The Relativistic Force-Norm and the Proof of E=mc2

[Matterwave]

In my derivation of the relativistic momentum (see my link above) I started from the original definition and got another result. How did you get your "modern notation"?



The question is, what m means.

I don't think we will come to a consensus. So I'll just drop the issue. You can take this as "you win" if you like.

 

[stevendaryl]

this results in the rocket equation
\dot v_R = \frac{{\dot m_R }}{{m_R }} \cdot v_{rel}

That's the correct equation of motion. But it doesn't seem consistent with F_R = \dot{p_R} = \dot{m_R} v_R + m_R \dot{v_R}

 

[DrStupid]

You are the one who needs to provide justification. I have already shown that it doesn't work, so it is your turn now.

Sorry, not this way! It was your claim that it doesn't work. Therefore you need to provide the corresponding justification. In regard to the rocket I disproved your claim instead of waiting for your proof. But that does not mean that I will do that again and again.

But I'm talking about classical mechanics

The thread is about E=mc². Therefore it makes no sense to limit the discussion to classical mechanics. This link shows what I'm talking about:

https://www.physicsforums.com/showpost.php?p=4737048&postcount=4

and there it needs to transform according to Galilean transformation and I've shown that it doesn't, so this is a 100% proof that $F=\dot p$ is inconsistent in classical mechanics

I have also shown that for closed systems (see my link above). Your proof for open systems is still missing. It's not sufficient just to claim it and wait for a disproof.

If you claim that a macroscopic theory gives the same results as a more fundamental microscopic one, you need to prove that. You can't simply postulate it and wait for others to disprove it.

As I didn't wrote something about macroscopic or microscopic theories I don't need to provide a corresponding proof. You raised that topic. You need to provide the proof.

Again, I'm only talking about classical mechanics here.

Again, that's off-topic.

And I'm doing so in order to prove that $F=\dot p$ requires justification and isn't valid in general, i.e. it is invalid in classical mechanics.

I'm still waiting for this proof.

The paper considers a function $m(t)$ that depends only on time. Thus the claim of the paper is completely valid.

I already explained why it is not valid. I will not repeat it. You can read it above.

PS: This discussion starts to get out of control and I'm not interested in meta discussions. Thus I will limit my replies to statements with justification and ignore everything else.

 

[stevendaryl]

Again, I'm only talking about classical mechanics here. And I'm doing so in order to prove that $F=\dot p$ requires justification and isn't valid in general, i.e. it is invalid in classical mechanics.

The one place where F = \dot{p} is preferable to F = m \dot{v} is in Lagrangian mechanics. We define the canonical momentum corresponding to a coordinate q to be p_q = \dfrac{\partial L}{\partial \dot{q}}, and we define the generalized force corresponding to coordinate [iteq]q[/itex] to be: F_q = \dfrac{\partial L}{\partial q}. In that case, the equations of motion are:

\dot{p_q} = F_q

not

m \ddot{q} = F_q

The latter is only true if the coordinate is cartesian and the force is due to a scalar potential.

 

[DrStupid]

That's the correct equation of motion. But it doesn't seem consistent with F_R = \dot{p_R} = \dot{m_R} v_R + m_R \dot{v_R}

This equation is part of the derivation. If it is not consistent with the result than there would be an error in the calculation. Do you see such an error?

 

[rubi]

Ok, I'm not going to waste any more time on this. Your claim in Post #18 was that $F=\dot p$ is to be favoured over $F=ma$ and I have provided several arguments that prove that the contrary is true, and you haven't addressed a single one of them. This makes the discussion entirely pointless.

 

[rubi]

The one place where F = \dot{p} is preferable to F = m \dot{v} is in Lagrangian mechanics. We define the canonical momentum corresponding to a coordinate q to be p_q = \dfrac{\partial L}{\partial \dot{q}}, and we define the generalized force corresponding to coordinate [iteq]q[/itex] to be: F_q = \dfrac{\partial L}{\partial q}. In that case, the equations of motion are:

\dot{p_q} = F_q

not

m \ddot{q} = F_q

The latter is only true if the coordinate is cartesian and the force is due to a scalar potential.

This is true, but a variable-mass system can't be described with a Lagrangian like this. The paper I posted above addresses this.

