What is Force? - A General Definition

[TuviaDaCat]

what is force anyway?
and i want the most general defenition.

 

[actionintegral]

This is a better question that you think! According to Bertrand Russell, the statement F=ma amounts to nothing more that a truism or circular definition.

 

[Hootenanny]

The most general definition of force is the rate of change of momentum;

\sum\vec{F} = \frac{d\vec{p}}{dt}

In classical physics (where the mass of a body in motion is constant) this can be expressed in a more familiar form;

\sum\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}

However, there are certain circumstances in classical physics such as rocketry where the above expression fails (also when considering relativistic speeds). The original 'general' expression of Newton's second law, however holds in all cases.

\sum\vec{F} = \frac{d\vec{p}}{dt}

 

[TuviaDaCat]

The most general definition of force is the rate of change of momentum;

\sum\vec{F} = \frac{d\vec{p}}{dt}

In classical physics (where the mass of a body in motion is constant) this can be expressed in a more familiar form;

\sum\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}

However, there are certain circumstances in classical physics such as rocketry where the above expression fails (also when considering relativistic speeds). The original 'general' expression of Newton's second law, however holds in all cases.

\sum\vec{F} = \frac{d\vec{p}}{dt}

you just defined the resultant force.
and even if it was called force, sigma f is a sum of forces, so such defenition is circular

 

[arildno]

The most general definition of force is the rate of change of momentum;

\sum\vec{F} = \frac{d\vec{p}}{dt}

In classical physics (where the mass of a body in motion is constant) this can be expressed in a more familiar form;

\sum\vec{F} = \frac{d\vec{p}}{dt} = \frac{d(m\vec{v})}{dt} = m\frac{d\vec{v}}{dt} = m\vec{a}

However, there are certain circumstances in classical physics such as rocketry where the above expression fail.

No, it doesn't. Where did you get that idea from?

 

[actionintegral]

Hi TuviaDaCat,

Whenever I am thinking about non-relativistice force, I mentally replace the word "force" with "acceleration". Opposing forces are just accelerations that cancel out.

 

[Andrew Mason]

No, it doesn't. Where did you get that idea from?

I think he meant F = mdv/dt fails if dm/dt \ne 0. The correct expression is, of course: F = dp/dt = mdv/dt + vdm/dt = ma + vdm/dt

AM

 

[lalbatros]

TuviaDaCat,

what is force anyway?
and i want the most general defenition

These demands are mutually exclusive.
Generality cannot tell you what a force is.
You better break the circle of circular definitions by recognizing as many practical examples as possible. Look in electromagnetism, fluid dynamics, elasticity, astronomy, ...

Hopefully you will forget about these all-out-of-nothing expectations from physics.

Michel

 

[pmb_phy]

what is force anyway?
and i want the most general defenition.

The (total) force on a particle equals the time rate of change of that particle's momentum.

This is a better question that you think! According to Bertrand Russell, the statement F=ma amounts to nothing more that a truism or circular definition.

If one assumes that F=dp/dt is a law of physics then then one is using circular logicic. However if one defines force as dp/dt the the circularity disappears. The error in login I mentioned is this - Newton's laws are said to hold only in an inertial frame, while an inertial frame is defined as any frame in which Newton's laws hold.In Eddington's words Every body continues in a state of rest or motion in sofar as it doesn't. A typical method today is to define the inertial frame in a way that has nothing to do with with Newton's first two laws, to define mass by Newton's third law, and to use the second as a definition of force.

For more on the details of this method please see

On force and the inertial frame, Robert W. Breheme, Am. J. Phys., 53(10, October 1985.

No, it doesn't. Where did you get that idea from?

That is a well known fact in both classical and relativistic mechanics. One need only turn to the Feynman Lectures to verify that fact. In volume I page 12-2, Feyman wrote

Newton also gave one rule rule about the force: that the forces between interacting bodies are equal and opposite - action equals reaction. In fact, the law F = ma is not exactly true; if it were a definition we should have to say that that it is always true; but it is not

Whenever I am thinking about non-relativistice force, I mentally replace the word "force" with "acceleration". Opposing forces are just accelerations that cancel out.

Ouch. In my humble opinion that's a bad habit.

