it suggests that the Earth is adjusting its force to the opposition that it may encounter; it sounds as if the Earth were acting like a radar, i.e. sending scouts around, getting feedback that the challenge is harder or easier, processing it and then sending back an appropriate action. But this is not true: as I said, the Earth creates a field in the surrounding space, which means that it sends out messages, but it does not care about "who" may be on the other side of the line and it does not wait for an answer.

To provide an alternative explanation, it comes useful to look at the factors that we had temporarily left aside: the constants "G" or "k". "G" is constant everywhere, irrespective of the medium through which gravity propagates. Instead the "k" for the electrostatic force depends on the characteristics of the medium, which may be more or less apt to propagate the force. This means that certain materials can act as an attenuating filter for electrostatic force and even as a barrier (that is the case of insulators). Instead, gravity cannot be screened or attenuated (other than by distance). 

This becomes important when the force reaches the target because its atoms can also be regarded as a medium. If an electrostatic influence reaches two equally charged objects, the charges in them will react in a similar manner, but such charges are incrusted in their respective bodies, which act as loads that must be dragged along. Thus the participants in an electrostatic interaction do "resist" in the proportion of their respective masses or, which amounts to the same, they get accelerated in inverse proportion to those masses. Instead gravitational force travels unimpeded through the body in question and reaches all its constituents without the need for mediation: transmission of the message happens "mouth to mouth". Thus the same message reaches all particles of the body, which are thus accelerated equally, without the need of any of them pulling from the others... (Well, for completeness, I should recall that we are always assuming a uniform field across the accelerated body: that the distance to all parts of the object in question is irrelevant and so there are no "tidal effects", which is what happens when a participant in the interaction is so big that the field ceases to be uniform. For example, the side of the Earth facing the Sun experiences a stronger field than the side facing away, which is farther away by a non-negligible distance... But that is not the case we are considering, so we leave it aside.) 

Hence the equivalence principle holds, not because heavier (i.e. more massive) objects resist more, but precisely because they do not resist: no matter if the object has many or few constituents, they are all reached by the gravitational field simultaneously and react in unison, without mutually interacting, so that is why they "free-fall" (in vacuum) side by side. Thus we have accounted for the acceleration version of the formula: acceleration of anything in the vicinity of a ball is m; acceleration of anything in the vicinity of the Earth is M. What about the force formula, how do you compose it...? Well, it is curious, because we have just ruled out "resistance", but it immediately returns through the back door: it may be that a body does not physically do anything to resist gravity, but anyhow it must suffer its effects in inverse proportion to the resistance (or inertia) that it does show up in other types of interaction. Explaining why requires a little excursus. 

If things don't resist gravity, then they must attract by their resistance capacity

Note that the "active gravitational parameter" (what causes gravity and goes into the gravitation formula) could theoretically be a factor having nothing to do with inertial mass.  This is what in fact happens with electrostatic attraction, where the strength of the interaction is given by the involved electric charges (Q and q), the distance r and the constant k, nothing else. It would be no problem if the same happened with gravity, i.e. if this force were created by a sort of gravitational charge being unrelated to mass. But what is indispensable is that the ensuing effect (acceleration) be distributed in inverse proportion to inertial masses: if we called such gravitational charge C and c, we would need acceleration by Cc/m and Cc/M, respectively. In other words, we need the objects to meet at the "center of mass" (COM) of the system that they jointly form. 

That is a must in all interactions. I will give you some other example that a guy told me about time ago in a discussion at a forum (I will even quote him):

• Two magnets, M1 and M2, are sitting on a frictionless surface with their opposite poles facing each other. Both magnets have the same inertial mass, but M1 produces a stronger magnetic field than M2. Now, will M2 have a higher acceleration than M1? Where will they meet? Of course, they will have the same acceleration and they will meet at their COM (which is determined by their inertial mass), which in this case is the mid-point.
• You and I are sitting in chairs on a frictionless surface. I reach out and pull you toward me. The result is that we both meet at our COM. Now we repeat the experiment, except this time you reach out and pull me toward you. The result is the same. We both meet at our COM. We repeat the experiment again, except this time we both pull simultaneously, me pulling with a slightly greater force than you. The result is exactly the same. We will always meet at our COM.

The same happens when the force in question is repulsive. If two charges are mutually repelled, the total effect will be dependent upon the charges (Qq) but it will be distributed in inverse proportion to the masses. Likewise when two balls collide: in the simple example where they have the same masses and crash against each other at the same speed, if the collision is perfectly elastic (no loss of energy), the ensuing acceleration will be evenly distributed: each ball will keep the same speed but in opposite direction, because the masses are equal.

