Newton's laws in variable mass systems

[Dickfore]

You can also find others that claim that Newton's laws, strictly speaking, apply only to particles, so it doesn't matter which definition you use.

This is the point. Please derive

$$\Sigma\mathbf{F}_{ex} = \frac{d \mathbf{P}}{d t}$$

for an (open) system with variable mass.

 

[D H]

Your point is fallacious. Specifically, it is your qualification "ex" (meaning external) on the force is that is fallacious.

So, without fallacies:

There is some larger, closed system of constant mass particles in which our somewhat arbitrary open system lives. Call the particles in our open system set A and the other particles set B. Whether a given particle is in set A or in set B varies with time. At some time t denote A(t) as the set of particles that are in set A at time t, B(t) as the set of particles in set B at time t. I'll split set B into two disjoint subsets, sets B_-(t) and ΔB(t) such that B(t) = B_-(t) ∪ ΔB(t). The reason for this partition of B(t) is that the subset ΔB(t) is about to join A(t): A(t+Δt) = A(t) ∪ ΔB(t) and B(t+Δt) = B_-(t).

The total momenta of the particles in sets A(t) and ΔB(t) are

$$\begin{aligned} P_{A(t)}(t) &= \sum_{i\in A(t)} p_i(t) \\ P_{\Delta B(t)}(t) &= \sum_{j\in\Delta B(t)} p_j(t) \end{aligned}$$

Assuming our Δt is small, the momentum of particle i at times t and t+Δt are approximately related by p_i(t+Δt) = p_i(t) + F_i(t)Δt, where F_i(t) is the net force acting on particle i at time t. Expanding that net force into the contributions from each particle,

$$\begin{aligned} P_{A(t)}(t+\Delta t) &\approx \sum_{i\in A(t)} \left( p_i(t) + \sum_{j\in A(t), j\ne i} F_ij(t) \Delta t + \sum_{j\in \Delta B(t)} F_ij(t) \Delta t + \sum_{j\in B(t)} F_ij \Delta t\right) \\ P_{\Delta B(t)}(t+\Delta t) &\approx \sum_{j\in \Delta B(t)} \left( p_j(t) + \sum_{i\in A(t)} F_ji(t) \Delta t + \sum_{i\in \Delta B(t), i\ne j} F_ji(t) \Delta t + \sum_{i\in B(t)} F_ji(t) \Delta t\right) \\ \end{aligned}$$

By Newton's third law, each of the following sums will vanish:

$$\begin{aligned} &\sum_{i\in A(t)}\sum_{j\in A(t), j\ne i} F_ij(t) \\ &\sum_{j\in \Delta B(t)}\sum_{i\in \Delta B(t), i\ne j} F_ji(t) \\ &\sum_{i\in A(t)}\sum_{j\in \Delta B(t)} F_ij(t) + \sum_{j\in \Delta B(t)}\sum_{i\in A(t)} F_ji(t) \end{aligned}$$

Using this, and summing to form the total momentum of particles A at time t+Δt yields

$$\begin{aligned} P_{A(t+\delta t)}(t+\Delta t) &= P_{A(t)}(t+\Delta t) + P_{\Delta B(t)}(t+\Delta t) \\ &\approx \sum_{i\in A(t)} \left( p_i(t) + \sum_{j\in B(t)} F_ij(t) \Delta t\right) + \sum_{j\in \Delta B(t)} \left( p_j(t) + \sum_{i\in B(t)} F_ji(t) \Delta t\right) \end{aligned}$$

The final term, $\sum_{j\in \Delta B(t)}\sum_{i\in B(t)} F_ji(t) \Delta t$, will be second order assuming that the set $\Delta B(t)\to\Phi\,\text{as}\,\Delta t\to 0$.

