B Gravitational Force acting on a massless body

[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.