Pete

 

[actionintegral]

Ouch. In my humble opinion that's a bad habit.
Pete

Ok - but why? What's the problem in setting m=1 everywhere? Or q=1? Or c=1? I'm not being a smart-aleck, I just don't like lugging around a lot of alphabetic luggage. If something is a constant, I set it =1 wherever possible and move on.

 

[Rach3]

Ok - but why? What's the problem in setting m=1 everywhere? Or q=1? Or c=1? I'm not being a smart-aleck, I just don't like lugging around a lot of alphabetic luggage. If something is a constant, I set it =1 wherever possible and move on.

I myself set \pi \equiv 1. But then I'm braver than most.

 

[rbj]

Ok - but why? What's the problem in setting m=1 everywhere? Or q=1? Or c=1?

there isn't. it's all a matter of defining units.

I'm not being a smart-aleck, I just don't like lugging around a lot of alphabetic luggage. If something is a constant, I set it =1 wherever possible and move on.

that's sort of what they're trying to do with Planck units.

 

[TuviaDaCat]

The (total) force on a particle equals the time rate of change of that particle's momentum.
If one assumes that F=dp/dt is a law of physics then then one is using circular logicic. However if one defines force as dp/dt the the circularity disappears. The error in login I mentioned is this - Newton's laws are said to hold only in an inertial frame, while an inertial frame is defined as any frame in which Newton's laws hold.In Eddington's words Every body continues in a state of rest or motion in sofar as it doesn't. A typical method today is to define the inertial frame in a way that has nothing to do with with Newton's first two laws, to define mass by Newton's third law, and to use the second as a definition of force.

For more on the details of this method please see

On force and the inertial frame, Robert W. Breheme, Am. J. Phys., 53(10, October 1985.

That is a well known fact in both classical and relativistic mechanics. One need only turn to the Feynman Lectures to verify that fact. In volume I page 12-2, Feyman wrote

Ouch. In my humble opinion that's a bad habit.

Pete

the equations described above were all about the sum of forces on a body, not the force on one...
and I am pretty much sure that the 2 classical forces, gravity and electricity must have a common base...

 

[Pythagorean]

Though I've been through the general physics courses, I'm going to try and avoid using a physics definition; it won't be too short, but I'm hoping the concept will be simple.

Force can be seen as change. Without force, there would be no physical events. Let's start with space. If you send something into space, you're putting a force on it by setting it into emotion, but the moment your hand leaves the object, and it's free-floating through space, there are eseentially no forces on it (in reality gravity from all the mass in the universe are pulling on it slightly, but not enough to matter) and it will continue on it's path indefinately, unless another force acts on it.

To take the example further, if all the forces in the universe suddenly stopped, here are some of the differences you'd notice (this is not a scientific explanation, but it should help understand the concept of force):

since there would be no electromagnetic force, big objects flying at each other wouldn't crash, they'd just simply pass through each other. The electromagnetic force is responsible for most (if not all) collisions you see on the macro level in your every day life. This is a result of the electrons from two objects repelling each other so greatly that they smash cars and faces and what not when two such objects meet each other.

Since there would be no gravitational force, at the moment all forces stopped acting everything would simply continue on the path it was headed for just before the forces stopped (and pass through other objects)

Because there's no nuclear forces, the same thing would happen with atoms, they'd simply fall apart, each particle of the atom (this is kind of sloppy, viewing the atoms as particles, but bare with me) would continue on whatever path it was on, there would be no turning and changing direction (that's actually considered acceleration, which is porportional to force) and the particles that make up the atom would simply drift off in their own directions.

And remember, since there's no electromagnetic force, none of these are interacting with each other in anyway. Too make a further ridiculous assumption on this already paradoxal model, I'd assume that statistics would allow all the particles that make up the universe would homogonize (assuming a closed universe). In that sense, forces (in addition to causing physical events) are responsible for a heterogenous planet, in which systems are separate (or 'closed') entities (like you and me and the Earth and a capped jug of wine).

My explanation may not be the ultimate one, but I'm hoping that if enough professonal physicists on this forum pick at it, you'll begin tod evelop a more accurate concept of force.

 

[Andrew Mason]

This is a better question that you think! According to Bertrand Russell, the statement F=ma amounts to nothing more that a truism or circular definition.

