Which formulation of Newton's second law is more fundamental?

[Joans]

Which formulation of Newton's second law is more fundamental??

Hi there!

I was Googling for interesting demonstrations in Physics, and I was lucky enough to find very interesting blog. Although it is completely new- the guy is writing it only for two weeks, but it already has a lot of nice physics demonstrations in it. All blog is about physics, so I am happy to share it with, although it is not the point of this discussion.
Here is the blog: http://triumphofmind.blogspot.com/


When I was reading that blog, I found that the guy writing it claimed that the formulation for second Newton's law of motion for constant mass object F=ma is not the fundamental one, and he emphasized that the right one is a=F/m. Although all sources including Wikipedia, HRW Principles of Physics, and other texbooks always write F=ma. And anyway is there any means of difference?

His argument is that acceleration is caused by the force, and not the force by acceleration as F=ma implies. But does causality matters here?

Also F=ma is direct result of fundamental formulation: F=dp/dt. And F=ma is mathematically more simple, because there is no risk of division by 0. Although for zero mass infinite acceleration souds reasonable...

Here is original post: http://triumphofmind.blogspot.com/2011/05/motion-of-matter-part-2.html

So what do you think?

 

[Histerize]

Hello,
I have just joined to this forum because I like physics. Although I am studying bioengineering, I had a course on physics and during the course the lecturer was not accepting such formulation of second Newton's law: The force is proportional to mass and acceleration of body. (F=ma) Simply because there is no causality consideration in formulation. But does it matter? Also in non-inertial systems, pseudo force is F=ma...

BTW, thanks for sharing the blog, I just had quick review of it, but I am going go through it tomorrow.

 

[espen180]

If the mass is zero, a classical treatment is insufficient. You need quantum theory and relativity theory to describe the particle.

 

[Joans]

How can you describe particle with no mass in theory of relativity or quantum??

I recognize that photon has no rest mass, but It is nonsense, because it does not exists at rest. And then it moves, it have mass from E=mcc...

What does it mean to have no mass? What it does not exist?

 

[vanhees71]

You should adapt to the modern notion of mass, and that's the invariant mass (as you pointed out yourself, the name "rest mass" in inappropriate for massless particles since these can not be at rest with respect to any inertial frame, but move with the speed of light). Setting $$c=1$$, the invariant mass of a particle is given by the four-momentum $$p$$ via the invariant equation $$p \cdot p=p_{\mu} p^{\mu}=E^2-\vec{p}^2=m^2.$$

 

[espen180]

If a particle has zero rest mass, it will travel at the speed of light in any inertial reference frame. The expression you mentioned, $E=mc^2$, is a special case for stationary objects. The general expression is $E^2=(mc^2)^2 + (pc)^2$. For the photon, this gives you $E=pc$. Quantum mechanics has the expression $E=hf$ for the photon, where h is Planck's constant and f is the frequency of the photon, so the correct momentum expression for a photon is $p=\frac{hf}{c}$.

Photons are, as you said, an example of a particle which has zero rest mass, but this does not imply that it doesn't exist!

 

[Joans]

That's hardcore... I presume then, that non-relativistic quantum mechanics which I am taught now is rubbish? However it is off-topic. So what about topic? Why does all textbooks and Wikipedia and other sources write rubbish?

 

[espen180]

Please give precise references. Without knowing exactly what your sources say I cannot possibly comment on them.

Nevertheless, you should never think that a physical theory exactly models reality. It is only an approximation. To say a theory is "rubbish" just because it is not applicable to all conceivable situations is a rather misguided stance.

 

[vanhees71]

That's hardcore... I presume then, that non-relativistic quantum mechanics which I am taught now is rubbish? However it is off-topic. So what about topic? Why does all textbooks and Wikipedia and other sources write rubbish?

