# Extending Newton's laws -- Is the concept of force still defined?

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• dextercioby
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In summary, it is argued that the correct interpretation of Newton's 2nd Law for one body of mass reads "The dynamics (i.e. vector sum of all external forces acting on the body = "all its interactions") dictates the kinetics (i.e. time derivative of the momentum vector = "motion")", under the assumption that the body's mass will not change during the action of the external forces and after that. However, if the effect of the external forces is to dictate the motion of the body by making it lose mass, then Newton's second law is no longer applicable. Dimensionally inconsistent, this definition implies a new definition of force. Forces may be defined only in the presence of minimum two bodies.

#### dextercioby

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Questions I came up with after reading Wikipedia's section on the Tsiolkovsky's equation.
It is argued that the correct interpretation of Newton's 2nd Law for one body of mass ##m## reads "The dynamics (i.e. vector sum of all external forces acting on the body = "all its interactions") dictates the kinetics (i.e. time derivative of the momentum vector = "motion")", under the assumption that the body's mass will not change during the action of the external forces and after that .

Now let us assume that the effect of the external forces is to dictate the motion of the body by making it lose mass, i.e. we step out of Newton's 2nd law's assumptions. We are also told that:

##\vec{F}_{\mbox{ext}}(t) = \frac{dm(t)}{dt}\vec{v}(t) + m\frac{d\vec{v}(t)}{dt} ## (1).

I have four questions: does (1) still make sense, i.e. is it correct? Does it imply a new definition of force replacing/enhancing the one from Newton's laws? Can the two terms of the RHS of (1), if the answer to the first question is "yes", be interpreted as forces? If the answer to the 3rd question is "yes", what would they represent physically?

Thank you!

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This depends on what you consider a ”force” on the system. If by ”force” you mean ”any momentum exchange” (including momentum lost or gained by mass entering or exiting your system) then Newton’s second law F = dp/dt holds as is.

dextercioby said:
We are also told that:

F→ext(t)=dm(t)dtp→(t)+mdp→(t)dt (1).
This is dimensionally inconsistent.

• topsquark
Orodruin said:
This depends on what you consider a ”force” on the system. If by ”force” you mean ”any momentum exchange” (including momentum lost or gained by mass entering or exiting your system) then Newton’s second law F = dp/dt holds as is.

This is dimensionally inconsistent.
I beg your pardon. V, not p, post corrected. No, I define a force as a primordial concept, i.e. the effect of any interaction a body may have. In other words, a force is defined only in the presence of minimum two bodies. If you define it "by the second" law, you already assume that the 2nd law is universal, i.e. it applies to systems of variable mass.

dextercioby said:
No, I define a force as a primordial concept, i.e. the effect of any interaction a body may have.
That is a bit vague as it is unclear what you would mean by ”any interaction”. An exchange of mass would be an interaction between an object and its surroundings, which is the only relevant thing once you have drawn the boundaries of the system you are considering.

dextercioby said:
If you define it "by the second" law, you already assume that the 2nd law is universal, i.e. it applies to systems of variable mass.
Yes, so? That is a feature of that definition, not something bad.

• vanhees71 and topsquark
I guess the must unusual scenario is if you have, say, a collection of marbles at rest in some inertial frame, and you define a body boundary to be shrinking (including fewer marbles). In this frame, there is no force, but in any other inertial frame there is a 'force'.

• topsquark
So https://en.wikipedia.org/wiki/Force. The force is a fundamental concept in Newtonian mechanics; it is the physical quantity that measures the effect of other bodies on a particular body. In other words, to define a force, you need to have at least 2 bodies (think of Coulomb's force. You must have a charged body and a test charge whose kinetics under the action of the Coulomb force you wish to determine/calculate). You cannot define a force with the second law's equality, because the 2nd law refers to the net (resultant) force acting on a particular body, so it automatically presupposes that effects of different bodies on a particular body (i.e. forces) can be summed as vectorial quantities.

Let me try to pose post 1 in another form then. Hypotheses once more: two bodies of masses ##m_1 (t)## variable which is acted on by a body of mass ##m_2## (constant) through a force we call it ##\vec{F}_{2,1} (t)##. The effect of the body ##m_2## is to make the other one both move and lose mass in time. Question: can we use the 3 known Newton's laws to calculate their motion (##\vec{x}_{1} (t), \vec{x}_{2} (t)##)? If yes, how? If not, why?

• topsquark
dextercioby said:
No, I define a force as a primordial concept, i.e. the effect of any interaction a body may have. In other words, a force is defined only in the presence of minimum two bodies.
Primordial?
Why is this a conundrum (at least in classical physics)? Start with the spirit of Newton: the vector momentum of the Universe is conserved. We arbitrarily partition our universe into two systems U' and S'. Any transfer of momentum beween is called a force. Isn't that the interpretation universally applied? How is this a problem?

