What if Force was not Mass times Acceleration?2019 Community Moderator Election Unicorn Meta Zoo #1: Why another podcast? Announcing the arrival of Valued Associate #679: Cesar Manara 2019 Moderator Election Q&A - QuestionnaireHow do we know that $F = ma$, not $F = k cdot ma$Why isn't it $E approx 27.642 times mc^2$?Physical meaning of force times areaHow did Newton find out force has something to do with acceleration?Mass, Acceleration, and ForceCalculating mass of an orbiting body with force and accelerationAcceleration due to gravitational forceHow force is mass times acceleration?Newton's second law with $F=kma$Force without accelerationHow force is equal to the product of mass and instantaneous acceleration?Force and center of mass acceleration
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What if Force was not Mass times Acceleration?
2019 Community Moderator Election
Unicorn Meta Zoo #1: Why another podcast?
Announcing the arrival of Valued Associate #679: Cesar Manara
2019 Moderator Election Q&A - QuestionnaireHow do we know that $F = ma$, not $F = k cdot ma$Why isn't it $E approx 27.642 times mc^2$?Physical meaning of force times areaHow did Newton find out force has something to do with acceleration?Mass, Acceleration, and ForceCalculating mass of an orbiting body with force and accelerationAcceleration due to gravitational forceHow force is mass times acceleration?Newton's second law with $F=kma$Force without accelerationHow force is equal to the product of mass and instantaneous acceleration?Force and center of mass acceleration
$begingroup$
We know that Force equals Mass times Acceleration, $F = ma$. But what if we had a law like $$F=m^2a$$
or
$$F = 2ma~?$$
newtonian-mechanics forces mass acceleration inertia
New contributor
pooja paliwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
We know that Force equals Mass times Acceleration, $F = ma$. But what if we had a law like $$F=m^2a$$
or
$$F = 2ma~?$$
newtonian-mechanics forces mass acceleration inertia
New contributor
pooja paliwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
6
$begingroup$
As we all know we live in a classical world where everything works on the basis on this law Ever heard of quantum mechanics?
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– Gert
6 hours ago
1
$begingroup$
Related, if not dupe, physics.stackexchange.com/q/104101/25301
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– Kyle Kanos
6 hours ago
3
$begingroup$
The proposed duplicate asks why there's no numerical constant out front (see also), but the question of why the constant of proportionality is "mass" rather than "mass squared" (or the like) is an interesting twist.
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– rob♦
5 hours ago
2
$begingroup$
Ultimately, this depends on what your particular definition of mass is. If by "mass" you mean "inertial mass," then the question is trivial - inertial mass is defined to be the proportionality constant between force and acceleration. So which definition of mass are we intended to use?
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– probably_someone
5 hours ago
$begingroup$
Without a clear definition of the terms mass and force the question is too broad for a reasonably confined answer.
$endgroup$
– GiorgioP
4 hours ago
add a comment |
$begingroup$
We know that Force equals Mass times Acceleration, $F = ma$. But what if we had a law like $$F=m^2a$$
or
$$F = 2ma~?$$
newtonian-mechanics forces mass acceleration inertia
New contributor
pooja paliwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We know that Force equals Mass times Acceleration, $F = ma$. But what if we had a law like $$F=m^2a$$
or
$$F = 2ma~?$$
newtonian-mechanics forces mass acceleration inertia
newtonian-mechanics forces mass acceleration inertia
New contributor
pooja paliwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
pooja paliwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited 2 hours ago
Qmechanic♦
108k122011253
108k122011253
New contributor
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asked 6 hours ago
pooja paliwalpooja paliwal
163
163
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pooja paliwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor
pooja paliwal is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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6
$begingroup$
As we all know we live in a classical world where everything works on the basis on this law Ever heard of quantum mechanics?
$endgroup$
– Gert
6 hours ago
1
$begingroup$
Related, if not dupe, physics.stackexchange.com/q/104101/25301
$endgroup$
– Kyle Kanos
6 hours ago
3
$begingroup$
The proposed duplicate asks why there's no numerical constant out front (see also), but the question of why the constant of proportionality is "mass" rather than "mass squared" (or the like) is an interesting twist.
$endgroup$
– rob♦
5 hours ago
2
$begingroup$
Ultimately, this depends on what your particular definition of mass is. If by "mass" you mean "inertial mass," then the question is trivial - inertial mass is defined to be the proportionality constant between force and acceleration. So which definition of mass are we intended to use?
$endgroup$
– probably_someone
5 hours ago
$begingroup$
Without a clear definition of the terms mass and force the question is too broad for a reasonably confined answer.
$endgroup$
– GiorgioP
4 hours ago
add a comment |
6
$begingroup$
As we all know we live in a classical world where everything works on the basis on this law Ever heard of quantum mechanics?
$endgroup$
– Gert
6 hours ago
1
$begingroup$
Related, if not dupe, physics.stackexchange.com/q/104101/25301
$endgroup$
– Kyle Kanos
6 hours ago
3
$begingroup$
The proposed duplicate asks why there's no numerical constant out front (see also), but the question of why the constant of proportionality is "mass" rather than "mass squared" (or the like) is an interesting twist.
$endgroup$
– rob♦
5 hours ago
2
$begingroup$
Ultimately, this depends on what your particular definition of mass is. If by "mass" you mean "inertial mass," then the question is trivial - inertial mass is defined to be the proportionality constant between force and acceleration. So which definition of mass are we intended to use?
