Talk:Newton's laws of motion/Archive 4

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Archive 6

Some other thoughts about this page

This is not related to the RFC and I hope this is not included there. I have questions about how this page is formatted and I did not want to jump in and edit here because there seems to be enough active editors here and too many cooks can spoil the broth.

I have some concerns that hopefully can help you improve the page.

• The quotes and the translations from Newton disrupt the flow of the content. Even the translation is often hard to understand since it uses a different scientific language. For example what exactly does Newton mean by motion in his second law (acceleration, velocity, position)? It is encyclopedic which is good. (It is not, nor should it be a textbook.) On the other hand, encyclopedias put such content as side boxes to separate it out from the main content.

• Often the history is mixed in with scientific content (in particular with the 1st law). This is another candidate for a box that people can skip or read separately if they so desire. At the least it should have its own section, like what is common on wikipedia.

• The alternate name for Newton's 1st law (law of inertia) is well established, but the alternate names for the other laws are not. It would not surprise me if one or more texts use them but for most people those names will be a distraction. (Try googling "Newton's second Law" vs "Law of resultant force" for instance).

• Have you considered splitting this article up into the main article plus separate articles covering more of the detail? The second and the third law in particular have the potential to be very long.

TStein (talk) 20:35, 11 July 2008 (UTC)

Boldface or arrows above?

The article mixes the two styles of symbols for vectors, but I think that, for consistency, we should pick one and use it throughout the article. Personally, I would prefer boldface, (as well as because I like it more) because it can be written without TeX, which don't align well with surrounding text when written inline (compare F with ${\vec {F}}$ or $\scriptstyle {\vec {F}}$). A similar issue is for the differential d which is written both in italics and in roman in this same article. What do you think? --A r m y 1 9 8 7 ! ! ! 22:50, 3 September 2008 (UTC)

I vote for boldface vectors for the same reasons. Also, my favorite textbooks use boldface.

As for the differential, I prefer to put d in italics, but scalar variable names in roman. You apparently prefer the inverse? MarcusMaximus (talk) 01:17, 4 September 2008 (UTC)

As for scalars, it is standard to write them in italics (it is the style that all browsers I've seen use to render the <var> tag, and all or almost all the textbooks and scientific papers I've ever seen use it); as for differentials, there are people using italics and people using roman; personally I prefer the latter for clarity, but italics seems to be far more common, at least on Wikipedia. MOS:MATH says "Both forms are correct; what is most important is to consistency within an article, with deference to previous editors." --A r m y 1 9 8 7 ! ! ! 10:06, 4 September 2008 (UTC)

Fg = mg vandalism

The user 125.24.228.54 likes to change force equations to read $F_{g}=mg$. I reverted the edits on this page. The same thing was done to the drag equation article. MarcusMaximus (talk) 18:33, 7 September 2008 (UTC)

Lack of third law in intro

Don't know how long it's been like this but the basic introductory part of the article, although titled three laws lack the third one. Just thought I'd point that out. —Preceding unsigned comment added by 64.231.185.133 (talk) 00:06, 10 September 2008 (UTC)

Thanks for calling this to our attention. I went back and found the edit where this was done, and it was quite a substantial act of vandalism by user 210.15.221.203. It had removed the entire third law, changed markup language that destroyed the references list, and changed random words throughout the article that made it nonsensical. I have done my best to restore it, but have been about about a dozen edits since then, so I can't be sure I got everything. MarcusMaximus (talk) 04:46, 10 September 2008 (UTC)

Requested semi-protection

This page has an obnoxiously high level of vandalism, so I requested semi-protection from unregistered IP addresses and new users. MarcusMaximus (talk) 04:59, 10 September 2008 (UTC)

Requested full protection

I put in a request for full protection for this page for an indefinite period of time. Since the semi-protection expired two days ago it has already been vandalized and reverted over a dozen times. It is vandalized constantly and unrelentingly, and there is a very good risk of losing valuable content. MarcusMaximus (talk) 07:30, 27 September 2008 (UTC)

Newton's Second Law

I was recently viewing this entry:

http://en.wikipedia.org/wiki/Newton%27s_laws_of_motion

and I noticed a common flaw in the definition of impulse which it critical in understanding Newton's Second Law. Impulse is not just the product of mass times the change in velocity, it is the product of both the change of velocity and the change in mass. If it were just m delta v it would be much more difficult to get a space vehicle into orbit. To prove this to yourself, push a cart with 100 pounds of potatos on it at a constant 5 MPH and then have your buddies add another 100 pound bag of potatos to it every second. You will soon see that while your delta v is 0 your delta m is what's going to stop you in your tracks! When I was in school the profs used to ask, "What is NSL?" The common answer was, "F=ma!" That is a simplification. —Preceding unsigned comment added by 75.70.62.142 (talk) 15:55, 30 December 2008 (UTC)

More to the point, Newton himself, (who you would have thought would have been the one to define the law) stated it (in latin) in terms of the rate of change of mass times velocity (i.e. rate of change of momentum).- (User) Wolfkeeper (Talk) 13:58, 5 June 2009 (UTC)

Page protection failure

This page was recently horribly vandalised by an anon editor because the Template:Classical mechanics template had been trashed. Seems like we might need to request protection for that also. SteveBaker (talk) 03:56, 7 January 2009 (UTC)

Question about Relativistic mass in Relativity section

The relativity section of this article is using relativistic mass in the derivation. This can be confusing for students in the US, at least. Here, the use of relativistic mass has been out of vogue for a while and has been replaced by relativistic Energy. In this understanding m always refers to the 'rest mass' and the relativistic momentum is p = γmv. I debated changing this section but I want to make sure that this notation is not prevalent in other fields and countries first. TStein (talk) 21:06, 23 January 2009 (UTC)

I come from the US, with a background in physics, so I offer no new perspective, but I say go ahead and change it. Rracecarr (talk) 22:36, 23 January 2009 (UTC)

Newton's Second Law video

I watched the video just now and I'm a bit uncertain as to its usefulness in describing Newton's second law. It says a lot of numbers and equations but doesn't explain them much. Whanhee 17:28, 25 January 2009 (UTC)

Non-constant mass in the second law.

I noticed a mistaken claim in the main section about the second law: that it only holds for systems of constant mass.

In fact, F = d(m v)/dt is always true, even if the mass is changing.

It's the simplified version F =m a that is only true for systems of constant mass.