This example serves to justify why it must be the same with gravity. Imagine that the two equal balls A and B are mutually attracted due to gravity. Then, if their "resistance" (what we could call their passive gravitational mass) were not proportional to inertial mass, but to something else, the two balls would not meet at the center of mass, but let us say closer to A. However, then they bounce against each other and now they must be accelerated in the opposite direction in proportion to inertial masses, that is to said equally. The outcome of this would be that the system as a whole would be accelerated in the direction of A. It would have self-accelerated (accelerated without any external force acting on it), which would violate Newton's Laws of motion. And now realize that the same would happen when the system is a compact object: if the constituents of a body were thrown against each other by gravity and not "resisting" in proportion to masses, the body would self-accelerate.    

So what is the solution? Well, if we want it both ways, if we want the equivalence principle to hold, i.e. all masses being attracted equally (out of a physical reason, i.e. "transparency") and we also want that the gravitational effect is distributed between the two sides in inverse proportion to inertial masses (out of respect for Newton's Laws of motion), we need that the active gravitational mass ("gravitational charge") be equal to inertial mass.

Translation into a mathematical derivation

I have the disadvantage that I need to express in many words what others see with a formula. But I have also the advantage of always trying to harmonize the two languages. So I will repeat now the whole story with formulae. 

I find it more convenient to reason on the basis of the so-called "reduced mass" reference frame because this handles the total acceleration between the two masses. This frame represents the perspective of one of the masses participating in the interaction (let us say it is M, because it is more massive). M is being accelerated itself, but we disregard that and we attribute all the motion to the other one (m). Based on a derivation that I will spare you, this means that the mass of the object that is deemed to move is a combination of the two masses, which takes this form: Mm/(M+m). This is called the "reduced mass" because it is a little less than the effective mass of the system (M+m) and it is usually referred to with the Greek letter μ. Likewise, the acceleration of m is in fact the "total acceleration" or sometimes called the "relative acceleration".

So let us start with an interaction where we do not pre-judge if the "charge" (the cause for the acceleration, here Q and q) is related or not to mass. In this case, in accordance with Newton's laws of motion, the relative acceleration will be the force F divided by the reduced mass, as follows:

Now let us do what is also usual: we have started with a reference frame where it is easier to reason, but let us move to the habitual one, which is an inertial reference, for example, the center of mass of the system, towards which the two bodies accelerate at their respective acceleration rates. For this purpose, what we have to do -always in accordance with Newton's laws- is distributing the total acceleration in inverse proportion to the masses, i.e. M takes a share of acceleration equal to m's share in the sum of the masses and vice versa, like this:

Now, if we asked no more, gravity would be something akin to electrostatic force, i.e. it would be caused by a sort of gravitational charge that is unrelated to inertial mass. But it cannot be so, we need that the red term vanishes out, so that the acceleration of any object (no matter its mass) is conditioned exclusively by the mass of the other object (equivalence principle). We need it because that is what matches experiments and what logic requires, in view of the also empirical fact that things are transparent to gravity. And the only way to obtain it is that Qq is equal to Mm, i.e. that the factors generating the gravitational effect are the same that determines its distribution between the objects in question.

Support forces: ground and fluids

Clarified this, we can move to the analysis of the other type of force, where inertial mass does play a passive level, i.e. it triggers actual resistance. Time to talk about contact forces. We will look in particular at support forces: those that are present when you are holding an object, say a steel ball, which is attracted to the Earth by gravity, but -without allowing it to fall- you gently lay it on the ground or on a fluid (water or air, for example). Let us consider each of these two cases.

The steel ball seeks support on the ground

Here we label the support force as "normal force", which is equal to the weight of the object, i.e. mg, but of the opposite direction. Because of this equality, students tend to make this mistake: they think that weight and normal force are the pair of forces referred to in Newton's third law: the object would exert weight (= - mg) on the ground (which would be the action) and the ground would exert a reaction force (= +mg). The next step is wondering: here the object is at rest, but if there is always, by definition, action and reaction of equal magnitude, how come that on other occasions objects move at all, how is it that, for example, a steel ball sinks through water?

The mistake is what I wrote in blue: we are not in the face of a Newton third law pair, because weight is not the force that the object exerts on the ground. Weight is the gravitational force that the Earth exerts on the object, whose 3rd law pair is the gravitational force that the object exerts on the Earth. This is so before Earth and object come into contact. If we now place the object (say the steel ball) on the ground, then it exerts contact force (which is basically electrostatic in nature, i.e. electron shells repel each other) on the ground, which in turn exerts a contact force on the object. Those two are also a Newton 3rd law pair. Thus we have four forces, ordered by pairs: each pair is of a given nature (either gravitational or electrostatic), of a given magnitude (same amount on both sides), but reciprocal (each element of the pair acts on its opponent). If you want to guess whether the ball moves or not, you need to look at forces acting on the same thing, by picking them out from each pair: here, for example, forces acting on the ball are gravity exerted by the Earth, pointing downward, and contact force exerted by the floor, pointing upward. In this special case, the support contact force always matches gravity force, so it does make the object motionless. But there may be other cases where the support exerts a stronger force and pushes it upward (as it happens with hot air being pushed up by cold air or cork being buoyed up to the surface of the water). Or it may happen the other way round (gravity is stronger than support force), as it happens when a steel ball sinks in water.