$endgroup$
– probably_someone
5 hours ago
$begingroup$
Without a clear definition of the terms mass and force the question is too broad for a reasonably confined answer.
$endgroup$
– GiorgioP
4 hours ago
6
6
$begingroup$
As we all know we live in a classical world where everything works on the basis on this law Ever heard of quantum mechanics?
$endgroup$
– Gert
6 hours ago
$begingroup$
As we all know we live in a classical world where everything works on the basis on this law Ever heard of quantum mechanics?
$endgroup$
– Gert
6 hours ago
1
1
$begingroup$
Related, if not dupe, physics.stackexchange.com/q/104101/25301
$endgroup$
– Kyle Kanos
6 hours ago
$begingroup$
Related, if not dupe, physics.stackexchange.com/q/104101/25301
$endgroup$
– Kyle Kanos
6 hours ago
3
3
$begingroup$
The proposed duplicate asks why there's no numerical constant out front (see also), but the question of why the constant of proportionality is "mass" rather than "mass squared" (or the like) is an interesting twist.
$endgroup$
– rob♦
5 hours ago
$begingroup$
The proposed duplicate asks why there's no numerical constant out front (see also), but the question of why the constant of proportionality is "mass" rather than "mass squared" (or the like) is an interesting twist.
$endgroup$
– rob♦
5 hours ago
2
2
$begingroup$
Ultimately, this depends on what your particular definition of mass is. If by "mass" you mean "inertial mass," then the question is trivial - inertial mass is defined to be the proportionality constant between force and acceleration. So which definition of mass are we intended to use?
$endgroup$
– probably_someone
5 hours ago
$begingroup$
Ultimately, this depends on what your particular definition of mass is. If by "mass" you mean "inertial mass," then the question is trivial - inertial mass is defined to be the proportionality constant between force and acceleration. So which definition of mass are we intended to use?
$endgroup$
– probably_someone
5 hours ago
$begingroup$
Without a clear definition of the terms mass and force the question is too broad for a reasonably confined answer.
$endgroup$
– GiorgioP
4 hours ago
$begingroup$
Without a clear definition of the terms mass and force the question is too broad for a reasonably confined answer.
$endgroup$
– GiorgioP
4 hours ago
add a comment |
7 Answers
7
active
oldest
votes
$begingroup$
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, $Fpropto a$.
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!
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6
$begingroup$
It would change something though as we wouldn’t have eg universal acceleration. Galileo would not have found the balls hit the ground at the same time when dropped from the tower ;)
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– innisfree
5 hours ago
6
$begingroup$
I disagree that it would be so simple. It would also have the consequence that when putting two objects together, we would have to add their masses squared instead of their masses, which seems pretty strange. Clearly what we now call mass is a better option than the others.
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– Javier
5 hours ago
1
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We’d also have to use some different units or fix the units somehow, eg introduce another dimensionful constant of nature or maybe just change the units of Newton’s constant $G$
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– innisfree
5 hours ago
3
$begingroup$
It would be nice if this answer (which is the accepted one) made mention of gravitational mass (and other non-inertial properties we expect mass to have, like additivity) and how they change.
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– jacob1729
4 hours ago
1
$begingroup$
If it was $m^2$ then it would definitely change something if the $m$ also refers to the mass associated with the force of gravity
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– Aaron Stevens
2 hours ago
|
show 1 more comment
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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 to be $v'=sqrt2 v$. 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 $mv$.
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$begingroup$
How would $g$ depend on $m$ here (for $F=km^2a$)?
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– KV18
1 hour ago
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@KV18 If you assumed that near the surface of the earth that the force of gravity is $mg$, then you just set up the equation: $km^2a=mg$ and solve: $a=g/km$ i.e. a more massive object would experience a smaller acceleration, which would make sense for this model of newtons second law
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– Aaron Stevens
1 hour ago
$begingroup$
But will not the force of gravity itself be $F=km^2g$?
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– KV18
1 hour ago
add a comment |
$begingroup$
"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.
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I do not understand the down votes to this answer. They only make clear that who gave them does not know anything about the point of view leaded by Kirchhoff, Mach, Hertz in the nineteenth century and nowadays present, anonimously, in several Mechanics textbooks (basically all of them which write that F=ma is a definition of F).
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– GiorgioP
4 hours ago
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@GiorgioP not only this one. It's strange that three answers had +2/-2 when I opened this page, and others 0/0.
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– Ruslan
2 hours ago
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@GiorgioP Down votes aren't only for answers that are incorrect or that people do not agree with
$endgroup$
– Aaron Stevens
2 hours ago
add a comment |
$begingroup$
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.
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add a comment |
$begingroup$
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 $a$.
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2
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Writing $k = m$ in $F = kma$ would change consequences of how things behave, at least in our current system, no?
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– JMac
5 hours ago
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No. If $a$ becomes $2a$, then we are basically using a "different" number system. I meant that the physics does not change, only a modification to the math - like using a shifted coordinate system where $X=x+2$ - kind of like that.
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– KV18
5 hours ago
1
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$k = m$ doesn't make $a$ become $2a$ though, it makes $F = m^2 a$.
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– JMac
5 hours ago
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I only meant $k$ for integers.
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– KV18
5 hours ago
1
$begingroup$
The question uses a non-integer $k$ as an example.
