# Newton's laws in variable mass systems

by cjl
Tags: laws, mass, newton, systems, variable
 Mentor P: 14,479 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 pi(t+Δt) = pi(t) + Fi(t)Δt, where Fi(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 $$\frac{d}{dt}P_{A(t)}(t) = F_{\text{ext}}(t) + \frac{dm_A(t)}{dt}v_e(t)$$
P: 3,015
 Quote by D H $$\frac{d}{dt}P_{A(t)}(t) = F_{\text{ext}}(t) + \frac{dm_A(t)}{dt}v_e(t)$$
So, you couldn't derive it. Nice.
 Mentor P: 14,479 Please do stop with the fallacious arguments.
Mentor
P: 14,479
 Quote by Dickfore 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.
P: 3,015
 Quote by D H 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?
 Mentor P: 14,479 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.
 P: 3,015 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.
 Mentor P: 14,479 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.
 P: 3,015 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.
 P: 16 This may be of use here: http://articles.adsabs.harvard.edu//...00227.000.html
 Sci Advisor HW Helper PF Gold P: 12,016 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: http://www.physicsforums.com/showthread.php?t=72176
Mentor
P: 14,479
 Quote by afallingbomb This may be of use here: http://articles.adsabs.harvard.edu//...00227.000.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.
 P: 3,015 Question: What form of Newton's Second Law is used in Hydrodynamics?
HW Helper
PF Gold
P: 12,016
 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.
P: 16
 Quote by Dickfore 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
HW Helper
PF Gold
P: 12,016
 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.
P: 3,015
 Quote by arildno 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:

 Quote by Dickfore 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.
P: 3,015
 Quote by afallingbomb 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.

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