Or in other words: One can't simply put $m=m(t)$ and use the same equations that are known to work for the constant mass case. Neither in Newtonian mechanics nor in the Lagrangian formulation nor in the Hamilton formalism. The resulting equations of motion don't describe experimentally testable facts in general.

 

[PAllen]

This is true, but a variable-mass system can't be described with a Lagrangian like this. The paper I posted above addresses this.

Or in other words: One can't simply put $m=m(t)$ and use the same equations that are known to work for the constant mass case. Neither in Newtonian mechanics nor in the Lagrangian formulation nor in the Hamilton formalism. The resulting equations of motion don't describe experimentally testable facts in general.

You could go all the way to using a Lagrangian density and the techniques of continuous systems. A bit of overkill, though.

 

[stevendaryl]

This equation is part of the derivation. If it is not consistent with the result than there would be an error in the calculation. Do you see such an error?

Well, I would say that the force on the rocket is

(1) F_R = -\dot{m}_F\ v_{rel}

rather than

(2) \tilde{F}_R = -\dot{m}_F\ v_F

And I would also say that

(3) F = m\ \dot{v}
rather than

(4) F = m\ \dot{v} + \dot{m}\ v

Interestingly, in this case, if you assume (1) and (3), (which I do), you get the same equations of motion as if you assume (2) and (4) (which you do):

F_R = m_R\ \dot{v}_R
\Rightarrow -\dot{m}_F\ v_{rel} = m_R\ \dot{v}_R

versus

\tilde{F}_R = m_R\ \dot{v}_R + \dot{m}_R\ v_R
\Rightarrow -\dot{m}_F\ v_F = m_R\ \dot{v}_R + \dot{m}_R\ v_R
\Rightarrow -\dot{m}_F\ v_F - \dot{m}_R\ v_R = m_R\ \dot{v}_R

Since \dot{m}_R = -\dot{m}_F and v_F = v_R + v_{rel}, the expression -\dot{m}_F\ v_F - \dot{m}_R\ v_R is equal to -\dot{m}_F\ v_R

So the question is: Is this two different ways of thinking about things, equally legitimate, or is it a matter of cancelling errors?

 

[DrStupid]

Well, I would say that the force on the rocket is

(1) F_R = -\dot{m}_F\ v_{rel}

Why would you say that? Can you derive this equation from fundamental laws or definitions?

And I would also say that

(3) F = m\ \dot{v}
rather than

(4) F = m\ \dot{v} + \dot{m}\ v

Force is defined as F=dp/dt and momentum as p=m·v. Thus F=m·a is valid for constant mass only. How do you justify the use of that equation for a system with variable mass?

So the question is: Is this two different ways of thinking about things, equally legitimate, or is it a matter of cancelling errors?

That depends on your justification for the equations you started with. Your calculation is acceptable if you know why you can do that. (I know it, but I will wait for your explanation )

 

[stevendaryl]

Why would you say that? Can you derive this equation from fundamental laws or definitions?

I would say so. Let me do a discrete approximation (which is appropriate, since matter is actually discrete).

So assume that you have a rocket of mass m_R initially traveling at speed v_R Some of the mass is made of fuel. Imagine a discrete amount of fuel \delta m being burnt and thrown out the back at relative velocity v_{rel} (this will actually be a negative value). Let \delta t be the time interval for this process.

To avoid the controversial step of considering variable mass, I will consider this system to be composed of two objects with constant mass (at least during time interval \delta t):

1. A rocket of mass m_R - \delta m.
   This has an initial velocity of v_R and a final velocity of v_R + \delta v_R. The change in momentum is \delta p_R = (m_R - \delta m)\ \delta v_R.
2. A quantity of fuel of mass \delta m.
   This has an initial velocity of v_R and a final velocity of v_R + v_{rel}. The change in momentum is \delta p_F = \delta m\ v_{rel}.

By conservation of momentum,

\delta p_R = -\delta p_F

So

\delta p_R = -\delta m\ v_{rel}

The average force on the rocket is \delta p_R/\delta t = -v_{rel}\ \delta m/\delta t.

In the continuum limit, F_R = - v_{rel} \dfrac{dm}{dt}

 

[OmCheeto]

It appears velo city has run away. If I were in high school, I'd probably run away too. I would answer the question, with evidence:

Binding Energy
...
In 2005, Rainville et al. published a direct test of the energy-equivalence of mass lost in the binding-energy of a neutron to atoms of particular isotopes of silicon and sulfur, by comparing the new mass-change to the energy of the emitted gamma ray associated with the neutron capture. The binding mass-loss agreed with the gamma ray energy to a precision of ±0.00004 %, the most accurate test of E=mc2 to date.