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– JMac
5 hours ago
add a comment |
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That is rather interesting, I've often thought of what would happen if you change a known relationship. If you doubled the mass, in your mass squared relationship, it would require four times the force to accelerate it the same. Another interesting change of relationship that happens in the real world is a rate of acceleration that increases or decreases over time. Try distance equals time cubed.
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add a comment |
$begingroup$
No matter the transformations you make into Newton's second law, as long as it's reverse transformation admits $rm a$ to be isolated.
A little bit of history
Since ancient times (at least Aristotle's time) people used to believe that for something to be at motion another thing must be acting upon it, otherwise, it should be at rest, i.e., forces were the cause of velocity. Newton's insight, -based on Galilee's observation that a straight uniform motion was as natural as rest- was to realize that the net result of actions on a body was proportional to the acceleration (second derivative of displacement), not the velocity (first derivative of displacement) as Aristotle thought.
The real deal
$F(r,x,t)$ in Newton's second law as expressed by equation
$$ F(r,x,t)=ma tag1$$
Is a very abstract concept that must be adapted to solve the observed motions in the universe. More importantly, $F(r,x,t)$ is an abstraction of some kind of position, time, and velocity function that satisfy the experimental observations. Since equation (1) can be written
$$F'(r,x,t)=a tag2$$
and $F'(r,x,t)$ has no form a priori, any transformation that can be reversed into equation (2), will give the same results as Newton's second law. However, now things like forces and masses do not necessarily behave the same as we expect.
For example, lets analyze the definition of gravitational force in the advent that Newton's second law was
$$F_G=2ma^2 tag3$$
In this particular case, the definition of $F_G$ here must satisfy (at least) the following observed facts:
- (By Galilee): “The acceleration of a falling body doesn’t depend on its mass”;
- (By Kepler): “The motion of planets around the sun are ellipses…”
The simplest definition of $rm F_G$ that satisfies both conditions is
$$F_G=fracsqrt2mr^4 tag4$$
However, with time, people would realize that $sqrtF_G$ was actually a better physical quantity, since more intuitive generalization of it, based on geometrical insights like Gauss law would arise. The same goes for the use of $rm m$ instead of $rm 2m$.
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add a comment |
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7 Answers
7
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oldest
votes
7 Answers
7
active
oldest
votes
active
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$begingroup$
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, $Fpropto a$.
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!
$endgroup$
6
$begingroup$
It would change something though as we wouldn’t have eg universal acceleration. Galileo would not have found the balls hit the ground at the same time when dropped from the tower ;)
$endgroup$
– innisfree
5 hours ago
6
$begingroup$
I disagree that it would be so simple. It would also have the consequence that when putting two objects together, we would have to add their masses squared instead of their masses, which seems pretty strange. Clearly what we now call mass is a better option than the others.
$endgroup$
– Javier
5 hours ago
1
$begingroup$
We’d also have to use some different units or fix the units somehow, eg introduce another dimensionful constant of nature or maybe just change the units of Newton’s constant $G$
$endgroup$
– innisfree
5 hours ago
3
$begingroup$
It would be nice if this answer (which is the accepted one) made mention of gravitational mass (and other non-inertial properties we expect mass to have, like additivity) and how they change.
$endgroup$
– jacob1729
4 hours ago
1
$begingroup$
If it was $m^2$ then it would definitely change something if the $m$ also refers to the mass associated with the force of gravity
$endgroup$
– Aaron Stevens
2 hours ago
|
show 1 more comment
$begingroup$
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, $Fpropto a$.
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!
$endgroup$
6
$begingroup$
It would change something though as we wouldn’t have eg universal acceleration. Galileo would not have found the balls hit the ground at the same time when dropped from the tower ;)
$endgroup$
– innisfree
5 hours ago
6
$begingroup$
I disagree that it would be so simple. It would also have the consequence that when putting two objects together, we would have to add their masses squared instead of their masses, which seems pretty strange. Clearly what we now call mass is a better option than the others.
$endgroup$
– Javier
5 hours ago
1
$begingroup$
We’d also have to use some different units or fix the units somehow, eg introduce another dimensionful constant of nature or maybe just change the units of Newton’s constant $G$
$endgroup$
– innisfree
5 hours ago
3
$begingroup$
It would be nice if this answer (which is the accepted one) made mention of gravitational mass (and other non-inertial properties we expect mass to have, like additivity) and how they change.
$endgroup$
– jacob1729
4 hours ago
1
$begingroup$
If it was $m^2$ then it would definitely change something if the $m$ also refers to the mass associated with the force of gravity
$endgroup$
– Aaron Stevens
2 hours ago
|
show 1 more comment
$begingroup$
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, $Fpropto a$.
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!
$endgroup$
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, $Fpropto a$.
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!
answered 5 hours ago
SteevenSteeven
28.1k866114
28.1k866114
6
$begingroup$
It would change something though as we wouldn’t have eg universal acceleration. Galileo would not have found the balls hit the ground at the same time when dropped from the tower ;)
$endgroup$
– innisfree
5 hours ago
6
$begingroup$
I disagree that it would be so simple. It would also have the consequence that when putting two objects together, we would have to add their masses squared instead of their masses, which seems pretty strange. Clearly what we now call mass is a better option than the others.
