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

[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:

$$L = \frac{m}{2} \dot{x}^i \dot{x}_i$$

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

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

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

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

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.

 

[haushofer]

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

Maybe if you consider the geodesic equation in which mass is not constant anymore? I didn't do the calculation, but don't you then get something like

$$\ddot{x}^{\rho} + \Gamma^{\rho}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} = \frac{\dot{m}}{m}\dot{x}^{\rho}$$

? The right hand side can then be considered as a force acting on the particle, pulling it out of geodesic motion.

 

[DrStupid]

It's an axiom, or assumption, not a definition.

It's an axiomatic definition.

But there is a physical difference between physical forces and the nonphysical quantity $\frac{dM}{dt} V$.

I am talking about forces as defined by Newton and not about "physical forces" (whatever that means).

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.

With limitation of the laws of motion to closed systems you need to divide the fuel into infinite pieces in order to apply forces. Without this limitation you just need two systems: the rocket and the exhaust. What is the benefit of the first option over the second?

 

[stevendaryl]

It's an axiomatic definition.

That's twisting the notion of "definition" to the point of meaninglessness.

With limitation of the laws of motion to closed systems you need to divide the fuel into infinite pieces in order to apply forces.

It seems straightforward to me: You consider two times: $t$ and $t+\delta t$. Let $M_{fuel}$ be the amount of fuel expended during time $\delta t$, and let $V$ be the velocity of the rocket at time $t$, and let $\delta V$ be the change in the velocity of the rocket, and let $U$ be the relative velocity of the fuel after it is expelled. Then conservation of momentum gives you:

$(M + M_{fuel}) V = M (V + \delta V) + M_{fuel} (V + U)$

Ignoring higher-order terms gives:

$M \delta V = - M_{fuel} U$

Since the mass of the rocket decreases as fuel is expended, we can write: $\frac{dM}{dt} = - M_{fuel}/\delta t$. So dividing through by $\delta t$ gives:

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

The quantity $\frac{dM}{dt} V$ just doesn't come into play, at all. This is just conservation of momentum. The force on the rocket is just

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

If you have an external force such as gravity, then the equations of motion are:

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

So:

$F_{rocket} = \frac{dM}{dt} U - M g$

The total force on the rocket is the gravitational force, $-Mg$ plus the contact force of the fuel against the rocket, $\frac{dM}{dt} U$.

If you want to call $\frac{dM}{dt} V$ a force, where does it come into play? It doesn't. You can certainly add and subtract it from both sides:

$M \frac{dV}{dt} + \frac{dM}{dt} V = \frac{dM}{dt} U - M g + \frac{dM}{dt} V$

Then you can say that the total force on the rocket has three parts:

1. $\frac{dM}{dt} U$: the contact force of the fuel pressing against the rocket
2. $-Mg$: the force of gravity
3. $\frac{dM}{dt} V$: the force due to changing mass

But term number three serves no purpose at all. It's just a term that "goes along for the ride" on both sides of the equal sign.

 

[vanhees71]

Maybe if you consider the geodesic equation in which mass is not constant anymore? I didn't do the calculation, but don't you then get something like

$$\ddot{x}^{\rho} + \Gamma^{\rho}_{\mu\nu}\dot{x}^{\mu}\dot{x}^{\nu} = \frac{\dot{m}}{m}\dot{x}^{\rho}$$

? The right hand side can then be considered as a force acting on the particle, pulling it out of geodesic motion.

How do you come to this equation? In GR, the EoM reads
$$\mathrm{D}_t p^{\mu}=\dot{p}^{\mu} + \Gamma^{\mu}_{\rho \sigma} p^{\rho} p^{\sigma} = K^{\mu},$$
where $p^{\mu}$ is the canonical (sic!) momentum and $K^{\mu}$ the four-force of all non-gravitational interactions acting on the particle.

 

[DrStupid]

It seems straightforward to me: [...]

Not for me. Without the limitation to closed systems I can directly apply the second law to rocket and exhaust:

$F_R = m_R \cdot \dot v_R + \dot m_R \cdot v_R$

$F_E = m_E \cdot \dot v_E + \dot m_E \cdot v_E$

The already exhausted reaction mass can be assumed to be zero (as if the rocked just started) because it does not interact with the rocket anymore:

$m_E = 0$

The relative exhaust velocity is

$u = v_E - v_R$

Conservation of mass results in

$\dot m_E = - \dot m_R$

And the third law says

$F_R = - F_E$

Everything together results in the differential rocket equation

$m_R \cdot \dot v_R = \dot m_R \cdot u$

without considering differential time steps, ignoring of higher-order terms or hand waving force equations. Additional forces can easily be included in the last condition. That means for gravity

$F_R = m_R \cdot g - F_E$

which results in

$m_R \cdot \dot v_R = g + \dot m_R \cdot u$

That is straightforward for me.