Refer to

H. Goldstein (1980). Classical Mechanics, p.2, eqn.(1-4), (1-5).

The section on open and closed systems appears to be somewhat in error. Tom Lougheed (talk) 02:55, 1 February 2009 (UTC)

This has been discussed a lot before, both on this page and on Talk:Isaac Newton. If you think of a rocket going at constant speed, and add two thrusters canceling each other exactly. Then the net force is 0, but the trusters are expelling mass and thus dp/dt is not 0! That is:

$\sum {\vec {F}}\neq {\frac {d{(m{\vec {v}})}}{d{t}}}$

So it's not valid for changing mass systems.
—Apis (talk) 20:39, 1 February 2009 (UTC)

Tom, refer to the footnotes in that section. They have solid citations backing up the current state of the article. MarcusMaximus (talk) 05:10, 26 February 2009 (UTC)

Just to be clear, and avoid sending readers looking for references from February 2009, the mistake in the two-thruster rocket example above is that it does not account for all the momentum changed by the applied forces. Instead, the equation should be written:

$\sum {\vec {F}}=m_{G}{\vec {a_{G}}}+{\dot {m_{of}}}{\vec {v_{of}}}+{\dot {m_{ob}}}{\vec {v_{ob}}}$

where ${\dot {m_{of}}}{\vec {v_{of}}}$ is the momentum of the mass outflowing forward and ${\dot {m_{ob}}}{\vec {v_{ob}}}$ is the momentum of the mass outflowing backwards. In this case, these last two terms cancel so $\sum {\vec {F}}=0$ and ${\vec {a_{G}}}=0$, as expected. -AndrewDressel (talk) 15:09, 5 June 2009 (UTC)

Andrew, it looks like you are not correctly using F=d(mv)/dt. Your equation seems to claim that the rocket's momentum is constant. We know its mass is decreasing, so it better have increasing velocity. Obviously under zero net force it can't be increasing its velocity, so there's a problem here. Most people would say that in order to prevent this error we just have to include all the mass, including the expelled propellant. But now we have a constant-mass system, just as the article and footnotes say you must use. MarcusMaximus (talk) 05:57, 8 June 2009 (UTC)

I didn't explain the source clearly enough. The author explains that the equation is valid only for an instant in time, hence ${\dot {m_{of}}}$ and ${\dot {m_{ob}}}$ but $m_{G}$, without the dot. -AndrewDressel (talk) 16:22, 8 June 2009 (UTC)

What does it mean for a dynamical equation to be valid at only an instant of time, but not for all time? MarcusMaximus (talk) 17:33, 8 June 2009 (UTC)

It means that "all the terms [must be] evaluated at time t". The equation does not say anything explicit about the rocket's momentum, let alone claim that the rocket's momentum is constant. Instead, the momentum at time t must be calculated as $(m_{G}(t_{0})-t({\dot {m_{ob}}}+{\dot {m_{of}}}))v_{G}(t_{0})$ if ${\dot {m_{of}}}$ and ${\dot {m_{ob}}}$ are constant. -AndrewDressel (talk) 18:22, 8 June 2009 (UTC)

Andrew, I see what you are referring to now. You're basically taking the speed of the rocket v_G times its current mass, assuming that both m(dot)'s and v_G are constant. I think these assumptions don't allow us to say anything about d(mv)/dt generally, because you almost have to assume your conclusion to justify your assumptions. The only question this equation can answer is "What is the momentum of a rocket with constant velocity and constant mass flowing out of opposing thrusters?" It doesn't tell us what we really want to know, which is, "What is the motion of this rocket, given that I know the mass flow rate?" MarcusMaximus (talk) 00:38, 9 June 2009 (UTC)

My intuition doesn't agree with this. If the second law holds good only for systems of constant mass, then the law will be F=ma only. Not rate of change of momentum. And it does explain the propulsion of single thruster rockets-the normal ones i mean. If newton's laws don't hold there, how are you able to say that the rocket will proceed with constant velocity? Sganesh 88 (talk) 06:49, 8 June 2009 (UTC)

Sganesh, the law does define the rate of change of momentum for systems of constant mass. However, the form F=d(mv)/dt does not provide the correct equation of motion for a rocket unless you include all the expelled mass in the system for the entire flight of the rocket--an impractical and unnecessary burden that actually turns the problem into a constant-mass system. This provides a counterexample to those who claim that F=d(mv)/dt applies to varying-mass systems. Anyway, a rocket's motion is correctly described using the law of conservation of momentum, or by using F=ma, and naturally putting the thrust in F and treating m as a variable. This is precisely how I simulate the motion of rockets at work, and it gives an answers that are verified by actual flight test data.

The main thing most people don't seem to consider is that when the mass of a system is allowed to vary, the mass that enters or leaves the system carries momentum with it equal to its mass times its velocity. By this mechanism, the momentum of the system increases or decreases independent of any external force. Therefore, to define the rate of change of momentum of a system only as F excludes the possibility of varying mass. A more general law would set d(mv)/dt equal to the sum of F and the rate of momentum transfer via incoming/outgoing mass. MarcusMaximus (talk) 07:48, 8 June 2009 (UTC)

That is true, but it doesn't really justify the claim that F=ma works for varying mass and F=dp/dt doesn't. Either can be applied to, say, a rocket, as long as you do it right. Either one will give bad results if applied incorrectly. It is basically semantic. Strictly, Newton's second law applies to an individual particle, and the mass of a particle never changes, classically. It is misleading to say that F = d(mv)/dt is general, and F = ma is a valid simplification when mass is constant. But F = dp/dt does have the advantage that it remains valid relativistically. Rracecarr (talk) 18:27, 8 June 2009 (UTC)

Rracecarr,

First, I think it's ok to forget about special relativity for the sake of clarity and pedantic effectiveness in the articles about classical mechanics. Talking about F=d(mv)/dt and varying mass tends to induce people to bring up rockets, not relativity, and rockets (in my experience) are handled wrong more often than not.

Second, you are correct that either equation can be misapplied to get the wrong answer. However, the d(mv)/dt form can only be applied to constant mass objects--it requires keeping track of all the entering/departing matter and calculating the velocity of the centers of mass of all the mass particles or bodies, including those that are no longer of interest (like expended fuel or jettisoned equipment). On the other hand, F=ma can be used correctly on a varying mass object without the bother of tracking all the expelled mass flow and separated bodies--one only needs to know the external forces on the object of interest and the forces exerted by the departing mass, and simply treat m as a variable. MarcusMaximus (talk) 00:38, 9 June 2009 (UTC)

Which method is more "bother" is up for debate. Conservation of momentum gives you v1-v0=ln(m0/m1) without the bother of integrating a variable acceleration over time. The reason I think variable mass should be left out of the article is that it is philosophically messy. Defining clearly what the total force on a particle is when what the "particle" is continually changes can be confusing. What you are actually doing, implicitly, when you apply F=ma to a rocket is applying it separately to each particle that makes up the rocket (none of which change in mass). You simply stop carrying out the calculation for a given fuel particle once it is separate from the rocket. Better to keep it simple and not cloud understanding of the basic physics. Rracecarr (talk) 01:13, 9 June 2009 (UTC)

I apologize if I seem argumentative, but I think it is very important to state clearly that F=d(mv)/dt does not apply to anything but systems of constant mass. There are all kinds of otherwise smart people who think their varying masses are taken care of by using the chain rule to get a term that contains m(dot). I've had debates about this with several physicists and I've even seen physics teachers on these talk pages argue that side. This glaring error even shows up on the NASA website! MarcusMaximus (talk) 02:57, 9 June 2009 (UTC)

