I ##F=\dot{p}=\dot{m}v+m\dot{v}## and Galilean invariance

[greypilgrim]

Hi.

In Newtonian physics, total mass is conserved, but open systems can obviously gain or lose mass, such as a rocket. But how can the term $\dot{m}v$ be Galilean invariant?

 

[A.T.]

In Newtonian physics, total mass is conserved, but open systems can obviously gain or lose mass, such as a rocket. But how can the term $\dot{m}v$ be Galilean invariant?

Why should it be conserved for an open system?

 

[tnich]

I don't think it is. Galilean invariance says that Newton's laws are invariant under a set of assumptions:
1) An absolute reference for space and time irrespective of motion of the observer
2) The velocity of an object in a frame moving at velocity $V$ relative to the absolute frame is
$\dot X' = \dot X - V$
where $\dot X$ is the velocity of the object in the absolute frame
This implies that $\ddot X' = \ddot X$
3) Mass is constant and invariant

If mass changes with time, then force is not invariant:

$F = \dot p = \dot m \dot X + m \ddot X$
$F' = \dot p' = \dot m \dot X' + m \ddot X'\\ = \dot m (\dot X - V) + m \ddot X\\ =F - \dot m V$

 

[greypilgrim]

Why should it be conserved for an open system?

What do you mean? I did write that mass doesn't have to be conserved in open systems.

Galilean invariance says that Newton's laws are invariant under a set of assumptions:

I have never seen those assumptions and all three of them seem odd to me.

1) An absolute reference for space and time irrespective of motion of the observer

Why would Galilean/Newtonian physics require an absolute reference frame?

2) The velocity of an object in a frame moving at velocity VV relative to the absolute frame is [...]

Those transformations for velocity and acceleration are derivatives of the Galilean transformation of the coordinates, not assumptions.

3) Mass is constant and invariant

Never seen this as an assumption. As far as I remember, the conservation of total mass in Newtonian physics can be derived from another conservation law (I think momentum).

 

[vanhees71]

Only closed systems can be Galileo invariant. E.g., the motion of a planet around the sun at a fixed point is an approximation for a very heavy (compared to the planet) sun, where you can neglect its motion due to the interaction with the planet. Only if you consider the system consisting of the Sun and the planet together, everything is Galileo invariant, and all 10 conservation laws (conservation of energy, momentum, angular momentum, center-of-mass velocity) hold true. Why mass is conserved in Newtonian is a very subtle issue, which as far as I know can only be explained from a careful analysis of the representation theory of the Galileo group in quantum mechanics (see Ballentine, Quantum Mechanics, for a very good reference about this).

In your example of a system with non-constant mass, you should consider a concrete example, like a rocket. The rocket alone, ejecting it's fuel to accerlate forward, is of course an open system, and thus Galileo invariance doesn't hold in this description. If you take into account both the rocket and the ejected fuel, Galileo invariance holds true and again the corresponding 10 conservation laws are fulfilled (+conservation of mass).

 

[greypilgrim]

Why mass is conserved in Newtonian is a very subtle issue

Isn't it a rather trivial implication from conservation of total linear momentum $P$ (symmetry under translations) and linear motion of the center of mass $V$ (symmetry under boosts) and $P=MV$?

 

[DrStupid]

Isn't it a rather trivial implication from conservation of total linear momentum $P$ (symmetry under translations) and linear motion of the center of mass $V$ (symmetry under boosts) and $P=MV$?

Only if the uniform motion of the center of mass is not derived from conservation of momentum and mass.

 

[greypilgrim]

Only if the uniform motion of the center of mass is not derived from conservation of momentum and mass.

It's the conserved quantity corresponding to symmetry under boosts.

 

[stevendaryl]

I'm not sure if this going off on a tangent, or not, but it is common to see people write that in cases of variable-mass systems (such as a rocket expending fuel), you should use $F = M \frac{dV}{dt} + \frac{dM}{dt} v$ instead of $F = M \frac{dV}{dt}$. I actually don't think that's a good way to think about variable mass systems.

Assume that initially you have a rocket of mass $M$ traveling at velocity $V$ and during a period $\Delta T$ it is acting on by a force $F$ and throws off a smaller mass $m$ at a relative velocity of $U$.

Then Newton's laws applied to the composite system gives:

$F \Delta T = \Delta P = (M-m) (V+\Delta V) + m (V+U) - MV = M \Delta V + m U - m \Delta V$

Assuming that $m$ and $\Delta V$ are both small, so that their product is negligible, this gives

$F \Delta T \approx M \Delta V + m U$

Since $\Delta M = -m$, we can write this as:

$F \Delta T \approx M \Delta V - \Delta M U$

Dividing through by $\Delta T$ gives (in the limit as $\Delta T \Rightarrow 0$

$F = M \frac{dV}{dt} - \frac{dM}{dt} U$

So the rule $F = M \frac{dV}{dt} + \frac{dM}{dt} V$ is only valid in the case $U = -V$. Remember, $U$ is the relative velocity between the rocket $M$ and the fuel $m$. So $U = -V$ is the case where the fuel has zero velocity after being expended (the total velocity of $m$ is $U+V$). That's not a realistic assumption about rockets. Certainly when a rocket first launches, $V \approx 0$ but $U < 0$ (the fuel is propelled backward).