Original article

I know this is a bit simplistic, but it was a simple question. Kind of like; "Give me a proof that the sun is shining". I don't think I'd even bother with such a question, much beyond, pointing at it, and claiming it does. Which is what Rainville et al, appear to have done, with E=mc².

 

[DrStupid]

To avoid the controversial step of considering variable mass, I will consider this system to be composed of two objects with constant mass (at least during time interval \delta t):
[...]
The average force on the rocket is \delta p_R/\delta t = -v_{rel}\ \delta m/\delta t.

This result is limited to the conditions above. You derived a force that would act on a "rocket" (it actually is a cannon) with constant mass during the time interval \delta t. As a rocket cannot change its momentum without changing its mass this condition cannot be fulfilled. If you consider the different mass at the begin and at the end of the time intervals you will get different equations.

However, you finally get the correct result because the error goes to zero for infinite small time intervals \delta t. If you forget the forces and return to the momentum you get

\frac{{\delta p_R }}{{\delta t}} = \left( {m_R - \delta m} \right) \cdot \frac{{\delta v_R }}{{\delta t}} = - \frac{{\delta p_F }}{{\delta t}} = - \frac{{\delta m}}{{\delta t}} \cdot v_{rel}

For infinite small steps this results in

\frac{{\delta v_R }}{{\delta t}} = - \frac{{\delta m}}{{\delta t}} \cdot \frac{{v_{rel} }}{{m_R }}

(due to \delta m \to 0)

Your use of the term "force" is at questionable but your balance of momentum is correct.


But there is a possibility to justify your calculation even for variable mass. With the original definition of Force (Lex 2) and momentum (Definition 2 in Newton's Principia) I initially get

F_R = m_R \cdot a_R + \dot m_R \cdot v_R
and
F_F = \dot m_F \cdot \left( {v_R + v_{rel} } \right)

In order to get your equations for the forces I just need to transform these equations into an inertial system with the same velocity of the rocket. In classical mechanics forces transform according to

F' = m' \cdot a' + \dot m' \cdot v' = m \cdot a + \dot m \cdot \left( {v - u} \right) = F - u \cdot \dot m

With u = v_R this results in

F'_R = m_R \cdot a_R
and
F'_F = \dot m_F \cdot v_{rel}

and with Lex 3 finally in

- \dot m_F \cdot v_{rel} = m_R \cdot a_R

This corresponds to your first calculation. As the resulting equation is frame-independent under Galilean transformation you may use it for the derivation of the rocket equation in all inertial systems. But you must not use the equations of force in other frames of reference or at different times.

 

[stevendaryl]

It appears velo city has run away. If I were in high school, I'd probably run away too. I would answer the question, with evidence:

This is a terminological issue. To me, there is a distinction between asking: "What is the empirical evidence that X is true?" and "What is the proof that X is true?"

For example, if someone asked for a proof that sin(45^o) = 1, it would not be sufficient to just construct a 45^o triangle and measure the lengths of the sides. Proof requires a logical or mathematical argument, or demonstration. Proof is more certain than empirical evidence, but that certainty comes at cost, which is that proof is always relative to a set of assumptions, and those assumptions could be wrong.

(Actually, empirical demonstrations also have hidden assumptions, but they are different kinds of assumptions.)

 

[DrStupid]

I know this is a bit simplistic

I wouldn't call it simplistic but basic because in physics experiments always have the final say. But it's always nice to have a theoretical background that predicts such experimental observations. That's where its starts to be complicate.

 

[stevendaryl]

This result is limited to the conditions above. You derived a force that would act on a "rocket" (it actually is a cannon) with constant mass during the time interval \delta t. As a rocket cannot change its momentum without changing its mass this condition cannot be fulfilled. If you consider the different mass at the begin and at the end of the time intervals you will get different equations.

What you're saying doesn't make a lot of sense to me. There is no difference between saying:

Description A
The rocket initially has mass M_R. The exhaust initially has mass 0. The rocket's mass decreases by \delta m and the exhaust's mass increases by \delta m

Description B
There are two objects, one of mass M_R - \delta m and another of mass \delta m. Initially, the two objects are traveling together, but the second object is burned and sent out the rear.