I guess I don't have any objection to pointing that out, if you think it's a common error. Honestly though, I don't know who could actually think the the velocity of matter entering or leaving the system could safely be ignored. I think my preference would be to phrase Newton's 2nd law as applying to a particle (where a particle comprises by definition a definite, unchanging amount of matter). Then in some section on "systems of particles" or something, rockets and fluid dynamics could be discussed, and it could be pointed out that you don't actually have to painstakingly apply Newton's 2nd law to every individual particle, but you do have to account for any momentum flux across the system boundary. Rracecarr (talk) 14:15, 10 June 2009 (UTC)

Wow, I just looked at the NASA link. That is amazingly bad. They go through all this rigamarole, with multiple errors (including moving to the rocket frame without accounting for pseudoforces), to come up with m dV = -v dm, which you can write down directly from conservation of momentum. Rracecarr (talk) 14:42, 10 June 2009 (UTC)

Well, I took a stab at fixing up the constant mass stuff and the open systems section. I found and added a good explicit reference about varying mass. I also removed that video. MarcusMaximus (talk) 07:24, 11 June 2009 (UTC)

Getting better, in my opinion. However, I think the first mention of "systems" is abrupt, awkward, and possibly misleading. Saying F=d(mv)/dt applies to systems of constant mass, when up to that point the discussion has focussed only on the effect of force on a "particle" or "object" or "body" is confusing. How do you even define "v" for a system of unconnected particles? Really, as soon as you start talking about systems, the equation needs to include summation symbols, and I think there should be at least a sentence smoothing the transition from talking about a single particle to a system of particles. Rracecarr (talk) 15:14, 11 June 2009 (UTC)

I dare say that MarcusMaximus studied Fluid Mechanics if he is working with rockets and will acknowledge that stating: F plus rate of momentum transfer equals d(mv) / dt is just a way of writting flow contributions on the opposite site of the standard practice in Fluid Mechanics (compare your claims with Navier-Stokes equations, that is to say: "Newton Second Law for continuous systems").

Laws of motion

1st law

Mass always changes its values unless it is in certain conditions but we ignore such changes and what about the radioactive elements which decay at every interval of time and thus reduces its mass while there is no external force acted upon it. Free neutron is the another example which is unstable and decay under 15 min. Similarly no external force was involved in creation of universe (bigbang).

2nd law

Let a body of mass of 1 kg on the surface of ground. We say F=W=mg. Is it possible for a body at rest to have acceleration “g” unit = m/s/s, Is there any rate of change of velocity”?

Nothing can travel at speed greater than speed of light c but if we put certain values of m and F in equation F=ma then we can get a speed viz greater than c provided there is no other attraction. e.g. F=150,000 N and m=0.5 kg. (a= rate of change of speed) but mass doesn’t remains constant if it travels close to the speed of light.

Similarly, if a body is moving in space with constant acceleration 1 m/s/s . It means that its speed increases with 1m/s at each second till it reaches to the speed of light. is it possible ?Further, gravitational time dilation is the effect of time passing at different rates in regions of different gravitational potential which was tested/ confirmed with difference of nanosecond recorded by atomic clocks at different altitudes. Thus if time does not remain constant how can we say that an object of mass is moving with v or a (in constant t).

3rd law

To every action there is equal but opposite reaction but in newton’s law of gravitation the resisting force of falling mass or reaction of gravitational pull is missing. Similarly in creation of the universe (bigbang) there is action but reaction is missing. 96.52.178.55 (talk) 04:59, 15 February 2009 (UTC) zarmewa khattak

This is newton's laws of motion, they don't hold at high speed or the special cases you describe. This was described by Einstein, see special relativity and general relativity.

Newton's laws refer to constant mass systems, if you want to model systems of changing mass, you have to break up the system into smaller "particles" so that you account properly for all mass.

In the case of a body resting on the ground there is a force F=mg acting on the body, but the ground is also exerting a force canceling out the one caused by gravity from earth. So the net force is 0, and thus the acceleration is 0.

In the case of a falling mass (e.g. an apple) the earth causes a force on the falling body, but the falling body causes an equal and opposite force on the earth. The body's mass (m) is small in this case, so the acceleration of it (F/m) is relatively large. However, earth's mass (M) is enormous in comparison, and thus the acceleration of the earth (F/M) is negligible when the force is of the same magnitude. If the other body has a mass comparable to that of the earth, the acceleration of the earth would be noticeable.

In the case of big bang, everything is pushing on each other and expanding, there is no problem there either. If you wonder what caused big bang, no one knows.
—Apis (talk) 11:24, 17 February 2009 (UTC)

You said net force = 0 and thus acceleration is 0. Then how we calculate weight of an object i.e. w=mg

The steel bar ruptures when all internally developed resisting forces fail to resist the applied force. Thus if we apply the same analogy to the paradox, the falling mass (Newton’s apple) should rupture first before falling on earth (bigger mass).

Similarly, you said that falling body causes an equal and opposite force on the earth. You also said g=a=F/m and g=a= F/M .This means that weight of 1 kg of sphere is decreases when it’s mass increases on the surface of earth.

Similarly, you also said acceleration of the earth would be noticeable if an object mass comparable to earth. So do we notice such accelerations in case of celestial bodies?

“in case of big bang everything is pushing each other and expanding”. This means someone is wrong either Newton (law of gravitation) or Sir Einstein. 96.52.178.55 (talk) 19:25, 22 February 2009 (UTC) K

You have some misunderstandings. I'll try to help with some of this: First, you are not quite right that "g=a". It is true that the acceleration due to gravity is equal to g, but that doesn't mean I'm not experiencing gravity unless I am accelerating. Almost to the contrary: if I were in an elevator accelerating down at 9.8 m/s², I would feel weightless within the elevator. I would still have the same mass but my weight with respect to the elevator would be zero.

We measure the weight of an object by measuring the reaction force on a scale. Gravity pulls me down with a force of mg; the scale pushes back (the reaction force); we measure the reaction force.

If nothing were under me, I would accelerate with an acceleration of g. The equal-and-opposite reaction would be the earth accelerating toward me at a rate of F/m_earth where the force on me and on the earth is equal (same amount of force) but opposite (we are pulled toward one another) but the accelerations are drastically different because the earth weighs so much.

As for "in the case of [the] big bang...", this is also not a problem. If you have two people on ice skates pushing on one another, they will accelerate away from each other; this is simple Newtonian mechanics. —Ben FrantzDale (talk) 15:05, 23 February 2009 (UTC)

For every action...