Define

$$\begin{aligned} F_{\text{ext}}(t) &\equiv \sum_{i\in A(t)} \sum_{j\in B(t)} F_ij(t) \\ \Delta m_{\Delta B(t)}(t) &\equiv \sum_{j\in \Delta B(t)} m_j \\ v_e(t) &\equiv \frac 1 {m(t)} \sum_{j\in \Delta B(t)} p_j(t) \end{aligned}$$

Note that by conservation of mass,

$$\Delta m_A(t) = \Delta m_{\Delta B(t)}(t)$$

With this the total momentum of particles A at time t+Δt becomes

$$P_{A(t+\delta t)}(t+\Delta t) \approx \sum_{i\in A(t)} p_i(t) + F_{\text{ext}}(t) \Delta t + \Delta m_{A(t)}(t) v_e(t)$$

Subtracting the momentum of particles A at time t, dividing by Δt and taking the limit Δt→0 yields

$$\frac{d}{dt}P_{A(t)}(t) = F_{\text{ext}}(t) + \frac{dm_A(t)}{dt}v_e(t)$$

 

[Dickfore]

$$\frac{d}{dt}P_{A(t)}(t) = F_{\text{ext}}(t) + \frac{dm_A(t)}{dt}v_e(t)$$

So, you couldn't derive it. Nice.

 

[D H]

Please do stop with the fallacious arguments.

 

[D H]

So, you couldn't derive it. Nice.

Of course not. Your equation is wrong in the context of F=dp/dt.

Why don't you derive your equation? State your assumptions and your definitions. Then use your equation to derive (a) the work performed on a variable mass system and (b) the time derivative (power) of the variable mass system's kinetic energy.

 

[Dickfore]

Of course not. Your equation is wrong in the context of F=dp/dt.

Why don't you derive your equation? State your assumptions and your definitions. Then use your equation to derive (a) the work performed on a variable mass system and (b) the time derivative (power) of the variable mass system's kinetic energy.

How can I derive a wrong equation?

 

[D H]

I don't know. You tell me. I've done math while all you have done is to use fallacious arguments. The equation in post #36 is yours, not mine. Define your terms and derive that result. Then use that result to calculate (a) the work performed on the variable mass system and (b) the time derivative of the system's kinetic energy.

 

[Dickfore]

It seems we have misunderstood each other. The equation I posted in #36 was in repsonse to your post #35. The point I tried to make is that that equation is incorrect for open systems, which you yourself demonstrated with the derivation in step #37.

In conclusion, you were wrong in post #14 where you stated a force transformation law.

 

[D H]

There is no error in post #14. I derived it in post #50. I challenged you to find the flaw and all you could do was use fallacious reasoning. I haven't the foggiest idea what you are going on about now. Please elaborate.

And use some math instead of fallacious arguments this time.

 

[Dickfore]

lol, what ever you say. I don't want to quarrel and get a ban. If someone reads this thread and knows Physics, they will draw a conclusion for themselves.

 

[afallingbomb]

This may be of use here:

http://articles.adsabs.harvard.edu//full/1992CeMDA..53..227P/0000227.000.html

 

[arildno]

The distinction between a geometric system, and a material system is crucial to understand when dealing with Newton's second law, and what forces act upon.

A geometric system is simply a mathematically defined region within space that we designate as our system, and particles/mass may well flow into, and out of that region.

Forces do NOT act upon a mere spatial region, they act upon material particles CONTAINED within that region.

Thus, F=dp/dt is perfectly valid, as long as we are talking about a material system, i.e, which is defined by consisting of the same particles throughout time (and in the Newtonian world, thus have fixed mass).

We may perfectly well formulate a Newton's 2.law for geometric systems, and it is highly useful, for example by calculating the reaction force on a tube section through which the fluid passes.

The rate of change of momentum within a geometric system consists not only of the effects of forces acting upon (momentarily) contained particles, but also that less momentum may flow into the region than leaves it, or vice versa (momentum itself being carried by massed particles).

The following thread goes into the details:
https://www.physicsforums.com/showthread.php?t=72176

 

[D H]

This may be of use here:

http://articles.adsabs.harvard.edu//full/1992CeMDA..53..227P/0000227.000.html

I already posted a link to the same article in [post= 2858679]post #14[/post].