What is the key to telling one case from the other? Here terminology may be a little confusing. We call "normal force" the one exerted by the ground and "buoyant force" the one exerted by the fluid. The reason for the word "normal" is that the plane pushes the ball perpendicularly. But actually "buoyant force" also acts perpendicularly, so we could have called them both "normal"... Better denominations would be those revealing the real difference, the factor accounting for why contact support force is sometimes equal, other times stronger or weaker than gravity, thus causing the ball to remain still or rise or fall... Obviously, the reason has to do with whether the objects involved are solids or fluids.

When we talk about solids (say our steel ball is resting on the rocky ground of the Earth), the support force automatically matches gravity because there is discipline in both armies, the army of the ball and the army of the ground. The defining characteristic of a solid is that cohesion among its particles is high (strong bonds), so when they are pushed, they hold tight, backing up each other, to avoid penetration. 

When applied to the ball, this means that it will transmit the force of gravity. Certainly, the idea that weight is the force exerted by the ball on the ground is not accurate, but it is not without any truth. Weight is the force suffered by the ball, but the ball "transmits" it to the ground. When I step on a ball, the ball conveys my push to the floor. Likewise, when the Earth pulls from the ball, the ball also conveys this pull, in the form of a push, to the ground. But it is important to note that this transmission is fully effective in this case only because all the constituents of the ball, when facing the opposition of the ground, find the collaboration of their colleagues and cooperate to avoid being disbanded.

Likewise, with the ground. The Earth is being accelerated by the ball, which means a tiny acceleration but of many constituents, matching exactly mg. But even if I helped the ball by jumping on it, the ground would resist, as it counts for this purpose with huge resources, which are fully committed in terms of cooperation, so the call will be answered by as many particles as are necessary to avoid penetration. (For this reason, "normal force" is called a constraint force, because its effect is banning a certain direction of progress for the object: something clashing against a firm ground can take any direction it pleases, except penetration. Tension, by the way, is another constraint force, whose role is avoiding that the object escapes away.) And how many are in particular needed? Well, I am not going to say that we can ascertain such a number. What we can guess is that it will be particles sufficiently committed to stop the progress of the ball. 

Thus it turns out that this time the support force exerted by the ground on the ball matches the gravity force on the same ball. But it does not need to be like that. It is not like that when...

The steel ball seeks support on water

Here the start point is identical. The ball is being pulled by the Earth, the ground (including the water) is being pulled against the ball. However, the interaction will be weaker because of the lack of firmness of the support. The water's army, certainly, may still count with many, many soldiers (think of the sea for example), but they show less solidarity: cohesion among troops is weaker in a liquid and even weaker in a gas. The sequence is as follows. The surface of our steel box touches the water. Steel is denser than water, which means that the first layer of steel soldiers brings into the interaction more mass than the first layer of water molecules. Unfortunately, the latter don't have strong bonds with their liquid medium, so when they are outnumbered, they slide over one another, giving way. This does not mean that the water soldiers abandon the battle: the whole army keeps attempting to recover the lost volume, from all directions, as they also "fall"  attracted by gravity: the water molecules placed between the surface and the bottom of the object push, like a piston, those below, which in turn bounce against the ground and push the ball upward;  those above push downward the top side; while the lateral ones neutralize each other. The net difference is that the object is always pushed upward by those water molecules that have been displaced by its volume, which is the Archimedes principle. Thus, unlike in the previous case, because the ground is a liquid, it does not involve unlimited resources: only those that I have just mentioned. That gives us the measure of the contact force applied by water on the ball (upward): mass of displaced water times g (for completeness, "g" makes sense because the particles pushing from the top are falling by gravity whereas the lower ones have been pushed by others also falling by gravity and have rebounded due to normal force exerted by the ground). 

This will also be the contact force to be exerted by the ball on the water (downward), not only because this follows from Newton's third law, but also because we know the reason: if the support water medium does not involve more soldiers in the battle, neither will the invading army of the ball. By this I do not mean that only a limited number of particles will participate on the side of the ball: I am not sure if all will participate or not, but in any case, only this level of "mass cooperation" will be requested. If you think of it, this is the same that happens with the Earth in the case of normal force: it only does as much as needed to avoid penetration.

In conclusion, we can guess the value of the support force exerted by a medium against (and also by) an object falling due to gravity, as per this rule of thumb: it is g times the mass that will cooperate/be involved in the interaction, which is 100% of the mass of the less massive object (when the support material is a solid) or the mass of the displaced fluid (when the support medium is a fluid).