As an amateur student of quantum physics (my PhD is is in cognitive nuerobilogogy) and also one who researches parapsychology, I have always had trouble getting my mind around this particular law. The quotation is indeed simplified but on the surface, in day to day life and in the field, would seem to be self defeating. It is already established that there are n absolutes in physics, that time is not linear and non local and that the actions we take now can effect the past, present and future. How does Newtons law fit in with quantum theory? Dr. K. M. —Preceding unsigned comment added by 74.13.83.108 (talk) 18:30, 17 April 2009 (UTC)

"It is already established ... that time is not linear and non-local and that the actions we take now can effect the past, present, and future," is an EXTREMELY QUESTIONABLE STATEMENT, and that it is probably best to dismiss it out-of-hand, especially the "already established" part. I don't think that you know what "alredy established" means, so you need to go back and work on that problem a lot more.98.67.166.234 (talk) 07:26, 12 November 2009 (UTC)

For every action...A Question?

I have an even more pressing point... isn't the third law incorrectly stated here? "If Body A exerts a force on body B, body B will exert a force equal and opposite in the same plane". But this isnt true... they work on different bodies.

Yes they do work on different bodies, as the passage you quote states: one force works on body B and the resulting reaction force works on body A. -AndrewDressel (talk) 21:50, 3 May 2009 (UTC)

Take the "weekly shopping" example: When you push a trolley forwards, you are exerting a forwards force on the ground, but it is the TROLLEY that exerts a backwards force on you... Woodyjojo (talk) 21:09, 3 May 2009 (UTC)

You are mixing two pairs of action/reaction forces. First, when you push on the ground, the ground pushes back with an equal and opposite force. Second, when you push on the cart, the cart pushes back with an equal and opposite force. -AndrewDressel (talk) 21:50, 3 May 2009 (UTC)

I think that it is good to emphasize that between the "action" and the "reaction" there is no time delay. The both happen simultaneously, and in cases that don't involve the Special Theory of Relativity, we know what "simultaneous" means. Simultaneous! Simultaneous! This is something about the subject that bothered me as a freshman college student - because the Professor did not emphasize the fact. (I found that at my university, Physics professors frequently did not put emphasis on what was important, and that left us students to struggle with it.98.67.166.234 (talk) 07:31, 12 November 2009 (UTC)

Proportionality changes to equality only by choice of units

The text correctly states the second law as "... the net force on a particle of constant mass is proportional to the time rate of change of its linear momentum" but incorrectly summarises it as F = d(mv)/dt.

Newton's proportionality only changes to the usual equality when units are particularly chosen to render the proportionality constant as unity, such as kilograms for mass and Newtons for force. Nh5h (talk) 04:37, 20 May 2009 (UTC) Charles

Correct! Indeed, F=kma, where k is a constant of value 1 is technically correct, and you are right about the choice of units causing the value of k to be 1. 1812ahill (talk) 15:32, 4 November 2009 (UTC)

Simpler introduction

I had a semester of quantum mechanics and a semester of relativity in college, but I've forgotten most of it. When I looked up this entry to refresh my memory, I couldn't get anything from this introduction.

Under WP:MTAA, the entire article should be understandable by a non-specialist. That isn't always possible, but certainly the introduction should be understandable.

It should explain what the three laws are in language that a non-specialist can understand. An intelligent layman, or a high school student taking science for the first time, is a non-specialist.

Does anybody have a suggestion for an easy-to-understand introduction? Does anybody object if I try? Nbauman (talk) 14:08, 31 May 2009 (UTC)

The proper formulation of 2d Newton law will be force to acceleration ratio stays the same if body is stiil the same (empiric law). It is definition of inertial mass. Another comment - this system of differential equations has no analytical solution in 99% of cases. —Preceding unsigned comment added by 140.168.71.18 (talk) 03:37, 1 June 2009 (UTC)

I attempted to make an improvement, and Xxanthippe deemed it unhelpful, but he didn't say why. I can perhaps see the appeal of stating the second law as simply F = ma and avoiding big words such as proportional, but I hope to gain consensus for replacing "action and reaction" with "forces always come in equal and opposite pairs" or the equivalent. The fact that "action and reaction" are archaic and misleading is demonstrated by the need always to clarify that Newton meant forces, as the current article does at least twice, and readily confirmed by external sources. - AndrewDressel (talk) 01:08, 2 June 2009 (UTC)

So, we currently have:

Newton's laws of motion are three physical laws that form the basis for classical mechanics. They are:

1. A body at rest stays at rest, and a body in motion stays in motion, unless it is acted on by an external force.

2. Force equals mass times acceleration (F = ma) (or alternately, force equals change in momentum), and

3. To every action there is an equal and opposite reaction.

I think 1 is fine. I would rather see a little more detail in 2: Force on what, mass of what, acceleration of what? My comments about 3 are above. I believe in the current form it is oversimplified to the point of being meaningless. The fact that force A on body 1 from body 2 has an equal and opposite counterpart, force B on body 2 from body 1 is crucial. -AndrewDressel (talk) 17:38, 7 June 2009 (UTC)

If you had too much detail it wouldn't be an introduction. I don't mind changing it, but under WP:MTAA, it has to be simple enough for a non-specialist to understand.

How do the introductory physics textbooks describe it? I suspect that this isn't the kind of thing a Wikipedia editor can just write out of his head. You have to see how the experts describe it. Nbauman (talk) 04:02, 8 June 2009 (UTC)

While that might prove fruitful, I wouldn't be surprised to find most introductory textbooks missing the mark. They have a captive audience and the rest of the chapter or text to straighten things out. I had to look through a few to find one that agrees with my point that expressing the 3rd law as "action and reaction" is not helpful. You are holding this article to a higher standard, which I think is a good thing. -AndrewDressel (talk) 13:02, 8 June 2009 (UTC)

One thing I've noticed after checking several texts is that I haven't seen an author try to explain them all at once. For example Halliday and Resnick introduce one law at a time and spend a page or two explaining it before moving on to the next. -AndrewDressel (talk) 16:34, 8 June 2009 (UTC)

Let's try to piece it together word by word:

1. A body at rest stays at rest, and a body in motion stays in motion, unless it is acted on by an external force.

a. mentions body to which law applies

b. mentions staying at rest

c. mentions staying in motion (maybe should also say "constant motion")

d. mentions external force (maybe should also say "net external force")

2. Force equals mass times acceleration (F = ma) (or alternately, force equals change in momentum), and

a. perhaps should use momentum first, since that's how Newton said it

b. probably has to include famous (infamous?) F = ma, but maybe only in parentheses along with corresponding text.

c. should mention body, which has mass, which experiences change in momentum or acceleration, and on which force acts.

d. should clarify that "force" means "external net force". "External" can't be too advanced if it gets mentioned in first law.

e. What about "constant mass"? I'm not sure.

f. perhaps should use the word proportional, as Newton did.

