Newton's laws in variable mass systems

D H

Another good one! You're just chock full of humor today, aren't you?

You can find tons of physics texts that define F as F=dp/dt, starting with Newton. That said, you can find others that define F as F=ma. 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.

The advantage of using F=dp/dt lies in the connection to the conservation laws. The disadvantage is the force acting on a mass-varying object is frame-dependent with this definition. The advantage of using F=ma is that whether mass is constant or varying, this definition makes force frame independent. The disadvantage is the lost connection with the conservation laws. Using F=ma in conjunction with work is a bad idea for variable mass systems. It will lead to incorrect results.

Dickfore

This is the point. Please derive

[tex]
\Sigma\mathbf{F}_{ex} = \frac{d \mathbf{P}}{d t}
[/tex]

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

[tex]\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}[/tex]

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,

[tex]\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}[/tex]

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

[tex]\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}[/tex]

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

[tex]\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}[/tex]

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

Define

[tex]\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}[/tex]

Note that by conservation of mass,

[tex]\Delta m_A(t) = \Delta m_{\Delta B(t)}(t)[/tex]

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

[tex]
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)
[/tex]

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

[tex]
\frac{d}{dt}P_{A(t)}(t) = F_{\text{ext}}(t) + \frac{dm_A(t)}{dt}v_e(t)
[/tex]

Dickfore

So, you couldn't derive it. Nice.

D H

Please do stop with the fallacious arguments.

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.

Dickfore

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//...00227.000.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:
http://www.physicsforums.com/showthread.php?t=72176

D H

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

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

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