The essence of f=ma is that for a given force, acceleration varies inversely as the mass. This is not circular. A 1 kg falling brick with a string on a pulley pulls a cart along a horizontal surface. The cart accelerates half as fast if I double the cart's mass; three times as fast if I remove 2/3 of its mass. So for a given size of falling brick, ma = constant. If I change the mass of the falling weight, I change the constant. We call that 'constant' the "force".

AM

 

[arildno]

I am shocked by the ignorance of Hootenanny, pmpphy and Andrew Mason (and possibly, Feynman's).
Learn the difference between MATRERIAL systems and geometrical systems.

A MATERIAL system (which consists of the SAME particles over time) is in the classical sense governed by two main laws:
1. Conservation of MASS.
2. A dynamical law known as Newton's 2.law that, due to 1., has two equivalent forms F=ma and F=dp/dt
(where "a" is the acceleration of the center of mass, m the total mass of the system, and p the total momentum of the system)
A GEOMETRICAL system does not contain the same particles over time, and is not governed by either 1 or either of the two forms of 2.


Don't apply laws on systems they are not valid for!

Read my tutorial:
https://www.physicsforums.com/showthread.php?t=72176

As for proof of my assertion that rocketry is, indeed, governed by F=ma, it suffices to say that a proper MATERIAL system is the rocket+fuel remaining+fuel ejected.

On this system, there are only internal forces working, hence the C.M of this system has zero acceleration.

Another proper material system is the following:
The rocket fuselage+the fuel that remains up to, and including time T.
That system S has mass m(T) (where T can be regarded as a parameter distinguishing between different material systems).
m(T) is a constant, and all particles contained within S accelerates equally with acceleration a(t) up to time t=T.

Let V be the exhaust velocity relative to S, and consider that at time T there is a particle P attached to S with mass dM.
In the interval T, T+dt, P is separated from S, experiencing momentum change dMV, that is, by Newton's 3.law, applying a force -dM/dtV on S at time T

For the time period t<=T, then, S obeys:
F(t)=m(T)a(t), F(T)=-\frac{dM}{dt}V
T was chosen arbitrarily.
Furthermore, by comparison of different material systems, we see we can introduce a mass function m(t) so fulfilling \frac{dm}{dt}=-\frac{dM}{dt}, we can formulate the proper law of motion for the NON-material system rocket+remaining fuel:
\frac{dm}{dt}V=ma, \frac{dm}{dt}&lt;0
This, of course, is neither the F=ma or F=dp/dt valid for a MATERIAL system, but that is beside the point, since our system isn't a material system in the first place.


F=ma is seen, however, to be equally valid for any particular MATERIAL system you're looking at.

 

[loseyourname]

That's a very misleading reading of Russell, actionintegral. You might want to provide a little context for that statement of his. He was not trying to say that physical equations are circular.

 

[pmb_phy]

Ok - but why? What's the problem in setting m=1 everywhere? Or q=1? Or c=1? I'm not being a smart-aleck, I just don't like lugging around a lot of alphabetic luggage. If something is a constant, I set it =1 wherever possible and move on.

Its possible to choose constants of nature such that c=1. One chooses a system of basic constants and when that is done then other quantities are then defined through those constants. You can choose the mass of a body to be 1 but if there are more than one particles then you have to allow for another particle having the value of two. Same with charge. If one thinks of force as being defined as F = dp/dt then and always remember that this is a definition rather than an equality then one is less likely to confuse F = ma with F = dp/dp. For example: you made the assertion that Opposing forces are just accelerations that cancel out. That is true if and only if you have the very special situation that the mass of each of the two particles are identical. In general what you said here is not true. If the masses are different then the accelerations will be different too.

Consider also the claims made by alrildno about our so-called ignorance of the subject. He states

A MATERIAL system (which consists of the SAME particles over time) is in the classical sense governed by two main laws:

Here he is defining a "material system" to be that definition such that his assertions are correct. In my extensive readings on the concepts of force and mass I've yet to see such a term used. He goes on to assume

1. Conservation of MASS.

This he states as a postulate of classical mechanics. However one simply does not need to define such a postulate since it can be reduced to a theorem. That theorem is stated as follows; If the total momentum of a system is conserved in all inertial frames of reference then it follows that the total mass of such a system is conserved. For a derivation of this theorem please see
http://www.geocities.com/physics_world/sr/conservation_of_mass.htm
As you can see from this derivation it is the postulate that momentum is conserved, not that mass is conserved. Conservation of momentum is a theorem which follows from Newton's second law as I recall. Actually mass is defined as the m such that mv is conserved. This is a definition based on observation, i.e. (loosely speaking) mass is defined such that momentum is conserved. Momentum is then defined as p = mv.