3. To every action there is an equal and opposite reaction.

a. mentions "equal and opposite", as it should

b. Should use "force" instead of "action", as laws 1 and 2 do, and as I try to explain above.

c. Should mention bodies that create the forces and that the forces act upon. Even Newton himself followed up in the same sentence with "corporum duorum". As Feynman explains in his lectures, Volume 1, Chapter 12, Section 1, What is a force?, "one of the most important characteristics of force is that it has a material origin."

d. The familiar "action reaction" expression can be provided in parentheses just as I suggest "F = ma" be for the 2nd law.

That leaves me with:

1. A body at rest stays at rest, and a body in motion stays in constant motion, unless it is acted on by a net external force.

2. The net external force on a body is proportional to its change in momentum, or more commonly, force equals mass times acceleration: F = ma.

3. The forces between two bodies are always equal and opposite, or more commonly, for every action there is an equal and opposite reaction.

Let the critiquing begin! -AndrewDressel (talk) 13:53, 8 June 2009 (UTC)

The first filter I use for a Wikipedia introduction is (on the authority of WP:MTAA) whether it is simple enough for a reader who is looking for an introduction; for example, a high school student studying physics for the first time.

If it's confusing to me, then it doesn't pass that filter.

If I have to stop and ask myself, "What exactly does that mean," then it doesn't pass that filter.

At Francois Monad's Nobel prize presentation, the speaker quoted Monad as saying, "In describing genetic mechanisms, there is a choice between being inexact and incomprehensible." and said, "In making this presentation, I shall try to be as inexact as conscience permits." [1]

An introduction has to be simple enough for the intended reader to understand it. If you try to be too exact, you can fail that test.

Sometimes you have to oversimplify in the introduction, and clarify it later. Nbauman (talk) 14:27, 8 June 2009 (UTC)

That might be all well and good, and I think I've read it before somewhere, but it doesn't say anything about the current proposal. -AndrewDressel (talk) 14:57, 8 June 2009 (UTC)

Overall I like your simplified version, Andrew. I think at this high level, though, it is even more effective to leave out words like "net" in front of "force", and possibly even "external", because I think the casual reader doesn't form a distinction between unmodified "force" and the highly qualified "net external force".

I think the third law is hard to understand because it deals with about 12 things in very short order: two bodies, two forces, exterted by which body, on which body, in which direction is each force, and of what magnitude? The bodies, directions, and magnitudes are fairly straightforward and are stated in the law. The biggest points of confusion seem to be the "by which" and "on which", because they are not clearly stated in the law. In practice this means students are unable to comprehend the difference between a 3rd law force that impedes all motion by giving a net force of zero (like pressing against a wall), and a 3rd law force that is commonly called an "inertial force" that is the natural companion of acceleration (like pushing a cart). Students know there are two forces that are equal and opposite somewhere in their free body diagram, but they don't know which body is exerting which force on what. MarcusMaximus (talk) 01:23, 9 June 2009 (UTC)

This sounds like WP:OR. Sources will be needed for such material in the main article. Xxanthippe (talk) 03:10, 9 June 2009 (UTC).

I'm just explaining here on the talk page what I think the challenges are to create a good front page article. MarcusMaximus (talk) 03:16, 9 June 2009 (UTC)

I agree with MarcusMaximus that adding terms like "net" and "external" make it confusing to the ordinary reader. What's the difference between "force" and "net force"? I don't know myself.

When I read an introductory paragraph for the first time, I ask myself what each word means. If I have to ask myself what "net" means, and what "external" means, it's much more difficult for me to understand, if I can understand it at all. Nbauman (talk) 17:23, 9 June 2009 (UTC)

I agree 100% with Nbauman. The whole article is simply a disaster in terms of intelligibility to the general reader. And it's not just a disaster because it's trying to be exact. It's a diaster because it's disorganized, internally inconsistent, and logically muddled.--76.167.77.165 (talk) 01:40, 8 October 2009 (UTC)

Newton's second law

(Copying this from my personal talk page)

Here's your WP:RS:

"According to Newton's second law, force is the time rate of change of the momentum"

[2]

Rracecarr (talk) 15:02, 19 June 2009 (UTC)

I saw that when I did a Google search. That's a blog. Blogs are not a WP:RS, according to WP:SPS. Who is Leonardo Motta? He's just some guy without any stated credentials who posts for free.

Besides, Motta himself also refers to it as "rate of change of momentum" in that link. So it's just as accurate to say, "rate of change of momentum" as "time rate of change of momentum", according to your own source, and it's simpler.

If you could find a well-known physicist or physics teacher using that phrase while addressing introductory students, I would accept it. Nbauman (talk) 15:26, 19 June 2009 (UTC)

NBauman, if you do a google search with "time rate of change" in quotes, you get hundreds of results from class lecture notes, homework problems, and textbooks in calculus, physics, and engineering, illustrating its widespread use to mean "the rate of change with respect to time". I didn't even think this would be a point of contention; it's in common usage.

[3][4][5][6][7][8][9]

And in particular, Richard Feynman uses it: [10]. Look in the Acceleration section on page 8 of that document to see the statement "Acceleration is defined as the time rate of change of velocity."

At the very least, it should be obvious that "rate of change" is ambiguous because it doesn't specify with respect to what. If you accept both as correct, and the rest of us only accept one, then we should go with the one that all of us accept. MarcusMaximus (talk) 22:30, 20 June 2009 (UTC)

First of all, blogs don't count, and neither do class notes WP:ELNO WP:RS. I am glad to find the chapters of Feynman online, since my copy is buried in storage right now.

I agree that it is accurate to say that force is the time rate of change of momentum in Newton's second law.

The issue is whether it is the best way to introduce the concept to an ordinary reader -- for example, a student studying physics for the first time.

There are lots of concepts in physics that are true but too complicated to explain to a beginner. A good teacher knows what to include and what to omit.

The idea that the change in momentum is the time change of momentum is true but it's not essential to understanding the relationship of force to momentum.

I think anyone reading this for the first time would understand it's about time. After all, velocity is the change in distance per unit time.

Feynman specifies the time rate of change because he's giving a precise definition -- and he's not explaining force to a student who's learning about it for the first time. He's emphasizing time. And apparently he got the point across to you.

I'm not disputing the facts. I'm arguing that there are only so many ideas you can put into an introductory sentence before a reader who is new to the topic gets confused. The question is, do you introduce the concept of time in the introduction, or do you point it out further down after the reader has understood the introductory concepts. It's a question of pedagogy.

I'm not going to change it. I'll wait to see if somebody else comes along and decides to to change it for the sake of clearer reading. Nbauman (talk) 00:14, 24 June 2009 (UTC)

Why do you think time rate of change is unclear? MarcusMaximus (talk) 01:21, 24 June 2009 (UTC)

Where can we apply first law?