$endgroup$
– Javier
5 hours ago
1
$begingroup$
We’d also have to use some different units or fix the units somehow, eg introduce another dimensionful constant of nature or maybe just change the units of Newton’s constant $G$
$endgroup$
– innisfree
5 hours ago
3
$begingroup$
It would be nice if this answer (which is the accepted one) made mention of gravitational mass (and other non-inertial properties we expect mass to have, like additivity) and how they change.
$endgroup$
– jacob1729
4 hours ago
1
$begingroup$
If it was $m^2$ then it would definitely change something if the $m$ also refers to the mass associated with the force of gravity
$endgroup$
– Aaron Stevens
2 hours ago
|
show 1 more comment
6
$begingroup$
It would change something though as we wouldn’t have eg universal acceleration. Galileo would not have found the balls hit the ground at the same time when dropped from the tower ;)
$endgroup$
– innisfree
5 hours ago
6
$begingroup$
I disagree that it would be so simple. It would also have the consequence that when putting two objects together, we would have to add their masses squared instead of their masses, which seems pretty strange. Clearly what we now call mass is a better option than the others.
$endgroup$
– Javier
5 hours ago
1
$begingroup$
We’d also have to use some different units or fix the units somehow, eg introduce another dimensionful constant of nature or maybe just change the units of Newton’s constant $G$
$endgroup$
– innisfree
5 hours ago
3
$begingroup$
It would be nice if this answer (which is the accepted one) made mention of gravitational mass (and other non-inertial properties we expect mass to have, like additivity) and how they change.
$endgroup$
– jacob1729
4 hours ago
1
$begingroup$
If it was $m^2$ then it would definitely change something if the $m$ also refers to the mass associated with the force of gravity
$endgroup$
– Aaron Stevens
2 hours ago
6
6
$begingroup$
It would change something though as we wouldn’t have eg universal acceleration. Galileo would not have found the balls hit the ground at the same time when dropped from the tower ;)
$endgroup$
– innisfree
5 hours ago
$begingroup$
It would change something though as we wouldn’t have eg universal acceleration. Galileo would not have found the balls hit the ground at the same time when dropped from the tower ;)
$endgroup$
– innisfree
5 hours ago
6
6
$begingroup$
I disagree that it would be so simple. It would also have the consequence that when putting two objects together, we would have to add their masses squared instead of their masses, which seems pretty strange. Clearly what we now call mass is a better option than the others.
$endgroup$
– Javier
5 hours ago
$begingroup$
I disagree that it would be so simple. It would also have the consequence that when putting two objects together, we would have to add their masses squared instead of their masses, which seems pretty strange. Clearly what we now call mass is a better option than the others.
$endgroup$
– Javier
5 hours ago
1
1
$begingroup$
We’d also have to use some different units or fix the units somehow, eg introduce another dimensionful constant of nature or maybe just change the units of Newton’s constant $G$
$endgroup$
– innisfree
5 hours ago
$begingroup$
We’d also have to use some different units or fix the units somehow, eg introduce another dimensionful constant of nature or maybe just change the units of Newton’s constant $G$
$endgroup$
– innisfree
5 hours ago
3
3
$begingroup$
It would be nice if this answer (which is the accepted one) made mention of gravitational mass (and other non-inertial properties we expect mass to have, like additivity) and how they change.
$endgroup$
– jacob1729
4 hours ago
$begingroup$
It would be nice if this answer (which is the accepted one) made mention of gravitational mass (and other non-inertial properties we expect mass to have, like additivity) and how they change.
$endgroup$
– jacob1729
4 hours ago
1
1
$begingroup$
If it was $m^2$ then it would definitely change something if the $m$ also refers to the mass associated with the force of gravity
$endgroup$
– Aaron Stevens
2 hours ago
$begingroup$
If it was $m^2$ then it would definitely change something if the $m$ also refers to the mass associated with the force of gravity
$endgroup$
– Aaron Stevens
2 hours ago
|
show 1 more comment
$begingroup$
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 to be $v'=sqrt2 v$. 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 $mv$.
$endgroup$
$begingroup$
How would $g$ depend on $m$ here (for $F=km^2a$)?
$endgroup$
– KV18
1 hour ago
$begingroup$
@KV18 If you assumed that near the surface of the earth that the force of gravity is $mg$, then you just set up the equation: $km^2a=mg$ and solve: $a=g/km$ i.e. a more massive object would experience a smaller acceleration, which would make sense for this model of newtons second law
$endgroup$
– Aaron Stevens
1 hour ago
$begingroup$
But will not the force of gravity itself be $F=km^2g$?
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
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 to be $v'=sqrt2 v$. 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 $mv$.
$endgroup$
$begingroup$
How would $g$ depend on $m$ here (for $F=km^2a$)?
$endgroup$
– KV18
1 hour ago
$begingroup$
@KV18 If you assumed that near the surface of the earth that the force of gravity is $mg$, then you just set up the equation: $km^2a=mg$ and solve: $a=g/km$ i.e. a more massive object would experience a smaller acceleration, which would make sense for this model of newtons second law
$endgroup$
– Aaron Stevens
1 hour ago
$begingroup$
But will not the force of gravity itself be $F=km^2g$?
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
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 to be $v'=sqrt2 v$. 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 $mv$.
$endgroup$
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 to be $v'=sqrt2 v$. 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 $mv$.
answered 2 hours ago
Aaron StevensAaron Stevens
15.7k42557
15.7k42557
$begingroup$
How would $g$ depend on $m$ here (for $F=km^2a$)?