 

[DrStupid]

Then Newton's laws applied to the composite system gives:
$F \Delta T = \Delta P = (M-m) (V+\Delta V) + m (V+U) - MV = M \Delta V + m U - m \Delta V$

How do you get this equation? Before the ejection of the reaction mass the rocket has the momentum $M \cdot V$ and if $\Delta V$ is the change of the velocity of the rocket, than it has the momentum $\left( {M - m} \right) \cdot \left( {V + \Delta V} \right)$ after the ejection. The resulting change of momentum is

F \cdot \Delta t = \Delta p = \left( {M - m} \right) \cdot \left( {V + \Delta V} \right) - M \cdot V = M \cdot \Delta V - m \cdot V - m \cdot \Delta V \approx M \cdot \Delta V - m \cdot V

which is consistent with $F = \dot p = M \cdot \frac{{dV}}{{dt}} + \frac{{dM}}{{dt}} \cdot V$. Could you please show me your full calculation, including the derivation of the first equation?

 

[vanhees71]

Isn't it a rather trivial implication from conservation of total linear momentum $P$ (symmetry under translations) and linear motion of the center of mass $V$ (symmetry under boosts) and $P=MV$?

The point is that you tacitly assume that $M$ is constant. There's no symmetry principle in classical mechanics that let's you derive this, and special relativity shows that mass is not separately conserved although in special relativity you have the same symmetries as in Newtonian mechanics (homogeneity and isotropy of space and boost invariance), however of course realized by a different symmetry group (Poincare group vs. Galileo group).

In quantum mechanics it turns out that mass occurs as a non-trivial central charge of the Galileo group's Lie algebra. This suggests a superselection rule according to which there are no superpositions of states with different masses, i.e., a non-relativistic quantum system has to live in a Hilbert space, where the total mass is fixed to one value.

 

[vanhees71]

How do you get this equation? Before the ejection of the reaction mass the rocket has the momentum $M \cdot V$ and if $\Delta V$ is the change of the velocity of the rocket, than it has the momentum $\left( {M - m} \right) \cdot \left( {V + \Delta V} \right)$ after the ejection. The resulting change of momentum is

F \cdot \Delta t = \Delta p = \left( {M - m} \right) \cdot \left( {V + \Delta V} \right) - M \cdot V = M \cdot \Delta V - m \cdot V - m \cdot \Delta V \approx M \cdot \Delta V - m \cdot V

which is consistent with $F = \dot p = M \cdot \frac{{dV}}{{dt}} + \frac{{dM}}{{dt}} \cdot V$. Could you please show me your full calculation, including the derivation of the first equation?

The save way to derive the rocket equation is to use the conservation of total momentum, which derives from first (symmetry) principles in Newtonian mechanics, i.e., you have
$$\vec{p}_{\text{rocket}} + p_{\text{fuel}}=\text{const}.$$
Now take the time derivative
$$\dot{\vec{p}}_{\text{rocket}}=-\dot{p}_{\text{fuel}}+\vec{F}.$$
Here, $\vec{F}$ is the total external force on the rocket (usually just the gravitational force of the Earth)
Assuming a given "exhaust of fuel" of $\mu$ (mass of fuel per unit time), you get
$$\dot{M} \vec{V}+M \dot{\vec{V}}=-\mu \vec{V}_{\text{fuel}}+\vec{F}.$$
Now we also assume (!) the conservation of total mass, i.e., $\dot{M}=-\mu$, and then you get
$$M \dot{\vec{V}} = \mu (\vec{V}-\vec{V}_{\text{fuel}})+\vec{F}.$$
Usually one assumes that the velocity of the fuel relative to the rocket $\vec{U}=\vec{V}_{\text{fuel}}-\vec{V}=\text{const}.$

Then we get
$$M \dot{\vec{V}}+\mu \vec{U}=\vec{F}.$$
Let's consider a simplified start of a rocket, which stays close to Earth and flies straight upward. Make the $z$ axis pointing upward. Then $\vec{F}=-M g \vec{e}_z$, and $\vec{U}=-U \vec{e}_z$ with $U>0$. Then we have
$$M \dot{V}_z+\dot{M} U=-M g$$
or
$$\dot{V}_z=-\frac{\dot{M}}{M} U - g.$$
With $U=\text{const}$ we can integrate wrt. $t$ from $t=0$ to $t$:
$$V_z(t)=-U \ln \left [\frac{M(t)}{M_0} \right]-g t.$$

 

[greypilgrim]

The point is that you tacitly assume that MM is constant.

Where do I assume that? From symmetry under translations it follows that the total momentum $P$ is constant, from symmetry under boosts that the center of mass moves at a constant velocity $V$. Since $P=MV$, the total mass $M$ must necessarily be constant as well.

 

[DrStupid]

The save way to derive the rocket equation is to use the conservation of total momentum [...]