In Description A, the objects change mass. In Description B, they don't. The only difference is how you divide the boundary around what counts as the objects. There is no conceptual difference between a rocket and a cannon--in both cases, you have a large mass splitting up into two unequal masses, and the smaller of the two is ejected.

You seem to be saying that the force depends on how you draw the boundary of two objects.

There is no fundamental reason in Newtonian physics to ever consider variable-mass objects. Since mass is conserved, you can always break the problem up into a many-particle problem involving tiny chunks of matter. So there is no need, ever, for a law of physics to describe what happens with variable mass. "Variable mass" is not a different situation, it is a different way of describing the same situation. So it's weird that the force would depend on how you describe things.

 

[DrStupid]

The only difference is how you divide the boundary around what counts as the objects. There is no conceptual difference between a rocket and a cannon--in both cases, you have a large mass splitting up into two unequal masses, and the smaller of the two is ejected.

There would be no conceptual difference if you would consider the transfer of momentum due to the change of the system boundaries after each step in Description B. But this is not included in your equations.

For the case that this explanation was too short:

In description A we start with a rocket of mass m_R and velocity v_R and no exhausted fuel (previus exhausted fuel can be neglected because it does not interact with the rocket) and we end up with a rocket of mass m_R-\delta m and velocity v_R + \delta v as well as with exhausted fuel with the mass \delta m and the velocity v_R + v_{rel}. The corresponding changes of momentum are

\delta p_R = \left( {m_R - \delta m} \right) \cdot \left( {v_R + \delta v_R } \right) - m_R \cdot v_R = m_R \cdot \delta v_R - \delta m \cdot \left( {v_R + \delta v_R } \right)

\delta p_F = \delta m \cdot \left( {v_R + v_{rel} } \right)Description B consits of 2 parts. In the first part the rocket with initial mass m_R and velocity v_R is split into the remaining part of the rocket with the mass m_R-\delta m and fuel with the mass \delta m, both traveling with the initial velocity of the rocket. This results in the following changes of momentum:

\delta p_R = \left( {m_R - \delta m} \right) \cdot v_R - m_R \cdot v_R = - \delta m \cdot v_R

\delta p_F = \delta m \cdot v_R

In the second step the velocity of the rocket is changed to v_R + \delta v and the velocity of the fuel to v_R + v_{rel}, leaving the mass of both parts constant. The corresponding changes of momentum are

\delta p_R = \left( {m_R - \delta m} \right) \cdot \left( {v_R + \delta v_R } \right) - \left( {m_R - \delta m} \right) \cdot v_R = \left( {m_R - \delta m} \right) \cdot \delta v_R

\delta p_F = \delta m \cdot \left( {v_R + v_{rel} } \right) - \delta m \cdot v_R = \delta m \cdot v_R

Both steps of description B result in the same equations as description A, but you described the second step of B only and neglected the first one. This is sufficient if you are interested in the acceleration only because the first step doesn't change the velocity. But as it changes the momentum this is not sufficient if you try to calculate the force.

 

[stevendaryl]

There would be no conceptual difference if you would consider the transfer of momentum due to the change of the system boundaries after each step in Description B. But this is not included in your equations.

Ah! Okay, I think I understand what you're saying.

It's the old continuity equation business. If you have a surface representing the boundary of an "object", there are two ways that momentum and energy can change:

(1) The particles within the boundary might change their energy or momentum.
(2) Particles may enter or leave through the boundary (taking energy and momentum with them).

So I think that this is just a terminological difference. I would not consider effect (2) to be a "force", at all. Effect (2) is present even for noninteracting particles. For example, suppose that I have a collection of noninteracting particles, initially confined within a region shaped like a cube (whose faces point in the x, y, and z directions). Half the particles have velocity zero, and the other have a nonzero velocity v in the x-direction. Initially, the stationary particles and the moving particles are both distributed uniformly within the cube.

If we consider the cube to be the object, then the object will be losing mass at a rate of \dfrac{dm}{dt} = -\rho v A where \rho is the density of the moving particles near the face pointing in the x-direction, v is the speed normal to the face, and A is the area of the face.

The particles that the cube is losing all have velocity v in the x-direction, so the momentum of the cube is changing with time according to

\dfrac{dp}{dt} = \dfrac{dm}{dt}\ v

I would not call this rate of change of momentum a "force" at all. To me, noninteracting particles by definition cannot exert forces.