2. A dynamical law known as Newton's 2.law that, due to 1.

If Newton's second law follows from "1" then it is not a law but a theorem. aldrino makes the false statement that (, has two equivalent forms F=ma and F=dp/dt). This is a totally invalid statement in that it does not correspond to what is observed in nature. It totally fails for a relativistic particle moving under a force (and fails for non-constant mass systems). (Notice how he avoids mention of relativistic particles? How convenient for him!)

A GEOMETRICAL system does not contain the same particles over time, and is not governed by either 1 or either of the two forms of 2.

I have to admit that I have no clue on what he means by this since the term is not defined in classical mechanics that I'm aware of and he does not define it here. He goes on to say

Don't apply laws on systems they are not valid for!

Newton's firtst two laws are valid under all circumstances and his third law fails when one gets into particles moving in fields such as the force between two charged particles. The reason being that the field has momentum.

I believe that aldrino's assertion

As for proof of my assertion that rocketry is, indeed, governed by F=ma, it suffices to say that a proper MATERIAL system is the rocket+fuel remaining+fuel ejected.

shows his misunderstanding of how forces is defined and what F = ma means. F = ma is not a definition. It is an equality under certain conditions. This equality fails under relativistic systems which aldrino fails to address.

I won't be addressing his comments any further on the forum (perhaps in PM if someone is really serious about his claims). I'm not interested any further in responding to comments about others "ignorance". It tells me that such a person is not willing to learn more than the assumptions he's already made. And anyone who claims that Feynman was ignorant of basic physics tells me that person has a lot to learn about basic classical mechanics. All those invalid assertions are just bad ju-ju.

Pete

ps - If anyone wishes to read the Am. J. Phys. article I referred to above then I'd be glad to e-mail it to them.

 

[pmb_phy]

the equations described above were all about the sum of forces on a body, not the force on one...
and I am pretty much sure that the 2 classical forces, gravity and electricity must have a common base...

When stated as F = dp/dt it is assumed that one understands "F" as what you refer to as "sum of forces" aka "total force."

The sum that you refer to is as follows: If F_21 is the force on body 2 due to body 1 in the abssence of all other forces and F_31 is the force on body one in the abssence of all other forces then the force (what you refer to as "sum of forces") is the quantity F_1 = F_21 + F_31. It must be understood that F_21 or F_31 refer only to what the force would be on object 1 in the absense of all other forces.

Pete

 

[Hootenanny]

Could somebody point me in the direction of a formal definition of both a material and geometrical system, for I have not come across these terms before.

 

[pmb_phy]

Could somebody point me in the direction of a formal definition of both a material and geometrical system, for I have not come across these terms before.

You shouldn't assume that such a definition exists which is universally accepted. Actually I've never seen those terms defined and I've been studying physics for 20 years (one never stops studying physics. One merely stops going to school. ) .

That said, one could take "material system" as one consisting of "material". But that requires that I define "material." Of course one could use the term "material" to mean the same thing as "matter" except that "matter" is not a well-defined quantity. Einstein defined the term to include the EM field whereas many people nowadays understand the term as referring to systems which are entirely composed of particles which have non-zero proper mass.

Note: F=dp/dt is not something I created out of nothing. This is the definition of "force" as given in almost all textbooks on classical mechanics. E.g. Feynman, Marion & Thornton, Corben $ Stehle, and in Jammer's account of the defintion of force as defined by Newton in Jammer's text "Concepts of Force" (i.e. Newton used the term "force" to mean F = dp/dt). A.P. French's text Newtonian Mechanics (page 166 Eq. 6-1) also explains that F dt = dp is how F is found in Newton's Principia. Nowhere in the Principia does F = ma appear. See also page 315 on French.

The other texts which don't use this definition define "force" as F = -grad U where U is the potential energy of the particle. This definition is used in treatments of analytical mechanics. E.g. Lanczos, Landau & Lif****z). You'll see the definition F = dp/dt in all relativistic treatments of mechanics since in that case

F = \frac{d(mv)}{dt} = \frac{d(\gamma m_0 v)}{dt}

In such case m will not be a constant in time but will vary as m = \gamma m_0. The relation F = ma willl not be valid in this case even when the proper mass remains constant in time.