Thank you for the derivation. But I already know that and I asked another question.

 

[vanhees71]

But I answered that (or at least I tried to). You cannot blindly use $\vec{F}=\dot{\vec{p}}$ in this example but you must go back to the fundamental law of momentum conservation for a closed system, and the closed system in this case consists of the rocket (including the unexhausted fuel) and the exhausted fuel. Of course @stevendaryl implicitly did exactly this. All I did was to reformulate his derivation in the hope it would clarify the argument.

For more examples, see Sommerfeld, Lectures on theoretical physics, vol. 1 (in the very beginning), where he treats all kinds of problems with "varying mass" like a chain falling from a table (both variants), a rain drop falling is vapour-saturated atmosphere and the like. As always, Sommerfeld is by at least an $\epsilon$ more comprehensive than many other textbooks!

 

[stevendaryl]

How do you get this equation? Before the ejection of the reaction mass the rocket has the momentum $M \cdot V$ and if $\Delta V$ is the change of the velocity of the rocket, than it has the momentum $\left( {M - m} \right) \cdot \left( {V + \Delta V} \right)$ after the ejection. The resulting change of momentum is

F \cdot \Delta t = \Delta p = \left( {M - m} \right) \cdot \left( {V + \Delta V} \right) - M \cdot V = M \cdot \Delta V - m \cdot V - m \cdot \Delta V \approx M \cdot \Delta V - m \cdot V

which is consistent with $F = \dot p = M \cdot \frac{{dV}}{{dt}} + \frac{{dM}}{{dt}} \cdot V$. Could you please show me your full calculation, including the derivation of the first equation?

You left out that the expelled mass, $m$ has momentum, also. The way I set things up:

Before: (separating the rocket from its fuel):

• Rocket has mass $M - m$
• Rocket has velocity $V$
• Fuel has mass $m$
• Fuel has velocity $V$
• Total momentum = $(M-m)V + mV = MV$

After:

• Rocket has mass $M-m$
• Rocket has velocity $V + \Delta V$
• Fuel has mass $m$
• Fuel has velocity $V+U$ ($U$ is the relative velocity between expended fuel and rocket, which is typically in the opposite direction from $V$)
• Total momentum = $(M-m) (V + \Delta V) + m (V+U) = MV + M \Delta V - mV - m \Delta V + mV + mU = MV + M \Delta V + m (U - \Delta V)$
• So the change in momentum is $MV + M \Delta V + m (U - \Delta V) - MV = M \Delta V + m (U - \Delta V)$
• In terms of $\Delta M = -m$: $\Delta P = M \Delta V - (\Delta M) U + \Delta M \Delta V \approx M \Delta V - (\Delta M) U$

It only reduces to $\Delta P = M \Delta V + (\Delta M)$ in the special case $U = -V$. Which might conceivably be correct in some circumstances, but not all. Probably not for an actual rocket.

 

[DrStupid]

You left out that the expelled mass, $m$ has momentum, also.

Of course I did. The expelled mass is not part of the rocket.

So the change in momentum is $MV + M \Delta V + m (U - \Delta V) - MV = M \Delta V + m (U - \Delta V)$

That's the change of the total momentum and as you do not include external forces it is always zero.

It only reduces to $\Delta P = M \Delta V + (\Delta M)$ in the special case $U = -V$.

You are confusing the force $M \cdot \dot V + \dot M \cdot V$ acting on the rocket with the force $M \cdot \dot V - \dot M \cdot U = 0$ acting on the total system of rocket and exhaust. Of course they are equal for U=-V only. This is the special case where the momentum of the rocket doesn't change.

 

[DrStupid]

But I answered that (or at least I tried to).

Than you didn't understand my question.

You cannot blindly use $\vec{F}=\dot{\vec{p}}$ in this example

Equations should never be blindly used - not only in this example.

but you must go back to the fundamental law of momentum conservation for a closed system

With $\vec{F}=\dot{\vec{p}}$ the conservation of momentum is included in the laws of motion.

 

[stevendaryl]

Of course I did. The expelled mass is not part of the rocket.

Yes, but for $F=Ma$, the external force goes into changing the momentum of both.

 

[DrStupid]

Yes, but for $F=Ma$, the external force goes into changing the momentum of both.

You neither included external forces nor did you talk about $F=Ma$.

 

[vanhees71]

Than you didn't understand my question.
Equations should never be blindly used - not only in this example.
With $\vec{F}=\dot{\vec{p}}$ the conservation of momentum is included in the laws of motion.

Of course you can derive the equations of motion from the fundamental conservation laws and vice versa for closed (!) systems, and that's what I did. If this doesn't answer your question, I didn't understand it, and if you want another clarification you have to tell me what's not clear with @stevendaryls and my derivation.

 

[stevendaryl]

You neither included external forces nor did you talk about $F=Ma$.

For the composite system rocket + fuel, the mass is unchanging, so we can we certainly use:

$F_{ext} = \frac{dP}{dt}$

where $P$ is the total momentum of rocket + fuel.