 

[haushofer]

How do you come to this equation? In GR, the EoM reads
$$\mathrm{D}_t p^{\mu}=\dot{p}^{\mu} + \Gamma^{\mu}_{\rho \sigma} p^{\rho} p^{\sigma} = K^{\mu},$$
where $p^{\mu}$ is the canonical (sic!) momentum and $K^{\mu}$ the four-force of all non-gravitational interactions acting on the particle.

It was a sophisticated guess if one makes the mass time dependent in the geodesic equation, but as I said, it could be wrong.

 

[stevendaryl]

Not for me. Without the limitation to closed systems I can directly apply the second law to rocket and exhaust:

$F_R = m_R \cdot \dot v_R + \dot m_R \cdot v_R$

$F_E = m_E \cdot \dot v_E + \dot m_E \cdot v_E$

I consider your derivation a little bogus. When you treat $M_E$ as zero, it means that you aren't considering the mass of the exhaust, but the incremental change in the mass of the exhaust. So you're actually doing the differential case that I was doing, but hiding that fact.

The mass of the exhaust isn't zero, except when the rocket first launches. But if you are really talking about a nonzero mass of the exhaust, then there is no single velocity to use in computing the term $V_E \frac{dM_E}{dt}$. Different parts of the velocity are moving at different speeds. The only times when it is meaningful to talk about an object having a single speed is when the object is rigid (so all the parts are moving at the same speed) or when the object is confined to a small region of space, so that all the pieces can be treated as approximately having the same velocity. But in those cases, you have well-defined objects with well-defined masses, so there is no point.

If you want to talk about a more complex case where you have extended objects that are not rigid, then to me, the way to do the analysis is like this:

$\frac{dP}{dt} = F_{ext} +$ the flux of momentum through the surface defining the object

where $F_{ext}$ is the external force acting on the object. The second term is not $V \frac{dM}{dt}$. It's a surface integral; something like this:

Flux = $\int_\mathcal{S} \rho \overrightarrow{V} (\overrightarrow{V} \cdot \overrightarrow{d \mathcal{S}})$

where $\rho$ and $V$ are the local mass density and velocity field, and $\mathcal{S}$ is the surface defining the object.

(I think that's right for a time-independent surface. If the definition of the surface changes with time, I'm not sure how to generalize it, right off the bat.)

My point is that there are almost no cases where $\frac{dM}{dt} V$ comes into play. If the situation is too complicated to analyze using fixed-mass objects, then that term is too primitive to be useful, anyway.

 

[vanhees71]

I don't get what you are after here. Take your example of the rocket. You clearly need the correct momentum balance equation. For the rocket with mass $M(t)$ you have (I just put the most simple 1D equation, for a rocket going straight up near the Earth, where gravity can be described by $g=\text{const}$). Then you have
$$\mathrm{d} p =\mathrm{d} t (M \dot{v} + \dot{M}(v-u))=-M g \mathrm{d} t.$$
Here, $u$ is the velocity of the center of mass of exhaust. The velocity $v-u=v_{\text{rel}}$ is usually taken as constant (it's the velocity of the exhaust in the rest frame of the rocket). Then the equation gets
$$M \dot{v}=-\dot{M} v_{\text{rel}}-M g.$$
Dividing by $M$ and integrating over $t' \in [0,t]$ gives
$$v(t)=-v_{\text{rel}} \ln \left (\frac{M(t)}{M_0} \right) - g t + v_0.$$

 

[DrStupid]

The mass of the exhaust isn't zero, except when the rocket first launches.

It makes no difference if the rocket first launches or not because the already ejected reaction mass doesn't interact with the rocket anymore. Therefore I can assume the rocket to be first launched in every point of the trajectory. I already explained that in my derivation above.

If you don't like this simplification you can include the exhaust. But that doesn't change anything. In absence of external forces all parts of the exhaust are moving with constant velocity. Therefore its total momentum changes by the mass transfer only, according to

$\dot p_T = \dot m_T \cdot v_T = F_T$

where $v_T$ is the velocity of the reaction mass crossing the system boundary between rocket and exhaust. This interaction doesn't change with inclusion of external forces as long as they do not affect the internal processes in the rocket engine.

My point is that there are almost no cases where $\frac{dM}{dt} V$ comes into play.

That would mean that there are almost no open systems. I do not agree.