• topsquark
hutchphd said:
Primordial?
Why is this a conundrum (at least in classical physics)? Start with the spirit of Newton: the vector momentum of the Universe is conserved. We arbitrarily partition our universe into two systems U' and S'. Any transfer of momentum beween is called a force. Isn't that the interpretation universally applied? How is this a problem?
Well, do you think it is fine for there to be force in one inertial frame and none in another, especially when no interaction of any kind is occurring? See my example above.

• topsquark
PAllen said:
especially when no interaction of any kind is occurring?
This depends on what you mean by "interaction". Arguably, exchanging mass between U' and S' is an interaction between the partitions U’ and S’.

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• topsquark, jbriggs444, Dale and 2 others
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• strangerep, topsquark and vanhees71
hutchphd said:
Primordial?
Why is this a conundrum (at least in classical physics)? Start with the spirit of Newton: the vector momentum of the Universe is conserved. We arbitrarily partition our universe into two systems U' and S'. Any transfer of momentum between is called a force. Isn't that the interpretation universally applied? How is this a problem?
It becomes a "problem" when interactions with mass-less particles occur (e.g. emission or absorption of photons). If the change in momentum is imparted by a photon, Newtonian mechanics don't really work. One cannot apply a force to accelerate a photon because they travel at only one speed, c. Newton's third law doesn't help because if one cannot impart a force to a photon, the photon cannot impart a force to a mass. One cannot really use ##\Delta p = F\Delta t## because photons do not experience time.

On the other hand, if one considers interactions as exchanges of energy, E, and used "force" as something that applied only to interactions between bodies of matter, we could define force as dE/ds.

AM

• topsquark
Andrew Mason said:
It becomes a "problem" when interactions with mass-less particles occur (e.g. emission or absorption of photons). If the change in momentum is imparted by a photon, Newtonian mechanics don't really work. One cannot apply a force to accelerate a photon because they travel at only one speed, c. Newton's third law doesn't help because if one cannot impart a force to a photon, the photon cannot impart a force to a mass. One cannot really use ##\Delta p = F\Delta t## because photons do not experience time.

On the other hand, if one considers interactions as exchanges of energy, E, and used "force" as something that applied only to interactions between bodies of matter, we could define force as dE/ds.

AM
Lucky then that photons don't exist in Newtonian theory but only in relativistic QED.

Furthermore, the entire "photons do not experience time" is an enormous popularisation that you hear way too often and should not really be used for physics argumentation.

• SammyS, apostolosdt, topsquark and 2 others
Orodruin said:
Furthermore, the entire "photons do not experience time" is an enormous popularisation that you hear way too often and should not really be used for physics argumentation.
An essential distinction between neutrinos and massless particles has been that neutrinos experience time since their properties can change as they travel. Massless particles, such as photons, don't experience change over time. How about: "photons cannot be associated with an inertial reference frame".

• topsquark
Andrew Mason said:
An essential distinction between neutrinos and massless particles has been that neutrinos experience time since their properties can change as they travel. Massless particles, such as photons, don't experience change over time. How about: "photons cannot be associated with an inertial reference frame".
Or just say light pulse. Your whole argument works classically with light pulses.

• topsquark
Andrew Mason said:
An essential distinction between neutrinos and massless particles has been that neutrinos experience time since their properties can change as they travel. Massless particles, such as photons, don't experience change over time.
Again, the "does not experience time" argument is way too popularised.

Also, neutrinos do not change properties as they travel. Neutrino oscillations are due to quantum interference between different propagation eigenstates and the propagation eigenstates not coinciding with the interaction eigenstates. Oscillations require neutrino masses to be different - not to be non-zero (i.e., it is perfectly possible for one eigenstate to be massless). The main difference is that photons do not mix with anything known. However, there are cases where photons do mix, such as with dark photons or with axions in an external magnetic field. In those cases, photons are very much subject to oscillations or similar effects so this argument does not really hold.

PAllen said:
Or just say light pulse. Your whole argument works classically with light pulses.
Not really, no. Light pulses generally are associated with a duration of at least one oscillation. In classical electromagnetism, one can very much impart momentum and energy into the electromagnetic field.

• vanhees71, hutchphd and topsquark
Orodruin said:
Not really, no. Light pulses generally are associated with a duration of at least one oscillation. In classical electromagnetism, one can very much impart momentum and energy into the electromagnetic field.
Classical in the sense of SR. @Andrew Mason argument was about the difficulties of handling this with Newtonian mechanics, especially trying to apply Newtonian momentum and force to light. This was solved with Maxwell's equations, but they don't work over a Galilean frame change, which, of course, was one of the progenitors of SR. For the difficulties with pre-SR classical mechanics applied to light, there was no need to bring in photons.