For that composite system, we have:

$F_{ext} = M \frac{dV}{dt} - U \frac{dM}{dt}$

which is not equal to $\frac{d}{dt} MV$ except in the special (unrealistic) case $U = -V$.

 

[DrStupid]

If this doesn't answer your question, I didn't understand it, and if you want another clarification you have to tell me what's not clear with @stevendaryls and my derivation.

stevendaryls already answered my question. Instead of the force acting on the rocket he derived the force acting on the total system of rocket and exhaust. Not being aware of the difference he erroneously concluded that $F = \dot p$ doesn't work because both forces are equal in the special case V=-U only.

For the composite system rocket + fuel, the mass is unchanging, so we can we certainly use:

$F_{ext} = \frac{dP}{dt}$

where $P$ is the total momentum of rocket + fuel.

For that composite system, we have:

$F_{ext} = M \frac{dV}{dt} - U \frac{dM}{dt}$

which is not equal to $\frac{d}{dt} MV$ except in the special (unrealistic) case $U = -V$.

You are still confusing the force $F_{ext}$ acting on the total system of rocket and exhaust with the force $\frac{d}{dt} MV$ acting on the rocket. Of course they are equal for V=-U only because that’s the special case where the exhaust leaves the rocket without momentum.

 

[greypilgrim]

Thanks for all your contributions. A question of mine remained unanswered, though:

Why mass is conserved in Newtonian is a very subtle issue

Isn't it a rather trivial implication from conservation of total linear momentum $P$ (symmetry under translations) and linear motion of the center of mass $V$ (symmetry under boosts) and $P=MV$?

What's so subtle about that?

 

[vanhees71]

Well, mass in non-relativistic theory is a quite complicated quantity. It comes into the theory when analysing the Galileo symmetry in quantum mechanics, i.e., you look for the unitary ray representations of the corresponding covering group. Then you realize that the proper unitary transformations do not lead to anything physically sensible. Simplified you can say that these proper representations would mean zero-mass particles, which do not make any sense in Newtonian physics nor in its quantized version. So you have to use also a non-trivial central extension of the Galileo Lie algebra, and the only non-trivial central extension is to introduce mass as a non-trivial central charge. Looked on mass from this very abstract point of view of extending classical Galileo symmetry to its quantum version, the fact that mass is a central charge implies that there is a socalled super selection rule, which forbids the superposition of states with different masses, i.e., if you have a system with a given mass no interaction can change this mass, because the unitary time evolution of quantum theory cannot connect Hilbert spaces with systems of different mass and thus mass is conserved in non-relativistic physics.

 

[greypilgrim]

That indeed sounds highly nontrivial. Is QM essential to understand mass, even in classical physics?

Anyway, what's wrong with my simple argument using symmetry under translations and under boosts and $P=MV$?

 

[vanhees71]

If you ask "why" in the natural sciences the answer can only be, "it's because of that or that fundamental principle". In classical mechanics mass is just a parameter for the inertia of matter. Its conservation is just postulated. So this doesn't answer your "why-question" very satisfactorially. The reason, why I answered with quantum-theoretical arguments is that the most fundamental theory we have today is quantum theory, and there is an argument from simpler principles, namely symmetry principles, which leads to the best answers to "why-questions" we can give today.

 

[stevendaryl]

You are still confusing the force $F_{ext}$ acting on the total system of rocket and exhaust with the force $\frac{d}{dt} MV$ acting on the rocket. Of course they are equal for V=-U only because that’s the special case where the exhaust leaves the rocket without momentum.

I think that the only sense in which $F = M \frac{dV}{dt} + V \frac{dM}{dt}$ is true is if you make it true by definition.

Consider a rocket, with no engine at all --- it's just drifting in space. But it's an old, crumbly rocket that is falling apart. So pieces of it are breaking off. That implies that $\frac{dM}{dt}$ is negative for the rocket. So that means that there is a force acting on the rocket? I think that's bizarre. Yes, you can say that it's true by definition, but it certainly isn't a force in the intuitive sense.

 

[Orodruin]

So that means that there is a force acting on the rocket? I think that's bizarre.

I do not think it bizarre. There is a change in momentum of the rocket because you are redefining what "the rocket" means. There is a transfer of momentum from "the rocket" to "not the rocket".

Yes, you can say that it's true by definition, but it certainly isn't a force in the intuitive sense.

There are many instances in physics where "intuition" leads you wrong. I think it is the reasonable definition, whether or not it makes "intuitive sense" or not. Intuitive sense is subjective.

 

[vanhees71]

There are many examples, where for a mechanical system the mass changes. There are some very illuminating exercises in Sommerfeld Lectures on Theoretical Physics.

E.g., there are two types of the problem of a chain falling from a table: (a) the chain is curled up and only the part hanging over the edge of the table moves or (b) the chain moves all the time as a whole. Another one is a raindrop falling in an oversaturated atmosphere and thus is gaining more and more water on its way down. Of course the rocket is among the standard problems with systems changing mass.

The fundamental laws are of course not the forces but the conservation laws and symmetries, i.e., here momentum conservation and translation symmetry.