However, I don't think it makes any difference. If you want to use the word "force" to describe the changes of momentum due to particles flowing across subjective boundaries, I don't care. It's a matter of accounting. I don't like it, because I think "force" should reflect something real, rather than an artifact of our modeling, but that's just aesthetics.

The nice thing about this boundary-dependent force is that it obeys Newton's laws of motion, just like "real" forces. So I'm convinced that it is harmless to consider it a force, although I wouldn't.

 

[DrStupid]

The particles that the cube is losing all have velocity v in the x-direction, so the momentum of the cube is changing with time according to

\dfrac{dp}{dt} = \dfrac{dm}{dt}\ v

I would not call this rate of change of momentum a "force" at all.

That's not a matter of taste. According to Newton's second law this is a force. The only choice you have is avoiding forces completely for open systems.

To me, noninteracting particles by definition cannot exert forces.

The particles cross a system boundary and carry momentum from one system to another. That actually is an interaction.

 

[stevendaryl]

Ok, I'm not going to waste any more time on this. Your claim in Post #18 was that $F=\dot p$ is to be favoured over $F=ma$ and I have provided several arguments that prove that the contrary is true, and you haven't addressed a single one of them. This makes the discussion entirely pointless.

I think I've gotten to the bottom of this dispute, and I'm fairly sure that it's just a terminological difference. Dr. Stupid's position simply amounts to calling the expression \dfrac{dm}{dt}\ \vec{v} a "force". There are no consequences to this choice. It's not a different theory, it's just a different way of accounting for things.

The way I prefer doing things is to reserve the word "force" to mean a physical interaction, a push or a pull exerted on particles. With this notion of force, Newton's equations of motion are:

(1) \vec{F}_{interaction} = m \dfrac{d\vec{v}}{dt}

Dr. Stupid's approach is to say that both the left-hand side and the right-hand side should be augmented:

\vec{F}_{total} = \vec{F}_{interaction} + \vec{F}_{boundary}

where \vec{F}_{boundary} reflects the rate at which momentum is passing across the object boundary.

In terms of \vec{F}_{total}, we have:

(2) \vec{F}_{total} = m \dfrac{d\vec{v}}{dt} + \dfrac{dm}{dt} \vec{v}

Assuming that (by definition?) \vec{F}_{boundary} = \dfrac{dm}{dt} \vec{v}, it's clear that equation (2) has exactly the same content as equation (1). You're just adding equal terms to both sides of an equation.

So in the case of the water balloon with equal leaks on each end, the Dr. Stupid approach would say that the force is nonzero, but that the force only contributes to \dfrac{dm}{dt} \vec{v}, not to m \dfrac{\vec{v}}{dt}

 

[stevendaryl]

That's not a matter of taste. According to Newton's second law this is a force.

Yes, it is purely a matter of taste. The equation of motion:

F_{physical} = m \dfrac{dv}{dt}

and the equation of motion

F_{total} = m \dfrac{dv}{dt} + \dfrac{dm}{dt} v

are EXACTLY equivalent, provided that

F_{total} = F_{physical} + \dfrac{dm}{dt} v

Adding the same quantity, \dfrac{dm}{dt} v, to both sides of an equation cannot possibly change the physical content of the equation.

If you had some independent way of computing F_{total}, then it would make sense to use the equation F_{total} = \dfrac{dp}{dt} even when the mass is variable. But there is no way to compute F_{total} other than adding up the known interaction forces and then adding a term \dfrac{dm}{dt} v. The only point of adding this term on the left side (the forces) is to cancel the same term appearing on the right side. Why not just leave it out, since it always cancels?

 

[DrStupid]

Dr. Stupid's approach is to say that both the left-hand side and the right-hand side should be augmented:

\vec{F}_{total} = \vec{F}_{interaction} + \vec{F}_{boundary}

where \vec{F}_{boundary} reflects the rate at which momentum is passing across the object boundary.

In classical mechanics I agree with this interpretation but this thread is primary about special relativity and that makes it different. In order to avoid confusions I better return to Newton's term "quantity of matter". In classical mechanics this property is identical with mass and can change in open systems only. But in special relativity it can also change in closed systems because it is frame dependent under Lorentz transformation. In that case both terms of the equation result from interactions.