There are three camps in classical physics regarding Newton's laws,

1. F=dp/dt=d(mv)/dt. With this definition of force, that F=m·dv/dt+v·dm/dt is a straightforward application of the chain rule. Dick: Does the chain rule not apply in physics?
2. F=ma, even for variable mass systems. The changes in state that result from mass variations are just another external force. With this definition the equations of motion fall straight out of Newton's 2nd law. An unfortunate side effect of this definition is the loss of connection to the conservation laws. Using this definition of force to compute work is invalid.
3. F=dp/dt, F=ma: Potato/patato. Newton's laws only apply to point masses with constant mass (e.g., Plastino & Muzzio). To this camp, there is no such thing as a variable mass system in the context of Newton's laws.

The last camp has withdrawn from the argument. The first two camps will arise at the same results if they are careful about their math.

 

[Dickfore]

Question: What form of Newton's Second Law is used in Hydrodynamics?

 

[arildno]

F=ma, even for variable mass systems. The changes in state that result from mass variations are just another external force. With this definition the equations of motion fall straight out of Newton's 2nd law. An unfortunate side effect of this definition is the loss of connection to the conservation laws. Using this definition of force to compute work is invalid.

This is a meaningless position.

Let us have an object O moving at constant velocity U; no part of it is subject to any force whatsoever.

I can perfectly well construct a system containing solely of parts of this object, yet that system will experience a constant increase in momentum, with nothing resembling of a force is acting upon any part of object O.

 

[afallingbomb]

Question: What form of Newton's Second Law is used in Hydrodynamics?

For a non-relativistic continuum (solid, fluid, etc) Newton's second law is simply the conservation of linear momentum expressed as:

http://en.wikipedia.org/wiki/Cauchy_momentum_equation

 

[arildno]

F=dp/dt=d(mv)/dt. With this definition of force, that F=m·dv/dt+v·dm/dt is a straightforward application of the chain rule. Dick: Does the chain rule not apply in physics?

As for this "camp":

Sure the chain rule work, as long as you use it properly.

Leibniz' rule of differentiation of an integral with moving boundaries is a fancy version of the chain rule, and that is the one to use here.

 

[Dickfore]

This is a meaningless position.

Let us have an object O moving at constant velocity U; no part of it is subject to any force whatsoever.

I can perfectly well construct a system containing solely of parts of this object, yet that system will experience a constant increase in momentum, with nothing resembling of a force is acting upon any part of object O.

I said the same thing:

YARLY! Imagine a collection of non-interacting balls traveling all uniformly relative to an inertial reference frame with a velocity V. If you mentally isolate a smaller and smaller subset of them, according to your formula, it seems there is a force acting on this subset. But, by definition, this force can not come from the neighboring points. Where does this force come from then?

but was ridiculed by D-H.

 

[Dickfore]

For a non-relativistic continuum (solid, fluid, etc) Newton's second law is simply the conservation of linear momentum expressed as:

http://en.wikipedia.org/wiki/Cauchy_momentum_equation

All i see on the lhs is mass per unit fluid volume element times acceleration of the said element.

 

[arildno]

I said the same thing:



but was ridiculed by D-H.

Well, DH is dead wrong on this issue, and that is why he turns to ridicule, since he doesn't have any solid arguments.

Sorry that I didn't see that part of the previous discussion.

 

[D H]

F=ma, even for variable mass systems. The changes in state that result from mass variations are just another external force. With this definition the equations of motion fall straight out of Newton's 2nd law. An unfortunate side effect of this definition is the loss of connection to the conservation laws. Using this definition of force to compute work is invalid.

This is a meaningless position.

Let us have an object O moving at constant velocity U; no part of it is subject to any force whatsoever.

I can perfectly well construct a system containing solely of parts of this object, yet that system will experience a constant increase in momentum, with nothing resembling of a force is acting upon any part of object O.

This is far from a meaningless position. Whether you want to call $\dot m\,dv/dt$ a force or not is mostly semantics. Whatever you call it, it has the units of force and affects the equations of motion just like a force. The equations of motion will be correct. The only gotcha here is that that calling this quantity a force represents a break between Newton's laws and the conservation laws.

This is the point of view taken by many, if not all, high-fidelity missile and spacecraft simulations. Those simulations don't care about the ultra high-fidelity details that can only be divined by resorting to CFD models. Those CFD-based simulations are too slow to use as the basis for long-term simulations or Monte Carlo testing. All those high-fidelity sims care about is developing the temporal history of the vehicle's state (or vehicles' states in the case of a multi-vehicle simulation).