• vanhees71
Massless particles don't make any sense in Newtonian physics. There's also no Newtonian quantum dynamics for massless particles, that makes physical sense, i.e., nothing observed in nature can be described as a massless classical or quantum Newtonian particle.

Even in the relativistic case, massless point particles are a bit delicate to handle. Well, even massive classical point particles are only approximately making sense. The best model one has is the Landau-Lifshitz approximation of the Lorentz-Abraham-Dirac equation.

Photons are never correctly described as classical massless point particles since there's no way to localize them and there's no proper position operator for massless quantum fields to begin with. The only correct description of a photon is as an asymptotic free Fock state of the electromagnetic quantum field.

The formulation "a photon doesn't experience time" is not even wrong. It's not making any sense even at a popular-science level. It's simply bad, superficial popular-science writing. Unfortunately such esoteric nonsense seems to sell better than popular-science books which describe the science as good as possible without using math, which is the only adequate language to really express the content of science.

• ForTheLoveOfPhysics, apostolosdt and Orodruin
Can we go back to the original post or the restatement at the end of post 7? Thank you!

The end of #7 is incompletely formulated, i.e., we don't know whether or not in which way the mass changes as a function of time. E.g., it could be like the usual rocket model in mechanics 1, where the fuel moves relative to the rocket body and you have thrust due to that (that's why we use rockets to begin with ;-)).

Another example is a drop of water falling in a saturated atmosphere getting heavier through condensation when otherwise freely falling.

In any case you get the EoM. by applying the conservation of total momentum applied to the closed system. It's nicely discussed in Sommerfeld, Lectures on Theoretical Physics, vol. 1 (mechanics).

• strangerep and hutchphd
vanhees71 said:
The formulation "a photon doesn't experience time" is not even wrong. It's not making any sense even at a popular-science level. It's simply bad, superficial popular-science writing. Unfortunately such esoteric nonsense seems to sell better than popular-science books which describe the science as good as possible without using math, which is the only adequate language to really express the content of science.
So are you saying that scientists should not even try to explain such things to the public (since the public is not going to understand the mathematics)?

Obviously photons are not conscious so the concept of them experiencing time is not a real one. But that is the nature of thought experiments - they only work in the imagination.

AM

Andrew Mason said:
So are you saying that scientists should not even try to explain such things to the public (since the public is not going to understand the mathematics)?
No, just that "photons don't experience time" is wrong, and the line of argument Siegel gives for it in that Forbes article is wrong. To accept the premise that there's any way to describe the experience of a photon is either to have contradicted oneself or to have accepted a lightcone coordinate basis as a basis for 'experience', which is so far outside what anyone would call 'experience' that it would effectively be a lie.

Asking if photons experience time is like asking if up is yellow. If you accept the implied premise that colour and direction have any relation, you've already gone wrong. Likewise, time just isn't a word you can use when talking about light unless you introduce stuff moving on timelike worldlines (i.e. normal matter that might have a clock) and talk about its definition of time.

• vanhees71
Andrew Mason said:
So are you saying that scientists should not even try to explain such things to the public (since the public is not going to understand the mathematics)?

Obviously photons are not conscious so the concept of them experiencing time is not a real one. But that is the nature of thought experiments - they only work in the imagination.

AM
The issue is that it's not possible to define the passage of time for a lightlike interval. If we describe this by saying light doesn't experience time we may be leaving the reader with the impression that zero time elapses.

Andrew Mason said:
So are you saying that scientists should not even try to explain such things to the public (since the public is not going to understand the mathematics)?
No, the opposite is what I'm saying. You have to explain such things correctly, and that's very difficult without math. To deliver some esoteric gibberish is less than writing nothing. It's an insult to any lay man interested in science.
Andrew Mason said:
Obviously photons are not conscious so the concept of them experiencing time is not a real one. But that is the nature of thought experiments - they only work in the imagination.

AM
This is not a thought experiment. It's empty gibberish!

• apostolosdt and hutchphd
dextercioby said:
So https://en.wikipedia.org/wiki/Force. The force is a fundamental concept in Newtonian mechanics; it is the physical quantity that measures the effect of other bodies on a particular body. In other words, to define a force, you need to have at least 2 bodies (think of Coulomb's force. You must have a charged body and a test charge whose kinetics under the action of the Coulomb force you wish to determine/calculate). You cannot define a force with the second law's equality, because the 2nd law refers to the net (resultant) force acting on a particular body, so it automatically presupposes that effects of different bodies on a particular body (i.e. forces) can be summed as vectorial quantities.