$endgroup$
– KV18
1 hour ago
$begingroup$
@KV18 If you assumed that near the surface of the earth that the force of gravity is $mg$, then you just set up the equation: $km^2a=mg$ and solve: $a=g/km$ i.e. a more massive object would experience a smaller acceleration, which would make sense for this model of newtons second law
$endgroup$
– Aaron Stevens
1 hour ago
$begingroup$
But will not the force of gravity itself be $F=km^2g$?
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
How would $g$ depend on $m$ here (for $F=km^2a$)?
$endgroup$
– KV18
1 hour ago
$begingroup$
@KV18 If you assumed that near the surface of the earth that the force of gravity is $mg$, then you just set up the equation: $km^2a=mg$ and solve: $a=g/km$ i.e. a more massive object would experience a smaller acceleration, which would make sense for this model of newtons second law
$endgroup$
– Aaron Stevens
1 hour ago
$begingroup$
But will not the force of gravity itself be $F=km^2g$?
$endgroup$
– KV18
1 hour ago
$begingroup$
How would $g$ depend on $m$ here (for $F=km^2a$)?
$endgroup$
– KV18
1 hour ago
$begingroup$
How would $g$ depend on $m$ here (for $F=km^2a$)?
$endgroup$
– KV18
1 hour ago
$begingroup$
@KV18 If you assumed that near the surface of the earth that the force of gravity is $mg$, then you just set up the equation: $km^2a=mg$ and solve: $a=g/km$ i.e. a more massive object would experience a smaller acceleration, which would make sense for this model of newtons second law
$endgroup$
– Aaron Stevens
1 hour ago
$begingroup$
@KV18 If you assumed that near the surface of the earth that the force of gravity is $mg$, then you just set up the equation: $km^2a=mg$ and solve: $a=g/km$ i.e. a more massive object would experience a smaller acceleration, which would make sense for this model of newtons second law
$endgroup$
– Aaron Stevens
1 hour ago
$begingroup$
But will not the force of gravity itself be $F=km^2g$?
$endgroup$
– KV18
1 hour ago
$begingroup$
But will not the force of gravity itself be $F=km^2g$?
$endgroup$
– KV18
1 hour ago
add a comment |
$begingroup$
"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.
$endgroup$
$begingroup$
I do not understand the down votes to this answer. They only make clear that who gave them does not know anything about the point of view leaded by Kirchhoff, Mach, Hertz in the nineteenth century and nowadays present, anonimously, in several Mechanics textbooks (basically all of them which write that F=ma is a definition of F).
$endgroup$
– GiorgioP
4 hours ago
$begingroup$
@GiorgioP not only this one. It's strange that three answers had +2/-2 when I opened this page, and others 0/0.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
@GiorgioP Down votes aren't only for answers that are incorrect or that people do not agree with
$endgroup$
– Aaron Stevens
2 hours ago
add a comment |
$begingroup$
"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.
$endgroup$
$begingroup$
I do not understand the down votes to this answer. They only make clear that who gave them does not know anything about the point of view leaded by Kirchhoff, Mach, Hertz in the nineteenth century and nowadays present, anonimously, in several Mechanics textbooks (basically all of them which write that F=ma is a definition of F).
$endgroup$
– GiorgioP
4 hours ago
$begingroup$
@GiorgioP not only this one. It's strange that three answers had +2/-2 when I opened this page, and others 0/0.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
@GiorgioP Down votes aren't only for answers that are incorrect or that people do not agree with
$endgroup$
– Aaron Stevens
2 hours ago
add a comment |
$begingroup$
"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.
$endgroup$
"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.
answered 5 hours ago
alephzeroalephzero
5,91521121
5,91521121
$begingroup$
I do not understand the down votes to this answer. They only make clear that who gave them does not know anything about the point of view leaded by Kirchhoff, Mach, Hertz in the nineteenth century and nowadays present, anonimously, in several Mechanics textbooks (basically all of them which write that F=ma is a definition of F).
$endgroup$
– GiorgioP
4 hours ago
$begingroup$
@GiorgioP not only this one. It's strange that three answers had +2/-2 when I opened this page, and others 0/0.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
@GiorgioP Down votes aren't only for answers that are incorrect or that people do not agree with
$endgroup$
– Aaron Stevens
2 hours ago
add a comment |
$begingroup$
I do not understand the down votes to this answer. They only make clear that who gave them does not know anything about the point of view leaded by Kirchhoff, Mach, Hertz in the nineteenth century and nowadays present, anonimously, in several Mechanics textbooks (basically all of them which write that F=ma is a definition of F).
$endgroup$
– GiorgioP
4 hours ago
$begingroup$
@GiorgioP not only this one. It's strange that three answers had +2/-2 when I opened this page, and others 0/0.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
@GiorgioP Down votes aren't only for answers that are incorrect or that people do not agree with
$endgroup$
– Aaron Stevens
2 hours ago
$begingroup$
I do not understand the down votes to this answer. They only make clear that who gave them does not know anything about the point of view leaded by Kirchhoff, Mach, Hertz in the nineteenth century and nowadays present, anonimously, in several Mechanics textbooks (basically all of them which write that F=ma is a definition of F).
$endgroup$
– GiorgioP
4 hours ago
$begingroup$
I do not understand the down votes to this answer. They only make clear that who gave them does not know anything about the point of view leaded by Kirchhoff, Mach, Hertz in the nineteenth century and nowadays present, anonimously, in several Mechanics textbooks (basically all of them which write that F=ma is a definition of F).