Pete

 

[Hootenanny]

That said, one could take "material system" as one consisting of "material". But that requires that I define "material." Of course one could use the term "material" to mean the same thing as "matter" except that "matter" is not a well-defined quantity. Einstein defined the term to include the EM field whereas many people nowadays understand the term as referring to systems which are entirely composed of particles which have non-zero proper mass.

Thanks for the info

Note: F=dp/dt is not something I created out of nothing. This is the definition of "force" as given in almost all textbooks on classical mechanics. E.g. Feynman, Marion & Thornton, Corben $ Stehle, and in Jammer's account of the defintion of force as defined by Newton in Jammer's text "Concepts of Force" (i.e. Newton used the term "force" to mean F = dp/dt). A.P. French's text Newtonian Mechanics (page 166 Eq. 6-1) also explains that F dt = dp is how F is found in Newton's Principia. Nowhere in the Principia does F = ma appear. See also page 315 on French.

I know, that is why I used it in my previous post as it is universally applicable. I didn't think that the definition using potential would be the 'most general' form so I omitted it, my mistake perhaps.

 

[pmb_phy]

I didn't think that the definition using potential would be the 'most general' form so I committed it, my mistake perhaps.

Its of limited use. It requires that there exists a function U such that F = -grad U. This is not always the case. The force of friction does not hace such a function associated with it. Velocity dependant forces also cannot be expressed as -grad U. For example; the magnetic force on a charged particle F = qvxB cannot not be written as -grad U since a particle moving in such a field moves with constant potential energy. The Lorentz force is written as

F = dp/dt = q[E + vxB]

The left equality is Newton's second law while the second equality is the Lorentz force law.

Pete

 

[pmb_phy]

actionintegral
Ok - but why? What's the problem in setting m=1 everywhere? Or q=1? Or c=1?

Speaking of units - From The Character of Physical Laws, Richard Feynman

For those of you who want some proof that physicists are human, the proof is in the idiocy of all the different units which they use for measuring energy

Pete

 

[rbj]

... Lif****z ...

i'd love to have a supposed obscenity embedded into my name so that when i introduce myself, i can emphasize the offending syllable and they can't bleep me.

...

Raquel Welsh: "Would you like to pet my kitty?"

Johnny Carson: "Sure, if you move the cat."

...

 

[Hootenanny]

Raquel Welsh: "Would you like to pet my kitty?"

Johnny Carson: "Sure, if you move the cat."

...

 

[superweirdo]

Somebody earlier said that if there would be no force, planets would go through each other. However, if there are no forces, I believe planets would just vanish b/c there would nothing holding it together. And I assume that electrons wouldn't go through each other, but when they would hit each other, what would happen? Would they pass their kenetic energy to its collisioner. If yes, then wouldn't it take some kind of force to pass the energy from one to another? What kind of force is it anyways?

 

[pmb_phy]

Somebody earlier said that if there would be no force, planets would go through each other.

Planets are bodies which are kept together by gravitational forces. If these forces didn't exist then planets wouldn't exist. The universe would simply consist of a gas of various particles.

However, if there are no forces, I believe planets would just vanish b/c there would nothing holding it together.

Precisely!

And I assume that electrons wouldn't go through each other, but when they would hit each other, what would happen?

It is unclear whether it is meaningfull to say that an electron can "go through" another electron. We don't know the precise structure of electrons and as such we are unable to determine what would happen if one electron passes through another. Especially when "center of electron" is not a well defined quantity in QM.

Would they pass their kenetic energy to its collisioner. If yes, then wouldn't it take some kind of force to pass the energy from one to another? What kind of force is it anyways?

You're now referring to "contact forces". Otherwise energy can be transferred to other particles through the field which they generate.

Pete

 

[Pythagorean]

Somebody earlier said that if there would be no force, planets would go through each other. However, if there are no forces, I believe planets would just vanish b/c there would nothing holding it together. And I assume that electrons wouldn't go through each other, but when they would hit each other, what would happen? Would they pass their kenetic energy to its collisioner. If yes, then wouldn't it take some kind of force to pass the energy from one to another? What kind of force is it anyways?

I'd imagine without force, the electrons would pass right through each other as well. I don't think interactions can happen without forces. Not my final answer though.

