Gravitational Force acting on a massless body

[Dale]

There is no physically interpretable dynamics of massless objects in Newtonian physics.

So then make that specific narrow claim, not your overly broad claim that there are no massless objects in Newtonian physics. There manifestly are such objects even if they fail some specific criterion that massive objects fulfill.

However, I disagree with even this more narrow claim. With a massless spring we determine the spring’s motion and internal energy. What dynamics are missing? What more dynamics do we get for a massive object?

 

[vanhees71]

I rest my case. It's obviously impossible to convince you about mathematical facts.

 

[ergospherical]

With a massless spring we determine the spring’s motion and internal energy. What dynamics are missing? What more dynamics do we get for a massive object?

This is pushing the analogy too far - if two particles are said to be connected by a massless spring it means only that they interact via a force which depends affine linearly on their separation. The actual details of how this force arises or the mechanism of energy storage is irrelevant to the problem. It's not an actual spring with its own dynamics, so much as a heuristic device to help you visualise the interaction.

 

[italicus]

@Dale

“So, an object attached to an ideal, massless spring”…

This is clearly an “ideal” situation, in which the spring mass is negligible, but not exactly zero. So negligible that you let it tend to zero, to simplify dynamics.
Any material body has a mass. If you apply a Force to a non-existing massless body, its acceleration would be infinite.

 

[Dale]

I rest my case. It's obviously impossible to convince you about mathematical facts.

@vanhees71 I have avoided making this personal, please do the same. If the only response you can do is to make it personal, then don’t respond at all.

Note, I have objected specifically to your overly broad claim that massless objects don’t exist in Newtonian physics. They clearly do as evidenced by textbooks that use them.

You have simply stated that they have no physically interpretable dynamics, which is a more narrow claim than your original. And since a massless spring’s motion and internal energy is often calculated, even this more narrow claim is unsupported.

 

[jbriggs444]

I am trying to understand the "debate" here and would prefer to see a claim cast in simpler terms, understandable to a first year physics student who is comfortable with the Newtonian formulation of classical mechanics.

Something like "for a massless particle subject to no external force other than gravity, all trajectories are consistent with the laws of Newtonian mechanics".

 

[ergospherical]

As per usual it's not even an interesting argument but rather trademark PF labouring over irrelevant minutiae

 

[Dale]

This is pushing the analogy too far - if two particles are said to be connected by a massless spring it means only that they interact via a force which depends affine linearly on their separation. The actual details of how this force arises or the mechanism of energy storage is irrelevant to the problem. It's not an actual spring with its own dynamics, so much as a heuristic device to help you visualise the interaction.

Even in the case where it is a stand in for an unspecified force the fact remains that said force is being analyzed by treating it as a massless spring. If you can use a theory to analyze object A by treating it as object B then it is rather a stretch to claim that the theory cannot handle object B or that somehow object B is excluded from the theory. In fact, it indicates the opposite, massless springs are so easy to handle with the theory that it is worthwhile to treat more difficult things as though they were massless springs.

 

[Dale]

It's a mathematical fact that there is no physically interpretable dynamics known for massless particles within Newtonian mechanics.

You have made this claim but with neither any explanation what you mean nor with any evidence to support it. A massless spring can be described in terms of its position, its velocity, and its internal energy, so specifically what physically interpretable dynamics are even missing?

Furthermore, in whatever sense you claim a massless spring doesn’t have dynamics is a far more narrow claim than your overly broad claim that massless objects simply don’t exist in Newtonian theory. They do. That is clearly demonstrated from the professional scientific literature.

 

[Dale]

This is clearly an “ideal” situation, in which the spring mass is negligible, but not exactly zero. So negligible that you let it tend to zero, to simplify dynamics.
Any material body has a mass. If you apply a Force to a non-existing massless body, its acceleration would be infinite.

Yes, obviously there are no real massless springs. The claim I am disputing is the overly broad claim from post 6 that there are no massless objects in Newtonian theory. I.e. that such objects cannot be used in Newtonian theory.