 

[haushofer]

That indeed sounds highly nontrivial. Is QM essential to understand mass, even in classical physics?

My 2 cents: no, this already plays at the classical level.

The action of a point particle with embedding coordinates $x^0(\tau),x^i(\tau)$ and wordlineparameter $\tau$ reads

<br /> S = \frac{m}{2} \int \frac{\dot{x}^i \dot{x}_i}{\dot{x}^0} d \tau<br />

This action is invariant under the infinitesimal Galilei transformations

<br /> \delta_H x^0 = \zeta^0, \ \ \delta_P x^i = \zeta^i, \ \ \delta_J x^i = \lambda^i{}_j x^i, \ \ \delta_G x^i = v^i x^0<br />

From this you can calculate the commutators. In particular, you'll find

<br /> [\delta_P, \delta_G] x^0 = [\delta_P, \delta_G] x^i = 0 \ \ \rightarrow [\delta_P, \delta_G] = 0<br />

But now comes the subtlety: the action is invariant, but the Lagrangian $L$ is only quasi-invariant under boosts, i.e. it transforms to a total derivative as you can check:
<br /> \delta_G L = \frac{d}{d\tau}\Bigl( m x_i v^i \Bigr)<br />

This has an important consequence: the Noether charge $Q_G$ of the boosts needs to be adjusted to this such that it remains conserved:

<br /> Q_G = p_i v^i x^0 - m x^i v_i<br />

If we now calculate the corresponding Poisson-brackets between the Noether charges $Q_P = p_i \zeta^i$ and $Q_G$, you will find due to this (signs depend on your conventions for these brackets)

<br /> \{Q_G, Q_P \}_{poisson} \sim m \zeta_i v^i<br />

where $m$ is the mass appearing in the action. This suggests that the Lie algebra of the Galilei group allows for a central extension (i.e. an extra generator $M$ which commutes with all the other generators), playing the role of mass:

<br /> [G,P] = M<br />

You can also check by the Jacobi-identities that this works out. Algebraically, in the non-relativistic limit $M$ replaces the Casimir operator $P_{\mu}P^{\mu}$ of the Poincare algebra which plays the role of mass. This centrally-extended Galilei algebra is called the Bargmann algebra. Interestingly, one can gauge this algebra and show that the corresponding gauge field of the generator $M$ roughly plays the role of the Newton potential. I like this result, because usually people tend to associate central extensions to quantization, but they already play an important role at the classical level.

By the way, this should also answer the question of the OP: in varying the action to obtain the EOM or symmetries, we treat the mass parameter $m$ as a constant. If you want to promote it to a function depending on $x^0$ or $x^i$, you should also vary it. I'm not sure if this makes sense, though, but it should be clear that the Galilei-symmetries are not apparent anymore.

 

[haushofer]

By the way, the fact that Galilei boosts play a subtle role in quantum mechanics can be seen by transforming the Schrodinger equation under them. You already see there easily that the wave function cannot be a simple scalar under Galilei boosts. The textbook by Ballentine has a very nice chapter on this.

 

[haushofer]

And, as a last fun-fact about the Bargmann algebra: the simplest $N=1$ supersymmetric extension of it necessarily break the relation between supertransformations and spacetime translations which is at the core of supersymmetry; the commutator of two supertransformations don't give a spacetime translation anymore in the non-relativistic case, but...the central element $M$.

Sorry, spend 4 years of my life on this, had to spit it out.

 

[stevendaryl]

I do not think it bizarre. There is a change in momentum of the rocket because you are redefining what "the rocket" means. There is a transfer of momentum from "the rocket" to "not the rocket".

Yes, I agree that momentum is changing. But the way you compute that is by using conservation of momentum, not by analyzing forces.

I think it is the reasonable definition, whether or not it makes "intuitive sense" or not.

What makes it a reasonable definition?

 

[stevendaryl]

What makes it a reasonable definition?

To answer my own question, a definition is reasonable if it does one or both of (1) provides insight, or (2) helps to organize information to calculate answers to problems. If it doesn't do either of those, then I don't think it's a reasonable definition.

As I said previously, the canonical case of variable mass is an object such as a rocket that throws mass behind it to propel forward. In this case, if there are no other forces (such as gravity) involved, then the equations of motion will be:

$M \frac{dV}{dt} - \frac{dM}{dt} U = 0$

where $U$ is the velocity of the exhaust relative to the rest of the rocket (for the rocket, both $\frac{dM}{dt}$ and $U$ will be negative, for a coordinate system in which the rocket is accelerating in the positive-x direction). The quantity $M \frac{dV}{dt} + V \frac{dM}{dt}$ doesn't come into play, at all. Now, of course you can come up with an equation for it:

$M \frac{dV}{dt} + V \frac{dM}{dt} = (U+V) \frac{dM}{dt}$

I guess that sort of makes sense: During a time $\delta t$, you're throwing off an amount of mass $- \frac{dM}{dt} \delta t$ and its velocity is $U+V$. So you're tossing out an amount of momentum equal to $-\frac{dM}{dt} (U+V) \delta t$, so the rocket's momentum must decrease by this amount by conservation of momentum. But I don't see how it helps to think of this in terms of forces, rather than conservation of momentum. It seems very awkward.

