What if Force was not Mass times Acceleration?

Question

We know that Force equals Mass times Acceleration, $F = ma$. But what if we had a law like $$F=m^2a$$ or $$F = 2ma~?$$

Answer 1

In contrast to the other answers, I am going to go a different direction. This is not a question of mathematics, where we can just say "oh yeah just redefine your variables." This is a physics question.

Going further, The OP is assuming $m$ stands for the mass of an object, not some arbitrary proportionality constant that ends up being the mass. Therefore, having a law like $F=km^2a$ ($k$ as just some constant for unit's sake) would not leave the universe unchanged. This is saying, for example, that if we double the "amount of stuff" associated with an object, that we would need $4$ times as much force to achieve the acceleration. This is certainly not true in our universe today, so we cannot say this law would be found in a similar universe.

There is also the question of the relation between inertial mass and gravitational mass, which we take to be equal. i.e. we know the force of gravity to be proportional to the mass of the object that gravity is acting on. Therefore, if forces behaved like $F=km^2a$, then we would find that, near the surface of the earth, that the acceleration due to gravity would actually depend on the mass of the object! Certainly this is not the same as our universe.

I understand the other answers' attempts to try and teach a lesson about definitions in physics, and how there are some things that are around purely by convention. But this is not one of those cases I believe. The OP is starting from $m$ means mass. So we should start there too and then see what that means for the proposed new "force laws".

This would be like me saying that it would be fine to take what we understand kinetic energy to be and rewrite it as $mv'$. Sure, I could say that I have "redefined" the velocity so that $v=\sqrt{2v'}$. And this is valid mathematically if we stay consistent with this in the rest of our equations. But if I start off by saying "let $v$ be the velocity of the object, i.e. the rate of change of displacement", then I cannot "redefine" my equations so that kinetic energy is $mv$.

Answer 2

It wouldn't change anything. We would just have defined mass differently. What is relevant about Newton's 2nd law is not $m$ but rather the fact that $F$ and $a$ are proportional, $F\propto a$.

The proportionality constant can then be called $m$ or $2m$ or $m^2$. That doesn't matter for the relationship that this law describes.

Newton's 2nd law defines mass (inertial mass) as the proportionality constant in $F=ma$. Had we decided upon $F=2ma$, then we would just have defined mass as half of the proportionality constant. It would change our language regarding mass (you would only weigh 40 kg rather than 80 kg i.e.), but it wouldn't change the way the world works.

If you on the other hand had asked about $F=ma^2$ or so, then you are messing with the world entirely with disastrous consequences that we can only try to guess as a result!

Answer 3

"Force" is just a word. You don't need it at all to do Newtonian mechanics, if you start from conservation of energy and momentum and the idea of gravitational potential energy. "Force" is then just a mathematical invention to keep track of how the momentum is transferred between different bits of stuff, when things move around.

The world doesn't change just because humans invent random new words.

Answer 4

One of the things that allowed us to invent and understand Newton's laws, which govern the motion of classical objects, is that they have the mathematical property of linearity. Linearity is a class of mathematical behaviors that people tend to be good at. If there are two or more forces acting on an object, the net force is the simple vector sum of the forces. If there is a system made of several masses, the total mass is the simple arithmetic sum of the individual masses.

For example, an Aristotelian misconception that was not properly refuted until the Renaissance is that "heavier objects fall faster." An argument against this, which I believe was due to Galileo, is a thought experiment, as follows. Imagine three identical cannonballs dropped from the same height: they should fall at the same rate because they all have the same mass. But tie two of them together with a low-mass connector, like glue or silk, and you have two objects whose masses are different by a factor of two. Can connecting two heavy masses with a silk thread change their fall time by a factor of two? It's a clever thought experiment.

In your formulation, two identical masses $m$ accelerating in tandem with the same acceleration $a$ would have the same amount of force,

$$ F = m^2 a, $$

acting on them. But if you tied them together so that you had a single object with mass $2m$ being acted upon by a force $2F$, then you would have

$$ 2F = (2m)^2 a $$

which predicts a different, smaller acceleration. Either you'd have to come up with an additional non-linear rule for adding forces, or you'd be describing a world that is different from the one where we live.

Answer 5

Then we would recall mass as $m' = m^2$ or $m' = 2m $. The whole idea of Newton's second law is that force and acceleration are proportional. We can then define inertial mass as the constant factor between them. So the world would look exactly the same.

Answer 6

Writing $F=kma$ and putting $k=1$ or $k=2,3...$ will not change the consequences of how things would actually behave.

All it does is put up a scaling factor to the value. But due to convenience, we agree on $k=1$ implying $ F=ma$.