Calling this a meaningless position is just wrong. You just have to be aware of the limitations of this position. This is not my typical take on this issue; I prefer the F=dp/dt point of view. (Although this is the approach I use when all I care about are the equations of motion).

 

[arildno]

And why should the momentum changing effect of volume expansion of your arbitrarily chosen control volume be called a "force"?

Why not call it by its proper name, namely..flux of momentum?

You might as well call a dog an apple, since both are organic compounds.


Classical mechanics, the mental framework we're in right now, is blind to other sources for change in momentum than forces acting upon particles contained within a system, along with whatever accretion of matter (carriers of momentum) to the system, for example through expansion of the volume of the geometric region we are talking about (for example having the interior region of ab expanding balloon as our control volume).

That other sources for momentum change exists as well is no good reason to make "force" into a flabby concept.

 

[D H]

And why should the momentum changing effect of volume expansion of your arbitrarily chosen control volume be called a "force"?

Because its a very useful concept?

Or maybe, aerospace engineers thinking that the thrust produced by a jet or a rocket is a meaningful quantity and testable quantity is just plain goofy. NOT.

 

[Dale]

Hi arildno, I remember seeing you present your definition of force before, but the textbooks I have used tend to agree with DH's definition. Do you have a supporting reference, or is this a personal approach?

 

[D H]

DH's definition.

Actually, I gave three:

1. F=dp/dt (e.g., Marion, Goldstein (just double-checked; Goldstein is rather insistent that F=dp/dt is the proper form)),
2. F=ma (e.g., Halliday&Resnick plus several engineering texts, with the mdot·v term denoted as a force)
3. F=dp/dt=ma is a potato/patato issue because Newton's laws apply only to constant mass point masses.

I generally prefer (1) F=dp/dt because of the connection with the conservation laws. I do use (2) when the driving concern is the equations of motion.

BTW, it is rather well-known that F=dp/dt is frame-dependent. I don't get what all the hub-bub is about.

 

[Dickfore]

When dealing with systems with variable mass, one needs to consider not only the "reactive forces", but also the shifting of the center of mass of a system due to its mass redistribution. Let us find an equation of motion for the center of mass of an open and not isolated system A.

In order that Newton's Laws be applicable, we must consider both the closure and isolation of this system A + B by identifying all the particles that flow out and into the system during a time interval $\Delta t$ as well as all particles that exert forces on the particles.

The acceleration of the center of mass of the system is given by:

$$\mathbf{a}^{(A)}_{\mathrm{C M}}(t) = \frac{d \mathbf{v}^{(A)}_{\mathrm{C M}}(t)}{dt}$$

where the velocity of the center of mass is by definition given by:

$$\mathbf{v}^{(A)}_{\mathrm{C M}}(t) \equiv \frac{\mathbf{P}^{(A)}(t)}{M^{(A)}(t)}$$

where:

$$\mathbf{P}^{(A)}(t) = \sum_{a \in A(t)}{\mathbf{p}_{a}(t)}$$

is the total momentum of the particles in the system A at the instant t and:

$$M^{(A)}(t) = \sum_{a \in A(t)}{m_{a}}$$

is their total mass. As was pointed out by D-H's derivation in post #37, these quantities can be time dependent in two different ways: Either because the quantities that enter the sum change with time, or because the set over which we sum changes itself, or both, of course.

According to the rules of Calculus, we may write:

$$\mathbf{a}^{(a)}_{\mathrm{C M}})(t) = \frac{1}{M^{(A)}(t)} \, \frac{d \mathbf{P}^{(A)}(t)}{d t} - \frac{\mathbf{P}^{(A)}(t)}{[M^{(A)}(t)]^{2}} \, \frac{d M^{(A)}(t)}{d t}$$

or:

$$M^{(A)}(t) \, \mathbf{a}^{(A)}_{\mathrm{C M}}(t) = \frac{d \mathbf{P}^{(A)}(t)}{d t} - \frac{\mathbf{P}^{(A)}(t)}{M^{(A)}(t)} \, \frac{d M^{(A)}(t)}{d t} = \frac{d \mathbf{P}^{(A)}(t)}{d t} - \mathbf{V}^{(A)}_{\mathrm{C M}}(t) \, \frac{d M^{(A)}(t)}{d t}$$