 

[rubi]

I think I've gotten to the bottom of this dispute, and I'm fairly sure that it's just a terminological difference. Dr. Stupid's position simply amounts to calling the expression \dfrac{dm}{dt}\ \vec{v} a "force". There are no consequences to this choice. It's not a different theory, it's just a different way of accounting for things.

I think you have analyzed the situation pretty beautifully and it's in the spirit of what i said in the middle of #46 and the end of #48 (although i wasn't aware of the fact that the difference between the two situations is entirely due to the boundary term). But I also think that this (together with the fact that $F=\dot p$ spoils Galilean invariance) is even more evidence in favour of the idea that the real fundamental second law is $F=ma$ instead of $F=\dot p$, since force is a real physical thing that can be measured and thus shouldn't depend on the boundaries that we draw mentally. Thus $F=\dot p$ seems like some kind of ensemble description really (much like the grand canonical ensemble in statistical mechanics).

But I'm still not convinced that $F=\dot p$ works everywhere. Think for example of a fuel consisting of ferromagnetic particles that are confined to a tank while inside the rocket whereas they can move freely as soon as they are exhausted. If there are any external magnetic fields, then it should make a real physical difference whether we count the fuel particles to the rocket tank or the freely moving fuel and it's not just a matter of drawing boundaries. I might be wrong of course, but it seems very plausible at last.

 

[DrStupid]

Yes, it is purely a matter of taste. The equation of motion:

F_{physical} = m \dfrac{dv}{dt}

You can define such a property but you should not call it force in order to avoid confusions. This term is reserved for dp/dt. Your F_{physical} and force can be equal (e.g. for all closed systems in classical mechanics) but they are not identical.

 

[mattt]

Well, obviously if you treat the system as a many-particle system (consisting of the empty balloon and the individual water molecules) with each of these "particles" evolving according to $F_i=m_i a_i$, then the sum of all momenta (the total momentum) is conserved ($\dot p_\mathrm{total} = 0$) or more generally, it is the sum of all the forces that act on the individual "particles". This just confirms the idea that in classical mechanics, only $F=ma$ is valid and every corresponding formula for variable-mass systems is just an effective version of $F=ma$.

This is exactly right.

You can't solve such problems without taking the relative exhaust velocity of the ejected matter into account. It should be obvious that it matters, because the resulting motion of the balloon/rocket/.. must depend heavily on both the direction and the magnitude of the relative exhaust velocity. The equation $F=\dot p$ can't take the relative exhaust velocity into account, since the term that shows up if you apply the product rule ($\dot m v$) doesn't depend on the relative exhaust velocity at all! Thus $F=\dot p$ must necessarily be wrong in almost all cases where $\dot m \neq 0$, except those in which $F_\mathrm{thrust}= \dot m v_\mathrm{rocket}$ happens to be true by coincidence.

Yes, you're right again.

There's really no way to save $F=\dot p$. It is just inconsistent.


The first law doesn't even matter. It follows already from a many-particle description of the situation (empty balloon + water molecules) that there is no force acting on the balloon and the water molecules that are still contained in the balloon, since they are moving at constant velocity and the fundamental description $F=ma$ of the situation really implies $F=0$. If $F=\dot p$ were consistent for a variable-mass system, then it would have to reproduce this fact.


Unfortunately there are some coincidental situations in which it works. That's the reason for why it is such a wide spread misconception that even some textbook authors get it wrong.

I know. People should be more careful when using "effective theories" (they should derive their formulas, in this case, from Newtonian mechanics of a system of many particles of constant mass).

It should be clear, that if we have a microscopic theory consisting only of particles of constant mass, then any effective description of a composite system like a variable-mass system requires a derivation from first principles, i.e it requires to be derived from a theory with only particles of constant mass. So even if it were true, we couldn't simply postulate $F=\dot p$ for $\dot m\neq 0$.

Totally agree with this.

 

[mattt]

Ah! Okay, I think I understand what you're saying.

It's the old continuity equation business. If you have a surface representing the boundary of an "object", there are two ways that momentum and energy can change:

(1) The particles within the boundary might change their energy or momentum.
(2) Particles may enter or leave through the boundary (taking energy and momentum with them).