 

[vanhees71]

And, as a last fun-fact about the Bargmann algebra: the simplest $N=1$ supersymmetric extension of it necessarily break the relation between supertransformations and spacetime translations which is at the core of supersymmetry; the commutator of two supertransformations don't give a spacetime translation anymore in the non-relativistic case, but...the central element $M$.

Sorry, spend 4 years of my life on this, had to spit it out.

Great that you did. Particularly in #31 you convincingly argued for mass conservation in Galilei-Newtonian spacetime from purely classical arguments. So one doesn't need the ray-representation argument of QT to argue for the extension of the Galilei algebra to the Bargmann-Wigner algebra in classical Newtonian mechanics!

 

[greypilgrim]

Sorry to ask again: After reading the last couple of responses, my

Isn't it a rather trivial implication from conservation of total linear momentum P (symmetry under translations) and linear motion of the center of mass V (symmetry under boosts) and P=MV?

seems far too simple, so what's wrong with it?

 

[vanhees71]

Well, in special relativity you have both symmetries but mass is not conserved!

 

[greypilgrim]

Ok, so this argument must clearly be false (probably even the non-relativistic version), where's the flaw?

 

[vanhees71]

The flaw is that in Newtonian mechanics the conservation of mass doesn't follow from the original Galilei symmetry but only from the extension of the Gailei-group algebra by a central charge named the Bargmann algebra (as I learned from #31 and the following very nice postings by @haushofer).

There's no flaw in the case of relativistic mechanics either. The reason here is that the Poincare algebra has no non-trivial central charges, i.e., the (proper orthochronous) Poincare group is the full symmetry group there is for SRT. Indeed obviously mass is not conserved. Take e.g., pair annihilation $\text{e}^+ + \text{e}^- \rightarrow \gamma + \gamma$, where the total mass on the left-hand side is $2 m_{\text{e}} \simeq 1.022 \; \text{MeV}$ and on the right-hand side vanishes.

 

[greypilgrim]

I still can't see what's wrong with this specific argument. Using Noether's theorem, we can derive from symmetry under translations that the total momentum $P$ is constant. We can also derive from symmetry under Galilean boosts that the velocity of the center of mass $V$ is constant. Since in Newtonian physics it's always $P=M\cdot V$, the total mass $M$ must also be also constant.

Would you mind pinpointing exactly which one of those three steps is wrong, and why?

By the way, I did not come up with this argument myself, I read it on Stack Exchange (answer by Malabarba).

 

[haushofer]

I still can't see what's wrong with this specific argument. Using Noether's theorem, we can derive from symmetry under translations that the total momentum $P$ is constant. We can also derive from symmetry under Galilean boosts that the velocity of the center of mass $V$ is constant. Since in Newtonian physics it's always $P=M\cdot V$, the total mass $M$ must also be also constant.

Would you mind pinpointing exactly which one of those three steps is wrong, and why?

By the way, I did not come up with this argument myself, I read it on Stack Exchange (answer by Malabarba).

Well, write down the action (I gave it some posts ago), promote the mass to be a function of the coordinates $\{x^0, x^i\}$ and back your statements up by calculation and we'll see. Let's take the static gauge $x^0 = \tau$ in the Lagrangian of the action I gave earlier:

<br /> L = \frac{m}{2} \dot{x}^i \dot{x}_i<br />

I'm not sure if this makes any sense, but just for the fun of it: If the mass can be a priori any function of the embedding coordinates, $m = m(x^0, x^i)$, the variation of the Lagrangian reads

<br /> \delta L = \frac{1}{2} \Bigl(\delta m \, \dot{x}^i \dot{x}_i + 2m \dot{x}^i \delta \dot{x}_i \Bigr)<br />

Let's now take a constant spatial translation $\delta x^i = \zeta^i$:

<br /> \delta L = \frac{1}{2} \Bigl(\frac{\partial m}{\partial x^i} \delta x^i \dot{x}^k \dot{x}_k + 0 \Bigr) = \frac{1}{2} \Bigl(\frac{\partial m}{\partial x^i} \zeta^i \dot{x}^k \dot{x}_k \Bigr)<br />

This is not zero in general, so I guess you see the subtlety here.

You also say: "We can also derive from symmetry under Galilean boosts that the velocity of the center of mass $V$ is constant." Can you show explicitly how? This whole business is a bit rusty for me, but aren't you mixing up equations of motion with the constancy of Noether charges? If so, the equations of motion you want to use already make use of the fact that mass is constant, isn't it? That makes it a bit of a circle reasoning.

As I said, this has been a while for me, but I don't get the Stack Exchange comment.

 

[DrStupid]

I think that the only sense in which $F = M \frac{dV}{dt} + V \frac{dM}{dt}$ is true is if you make it true by definition.

I don't need to make it true by definition because it already is true by definition - since more than three centuries.