$endgroup$
– GiorgioP
4 hours ago
$begingroup$
@GiorgioP not only this one. It's strange that three answers had +2/-2 when I opened this page, and others 0/0.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
@GiorgioP not only this one. It's strange that three answers had +2/-2 when I opened this page, and others 0/0.
$endgroup$
– Ruslan
2 hours ago
$begingroup$
@GiorgioP Down votes aren't only for answers that are incorrect or that people do not agree with
$endgroup$
– Aaron Stevens
2 hours ago
$begingroup$
@GiorgioP Down votes aren't only for answers that are incorrect or that people do not agree with
$endgroup$
– Aaron Stevens
2 hours ago
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 5 hours ago
Ollie113Ollie113
777
777
add a comment |
add a comment |
$begingroup$
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 $a$.
$endgroup$
2
$begingroup$
Writing $k = m$ in $F = kma$ would change consequences of how things behave, at least in our current system, no?
$endgroup$
– JMac
5 hours ago
$begingroup$
No. If $a$ becomes $2a$, then we are basically using a "different" number system. I meant that the physics does not change, only a modification to the math - like using a shifted coordinate system where $X=x+2$ - kind of like that.
$endgroup$
– KV18
5 hours ago
1
$begingroup$
$k = m$ doesn't make $a$ become $2a$ though, it makes $F = m^2 a$.
$endgroup$
– JMac
5 hours ago
$begingroup$
I only meant $k$ for integers.
$endgroup$
– KV18
5 hours ago
1
$begingroup$
The question uses a non-integer $k$ as an example.
$endgroup$
– JMac
5 hours ago
add a comment |
$begingroup$
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 $a$.
$endgroup$
2
$begingroup$
Writing $k = m$ in $F = kma$ would change consequences of how things behave, at least in our current system, no?
$endgroup$
– JMac
5 hours ago
$begingroup$
No. If $a$ becomes $2a$, then we are basically using a "different" number system. I meant that the physics does not change, only a modification to the math - like using a shifted coordinate system where $X=x+2$ - kind of like that.
$endgroup$
– KV18
5 hours ago
1
$begingroup$
$k = m$ doesn't make $a$ become $2a$ though, it makes $F = m^2 a$.
$endgroup$
– JMac
5 hours ago
$begingroup$
I only meant $k$ for integers.
$endgroup$
– KV18
5 hours ago
1
$begingroup$
The question uses a non-integer $k$ as an example.
$endgroup$
– JMac
5 hours ago
add a comment |
$begingroup$
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 $a$.
$endgroup$
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 $a$.
edited 5 hours ago
answered 5 hours ago
KV18KV18
1,294617
1,294617
2
$begingroup$
Writing $k = m$ in $F = kma$ would change consequences of how things behave, at least in our current system, no?
$endgroup$
– JMac
5 hours ago
$begingroup$
No. If $a$ becomes $2a$, then we are basically using a "different" number system. I meant that the physics does not change, only a modification to the math - like using a shifted coordinate system where $X=x+2$ - kind of like that.
$endgroup$
– KV18
5 hours ago
1
$begingroup$
$k = m$ doesn't make $a$ become $2a$ though, it makes $F = m^2 a$.
$endgroup$
– JMac
5 hours ago
$begingroup$
I only meant $k$ for integers.
$endgroup$
– KV18
5 hours ago
1
$begingroup$
The question uses a non-integer $k$ as an example.
$endgroup$
– JMac
5 hours ago
add a comment |
2
$begingroup$
Writing $k = m$ in $F = kma$ would change consequences of how things behave, at least in our current system, no?
$endgroup$
– JMac
5 hours ago
$begingroup$
No. If $a$ becomes $2a$, then we are basically using a "different" number system. I meant that the physics does not change, only a modification to the math - like using a shifted coordinate system where $X=x+2$ - kind of like that.
$endgroup$
– KV18
5 hours ago
1
$begingroup$
$k = m$ doesn't make $a$ become $2a$ though, it makes $F = m^2 a$.
$endgroup$
– JMac
5 hours ago
$begingroup$
I only meant $k$ for integers.
$endgroup$
– KV18
5 hours ago
1
$begingroup$
The question uses a non-integer $k$ as an example.
$endgroup$
– JMac
5 hours ago
2
2
$begingroup$
Writing $k = m$ in $F = kma$ would change consequences of how things behave, at least in our current system, no?
$endgroup$
– JMac
5 hours ago
$begingroup$
Writing $k = m$ in $F = kma$ would change consequences of how things behave, at least in our current system, no?
$endgroup$
– JMac
5 hours ago
$begingroup$
No. If $a$ becomes $2a$, then we are basically using a "different" number system. I meant that the physics does not change, only a modification to the math - like using a shifted coordinate system where $X=x+2$ - kind of like that.
$endgroup$
– KV18
5 hours ago
$begingroup$
No. If $a$ becomes $2a$, then we are basically using a "different" number system. I meant that the physics does not change, only a modification to the math - like using a shifted coordinate system where $X=x+2$ - kind of like that.
$endgroup$
– KV18
5 hours ago
1
1
$begingroup$
$k = m$ doesn't make $a$ become $2a$ though, it makes $F = m^2 a$.