So I think that this is just a terminological difference. I would not consider effect (2) to be a "force", at all. Effect (2) is present even for noninteracting particles. For example, suppose that I have a collection of noninteracting particles, initially confined within a region shaped like a cube (whose faces point in the x, y, and z directions). Half the particles have velocity zero, and the other have a nonzero velocity v in the x-direction. Initially, the stationary particles and the moving particles are both distributed uniformly within the cube.

If we consider the cube to be the object, then the object will be losing mass at a rate of \dfrac{dm}{dt} = -\rho v A where \rho is the density of the moving particles near the face pointing in the x-direction, v is the speed normal to the face, and A is the area of the face.

The particles that the cube is losing all have velocity v in the x-direction, so the momentum of the cube is changing with time according to

\dfrac{dp}{dt} = \dfrac{dm}{dt}\ v

I would not call this rate of change of momentum a "force" at all. To me, noninteracting particles by definition cannot exert forces.

Exactly. You have described the issue perfectly here. It is just "another terminology" that, in my opinion, only complicate things (after all, there is no cube; in reality there are only particles in your example).

It can be harmless or it can be dangerous (it is harmless if you know what is going on in reality, but if someone really thinks that in your example there is "a force" acting on your "vanishing cube"...that can be dangerous, for his undertanding).

 

[PAllen]

Note, that if you define all systems as consisting or particles, or all systems as describe by continuous elements of ρdv, then F=dp/dt holds precisely, with no anomalies or confusions. As has been noted, confusions originate in defining composite opjects, for which dm/dt is presumed to apply.

Note, that modern approaches to relativity do not introduce dm/dt. Instead, mass/density are invariant scalars, and momentum is 4-vector (covector if you care about the Lagrangian), with appropriate transformation properties.

 

[DrStupid]

Note, that modern approaches to relativity do not introduce dm/dt. Instead, mass/density are invariant scalars, and momentum is 4-vector (covector if you care about the Lagrangian), with appropriate transformation properties.

That's why it's better to start with Newton's "quantity of matter" as a parameter that is not specified and than derive the properties of this parameter from basic definitions and the corresponding transformation.

 

[rubi]

Note, that modern approaches to relativity do not introduce dm/dt. Instead, mass/density are invariant scalars, and momentum is 4-vector (covector if you care about the Lagrangian), with appropriate transformation properties.

This is right. And in relativity, the time parameter in derivatives should really be replaced by the proper time in order to make things Lorentz invariant and in this case, the equation of motion in relativity is $f^\mu = m a^\mu$ as well, where $m$ is the (constant) rest mass of the particle, much in accordance to what happens in classical mechanics. It also generalizes beautifully to non-inertial frames, where this equation becomes $m\left(\frac{\mathrm d^2 x^\mu}{\mathrm d \tau^2} + \Gamma^\mu_{\rho\sigma} \frac{\mathrm d x^\rho}{\mathrm d \tau} \frac{\mathrm d x^\sigma}{\mathrm d \tau}\right) = f^\mu$. Making the $m$ parameter dependent on the velocity would spoil pretty much all of the coordinate-invariant notions of relativity.

 

[PAllen]

This is right. And in relativity, the time parameter in derivatives should really be replaced by the proper time in order to make things Lorentz invariant and in this case, the equation of motion in relativity is $f^\mu = m a^\mu$ as well, where $m$ is the (constant) rest mass of the particle, much in accordance to what happens in classical mechanics. It also generalizes beautifully to non-inertial frames, where this equation becomes $m\left(\frac{\mathrm d^2 x^\mu}{\mathrm d \tau^2} + \Gamma^\mu_{\rho\sigma} \frac{\mathrm d x^\rho}{\mathrm d \tau} \frac{\mathrm d x^\sigma}{\mathrm d \tau}\right) = f^\mu$. Making the $m$ parameter dependent on the velocity would spoil pretty much all of the coordinate-invariant notions of relativity.

Well, except that in SR, p=mU, so F=dp/d\tau is exactly equivalent. Further, it is more accurate because you really want 4-velocity and 4-acceleration to be be vectors, while momentum and force are covectors, for a Lagrangian treatment. Thus, p = mU<lower index with metric>, and force is dp/d\tau

 

[rubi]

Well, except that in SR, p=mU, so F=dp/dt is exactly equivalent. Further, it is more accurate because you really want 4-velocity and 4-acceleration to be be vectors, while momentum and force are covectors, for a Lagrangian treatment. Thus, p = mU<lower index with metric>, and force is dp/d\tau

This is right and I didn't want to deny that. Of course $F=ma$ and $F=\dot p$ are always equivalent if $m$ is a constant and $p=mv$. I'm just saying that $m=\mathrm{const}$ is conceptually better, because it makes things Lorentz invariant and even allows for coordinate-free descriptions. (Of course you're also right about the vector/covector distinction that needs to be taken seriously in general. I have just neglected it here in order to not be too technical.)