Obviously, I and everyone else understands that such objects are used as approximations to real objects, and the real object has mass. But the point is that such massless objects are compatible with the theory, as evidenced by their frequent use in introductory textbooks on the subject and as evidenced by the fact that we frequently treat other objects or forces as though they were massless springs for theoretical calculations.

To make a blanket claim that there are no massless objects in Newtonian physics is going too far.

 

[George Jones]

None of this examples treat the dynamics of a massless system. All are in the category I mentioned from the first posting on this subject on: You neglect the mass of parts of the setup (here the springs) against other masses. Again: There is no physically interpretable dynamics of massless objects in Newtonian physics.

I, too, say that massless objects do not exist in Newtonian mechanics.

Nonsense. In the many textbooks where massless strings and springs are used they are indeed describing theoretical objects which are theoretically assigned zero mass and then treated according to standard Newtonian theory.

I would agree that massless strings don’t exist in reality, but they are commonplace in Newtonian theory. To state otherwise is to contradict most introductory physics textbooks.

While it is true that first-year texts abound with examples that use "massless" strings/springs, it is also true that these same texts point out that out that the strings/springs are not massless. A random sample of three popular first-year texts (itailcs and bold below are used in the texts):

Halliday and Resnick "In the special case in which the weight of the spring is negligible ..."

Serway and Jewett "In problem statements, the synonymous terms light and of negligible mass are used to indicate that a mass is to be ignored when you work the problems."

Knight: "Often in physics and engineering problems the mass of the string or rope is much less than the masses of the objects that it connects. In such cases, we can adopt the massless string approximation. In the limit $m_s \rightarrow 0$, Equation 7.8 becomes ..."

When I teach first-year physics, I use the term "negligible" before taking $m=0$ in the first few such examples that I present.

 

[Dale]

While it is true that first-year texts abound with examples that use "massless" strings/springs, it is also true that these same texts point out that out that the strings/springs are not massless.

Completely agreed, as mentioned earlier. Real springs are not massless and it is just an approximation.

Nonetheless, in the theory they are in fact treated as having 0 mass, and nothing about the theory breaks when that is done.

 

[gmax137]

Nonetheless, in the theory they are in fact treated as having 0 mass, and nothing about the theory breaks when that is done.

As long as they have exactly zero net force, right? Other wise you get a "divide by zero" error.

I know you know that, so I'm having a hard time deciphering the point of this discussion.

 

[italicus]

As long as they have exactly zero net force, right? Other wise you get a "divide by zero" error.

I know you know that, so I'm having a hard time deciphering the point of this discussion.

Let’s consider a mathematical pendulum. The string of suspension is supposed to have negligible mass, which we pose equal to zero to solve the problem.
Nonetheless, we’re perfectly able to calculate which is the variable tension in the string when the mass swings.
So it seems a little strange that a really massless string can transmit a force to the pin of suspension, doesn’t it?
Many assumptions of basic physics are similar : think of a pulley “ without mass” so without inertia; or a “perfectly flexible, inextensible and massless rope “ that lifts a massive body! Or the definition of “ rigid body”, a plane without any friction …

All useful abstractions, but in real mechanics things are quite different.
I received a lot of observations by people , that weren’t able to grasp the point. The best thing to do Is to clarify that they’re assumptions only, at the very beginning.

 

[gmax137]

I promised myself I would not get drawn into this. But

Let’s consider a mathematical pendulum. The string of suspension is supposed to have negligible mass, which we pose equal to zero to solve the problem.

No, my high-school physics books did not write the equations of motion for the string and then set Mstring=0.

 

[Dale]

So it seems a little strange that a really massless string can transmit a force to the pin of suspension, doesn’t it?

Certainly, but nothing breaks in the theory from it.

 

[gmax137]

So it seems a little strange that a really massless string can transmit a force to the pin of suspension, doesn’t it?

Not much stranger than the Earth transmitting the force of gravity (mg) to the pendulum bob with no string at all.