It is practically impossible for a single mass to exist in universe and all other masses are always under the influence of gravitation (external force) unequivocally 96.52.178.55 (talk) 06:13, 10 July 2009 (UTC)Quoth Khattak #1

That is a good question. It is primarily used as a qualitative statement of principle, which allows us to say that "the only things causing this object's motion to change are these forces". There are also many situations where gravity is negligible compared to other forces. MarcusMaximus (talk) 01:07, 11 July 2009 (UTC)

Reordered discussion of 2nd Law

I changed the order of the sentences in the discussion of Newton's 2nd law in so it is introduced in a more logical way, at least in my opinion. I'd like to hear feedback on what other editors think. MarcusMaximus (talk) 03:30, 13 July 2009 (UTC)

1- Change in motion is proportional to force impressed and impulse impressed can only produce motion if greater than the resisting force of an object. So is this initial amount of impressed force which was used to overcome the resisting force, excluded or included in the mentioened impressed force viz proportional to change in motion? What is true net picture of this impulse impressed, should it be subtracted if included or added in case excluded or otherwise should be justified clearly if remains the same?

2- Here is the sentence in the article to which I want to draw your attention “The product of the mass and velocity is momentum (which Newton himself called "quantity of motion"). Therefore, this equation expresses the physical relationship between force and momentum.”

Momentum (mxv) and motion are two different things. Momentum is the phenomenon which occurs after the cessation of impulse impressed or it is a potential control exists in a body while in the article the change in motion requires an impulse impressed. Thus "momentum" which is mentioned again and again doesn’t jibe with "motion" lucidly in the article. Since this is caption of Newton’s second commandment of Motion not momentum therefore either momentum should be culled or it needs more careful revising.

I hope I have explained things clearly enough that you can understand what I mean.96.52.178.55 (talk) 05:57, 16 July 2009 (UTC) Khattak #1

Editors who don't read the article

Two editors criticised the article, one with a 'clarify me', and the other with a peremptory hiding of some of the material.

There are signs that neither of these editors actually read even the relevant parts of the article, let alone the whole thing, nor its existing inline references (which already included an online link to relevant source immediately available for download with a few clicks of the mouse).

The 'clarify me' edit modified a line that said "A more direct translation is:", and added "Clarifyme date=February 2009", asking "compared to what?" But on the lines immediately above this comment, can be seen the Latin of Law III and its English translation. Duh!!! It can't have been too hard for the editor to actually read the lines that went immediately before, and to notice that this was the translation referred to.

Another editor (earlier today) thought it was appropriate to give this order: "Unless someone can find out where this translation came from, this whole area will be hidden from view until further notice." This was accompanied by another order about the required format of the reference.

The editor then hid a lengthy passage of the article. This action made the immediately-following paragraph nonsensical, because the article went on to explain Newton's use of the expression 'motion', and this usage occurs in the hidden paragraph, but nowhere else that was immediately relevant. In this way, the edit mutilated the article, depriving it of part of the sense that it previously had, apart from the material effectively deleted. The editor clearly did not read the next-following paragraph(s) to check whether the proposed deletion would create a need for consequential amendments to keep the article making sense after the edit.

Also, an existing earlier inline reference already and clearly gives a whole source for the Principia English translation of Andrew Motte, available online. The challenged translation is to be found as part of that. The given source is available in full online, with a link that was already supplied, and the source does actually include a 'contents' page. There was nothing to stop the editor who made the effective deletion from actually reading the lead paragraph of the article to be edited, clicking on the link to the given reference, and checking the content of the reference already supplied before presuming a lack of source and effectively cutting the visible text.

Now there's no ownership on WP, but constructive editing does mean trying to improve the articles. If an editor thinks that some of the material in the article is significantly defective and needs deletion, but has no time to verify anything, and no time to make the revised text consistent and meaningful, it seems reasonable to leave it to somebody else to do the checking; but in that case, it would also be constructive editing, as well as considerate towards other editors and users, not just to immediately cut and slash, but first to make a note of the possible problem on the talk page.

In response to the first comment, I edited just now by spelling out the obvious place where the comparison translation is to be found. In reponse to the hiding, I put in some slightly more convenient links, but the content pointed to is already there in the online source provided by the pre-existing lead paragraph. Terry0051 (talk) 14:15, 6 September 2009 (UTC)

I was actually quite pleased that that paragraph was hidden. Why do we need lengthy, opaque passages from the Principia? Strad (talk) 20:06, 6 September 2009 (UTC)

[From Terry0051] That's a different question entirely. By all means, shorten the description to improve it by leaving it clean, coherent and with all its essential facts. That's not what happened. What happened was slash and cut for a reason not completely articulated but to the extent articluated not applicable, leaving references hanging and a lack of meaning and clarity. Terry0051 (talk) 01:14, 7 September 2009 (UTC)

Yes, I agree. It doesn't make a lot of sense to remove good but unsourced material from an article, especially when finding the source is as easy as pasting a passage of the quotation into Google. It's not responsible to remove material and render the remaining material nonsensical. Strad (talk) 01:38, 7 September 2009 (UTC)

problems

There are a lot of problems with this article:

1. It needlessly presents each law in a bunch of different forms. E.g., the first law is presented at least four times, depending on how you count: "In the absence of..., " "There exists a set of ...," "Corpus omne perseverare...," "Every body persists ..." This is just sheer sloppiness and disorganization (and maybe showboating, in the case of the Latin).
2. Almost no common, everyday examples are given.
3. A vast amount of attention is given to relatively unimportant issues, such as how to deal with varying mass.
4. Like a lot of WP articles on math and science, it reads like it was written by a grad student who wanted to show off how smart he was, rather than by someone actually trying to communicate with the typical reader of WP.
5. There are errors, e.g., at "There is a class of frames..."
6. The article is strewn with statements to the effect that xyz is true, except in relativity. It doesn't need to say this in every single place where a statement is made that fails in relativity. Also, the article presumes the interpretation of relativity in which mass varies; it's much more common these days to take mass as a constant (a Lorentz scalar, equal to what used to be called the "rest mass"), and construct four-vectors like p=mγv. And in any case, it's naive and/or wrong for the article to present relativity as a bunch of correction factors to be thrown in to equations like Newton's laws; what's really different about relativity is its depiction of the non-absolute nature of spacetime, which is completely different from the Newtonian concept.
7. There is virtually no discussion of the empirical evidence that would have led Newton to form the laws, or that verifies their validity (within their realm of applicability).

--76.167.77.165 (talk) 01:30, 8 October 2009 (UTC)

Be bold. Strad (talk) 02:09, 8 October 2009 (UTC)

[From Terry0051] Some of the solutions advocated by 76.167.77.165 do not seem to be mutually consistent. For example, calling it "showing off" (or similar) to give the original form of the laws, but at the same time asking for "the empirical evidence that would have led Newton to form the laws, or that verifies their applicability". The latter really calls for the original form of the laws to be given too, otherwise anachronism and confusion are likely to result. Terry0051 (talk) 08:07, 8 October 2009 (UTC)

Spelling error - "ovelaping" ---> overlapping

That's it. Easily found using Ctrl+F. Cheers :-) —Preceding unsigned comment added by 118.93.95.162 (talk • contribs)

Fixed. Mindmatrix 12:51, 14 October 2009 (UTC)

Strange wording of 1st law.