 

[PeterDonis]

$m=\mathrm{const}$ is conceptually better, because it makes things Lorentz invariant

$m$ can be a function of proper time and still be a Lorentz scalar and preserve covariance and transformation properties.

 

[rubi]

$m$ can be a function of proper time and still be a Lorentz scalar and preserve covariance and transformation properties.

This is correct as well, but i was talking about velocity dependence in that post. I should have made that clear.

I'm not sure what the effects of a $\tau$-dependent mass would be in relativity. I suspect it would introduce the same difficulties with boundary terms (this time 4-dimensional) that stevendaryl explained in his posts. But point-particle interactions in relativity are problematic anyway, so I'm not sure whether that situation is relevant anyway.

 

[PAllen]

$m$ can be a function of proper time and still be a Lorentz scalar and preserve covariance and transformation properties.

Except then, if you take force as dp/d\tau, you get the result that the 4-force does not transform properly - its magnitude is frame dependent. I think for simplicity and consistency you need to treat either a system of particles where m is constant for each, or a continuous system based on \rhodv.

 

[PeterDonis]

Except then, if you take force as dp/d\tau, you get the result that the 4-force does not transform properly - its magnitude is frame dependent.

Can you give an example? Bear in mind I'm not really thinking of something like a rocket; for that example I agree that the system should be modeled as a bunch of objects each with a constant rest mass: the payload plus the individual packets of fuel/reaction mass that get ejected. Each fuel/reaction mass packet just gets ejected at a slightly different proper time along the payload's worldline (i.e., it follows the same worldline as the payload up until some proper time $\tau$, and then gets ejected).

I'm thinking of something more like this: I have an object that gets heated by laser beams coming at it isotropically, so the net 3-force on it is zero in its rest frame; the only thing that happens to it in its rest frame is that it gets heated, meaning its rest mass increases. That means the 4-force on the object is *not* zero, even though the 3-force in the rest frame is; the 4-force $dp_{\mu} / d \tau$ includes a "timelike" term for the heating, which then transforms into "spacelike" terms in other frames (corresponding to the fact that the 3-force in those other frames is not zero even though the object's velocity in an inertial frame in which it is moving does not change--its 3-acceleration is zero in any frame).

 

[PAllen]

Can you give an example? Bear in mind I'm not really thinking of something like a rocket; for that example I agree that the system should be modeled as a bunch of objects each with a constant rest mass: the payload plus the individual packets of fuel/reaction mass that get ejected. Each fuel/reaction mass packet just gets ejected at a slightly different proper time along the payload's worldline (i.e., it follows the same worldline as the payload up until some proper time $\tau$, and then gets ejected).

This is what I was thinking of.

I'm thinking of something more like this: I have an object that gets heated by laser beams coming at it isotropically, so the net 3-force on it is zero in its rest frame; the only thing that happens to it in its rest frame is that it gets heated, meaning its rest mass increases. That means the 4-force on the object is *not* zero, even though the 3-force in the rest frame is; the 4-force $dp_{\mu} / d \tau$ includes a "timelike" term for the heating, which then transforms into "spacelike" terms in other frames (corresponding to the fact that the 3-force in those other frames is not zero even though the object's velocity in an inertial frame in which it is moving does not change--its 3-acceleration is zero in any frame).

I agree, an this is a great example of why dp/d\tau is fundamental. You have 4-force on the body, with the time component nonzero in its rest frame. This leads to change in the norm=invariant mass of the body.

 

[PeterDonis]

You have 4-force on the body, with the time component nonzero in its rest frame. This leads to change in the norm=invariant mass of the body.

Yes, and the magnitude of the force--meaning the norm of the 4-force vector (or covector if we're being precise) is a Lorentz scalar, just as it should be. It's not frame-dependent. The magnitude of the *3-force* is frame-dependent, of course; but we would expect it to be since it's not a 4-vector to begin with.