How do you come to this conclusion? There's a lot to learn in non-relativistic quantum mechanics and it's a great theory within it's realm of applicability. Any physical theory has a certain validity range, and a lot can be described with classical (Newtonian or relativistic) mechanics, classical field theory (Maxwellian electromagnetics), but certain aspects of nature can only be described by quantum theory, and the non-relativistic theory has a wide range of successful applications! A lot of atomic, molecular, condensed-matter physics, even part of nuclear physics can be very successfully described with non-relativistic quantum mechanics. Here, the electromagnetic field enters in the sense of a classical field, and not necessarily as a quantum field, and even then it can make perfect sense to use a non-relativistic quantum description of the matter part (e.g., electrons in not too heavy atoms or molecules and condensed matter).

Of course, one cannot describe photons with a non-relativistic theory. As you shall perhaps learn at a more advanced level, zero-mass particles do not make sense in non-relativistic quantum theory. That's all related to the representation theory of the Galileo group, i.e., the symmetry group of Newtonian space-time. For massless particles you have to deal necessarily with the Minkowski space of the special theory of relativity and thus with help of the representations of the Poincare group, i.e., the symmetry group of Minkowski space. Relativistic quantum theory is also more difficult to learn since you have to deal necessarily with a many-body theory and thus quantum field theory. In non-relativistic quantum theory you can get quite far with a fixed number of particles or even with single-particle pictures.

As little sense it makes to learn quantum mechanics without a solid background in classical mechanics and electrodynamics it makes no sense to try to learn relativistic quantum field theory without a solid knowledge about the non-relativistic theory.

 

[AlephZero]

Newton's second law as stated by Newton in "Principia" was F = dp/dt.

End of story, IMO.

 

[Jedi_Sawyer]

I don't think that either formulation is correct. Both F=ma and F= dp/dt assume that all change in impulse, (mv) , is a function of just force considerations and I don't think that is true.

 

[vanhees71]

I don't think that either formulation is correct. Both F=ma and F= dp/dt assume that all change in impulse, (mv) , is a function of just force considerations and I don't think that is true.

Can you elaborate a bit more? Of course, Newton got it right in the first shot: The correct form is F=dp/dt, and F=m a is just a special case for bodies with a constant mass in non-relativistic motion. That's an example for great intuition of a genius. Newton has of course never thought about relativity, which became an issue only much later when electromagnetism has been understood by the works of Faraday and Maxwell, and the Maxwellians started to analyse this theory.

At the end of course, the experiment has the last word, and I think Newton's 2nd Law is very well established for situations, where classical dynamics is applicable. So, where do you see some flaw?

 

[Jedi_Sawyer]

Suppose that the description of impulse has another component relating the change in energy of the system per change in velocity of that energy, or
I = ΔP = ∫ F dt + + ΔU/Δv. The subscripts didn't copy over but in the above equation the term U refers to the internal energy of a thermodynamic system and is a scalar and is dependant on the velocity, and the v vector is the directional velocity of that energy and is dependant on the temperature. A dimensional analysis reveals that the extra term has the correct units for the impulse, Mass ∙ velocity. In order to utilize the extra term we need a system whose Internal Energy is affected by the vectored velocity of that energy, and the vectored velocity is affected by the temperature. The gaseous exhaust or the expelled propellant of a rocket is very possibly an example of such a system. Since for a gas momentum is proportional to pressure, we conclude that if we reduce the pressure of a gas say by reducing the temperature we are reducing momentum. Conversely if he increase the pressure say by heating a gas we are increasing the momentum. You know that a pressurized balloon with a certain amount of gas will propel a balloon a lot farther than the same amount of gas in a less pressurized balloon.

 

[Joans]

How do you come to this conclusion?

I asked the question what should I say to children, F=ma or a=F/m, but almost all replies were that I should adopt different notion of mass and go to the relativistic quantum. I definitely recognize the applicability of theories, it was irony which was hardly noticed. Classical physics is mostly of importance in our lives. And post was made under Classical Physics section, which also says something.

Please give precise references. Without knowing exactly what your sources say I cannot possibly comment on them.