In the 'The three laws' section below the introduction, it says: "A body persists its state of rest or of uniform motion unless acted upon by an external unbalanced force." I presume this is a typo and will ammend to 'persists in a state'. 1812ahill (talk) 15:43, 4 November 2009 (UTC)

Metric simplification

I see there is a past discussion of this subject in archives 1 but the issue still holds.

Stating that the second law can be expressed F = ma is an implicit simplification. Newton did not say this, he defined a relationship in terms one of proportionality, not equality. That is to say, F ∝ ma or if you prefer F = kma where k is some constant. F = ma is not some fundamental physical truth but a trick of units (the Newton is defined in terms of this equation making the constant 1). However, this is only holds in the metric system.

This is a point worth making since I have seen many people confused by it, for example using pounds for mass and pound-force for force and asking why they get nonsensical results. In my view we don't need to labour this point and we certainly don't need to start littering equations with constants all the way through the article, but we should at least acknowledge that missing step to get to F = ma and that it is a trick of units rather than some fundamental physical truth. CrispMuncher (talk) 14:53, 30 November 2009 (UTC)

I wouldn't call it a trick of units nor does it only hold true for the metric system. F = ma is true independent of units. To use the formula in numerical calculations we have to use coherent units. The SI gives us coherent units but we could also use pounds mass, poundals and feet per second per second or slugs, pounds force and feet per second per second. To use pounds mass and pound force acceleration must be in "g"s. JIMp _talk·cont 20:02, 11 January 2010 (UTC)

Verbatim?

Are the laws here verbatim to what Newton stated? IceBlade710 (talk) 04:11, 7 January 2010 (UTC)

Well, as far as I can see in the present text of this article, each of the laws is stated in a number of ways, including Newton's original. But if you want to read more directly how he put the laws, including his comments on each of them, you can go here, to the original English translation (of Newton's Latin). (If you did really want the original Latin, there are links in the article on the 'Principia' :) .) Terry0051 (talk) 22:30, 11 January 2010 (UTC)

Should momentum (P) be upper case or lower case?

I made recent change introducing the variable "P" as momentum and also making it upper case as indicated in the [Momentum] article: "In physics, the usual symbol for momentum is a uppercase bold P". This change was immediately reverted with an indication that I was simply wrong. Upon review, I have noticed that the momentum article is apparently not internally consistent. However, the particular fact that I was referencing from the Momentum article is sourced (apparently by a printed textbook). A textbook that I have at home uses a stylized lowercase "p" (possibly Greek, but I can't tell).

What would be the definitive source for this detail?

1. (cur) (prev) 05:25, 14 September 2009 Xxanthippe (talk | contribs) (33,918 bytes) (Reverted good faith edits by Nobar; Does not seem to be the case. (TW)) (undo)
2. (cur) (prev) 03:51, 14 September 2009 Nobar (talk | contribs) (33,932 bytes) (→Newton's second law: Momentum (P) should be represented in upper case: see Momentum) (undo)

--Nobar (talk) 06:34, 14 September 2009 (UTC)

Feynman, Kleppner & Kolenkow, and Serway & Faughn use p for the momentum of a particle. Halliday & Resnick probably do as well (the 1970 edition of their Fundamentals of Physics uses p, at least; I don't have access to Physics). Halliday & Resnick (1970 Fundamentals) and Kleppner & Kolenkow specifically reserve P to mean the net momentum of a system of particles. Strad (talk) 04:42, 16 September 2009 (UTC)

I confirm the correctness of the above comment. Lower case p is used generally for the momentum of an individual particle see, for example, Dirac's book on Quantum Mechanics. WP editors should not use WP as a source for their WP edits. Xxanthippe (talk) 05:15, 16 September 2009 (UTC).

There is simply a general tendency to use lowercase letters rather than uppercase ones. Therefore you see p more often than P. There is no standard, widely recognized distinction between a lowercase p meaning one thing and an uppercase P meaning another thing.--76.167.77.165 (talk) 01:38, 8 October 2009 (UTC)

No, lower-case "p" is the standard widely-recognised symbol for momentum. JIMp _talk·cont 21:15, 14 January 2010 (UTC)

Laws

Maybe and easier way to describe the laws would be:

Newtons 1st law: Every object continues in its state of rest, or uniform motion in a straight line unless acted upon by and external force.

Newtons 2nd law: The external force (F) acting on a body is proportional to the product of its mass (M) and the acceleration (A) produced by that force. F=ma

Newtons 3rd law: Every action has an equal and opposite reaction. —Preceding unsigned comment added by Pricipalsofflight (talk • contribs) 03:34, 21 March 2010 (UTC)

Newton's 3rd law for bodies in motion

Does Newton's third law apply to bodies in motion ? (not in a state of equilibrium).

SPECIFICALLY:

A 1 ton box falling in a column of almost empty air, containing only a few air molecules ?

The box will fall under gravity and exert a force (equal to its weight) on the air molecule.

The air molecule will exert the SAME force back on the box.

By newton's third law.

Therefore the box should not move at all.

UNLESS the air molecule moves.

But then the system is not in equilibrium and newtons' third law cannot really be held to apply to this system as a whole, correct ?

A body in motion is in equilibrium, it is impossible to distinguish between a body in motion and a body that is stationary as speed can only be determined relative to another body.

The force between the box and the air molecules will only be equal to the weight of the box when the box reaches its terminal velocity, at this speed the forces between the box and the air will equal the weight of the box. The box will then stop accelerating and continue falling at a constant speed. The air molecules will move, the total force on them causing them to accelerate will equal the total force on the box. Rolo Tamasi (talk) 08:36, 22 February 2008 (UTC)

Newton's 3rd Law still holds. Newton's 3rd Law refers to pairs of forces: if object is exerting a force on object 2, object 2 is exerting a force on object 1 that is equal in magnitude and opposite in direction. The presence of acceleration does not alter this. In your example, when the falling box hits an air molecule it exerts a force on the air molecule (that is not equal to the weight of the box) and the air molecule exerts the same force back on the box. This does not result in zero acceleration though since the two forces are not acting on the same object. As for the box, there is the force of gravity acting down and the force due to the air molecule acting up. The mistake in your example was assuming that the force of the box on the air molecule equals its weight. PhySusie (talk) 11:57, 22 February 2008 (UTC)

Except, at terminal velocity the force does equal the weight of the box and 3rd law still applies. 128.91.26.30 was therefore not wrong to say the force equals the weight as long as it is appreciated that this is a specific not a general occasion. I think the error (in addition to the error that a body in motion is not in equilibrium) was to say that when this happens the box should not move at all. That is wrong, when the force equals the weight the box stops accelerating it does not stop moving. Rolo Tamasi (talk) 20:21, 22 February 2008 (UTC)

PhySusie said: "The mistake in your example was assuming that the force of the box on the air molecule equals its weight". My question: Why is the force on the air molecule not equal to the weight of the box ?