As far as I know, it's the electromagnetic force that passes kinetic energy in collisoins.

 

[Mickey]

what is force anyway?

That's a question that physicists work on as part of the job description, I think.

It's a general case of questions like "what is gravity?" and "what is magnetism?" and so on. Looking for an answer to the question has given us many discoveries, namely that some forces are the same.

The greatest of Newton's contributions may be that he gave us such an interesting question.

 

[rbj]

Raquel Welsh: "Would you like to pet my kitty?"

Johnny Carson: "Sure, if you move the cat."

...

well, i screwed up. first it was supposed to be Zsa Zsa Gabor not Raquel Welsh (Sophia Loren, etc. i get all of those '60s sex-bombs mixed up). second, it's a false urban legend:

http://www.snopes.com/radiotv/tv/zsazsa.htm

see a video clip where this is discussed (Johnny and Jane Fonda):

http://www.snopes.com/radiotv/video/zsazsav.rm

still worth a giggle.

 

[pmb_phy]

That's a question that physicists work on as part of the job description, I think.

It's a general case of questions like "what is gravity?" and "what is magnetism?" and so on. Looking for an answer to the question has given us many discoveries, namely that some forces are the same.

The greatest of Newton's contributions may be that he gave us such an interesting question.

At this point I believe that the OP should clarify what he is asking. Is he seeking a defintion of force or the mechnism of interacting bodies which produces accelerations and for which there is a non-zero force on the body.

Jammer addresses Newton's second law in his book Concepts of Force. On page 124 Jammer writes

The second law, likewise, has two possible interpretations: it may serve as a quantitative definition of force or as a generalization of emperical facts. In modern notation, according to Netwon, asserts F ~ \Delta(mv).
...
Force, for Newton, was a concept given a priori, intuitively and ultimately in analogy to human musclular force.

Newton took the second law to refer to emperical facts. However the modern view is the second law is to be taken as a definition of force.

Pete

 

[Andrew Mason]

Newton took the second law to refer to emperical facts. However the modern view is the second law is to be taken as a definition of force.

How modern? Feynman was quite clear that F=ma was more than a definition as the concept of force is based on empirical fact. Tension in a string; extension of a spring; stress/strain on a beam: F=-kx or F=mg or F = kq^2/r^2. These are real phenomena and more than definitions.

AM

 

[pmb_phy]

How modern?

Nothing has really changed since Newton published his Principia. Nowadays physicists know that F = dp/dt isn't just a convinient for byt a neccesary form for the definition of force. The first relativistic instance I know of this was in an article written by Max Plank in which he stated that the Lorentz force equation can be written as

\frac{dp}{dt} = q[E + vxB]

where p = \gamma m_0[/tex]. This form remained to this day. I doubt that you could find and article or a upper level undergrad textbook published after 1970 that would indicate otherwise.<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Feynman was quite clear that F=ma was more than a definition as the concept of force is based on empirical fact. </div> </div> </blockquote><img src="https://cdn.jsdelivr.net/joypixels/assets/8.0/png/unicode/64/1f615.png" class="smilie smilie--emoji" loading="lazy" width="64" height="64" alt=":confused:" title="Confused :confused:" data-smilie="5"data-shortname=":confused:" /> clear in what sense? Where did you get that opinion from? What Feynman actually wrote was (as posted above).<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Newton also gave one rule rule about the force: that the forces between interacting bodies are equal and opposite - action equals reaction. In fact, the law F = ma is not exactly true; if it were a definition we should have to say that that it is <i>always</i> true; but it is not. </div> </div> </blockquote>If you looked at the context of that that statement made by Feynmen then you&#039;ll see that he means that F = dp/dt<br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> Tension in a string; extension of a spring; stress/strain on a beam: F=-kx or F=mg or F = kq^2/r^2. These are real phenomena and more than definitions. </div> </div> </blockquote>In the cases you give, the equation of motion will be<br /> <br /> dp/dt = -kx<br /> dp/dt = mg<br /> dp/dt = kq^2/r^2<br /> <br /> It has been emphsized in almost all my posts on this that in non-relativistic cases and when the mass is not a function of time then F = ma is an <i>equality</i>, but not a <i>definition</i>.<br /> <br /> Pete

 

[Gokul43201]

what is force anyway?
and i want the most general defenition.

A force is that thing, which when acting individually on a single (massive) particle, causes it to accelerate.