$endgroup$
– JMac
5 hours ago
$begingroup$
$k = m$ doesn't make $a$ become $2a$ though, it makes $F = m^2 a$.
$endgroup$
– JMac
5 hours ago
$begingroup$
I only meant $k$ for integers.
$endgroup$
– KV18
5 hours ago
$begingroup$
I only meant $k$ for integers.
$endgroup$
– KV18
5 hours ago
1
1
$begingroup$
The question uses a non-integer $k$ as an example.
$endgroup$
– JMac
5 hours ago
$begingroup$
The question uses a non-integer $k$ as an example.
$endgroup$
– JMac
5 hours ago
add a comment |
$begingroup$
That is rather interesting, I've often thought of what would happen if you change a known relationship. If you doubled the mass, in your mass squared relationship, it would require four times the force to accelerate it the same. Another interesting change of relationship that happens in the real world is a rate of acceleration that increases or decreases over time. Try distance equals time cubed.
$endgroup$
add a comment |
$begingroup$
That is rather interesting, I've often thought of what would happen if you change a known relationship. If you doubled the mass, in your mass squared relationship, it would require four times the force to accelerate it the same. Another interesting change of relationship that happens in the real world is a rate of acceleration that increases or decreases over time. Try distance equals time cubed.
$endgroup$
add a comment |
$begingroup$
That is rather interesting, I've often thought of what would happen if you change a known relationship. If you doubled the mass, in your mass squared relationship, it would require four times the force to accelerate it the same. Another interesting change of relationship that happens in the real world is a rate of acceleration that increases or decreases over time. Try distance equals time cubed.
$endgroup$
That is rather interesting, I've often thought of what would happen if you change a known relationship. If you doubled the mass, in your mass squared relationship, it would require four times the force to accelerate it the same. Another interesting change of relationship that happens in the real world is a rate of acceleration that increases or decreases over time. Try distance equals time cubed.
edited 4 hours ago
answered 4 hours ago
James MontagneJames Montagne
114
114
add a comment |
add a comment |
$begingroup$
No matter the transformations you make into Newton's second law, as long as it's reverse transformation admits $rm a$ to be isolated.
A little bit of history
Since ancient times (at least Aristotle's time) people used to believe that for something to be at motion another thing must be acting upon it, otherwise, it should be at rest, i.e., forces were the cause of velocity. Newton's insight, -based on Galilee's observation that a straight uniform motion was as natural as rest- was to realize that the net result of actions on a body was proportional to the acceleration (second derivative of displacement), not the velocity (first derivative of displacement) as Aristotle thought.
The real deal
$F(r,x,t)$ in Newton's second law as expressed by equation
$$ F(r,x,t)=ma tag1$$
Is a very abstract concept that must be adapted to solve the observed motions in the universe. More importantly, $F(r,x,t)$ is an abstraction of some kind of position, time, and velocity function that satisfy the experimental observations. Since equation (1) can be written
$$F'(r,x,t)=a tag2$$
and $F'(r,x,t)$ has no form a priori, any transformation that can be reversed into equation (2), will give the same results as Newton's second law. However, now things like forces and masses do not necessarily behave the same as we expect.
For example, lets analyze the definition of gravitational force in the advent that Newton's second law was
$$F_G=2ma^2 tag3$$
In this particular case, the definition of $F_G$ here must satisfy (at least) the following observed facts:
- (By Galilee): “The acceleration of a falling body doesn’t depend on its mass”;
- (By Kepler): “The motion of planets around the sun are ellipses…”
The simplest definition of $rm F_G$ that satisfies both conditions is
$$F_G=fracsqrt2mr^4 tag4$$
However, with time, people would realize that $sqrtF_G$ was actually a better physical quantity, since more intuitive generalization of it, based on geometrical insights like Gauss law would arise. The same goes for the use of $rm m$ instead of $rm 2m$.
$endgroup$
add a comment |
$begingroup$
No matter the transformations you make into Newton's second law, as long as it's reverse transformation admits $rm a$ to be isolated.
A little bit of history
Since ancient times (at least Aristotle's time) people used to believe that for something to be at motion another thing must be acting upon it, otherwise, it should be at rest, i.e., forces were the cause of velocity. Newton's insight, -based on Galilee's observation that a straight uniform motion was as natural as rest- was to realize that the net result of actions on a body was proportional to the acceleration (second derivative of displacement), not the velocity (first derivative of displacement) as Aristotle thought.
The real deal
$F(r,x,t)$ in Newton's second law as expressed by equation
$$ F(r,x,t)=ma tag1$$
Is a very abstract concept that must be adapted to solve the observed motions in the universe. More importantly, $F(r,x,t)$ is an abstraction of some kind of position, time, and velocity function that satisfy the experimental observations. Since equation (1) can be written
$$F'(r,x,t)=a tag2$$
and $F'(r,x,t)$ has no form a priori, any transformation that can be reversed into equation (2), will give the same results as Newton's second law. However, now things like forces and masses do not necessarily behave the same as we expect.
For example, lets analyze the definition of gravitational force in the advent that Newton's second law was
$$F_G=2ma^2 tag3$$
In this particular case, the definition of $F_G$ here must satisfy (at least) the following observed facts:
- (By Galilee): “The acceleration of a falling body doesn’t depend on its mass”;
- (By Kepler): “The motion of planets around the sun are ellipses…”
The simplest definition of $rm F_G$ that satisfies both conditions is
$$F_G=fracsqrt2mr^4 tag4$$
However, with time, people would realize that $sqrtF_G$ was actually a better physical quantity, since more intuitive generalization of it, based on geometrical insights like Gauss law would arise. The same goes for the use of $rm m$ instead of $rm 2m$.