 

[stevendaryl]

I don't need to make it true by definition because it already is true by definition - since more than three centuries.

No, it's not true by definition. I think that's a misinterpretation of Newton's laws. For one thing, Newton's original second law was not expressed in terms of change of momentum, it was expressed in terms of acceleration. It's not explicitly stated, but I think that his laws really implicitly assume constant-mass objects, so the distinction between $F = m \frac{dv}{dt}$ and $F = \frac{d}{dt} mv$ is moot. But I think it's a misinterpretation of Newton's laws to think that the second law is a definition of force. For one thing, there can be forces at work on a system where there is no acceleration. There's a whole branch of physics called "statics" which is about computing forces in situations where there is no motion at all. So forces are not identified with acceleration in Newtonian physics.

Secondly, Newton's third law (equal and opposite forces) is definitely not true by definition.

I think that the best way to think about Newton's laws is to consider forces to be primitive, not defined quantities. Between any two objects, $A$ and $B$, there is potentially two forces: $F_{AB}$, the force of $A$ on $B$, and $F_{BA}$, the force of $B$ on $A$. These are vector quantities. Then Newton's second law should be understood as:

$\sum_{j} F_{ji} = M_i \frac{dV_i}{dt}$

Newton's third law is understood as the assumed symmetry

$F_{ij} = - F_{ji}$

Newton's laws are really about constant-mass objects. In light of the fact that mass is conserved in Newtonian mechanics, we can always derive whatever we want to know about objects of changing mass by redrawing our object boundaries so that the problem at hand deals with constant-mass objects. So there is no logical need to have a rule about variable-mass objects.

 

[DrStupid]

For one thing, Newton's original second law was not expressed in terms of change of momentum, it was expressed in terms of acceleration.

Newton's original second law was expressed in terms of change of motion (see http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/46) and motion is defined as the product of mass and velocity (see http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/26) which is nothing else than momentum.

It's not explicitly stated, but I think that his laws really implicitly assume constant-mass objects, so the distinction between $F = m \frac{dv}{dt}$ and $F = \frac{d}{dt} mv$ is moot.

I don't see any reason for such a restriction. The wording of the second law doesn't suggest something like that and the second and third law work quite well for variable mass systems. Just the first law works for closed systems only (which is therefore the best interpretation of Newton's term "body" in this particular case).

Secondly, Newton's third law (equal and opposite forces) is definitely not true by definition.

1. For systems with at least three bodies pair wise interactions are just one of infinite different possibilities to divide the net forces into components. The definition that forces always act pair wise and equal in both directions is the most convenient method to aply conservation of momentum but not the only possible.

2. The third law also distinguishes forces from factious forces - again simply by definition. Newton's attempt to change that in order to generalize the laws of motion to non-inertial systems (see http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/49) didn't became accepted back then. However, today it is quite normal to use factious forces, which doesn't comply with the third law. It is just a matter of definition that they are no forces.

Both aspects of the third law are just true by definition.

So there is no logical need to have a rule about variable-mass objects.

Yes, that's correct, because variable-mass systems are already covered by the laws of motion and there is no logical need to invent an artificial restriction to systems with constant mass.

 

[stevendaryl]

Newton's original second law was expressed in terms of change of motion (see http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/46) and motion is defined as the product of mass and velocity (see http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/26) which is nothing else than momentum.

But I don't think that there was any indication that he considered variable-mass objects.

1. For systems with at least three bodies pair wise interactions are just one of infinite different possibilities to divide the net forces into components. The definition that forces always act pair wise and equal in both directions is the most convenient method to aply conservation of momentum but not the only possible.

My point is that "forces act pair wise" doesn't make any sense if you think of force as being defined as the rate of change of momentum.

2. The third law also distinguishes forces from factious forces - again simply by definition. Newton's attempt to change that in order to generalize the laws of motion to non-inertial systems (see http://cudl.lib.cam.ac.uk/view/PR-ADV-B-00039-00001/49) didn't became accepted back then. However, today it is quite normal to use factious forces, which doesn't comply with the third law. It is just a matter of definition that they are no forces.

To me, the better way to make sense of forces in a noninertial coordinate system is to understand it as a vector equation:

$\overrightarrow{F} = M \frac{d}{dt} \overrightarrow{v}$

If you understand the right side in terms of parallel transport (yeah, I know Newton didn't think of it that way, because differential geometry had not been invented yet) then it applies in non-cartesian coordinate systems, as well as rectangular:

$\frac{d}{dt} \overrightarrow{v} = \sum_j e_j (\frac{d v^j}{dt} + \Gamma^j_{kl} v^k v^l)$

This works for non-cartesian coordinates, but not for non-inertial coordinates (whose translations to inertial, cartesian coordinates are time-dependent). But even the noninertial case can be understood as $F^\mu = M (\frac{dv^\mu}{ds} + \Gamma^\mu_{\nu \lambda} v^\nu v^\lambda)$ if you take the 4-dimensional view of considering $t$ as a coordinate like the spatial coordinates.