 

[robphy]

When I introduce "force" to my class, I say:

a "force" is a push or a pull on an object due to another object,
which can be added [and scalar multiplied] like a vector

later,
\sum \vec F_i \equiv \vec F_{net} \stackrel{\mbox{Newton\ II}}{=} \frac{d\vec p}{dt}

 

[Andrew Mason]

clear in what sense? Where did you get that opinion from? What Feynman actually wrote was (as posted above).

Feynman, Lectures, vol 1, ch 12, p. 12-3:

"In the same way, we cannot just call f=ma a definition, deduce everything purely mathematically, and make mechanics a mathematical theory, when mechanics is a description of nature. By establishing suitable postulates it is always possible to make a system of mathematics, just as Euclid did, but we canot make a mathematics of the world, because sooner or later we have to find out whether the axioms are valid for the object of nature. "​

AM

 

[pmb_phy]

When I introduce "force" to my class, I say:

a "force" is a push or a pull on an object due to another object,
which can be added [and scalar multiplied] like a vector

later,
\sum \vec F_i \equiv \vec F_{net} \stackrel{\mbox{Newton\ II}}{=} \frac{d\vec p}{dt}

The reason I refer to F_net as simply F is due to things like the electric field E at a point P due to charges in its vicinity. Those charges produce a (net) force on a test particle placed at P. Normally we say that the force F at P is given by F = qE. I don't recall anyone writing F_net = qE. I also keep with many texts such with Goldstein when I write F = dp/dt. But that is my personal preference. If one merely observers the test particle and what it is doing then what one measures is dp/dt of the particle.

Pete

 

[robphy]

The reason I refer to F_net as simply F is due to things like the electric field E at a point P due to charges in its vicinity. Those charges produce a (net) force on a test particle placed at P. Normally we say that the force F at P is given by F = qE. I don't recall anyone writing F_net = qE. I also keep with many texts such with Goldstein when I write F = dp/dt. But that is my personal preference. If one merely observers the test particle and what it is doing then what one measures is dp/dt of the particle.

Pete

I would say F_electric = qE since it is but one of the many forces that can appear in "the vector sum of the forces".

 

[pmb_phy]

I would say F_electric = qE since it is but one of the many forces that can appear in "the vector sum of the forces".

In that case, to be consistent, why not write F_electric = qE_net?

Also, in analytical mechanics one is given the Lagrangian and unless one knows the reason for the potential then one derives the (canonical) force from which it is impossible to determine if there are more than one sources.

Better yet let me ask you these two questions: (1) Under what circumstances would you call F a net sum when this force is derived from a Lagrangian that I give you? (2) under what conditions do you use the term E_net rather than E?

Pete

 

[TuviaDaCat]

im wondering about wether force must be defined using time...
after all, i can measure a force, without watching a body acclerate in a system which one force far greater all other forces...

 

[robphy]

In that case, to be consistent, why not write F_electric = qE_net?

E is the value of the vector field at a point. Implicitly, it is the vector sum of those E-fields from other objects that contribute to it. You could write "net", if you wish. If I really want to emphasize things, I use a lot of "decorations" and write:
F_{electric on q} = qE_{[net] at q's location}

I think the point of emphasizing "net" in F_net is that it is the net-force that is related to the acceleration of the particle, and not that each force somehow contributes an acceleration to the particle. This should remind someone that forces need to be added vectorially first... then one can use Newton-II.

In addition, many problems say "one applies a force F"... but that is not necessarily the net force.

Also, in analytical mechanics one is given the Lagrangian and unless one knows the reason for the potential then one derives the (canonical) force from which it is impossible to determine if there are more than one sources.

Better yet let me ask you this question: Under what circumstances would you call F a net sum when this force is derived from a Lagrangian that I give you?

Pete

Let me think more about this.
But let me say this much...
the point of "net" is that forces superpose like vectors... in particular, the forces from various external objects. Similarly, if you have numerous potentials, then the potentials are additive [and so the derived forces are additive as vectors]. Certainly, if there is just one external object (or potential), then the sum is trivial.

 

[pmb_phy]

But let me say this much...
the point of "net" is that forces superpose like vectors... in particular, the forces from various external objects. Similarly, if you have numerous potentials, then the potentials are additive [and so the derived forces are additive as vectors]. Certainly, if there is just one external object (or potential), then the sum is trivial.