$endgroup$
add a comment |
$begingroup$
No matter the transformations you make into Newton's second law, as long as it's reverse transformation admits $rm a$ to be isolated.
A little bit of history
Since ancient times (at least Aristotle's time) people used to believe that for something to be at motion another thing must be acting upon it, otherwise, it should be at rest, i.e., forces were the cause of velocity. Newton's insight, -based on Galilee's observation that a straight uniform motion was as natural as rest- was to realize that the net result of actions on a body was proportional to the acceleration (second derivative of displacement), not the velocity (first derivative of displacement) as Aristotle thought.
The real deal
$F(r,x,t)$ in Newton's second law as expressed by equation
$$ F(r,x,t)=ma tag1$$
Is a very abstract concept that must be adapted to solve the observed motions in the universe. More importantly, $F(r,x,t)$ is an abstraction of some kind of position, time, and velocity function that satisfy the experimental observations. Since equation (1) can be written
$$F'(r,x,t)=a tag2$$
and $F'(r,x,t)$ has no form a priori, any transformation that can be reversed into equation (2), will give the same results as Newton's second law. However, now things like forces and masses do not necessarily behave the same as we expect.
For example, lets analyze the definition of gravitational force in the advent that Newton's second law was
$$F_G=2ma^2 tag3$$
In this particular case, the definition of $F_G$ here must satisfy (at least) the following observed facts:
- (By Galilee): “The acceleration of a falling body doesn’t depend on its mass”;
- (By Kepler): “The motion of planets around the sun are ellipses…”
The simplest definition of $rm F_G$ that satisfies both conditions is
$$F_G=fracsqrt2mr^4 tag4$$
However, with time, people would realize that $sqrtF_G$ was actually a better physical quantity, since more intuitive generalization of it, based on geometrical insights like Gauss law would arise. The same goes for the use of $rm m$ instead of $rm 2m$.
$endgroup$
No matter the transformations you make into Newton's second law, as long as it's reverse transformation admits $rm a$ to be isolated.
A little bit of history
Since ancient times (at least Aristotle's time) people used to believe that for something to be at motion another thing must be acting upon it, otherwise, it should be at rest, i.e., forces were the cause of velocity. Newton's insight, -based on Galilee's observation that a straight uniform motion was as natural as rest- was to realize that the net result of actions on a body was proportional to the acceleration (second derivative of displacement), not the velocity (first derivative of displacement) as Aristotle thought.
The real deal
$F(r,x,t)$ in Newton's second law as expressed by equation
$$ F(r,x,t)=ma tag1$$
Is a very abstract concept that must be adapted to solve the observed motions in the universe. More importantly, $F(r,x,t)$ is an abstraction of some kind of position, time, and velocity function that satisfy the experimental observations. Since equation (1) can be written
$$F'(r,x,t)=a tag2$$
and $F'(r,x,t)$ has no form a priori, any transformation that can be reversed into equation (2), will give the same results as Newton's second law. However, now things like forces and masses do not necessarily behave the same as we expect.
For example, lets analyze the definition of gravitational force in the advent that Newton's second law was
$$F_G=2ma^2 tag3$$
In this particular case, the definition of $F_G$ here must satisfy (at least) the following observed facts:
- (By Galilee): “The acceleration of a falling body doesn’t depend on its mass”;
- (By Kepler): “The motion of planets around the sun are ellipses…”
The simplest definition of $rm F_G$ that satisfies both conditions is
$$F_G=fracsqrt2mr^4 tag4$$
However, with time, people would realize that $sqrtF_G$ was actually a better physical quantity, since more intuitive generalization of it, based on geometrical insights like Gauss law would arise. The same goes for the use of $rm m$ instead of $rm 2m$.
answered 6 mins ago
J. ManuelJ. Manuel
1,020222
1,020222
add a comment |
add a comment |
pooja paliwal is a new contributor. Be nice, and check out our Code of Conduct.
pooja paliwal is a new contributor. Be nice, and check out our Code of Conduct.
pooja paliwal is a new contributor. Be nice, and check out our Code of Conduct.
pooja paliwal is a new contributor. Be nice, and check out our Code of Conduct.
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6
$begingroup$
As we all know we live in a classical world where everything works on the basis on this law Ever heard of quantum mechanics?
$endgroup$
– Gert
6 hours ago
1
$begingroup$
Related, if not dupe, physics.stackexchange.com/q/104101/25301
$endgroup$
– Kyle Kanos
6 hours ago
3
$begingroup$
The proposed duplicate asks why there's no numerical constant out front (see also), but the question of why the constant of proportionality is "mass" rather than "mass squared" (or the like) is an interesting twist.
$endgroup$
– rob♦
5 hours ago
2
$begingroup$
Ultimately, this depends on what your particular definition of mass is. If by "mass" you mean "inertial mass," then the question is trivial - inertial mass is defined to be the proportionality constant between force and acceleration. So which definition of mass are we intended to use?
$endgroup$
– probably_someone
5 hours ago
$begingroup$
Without a clear definition of the terms mass and force the question is too broad for a reasonably confined answer.
$endgroup$
– GiorgioP
4 hours ago