The most popular textbook in freshman's courses is Halliday, Resnick, Jearl Walker Principles of Physics or Fundamentals of Physics. Now I quote how they present Newton's second law:
F=ma - the net force on a body is equal to the product of the body's mass and it's acceleration.
Now let's go to another most popular source, Wikipedia:
F=dp/dt since the law is valid only for constant-mass systems, the mass can be taken outside the differentiation. Thus F=ma where F is the net force applied, m is the mass of the body, and a is the body's acceleration. Thus, the net force applied to a body produces a proportional acceleration. Although verbal formulation seems to be correct, but formula do not reflect this? Should it be a=F/m?
Another reference: Prof Walter Lewin from MIT OCW states the same.

Newton's second law as stated by Newton in "Principia" was F = dp/dt.
End of story, IMO.

Surely you are right, but if m is constant, the law without loss of generality becomes F=ma, although a=F/m seems to be more reasonable to be taught at schools or am I missing something?

 

[vanhees71]

I've not read the complete thread and thought you were asking about mass in relativity. Zero mass simply makes no sense in non-relativistic physics. The deep mathematical reason for this is the structure of the Galileo group, which is the basic foundation of non-relativistic quantum mechanics.

 

[espen180]

They are equivalent. Whether you use F=ma or A=F/m makes absolutely no difference, although the correct expression is F=dp/dt.

 

[D H]

Newton's second law as stated by Newton in "Principia" was F = dp/dt.

End of story, IMO.

Newton said nothing of the sort. Newton's second law as stated by Newton in "Principia" is "Mutationem motus proportionalem esse vi motrici impressae, et fieri secundum lineam rectam qua vis illa imprimitur."

In English, "The alteration of motion is ever proportional to the motive force impressed; and is made in the direction of the right line in which that force is impressed."

Our modern metric system of units that let's us now say that F=ma postdates Newton by about a century. Our modern notation that let's us say that $\vec F = m \vec a$ postdates Newton by a couple of centuries.

They are equivalent. Whether you use F=ma or A=F/m makes absolutely no difference, although the correct expression is F=dp/dt.

They are equivalent only if the mass is constant. In variable mass situations they are not equivalent. The form F=ma means the force is the same in all inertial frames. This is not true with F=dp/dt. Now force is not invariant. It is a frame-dependent quantity instead.

People who deal with variable mass systems use F=ma, not F=dp/dt.

 

[vanhees71]

Newton's physics is unchanged since its formulation by Newton himself. Only the form to express it has changed over the centuries. The most fundamental and correct version of the 2nd Law since Newton's Principia is

$$\frac{\mathrm{d} \vec{p}}{\mathrm{d} t}=\vec{F}.$$

You have to read the definitions carefully. What Newton calls "motion" indeed is "momentum" in modern terminology.

For the special case of bodies with constant mass, m, you have of course

$$m \vec{a}=\vec{F}$$

due to (Newton's!) definition of momentum

$$\vec{p}=m \vec{v}=m\frac{\mathrm{d} \vec{x}}{\mathrm{d}t}.$$

 

[D H]

Newton's physics is unchanged since its formulation by Newton himself.

Wrong. Newton formulated his laws geometrically and verbosely. The algebraic interpretations came later. The interpretation as vectors came later. Newton's fluxions were geometrically defined. The algebraic interpretation of derivatives and the very concept of what exactly a derivative is underwent massive changes with Weierstrass.

Newton's second law, in Newton's terms, applies to point masses with constant mass only. For systems of constant mass, arguing whether the correct form is F=ma or F=dp/dt is akin to arguing over how many angels can dance on the point of a pin. Extending Newton's laws to systems of particles yields F=ma or F=dp/dt, you pick, if the system as a whole has constant mass.

There are three choices for a variable mass system:

• Newton's laws don't apply to variable mass systems. Yes, some people do say that.
• F=ma is the correct form. This choice makes force a frame-invariant quantity, but leads to a disconnect between force and momentum.
• F=dp/dt is the correct form. This choice makes force a frame-dependent quantity, but obviously does maintain a connection between force and momentum.

With rockets at least, the consensus is to use F=ma.