I fully understand this view and wouldn't argue against it. I don't use it myself. But I have my reasons, some of which I posted here.

Pete

ps - I have scanned in the Am. J. Phys. article Force and the inertial frame by Robert W. Brehme and can e-mail it to those who are interested. The topic of the paper is identical to the topic of this thread. This link may work

www.geocities.com/physics_world/gr/bondi_1957.pdf

 

[Andrew Mason]

Just a further comment on Feynman, Vol1, Ch. 12 of Lectures. Feynman maintains that force as a physical concept has a meaning that exists independently of F=ma or F=dp/dt. He says that if it were merely a definition (he uses the example of 'gorce', which has no physical meaning) it would have no use.

AM

 

[robphy]

Just a further comment on Feynman, Vol1, Ch. 12 of Lectures. Feynman maintains that force as a physical concept has a meaning that exists independently of F=ma or F=dp/dt. He says that if it were merely a definition (he uses the example of 'gorce', which has no physical meaning) it would have no use.

AM

My post #36 is consistent with this.
In fact, as a prelude to Newton's Law of Motion F=dp/dt (F=ma to the algebra-based class), I discuss what could be called "Aristotle's Law of Motion" F=mv. (Of course, Aristotle does not have a correct law of inertia.)

 

[pmb_phy]

Andrew Mason
Just a further comment on Feynman, Vol1, Ch. 12 of Lectures. Feynman maintains that force as a physical concept has a meaning that exists independently of F=ma or F=dp/dt. He says that if it were merely a definition (he uses the example of 'gorce', which has no physical meaning) it would have no use.

As it is obvious in that chapter, Feynman took F = ma (actually dp/dt) as a law of nature. As I mentioned earlier it is more common nowadays to take F = dp/dt as a definition of F. Otherwise you run into circular logic. That Feyman held F to be F = dp/dt is evident on page 15-9 where he writes

To see the consequences of Einstein's modification of m we start with Newton's law is the rate of change of momentum, or

F = d(mv)/dt

...

Force now has as much of a meaning as tourque does since tourqe is a defined quantity. I.e. in the case of force it is used to simplify laws, such as the Lorentz force law dp/dt = q[E + vB]. Same idea holds for things like momentum too.

My post #36 is consistent with this.
In fact, as a prelude to Newton's Law of Motion F=dp/dt (F=ma to the algebra-based class), I discuss what could be called "Aristotle's Law of Motion" F=mv. (Of course, Aristotle does not have a correct law of inertia.)

Note: F = ma was never written by Newton in his Principia. F = ma comes from Euler, wyhich, of course, is not true in all cases (I.e. for relativistic particles moving in a field)

Pete

 

[Andrew Mason]

As it is obvious in that chapter, Feynman took F = ma (actually dp/dt) as a law of nature. As I mentioned earlier it is more common nowadays to take F = dp/dt as a definition of F. Otherwise you run into circular logic.

I understand what you are saying, but I don't agree that force must be 'defined' in order to avoid circular logic. You cannot question the validity of a definition. But you can question the validity of Newton's second law. You can do experiments to see if it is true.

If we were to measure the acceleration of a given mass with a given force applied to it and then measure the acceleration with double the force applied and found the acceleration to be something other than double, we would have to question the validity of F=ma.

AM

 

[robphy]

I understand what you are saying, but I don't agree that force must be 'defined' in order to avoid circular logic. You cannot question the validity of a definition. But you can question the validity of Newton's second law. You can do experiments to see if it is true.

If we were to measure the acceleration of a given mass with a given force applied to it and then measure the acceleration with double the force applied and found the acceleration to be something other than double, we would have to question the validity of F=ma.

AM

...or ask if we are working in an inertial frame.

 

[robphy]

Note: F = ma was never written by Newton in his Principia. F = ma comes from Euler, wyhich, of course, is not true in all cases (I.e. for relativistic particles moving in a field)

Pete

Yes, I am aware... but that's why I put that in parenthesis... for the algebra-based (i.e. non-calculus) students that I may teach.

 

[pmb_phy]

Yes, I am aware... but that's why I put that in parenthesis... for the algebra-based (i.e. non-calculus) students that I may teach.

Ok. At this point I think anything else I have to contribute would be mere semantics since all that needed to be said was said.

Pete