Let us find the derivatives on the rhs of this equality. In doing so, we will adopt the following notation. Let the index a enumerate the particles that were inside the system A at time t; let the index i enumerate the particles that exited system A into system B and let j enumerate the particles that entered from B to A during the time interval $\Delta t$. Then we have:

$$\mathbf{P}^{(A)}(t + \Delta t) = \sum_{a}{\mathbf{p}_{a}(t + \Delta t)} - \sum_{i}{\mathbf{p}_{i}(t + \Delta t)} + \sum_{j}{\mathbf{p}_{j}(t + \Delta t)}$$

We will keep quantities up to order $O(\Delta t)$ in the above sum. It is important to realize that the number of particles that enter or exit (and, therefore, both the mass and momentum they carry with them) is a quantity of the order $O(\Delta t)$. That is why we can exchange the argument from $t + \Delta t$ to $t$ in the last two sums. Similarly, we can write:

$$p_{a}(t + \Delta t) = p_{a}(t) + \Delta t \, \sum_{b \in A + B, b \neq a}{\mathbf{F}_{b a}(t)} + o(\Delta t) = p_{a}(t) + \Delta t \, \left[\sum_{a' \in A, a' \neq a}{\mathbf{F}_{a' a}(t)} + \sum_{b \in B}{\mathbf{F}_{b a}(t)}\right] + o(\Delta t)$$

Collecting everything together, we may write:

$$\mathbf{P}^{(A)}(t + \Delta t) = \mathbf{P}^{(A)}(t) + \Delta t \, \left[\sum_{a, a' \in A, a' \neq a}{\mathbf{F}_{a' a}(t)} + \sum_{a \in A, b \in B}{\mathbf{F}_{b a}(t)}\right] - \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i} \mathbf{v}_{i}(t)} + \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j} \mathbf{v}_{j}(t)} + o(\Delta t)$$

$$\frac{d\mathbf{P}^{(A)}(t)}{d t} = \sum_{a, a' \in A, a' \neq a}{\mathbf{F}_{a' a}(t)} + \sum_{a \in A, b \in B}{\mathbf{F}_{b a}(t)} - \lim_{\Delta t \rightarrow 0}{\left[\frac{1}{\Delta t} \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i} \, \mathbf{v}_{i}(t)} \right]} + \lim_{\Delta t \rightarrow 0}{\left[\frac{1}{\Delta t} \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j} \, \mathbf{v}_{j}(t)} \right]}$$

The first term in this equation is exactly equal to zero due to Third Newton's Law (the summation is over the same system A and $\mathbf{F}_{a a'} = -\mathbf{F}_{a' a}$) The second term is the sum of all the instantaneous external forces that act on the system $\sum{\mathbf{F}_{\mathrm{ext}}(t)}$. The third and the fourth term are the outgoing and incoming momentum flux, respectively, these are frame dependent quantities (because velocity is a frame dependent quantity). Although expressed as limits, these are actually finite quantities (if the particle flux is finite).

The other derivative is:

$$M^{(A)}(t + \Delta t) = \sum_{a}{m_{a}} - \sum_{i}{m_{i}} + \sum_{j}{m_{j}} = M^{(A)}(t) - \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i}} + \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j}}$$

$$\frac{d M^{(A)}(t)}{d t} = -\lim_{\Delta t \rightarrow 0}{\left \frac{1}{\Delta t} \, \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i}}\right]} + \lim_{\Delta t \rightarrow 0}{\left \frac{1}{\Delta t} \, \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j}}\right]}$$

Again, both of these terms are finite.