To Rolo: I am SPECIFICALLY talking about 1 SINGLE air molecule in my example, so why are you bringing up "air" and terminal velocity ?

The reason I am talking about terminal velocity is that is the only time when the force on the box from impact with the air is equal to the weight of the box, which was the condition stated.

But now I understand your scenario. We are not really talking a bout a box in air at all. We are talking about a single collision between two objects of hugely different mass.

As this is a momentary incident the fact that the box is being accelerated by gravity is also irrelevant. The only relevant issues are that the box and the molecule have a speed differential and a big mass differential and that they collide.

The question is what is the force of that impact? There is little reason why it would be equal to the weight of the box (or, equally, the weight of the molecule).

The calculation of the force of the impact is not easily determined because it depends upon the nature of the collision, specifically the time it takes. If it is zero time the force is infinite but for a zero time – pretty unhelpful stuff!

What we do know is the force will be applied equally to the two objects and thus the effect of the impact on their speeds will be inversely proportional to their relative masses. If we knew the speed differential and the mass relativities and assumed no energy was absorbed by the impact we could calculate the speed changes.

Thus we can see that the molecule will “ping” off the surface of the box experiencing a substantial acceleration while the change of speed of the box will be almost undetectable.

Hope that helps. Rolo Tamasi (talk) 10:49, 23 February 2008 (UTC)

24.125.237.100 (talk) 00:48, 17 April 2010 (UTC)this page doesnt help in 8th grade physical science24.125.237.100 (talk) 00:48, 17 April 2010 (UTC)

Rolo: Thanks for that very illuminating post. I think I get it now ! —Preceding unsigned comment added by 71.242.39.174 (talk) 23:41, 23 February 2008 (UTC)

Your example appears to confuse force with acceleration. Consider the second law F=ma. If the forces on the air molecule and the box are equal and opposite, then the net change in the acceleration of the box will be negligible. You can just check this with some plausible numbers if you don't believe it. Silly rabbit (talk) 03:47, 23 February 2008 (UTC)

THERE'S A BASIC PROBLEM IN YOUR INTERPRETATION! There are 2 pairs of 3rd law forces (which ARE equal and have opposite directions): 1) Weight exerted by earth on box vs force exerted by box on earth (according to gravitational law) 2) Draw force exerted by molecules on box vs force exerted by box on molecules (which makes them move turbulently) THE DRAG FORCE IS NOT RELATED TO THE WEIGHT AND DON'T HAVE TO BE EQUAL!!!! THE ARE BOTH EXERTED ON THE SAME OBJECT SO THEY CANNOT BE A 3RD LAW PAIR! This confussion is very common. Even our dear friend from NASA (see video on 3rd law) makes this mistake. —Preceding unsigned comment added by Andinosa (talk • contribs) 09:50, 16 April 2009 (UTC)

Edit request from Elobroxium, 3 May 2010

{{editsemiprotected}}

I would suggest adding a sentence to the end of the section on Newton's first law. Following "....and Newton's second law does not hold in the form F = ma.[12]." I suggest adding

"Newton's first law is consistent with his second law but is not a consequence of it."

The purpose of this addition is to make it clear that the two laws are independent axioms of Newtonian mechanics. I have noticed that the question sometimes arises in physics forums as to whether the first law is not a consequence of the second. Also some academic publications suggest this. Standard physics textbooks avoid answering the question directly by discussing the concept of inertial frames. So the question is never fully answered.

One reference on the internet that discusses the problem thoroughly is http://www.quartets.de/acad/firstlaw.html and I would suggest your viewing this article when considering my suggestion. Elobroxium (talk) 10:21, 3 May 2010 (UTC)

This will need a check from an expert; I will get one here ASAP. For now, I will cancel outr the {{editsemiprotected}}, but I'll check back here. Thanks for the suggestion, more soon.  Chzz  ►  13:43, 3 May 2010 (UTC)

No, I don't think the addition is needed, and I think the one sentence addition proposed would only confuse people more. There are several ways of resolving the apparent paradox but, I must admit, it is not a problem I've ever come across in teaching Newton's laws to teenagers, which I did for several years. Newton's first law is only "Newton's" law in that he was the first to state it clearly and succinctly, and also the first to realize certain consequences of it: as the article (and Newton himself) point out, it had already been stated allegorically, not least by Galileo. If you don't accept the first law (say, because you're an Aristotelian), you won't accept any of the rest. The second law goes a lot further: it says that the acceleration is proportional to the force (rather than to the square of the force, for example) and that it is in the same direction as the force. The second law could be disproved without affecting the first law: in that sense, it is the more fundamental (even if General Relativity poses some interesting questions as to "what is a straight line?") Physchim62 (talk) 15:43, 3 May 2010 (UTC)

I don't see how this addition is necessary, and frankly I don't see its relevance to the aritcle. And as Physchim62 points out, I think this would only serve to confuse a the lay reader. Also, I am thinking that this is an obscure issue, in other words, not really a mainstream view. However, whether or not it is an issue that comes up from time to time, I think this is an anecdote that has only historical value, and belongs with one of the history articles. This is not a history article. Also, in order to add it to a history article I think some verifiable sources would have to be provided. In any case, thanks for being willing to contribute. ----Steve Quinn (formerly Ti-30X) (talk) 02:21, 4 May 2010 (UTC)

I think this would make a fine addition, though not necessarily in this form, if we can find a solid reference that demonstrates notability. -AndrewDressel (talk) 18:59, 4 May 2010 (UTC)

As I wrote on your talk page, thanks for fixing that link. Now, maybe, I can get a better understanding of the issue. ----Steve Quinn (formerly Ti-30X) (talk) 00:50, 5 May 2010 (UTC)

I think that the suspicion that Newton's first law is merely a consequence of the second is more widely held than Physchim62 suggests. http://www.quartets.de/acad/firstlaw.html quotes "The Cambridge Companion to Newton" as one instance of this mistake. In that book the redundancy of the first law is not discussed but rather accepted. I also found a very lengthy discussion of the question in "The Philosophy of Science" by J. Earman and M. Friedman (published by University of Chicago, starting on page 345) in the philosophy section of the library. In the mathematics section I found a reference to the problem (but no resolution of it) in a footnote on page 49 of Louis Jagerman's "The Mathematics of Relativity for the Rest of Us". In a physics forum I found a thread discussing the problem in http://www.physicsforums.com/archive/index.php/t-165100.html.

Physics textbooks avoid even raising the question because the modern approach is to treat the first law as a definition of an inertial frame, but for people who are not professional physicists the argument, given on the webpage of "quartets" and which all these references above assume, is quite convincing.

I think that your article in Wikipedia should, even if only in a footnote, mention the problem of the apparent redundancy of the first law and give its resolution.

Elobroxium

Elobroxium (talk) 10:01, 6 May 2010 (UTC)