Let me try to pose post 1 in another form then. Hypotheses once more: two bodies of masses ##m_1 (t)## variable which is acted on by a body of mass ##m_2## (constant) through a force we call it ##\vec{F}_{2,1} (t)##. The effect of the body ##m_2## is to make the other one both move and lose mass in time. Question: can we use the 3 known Newton's laws to calculate their motion (##\vec{x}_{1} (t), \vec{x}_{2} (t)##)? If yes, how? If not, why?
If your goal is to provide a general definition of Newtonian force, why is the effect of the body ##m_2## to make ##m_1 (t)## both move and lose mass in time? In general, forces just cause bodies to change their motion, not their mass.

AM

Andrew Mason said:
In general, forces just cause bodies to change their motion, not their mass.
This depends on your definition of force.

The first law states what a force is but does not quantify it: it is that which causes a body to change its motion. If a body loses mass, it is no longer the same body. So it seems to me that we first have to define what a body is.

AM

Andrew Mason said:
The first law states what a force is but does not quantify it: it is that which causes a body that change its motion. If a body loses mass, it is no longer the same body. So we first have to define what a body is.
Again, a question of definition.

Andrew Mason said:
So it seems to me that we first have to define what a body is.
That would seem prudent regardless. If one desires, the ejected mass can be thought of as yet another body. This argument seems fruitless. We all mean what we say, but each has to say exactly what he means.
PedagogicalIy I prefer the momentum emphasis.

• Orodruin
hutchphd said:
That would seem prudent regardless. If one desires, the ejected mass can be thought of as yet another body. This argument seems fruitless. We all mean what we say, but each has to say exactly what he means.
PedagogicalIy I prefer the momentum emphasis

I would propose to restrict the definition of "body" to baryonic matter (using it in the general sense to include electrons in atoms). After all, the concept of force originated with macroscopic objects containing baryonic matter. I realize that this may not work if you want to include as bodies such objects as dark matter, neutrinos, free electrons, black holes etc. However, the concept of force is not particularly useful, if it can apply at all, to such objects.

How about: "A body is an object comprised of baryonic matter, the quantity and identity of which does not change with time".

If you don't like using the term "baryonic matter" to include electrons, we could also define a body in terms of its constituent particles: "A body is a collection of protons, neutrons and electrons, the quantity and identity of which remain constant over time".

We could then define "force" as a physical interaction with the body that changes the motion of a body. We could also define the direction of the force as the same as the direction of the resulting change in motion of the body.

The next step would be to define the magnitude of a change in motion. That is when you get into the Newton's second law, not as a definition of force but as a way of defining its magnitude.

This approach avoids the problem of using mass or inertia or momentum in the definition of what a force is and avoids the circular reasoning in using the second law to define force.

AM

I won't stop you.

Andrew Mason said:
I would propose to restrict the definition of "body" to baryonic matter (using it in the general sense to include electrons in atoms). After all, the concept of force originated with macroscopic objects containing baryonic matter. I realize that this may not work if you want to include as bodies such objects as dark matter, neutrinos, free electrons, black holes etc. However, the concept of force is not particularly useful, if it can apply at all, to such objects.

How about: "A body is an object comprised of baryonic matter, the quantity and identity of which does not change with time".

Baryonic matter appears in relativistic quantum field theories. It has very little to do with classical forces and classical forces are not very useful in the Standard Model either. Already with baryonic matter you run into issues of mass non-conservation.

Andrew Mason said:
We could then define "force" as a physical interaction with the body that changes the motion of a body.
If we accept your definition of "body" as baryonic matter with fixed constituents, you are already out of the constant mass case - as required by relativistic quantum theory.

Either way, your post is full of "we could define" and similar. Even if there were no caveats in the definition of what constitutes a "body", you could also do it in different ways that all capture the essence of Newton's laws of motion when applied to the constant mass case.

• vanhees71
Orodruin said:
If by ”force” you mean ”any momentum exchange” (including momentum lost or gained by mass entering or exiting your system) then Newton’s second law F = dp/dt holds as is.
As a matter of personal preference I would define force as the rate of change of momentum with respect to time. That makes it easy to generalize to relativity where the rate of change of the four momentum with respect to proper time is a tensor.

Dale said:
As a matter of personal preference I would define force as the rate of change of momentum with respect to time. That makes it easy to generalize to relativity where the rate of change of the four momentum with respect to proper time is a tensor.
It is certainly the case that force=dp/dt. But if one defines force as dp/dt then one has to define momentum. If momentum is defined as mass x velocity, one has to define mass in a quantifiable way. But the only way to define mass in a quantifiable way (without resorting to something based on the number of protons+electrons + neutrons) is by its inertia ie.: m=(dp/dt)/(dv/dt) and everything becomes circular. I think the OP was trying to avoid that.

AM

• Dale
Andrew Mason said:
If momentum is defined as mass x velocity,
Then momentum can be defined through Noether’s theorem.

• vanhees71