Combining everything together, we can finally write:

$$M^{(A)}(t) \, \mathbf{a}^{(A)}(t) = \sum{\mathbf{F}_{\mathrm{ext}}}(t) - \mathbf{\Pi}_{\mathrm{out}}(t) + \mathbf{\Pi}_{\mathrm{in}}(t)$$

where $\mathbf{\Pi}_{\mathrm{in/out}}(t)$ is the incoming (outgoing) momentum flux in the center-of-mass reference frame (a frame independent quantity, because relative velocities are Galilean invariants):

$$\mathbf{\Pi}_{\mathrm{out}}(t) = \lim_{\Delta t \rightarrow 0}{\left[\frac{1}{\Delta t} \, \sum_{i \in \Delta N_{\mathrm{out}}}{m_{i} \, \mathbf{u}_{i}}\right]}, \; \mathbf{u}_{i} = \mathbf{v}_{i} - \mathbf{v}^{(A)}_{\mathrm{C M}}$$

$$\mathbf{\Pi}_{\mathrm{in}}(t) = \lim_{\Delta t \rightarrow 0}{\left[\frac{1}{\Delta t} \, \sum_{j \in \Delta N_{\mathrm{in}}}{m_{j} \, \mathbf{u}_{j}}\right]}, \; \mathbf{u}_{j} = \mathbf{v}_{j} - \mathbf{v}^{(A)}_{\mathrm{C M}}$$

By specifying the number of particles that leave (enter) the system in unit time $\varphi_{\mathrm{out}/\mathrm{in}}(m, \mathbf{u}, t)$ with a relative velocity in a unit velocity-space volume around $\mathbf{u}$ and in unit interval around the mass $m$, we can express the above quantities as:

$$\mathbf{\Pi}_{\mathrm{out}/\mathrm{in}}(t) = \int{m \, \mathbf{u} \, \varphi_{\mathrm{out}/\mathrm{in}}(m, \mathbf{u}, t) \, d^{3}u \, dm}$$

Also, the mass rate of change is:

$$\frac{d M^{(A)}(t)}{d t} = -\mu_{\mathrm{out}}(t) + \mu_{\mathrm{in}(t)}$$

where the incoming (outgoing) mass flux rate is given by:

$$\mu_{\mathrm{out}/\mathrm{in}}(t) = \int{m \varphi_{\mathrm{out}/\mathrm{in}}(m, \mathbf{u}, t) \, d^{3}u \, dm}$$

 

[pallidin]

Hmmm... I did not know that the "shifting of the center of mass" is physically allowable in a closed system. Any examples?

 

[Dickfore]

Hmmm... I did not know that the "shifting of the center of mass" is physically allowable in a closed system. Any examples?

But, no one said it's a closed system.

 

[pallidin]

There is no such thing as an open system.
All systems are closed by default.
Any examples on a totally open system?

 

[pallidin]

Let's get real here. Has a totally open system EVER been observed beyond mathematical construct in true reality?
Can't think of a single one...

 

[Dickfore]

There is no such thing as an open system.
All systems are closed by default.
Any examples on a totally open system?

Your body.

 

[D H]

Hmmm... I did not know that the "shifting of the center of mass" is physically allowable in a closed system. Any examples?

A rocket that in the ten minutes or so that it takes to get from the ground to insertion orbit spews out 90% of of its mass. The center of mass does shift, and this shift does indeed impact the dynamics. Ignoring that effect results in an error on the order of 10s (maybe 100?) of meters. (It's been a while since I attacked this problem.)

Even worse, at least as far as modeling is concerned, is the impact on rotation. For translational state, all that matters is the rate at which the center of mass is moving within the vehicle. The rotational analog of a shifting center of mass is a shifting inertia tensor. However, knowing the rate at which the inertia tensor is changing is not enough to solve the rotational problem. You get a nasty line integral; the path taken by the flowing fuel / exhaust gases inside the vehicle also come into play. Normally this effect is very tiny (a launch vehicle turns about 9 degrees/minute during the climb to insertion burn). There are some other circumstances where this effect can be significant.

 

[D H]

There is no such thing as an open system.
All systems are closed by default.
Any examples on a totally open system?

Jets, rockets, turbines, ...

 

[pallidin]

Jets, rockets and turbines can not perform without an external environment of some sort, even if it's just gravity itself.
The external environment envelopes and defines the nature of the thrust.
Being thus intertwined, the system is NOT closed.

 

[Dickfore]

Jets, rockets and turbines can not perform without an external environment of some sort, even if it's just gravity itself.
The external environment envelopes and defines the nature of the thrust.
Being thus intertwined, the system is NOT closed.

And that's what we said: the system is open.