Both aspects of the third law are just true by definition.

I would say not. It doesn't actually matter, I guess. But as I said in another comment, the two considerations for whether definitions are useful are:

1. Does it lead to better insights?
2. Does it help to organize facts in solving problems?

I don't see how the view that $\overrightarrow{F} = \frac{dM}{dt} \overrightarrow{V} + M \frac{d\overrightarrow{V}}{dt}$ is useful by either criteria. You have to add more assumptions about the distinction between real forces and fictitious forces, and you have to make up the distinction between "net force" and component forces (the latter can be nonzero even when the former adds up to zero), which makes little sense if force is by definition the change in momentum. (You could define it counterfactually, I suppose, as a force is by definition, the change in momentum that would result if it were the only force at work).

The whole thing seems like very bad pedagogy, to me. It doesn't help in solving problems or in understanding what's going on.

 

[DrStupid]

But I don't think that there was any indication that he considered variable-mass objects.

And I don't think that there is any indication that he excluded variable-mass systems.

My point is that "forces act pair wise" doesn't make any sense if you think of force as being defined as the rate of change of momentum.

Forces are defined as proportional to the rate of change of momentum - not to be identical with the rate of change of momentum. You may see them as the momentum flow from one system to another. In analogy to thermodynamics momentum would be the state function and force the corresponding process function. In this case it is quite obvious that in a closed system consisting of two bodies A and B the flow from A to B is equal but opposite to the flow from B to A (due to conservation of momentum). In case of more than two bodies it must bedefinied that the transfer of momentum is always pair wise between each two of them.

To me, the better way to make sense of forces in a noninertial coordinate system is to understand it as a vector equation:

$\overrightarrow{F} = M \frac{d}{dt} \overrightarrow{v}$

This is again a limitation of Newton's second law to systems with constant mass and I still do not see a corresponding justification. The second law obviously works for open systems. Why shouldn't it be used for them? What is the benefit of a restriction to closed systems?

I don't see how the view that $\overrightarrow{F} = \frac{dM}{dt} \overrightarrow{V} + M \frac{d\overrightarrow{V}}{dt}$ is useful by either criteria.

And I do not see how a restriction of the Newtonean mechanis to closed systems is useful by any means. It seems we are not going to come to an agreement.

 

[vanhees71]

Well, as I stated above there are some problems where one needs to take the "changing mass" into account, e.g., rockets, a falling raindrop in a satisfied atmosphere gaining more and more water on its way falling down, a chain moving from a table etc. All these phenomena are most clearly treated, using momentum conservation of closed systems, and indeed Newton got the law right from the very start.

 

[stevendaryl]

Forces are defined as proportional to the rate of change of momentum - not to be identical with the rate of change of momentum.

I think you're mixing up terminology. Something is "defined" to be proportional to something else. It's an axiom, or assumption, not a definition. Newton's laws are introducing a concept called "force", which is assumed to have certain properties:

1. It is a vector quantity
2. For any pair of objects, there is a force of one object on another
3. The force of one object on the other is the negative of the force of second object on the first
4. The vector sum of the forces acting on one object is proportional to the rate of change of momentum of that object

To me, #4 is an assumption, or axiom, about the way forces work. It's not a definition.

You may see them as the momentum flow from one system to another.

Yes, momentum flow makes sense. But there is a physical difference between physical forces and the nonphysical quantity $\frac{dM}{dt} V$. The latter quantity can be zero or nonzero depending on how you define your "object". It's not a physical effect, it's a modeling effect. Just to go down the list:

1. Gravitational forces don't cause $M$ to change.
2. Electrical forces don't cause $M$ to change.
3. Springs attached to an object don't cause $M$ to change.

What causes $M$ to change is a modeling choice made by the physicist.

This is again a limitation of Newton's second law to systems with constant mass and I still do not see a corresponding justification. The second law obviously works for open systems. Why shouldn't it be used for them? What is the benefit of a restriction to closed systems?

Given that mass is conserved (and more specifically, obeys a continuity equation), you can always solve problems by considering forces on constant-mass objects. So it's certainly not a limitation on what can be calculated. If you want to compute the motion of a rocket, you consider the rocket plus the fuel together.

As I said, there are two reasons for choosing one definition over another: (1) Does it provide insight? (2) Does it help in solving problems?

I think that the application of $F = M \frac{dV}{dt} + \frac{dM}{dt} V$ to variable mass systems fails in both respects. It doesn't provide insight, because it mixes up modeling choices (how to draw the boundary affects this notion of "force") with physical fields (electrical forces, gravitational forces). And it doesn't help solve problems. It seems to me to be a useless way to look at things.

And I do not see how a restriction of the Newtonean mechanis to closed systems is useful by any means.

The advantage to considering constant-mass definitions is that then the actual physical forces (you know, gravity, electric forces, contact forces, etc.) are brought to the fore. Whereas in variable-mass systems, the forces are artifacts of how you draw the object boundary.

 

[stevendaryl]

Once again, I would like to see a single example where considering $\frac{dM}{dt} V$ a force helps to understand a problem.