EDIT: sorry, that's a distraction from the thread.

 

[italicus]

Not much stranger than the Earth transmitting the force of gravity (mg) to the pendulum bob with no string at all.

EDIT: sorry, that's a distraction from the thread.

It’s not a distraction. But you should know that the bob is in the Earth gravitational field. Newton law of universal gravitation applies here. Remember that he was not able to explain the “action at a distance “ of his law: “Hypothesis non fingo” …until a scientist came at the beginning of last century, who changed the point of view radically.
This is a progress of science., that happens not only when we discover and learn something new, but also when we look differently at old questions, and find new and more adequate solutions.

 

[weirdoguy]

Ok, so what EXACTLY breaks down when we cosider massles things in Newtonian mechanics? Any particular example?

 

[italicus]

Ok, so what EXACTLY breaks down when we cosider massles things in Newtonian mechanics? Any particular example?

Simple example: the second law of dynamics.

 

[Dale]

Simple example: the second law of dynamics.

Massless objects don’t violate Newton’s 2nd law

 

[italicus]

Massless objects don’t violate Newton’s 2nd law

Please justify your opinion by physical issues. Here we are speaking of what Physics says, not you or me.

 

[Dale]

Please justify your opinion by physical issues. Here we are speaking of what Physics says, not you or me.

I already showed what “Physics says” by citing several different textbooks. But to be clear the equation $\Sigma F=ma$ for $m=0$ simply means that $\Sigma F=0$ regardless of $a$. $0=0a$ is perfectly valid.

 

[jbriggs444]

I already showed what “Physics says” by citing several different textbooks. But to be clear the equation $F=ma$ for $m=0$ simply means that $F=0$ regardless of $a$. $0=0a$ is perfectly valid.

One can argue that you lose predictivity. Knowing that force is zero and that mass is zero does not allow one to predict acceleration. As you say, this is perfectly compatible with Newton's 2nd law.

 

[italicus]

One can argue that you lose predictivity. Knowing that force is zero and that mass is zero does not allow one to predict acceleration. As you say, this is perfectly compatible with Newton's 2nd law.

When F=0, any mass, as big as you want, has zero acceleration , as stated by the first principle of dynamics, which exists indipendently by the second. Remember the story of the Newton’s “Principia mathematica “.
Anyway, a = 0/0 has no mathematical or physical meaning.
You lose not only predictivity , you lose dynamics principles posing m=0 ; the correct assumption to be made is that, when mass is so small compared to other masses involved in the problem, you “assume “ it to be zero, sic and simpliciter .

 

[valenumr]

Yes, I believe that this is wrong.

So the notion is that one adopts the [incorrect] model of a photon as a little bullet, takes the [unconventional] notion of mass as relativistic mass, $\frac{E}{c^2}$ and the [well accepted] notion of momentum as $p=\frac{E}{c}$. Then one applies the Newtonian notion of gravitational force and computes the radius of curvature required so that [If I have understood correctly]:$$F = G\frac{m_1 m_2}{r^2} = \frac{dp}{dt} = m_1 \frac {dv}{dt} = m_1 v \frac{d\theta}{dt} = m_1 \frac{v^2}{r} = m_1 \frac{c^2}{r}$$Solving for r:$$r = \frac{Gm_2}{c^2}$$So yes, that leads to a prediction. But not to a prediction that depends on wavelength.

Yeah, my last response was overly sarcastic. My train of thought was that a relativistic observer would see both frequency shift and length contraction, the latter of which would cause apparent angles to change, this led me to wonder if the lensing would be frequency dependent. So I tried to kind of treat it classically, with much abuse, and couldn't make it work out.

 

[vanhees71]

Even in the case where it is a stand in for an unspecified force the fact remains that said force is being analyzed by treating it as a massless spring. If you can use a theory to analyze object A by treating it as object B then it is rather a stretch to claim that the theory cannot handle object B or that somehow object B is excluded from the theory. In fact, it indicates the opposite, massless springs are so easy to handle with the theory that it is worthwhile to treat more difficult things as though they were massless springs.