Writing $a \to 2a$ instead does not change anything but the convention you have to abide by, for it is the same argument as writing $kma=F$.

If $k=m$ as you have included ($F=m^2a$), things would be the same if you assume $m^{'}=m^2$ and assume that $m^{'}$ is the actual mass.

Besides, $m^{'}$ is meant to be a constant of proportionality, and so it makes no difference to the consequences of physics. $F=ma$ is based on the understanding that $F$ is proportional to $a$.

Answer 7

I have not read all the answers but most of them are conceptually wrong. Newton's second law does not state that force is proportional to acceleration. Mass is not just a proportionality constant.

Newton's second law of motion states that

The rate of change of motion (momentum) is directly proportional to the motive force and the direction of the change is parallel to the force.

Hence instead of $F\propto a$ we have $\vec{F}\propto \frac{d\vec{p}}{dt}$. Changing the proportionality into equality $$\vec{F} = k\frac{d(m\vec{v})}{dt}$$ Solving through $$\vec{F} = km\frac{d\vec{v}}{dt} + k\vec{v}\frac{dm}{dt}$$ (This equation is actually wrong, but that is not the point here). In most cases mass of system is constant and hence the second term becomes zero, but the most general form of second law is as given above.

Now returning to the question. What if $k$ is something other than $1$. That will not change the physics. The only difference will be that accelerating objects will be harder (if $k>1$).

The other question is invalid because $m²$ is not possible since $\vec{p}=m\vec{v}$. One can again ask why is momentum not $m²\vec{v}$? You can consider momentum as $m²\vec{v}$ just don't call $m$ as the mass, since momentum is defined that way.

Answer 8

No matter the transformations you make into Newton's second law, as long as it's reverse transformation admits $\rm a$ to be isolated.

$F(r,v,t)$ in Newton's second law as expressed by equation $$ F(r,v,t)=ma \tag{1}$$

Is a very abstract concept that must be adapted to solve the observed motions in the universe. More importantly, $F(r,v,t)$ is an abstraction of some kind of position, time, and velocity function that satisfy the experimental observations. Since equation (1) can be written

$$F'(r,v,t)=a \tag{2}$$

and $F'(r,v,t)$ has no form a priori, any transformation in (1) that can be reversed into (2), will give the same results as Newton's second law. However, What one would call force is not necessarily what we currently call it but they should be easily converted into one another.

For example, lets analyze the definition of gravitational force in the advent that Newton's second law were

$$F(x,v,t)=2ma^2 \tag{3}$$

In this particular case, the definition of gravitational force $F_G$ here must satisfy (at least) the following observed facts:

1. (By Galilee): “The acceleration of a falling body doesn’t depend on its mass”;
2. (By Kepler): “The motion of planets around the sun are ellipses…”

The simplest definition of $F_G$ that satisfies both conditions is

$$F_G=\frac{(2m)^2(2M)^2}{r^4} \tag{4}$$

Which leads to the exactly same observations as current definition do. In other words, the core idea at Newton’s second law, is that “the amount of pushes and pulls from everything else in the universe into a body (also known as force)” is the cause of this body’s acceleration, and not it’s velocity as previously thought. Nonetheless, it is important to notice that any function of $a$ would satisfy this principle, yet equation $(2)$ is the simplest choice though.

The problem is, that by plugging his theory’s “void” function or “experimental part” function $F$ in left with a given and unmutable function of acceleration $f(a)≡a$ at right, Newton’s theory cannot be used in setups where the definition of space and time are no longer as simple as the ones given in Galilee’s principle of relativity, which are: “time is universal” and “rods have the same length to everyone”.

The big advantage of the so called Lagrangian mechanics is that instead of fixing an universal $F$ to an $f(a)$, it fixes an universal relation between then. Lagrangian mechanics introduces a single “void” function $L$ that simultaneously contains a void $F$ and a void $f(a)$, given by $$F=\frac{\partial L}{\partial x} \tag{5}$$ And $$f(a)=\frac{d}{dt} \frac{\partial L}{\partial \dot{x}} \tag{5}$$

$F$ and $f(a)$ in $L$ sometimes become so entangled that it is difficult to perceive which is what. However, this makes its easily generalizable to any known mechanical theory, from Newtonian to relativistic, and with a somehow different interpretation, into quantum mechanics. A common misconception of Lagrangian mechanics is that, since $$L=\frac{1}{2}mv^2 – U \tag{6}$$ It is just Newtonian mechanics in disguise. This is wrong. The form given in $(6)$ is just the one that recovers Newtonian mechanics. $L$ is not given a priori. It needs to be found depending on the specific problem and conditions. For example, the same Lagrangian theory works fine in restrict and general relativity by using the appropriate form form $L$.