Again: Such approximate descriptions do not treat the dynamics of massless objects but are approximations where the mass of parts of the system is neglected to simplify the description. As I stressed above, there is a well-known mathematical no-go theorem for massless representations of the Galilei group (or its quantum mechanical extension).

 

[vanhees71]

Not much stranger than the Earth transmitting the force of gravity (mg) to the pendulum bob with no string at all.

EDIT: sorry, that's a distraction from the thread.

This entire nonsensical discussion about fictions that are mathematically disproven to exist is a distraction from the thread!

 

[vanhees71]

I, too, say that massless objects do not exist in Newtonian mechanics.
While it is true that first-year texts abound with examples that use "massless" strings/springs, it is also true that these same texts point out that out that the strings/springs are not massless. A random sample of three popular first-year texts (itailcs and bold below are used in the texts):

Halliday and Resnick "In the special case in which the weight of the spring is negligible ..."

Serway and Jewett "In problem statements, the synonymous terms light and of negligible mass are used to indicate that a mass is to be ignored when you work the problems."

Knight: "Often in physics and engineering problems the mass of the string or rope is much less than the masses of the objects that it connects. In such cases, we can adopt the massless string approximation. In the limit $m_s \rightarrow 0$, Equation 7.8 becomes ..."

When I teach first-year physics, I use the term "negligible" before taking $m=0$ in the first few such examples that I present.

Yes, that's fully justified and the correct formulation! For advanced undergrads you can demonstrate within quantum mechanics that massless representations of the Galilei group don't lead to useful dynamics. That's why in standard QM the Galilei group is represented by a ray representation or, equivalently by a central extension of the covering group of the Galilei group with mass as a central charge of the corresponding Bargmann-Wigner group, leading also to a mass superselection rule.

Also in the classical formulation, I don't see, how you can have massless objects in Newtonian mechanics with a useful dynamics. At least a hint is that the limit $m \rightarrow 0$ in the Hamiltonian formulation of the action principle in Newtonian physics doesn't make sense (while in relativistic physics it does).

 

[Dale]

Anyway, a = 0/0 has no mathematical or physical meaning.

Absolutely, $a=\Sigma F/m$ is not valid for $m=0$, but that is not Newton's 2nd law. Newton's 2nd law is $\Sigma F = m a$ which is valid and meaningful for $m=0$.

One can argue that you lose predictivity.

Actually, you often don't lose predictivity. Instead of using Newton's 2nd law to determine the acceleration you simply use Newton's 2nd law to determine the net force. You still have determined one value from the equation.

What you do lose is a certain amount of flexibility. Instead of being able to use it to determine either force or acceleration, you can only use it to determine force. As long as that is all you need in a specific scenario then you have not lost any predictivity. Usually wherever massless objects are used, all that is needed is the force, so they are predictive.

 

[Dale]

Again: Such approximate descriptions do not treat the dynamics of massless objects but are approximations where the mass of parts of the system is neglected to simplify the description.

And such simplifications do not break the theory.

within quantum mechanics that massless representations of the Galilei group don't lead to useful dynamics

So make that claim instead of the one that you did make.

 

[vanhees71]

And such simplifications do not break the theory.

So make that claim instead of the one that you did make.

No these simplifications don't break the theory, and I never claimed so. My claim is that you cannot give physical meaning to massless objects (point particles or extended objects of any kind) within Newtonian physics.

Your claim that $F=ma$ describes anything for $m=0$ doesn't make sense, because then it says $F=0$, but it doesn't define an equation of motion for a massless point particle, because there are no kinematical quantities in this equation.

 

[Interested bystander]

I'm no physicist but even I can see that ignoring the mass of something in a system for the sake of simplicity, does not equate to a mathematical description of a massless version of that thing. None of the examples given purport to describe massless objects. They are simply approximations of real systems with some detail omitted for simplicity.