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Sloppiness with F=dP/dt

  1. Jan 28, 2005 #1

    arildno

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    Many persons think that classical systems with varying mass obey the fundamental law:
    [tex]\vec{F}=\frac{d\vec{P}}{dt}[/tex]
    This is simply wrong!

    Every classical system in which the mass is variable, is a GEOMETRIC system, meaning that it does not consist of the same material particles over time.

    The opposite is called a MATERIAL system, in which the system does consist of the same material particles over time.
    In the classical approximation, a material system DO obey F=dP/dt; however, at the same time, a classical, material system is characterized by CONSTANT MASS (this is in actuality wrong; due to relativistic mass increase).

    So, what law does a geometric system obey?

    The forces acting upon a set of material particles enclosed by a geometric control volume obeys:
    [tex]\vec{F}=\frac{d\vec{P}}{dt}+\dot{\vec{M}}[/tex]
    where [tex]\frac{d\vec{P}}{dt}[/tex] is the change of momentum enclosed by the control volume, and [tex]\dot{\vec{M}}[/tex] is the momentum flux out of the geometric control surface.

    The interpretation of the momentum flux is quite simple:
    Material particles may choose to leave (or enter) our chosen control volume, carrying their own momentum with them.
    This is a mechanism for momentum change within the control volume which is not necessarily the result of some force acting upon the enclosed material!

    Without including the momentum flux you will end up with wrong answers, except in a few lucky special cases.
     
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  3. Jan 28, 2005 #2

    dextercioby

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    So your formula in which branch of theoretical classical dynamics is included?? :confused:

    Daniel.

    EDIT:Give an example
     
  4. Jan 28, 2005 #3

    arildno

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    It has nothing to do with "theoretical" classical dynamics; it has to do with the simple difference between a geometrical and material system, and it pertains to ALL examples in classical mechanics in which the mass is variable.

    Examples:
    1. A silly one:
    Consider a plate P with density [tex]\rho[/tex] which moves with uniform velocity [tex]U\vec{i}[/tex] and which at t=0 coincides with the square [tex]I=[0,L]\times[0,L][/tex]
    As time goes, more and more of the plate leaves I, that is, the amount of momentum enclosed in I changes with time (I is a geometric control volume at rest with respect to the inertial reference frame in which P has velocity U).

    In fact:
    At times [tex]0<=t<=\frac{L}{U}[/tex], the momentum enclosed in I is:
    [tex]\vec{P}=\rho(L^{2}-LUt)U\vec{i}[/tex]
    That is, the rate of change of momentum within I is:
    [tex]\frac{d\vec{P}}{dt}=-\rho{L}U^{2}\vec{i}[/tex]

    However, since each part of P moves with uniform velocity, there certainly cannot be a net external force acting upon parts of P still enclosed in I!

    The resolution is found by the momentum flux, which, for a control volume at rest, in (3-D) integral form is given by:
    [tex]\dot{\vec{M}}=\int_{S}(\rho\vec{v})(\vec{v}\cdot\vec{n})dA[/tex]
    where S is the enclosing surface, [tex]\vec{v}[/tex] is the velocity profile of material particles at the boundary, and [tex]\vec{n}[/tex] is the outward, unit normal vector.
    In our case then, the only contribution to the momentum flux comes from the x=L side:
    [tex]\dot{\vec{M}}=LU^{2}\vec{i}[/tex]
    Thus,
    [tex]\frac{d\vec{P}}{dt}+\dot{\vec{M}}=-LU^{2}\vec{i}+LU^{2}\vec{i}=\vec{0}[/tex]
    as it should be.
    I'll post some less sillier examples later.
     
    Last edited: Jan 28, 2005
  5. Jan 28, 2005 #4

    arildno

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    2. A not so silly example from fluid mechanics:
    Consider a U-tube with its open sections normal to the horizontal.
    Let a jet of fluid (of constant density) enter the U-tube with velocity [tex]V\vec{i}[/tex] and leave with velocity [tex]-V\vec{i}[/tex]
    The cross-sections have area A
    That is, we assume that at the in/out-lets, the fluid velocity is constant across the cross-section.
    We also assume stationary flow.

    From the stationary flow, it follows that the momentum enclosed within the (U-tube+in/outlet-sections) is constant in time.
    But the fluid has turned, there certainly acts a force upon it from the U-tube!
    We have therefore:
    [tex]\vec{F}=\int_{S}(\rho\vec{v})(\vec{v}\cdot\vec{n})dA[/tex]
    At the inlet, we have [tex]\vec{n}_{i}=-\vec{i}[/tex]; at the outlet [tex]\vec{n}_{o}=-\vec{i}[/tex]
    At the U-tube surface, we have [tex]\vec{v}\cdot\vec{n}=0[/tex]
    Hence, we get:
    [tex]\vec{F}=-2\rho{A}V^{2}\vec{i}[/tex]

    The external forces acting upon the enclosed fluids are (neglecting gravity):
    1. The pressure forces from the fluid acting upon the inlet and outlet sections.
    2. The force from the U-tube.
    For simplicity then, if we disregard the pressure forces as negligible, it follows from Newton's 3.law that the fluid exerts upon the tube a reaction force [tex]\vec{R}[/tex] :
    [tex]\vec{R}=2\rho{A}U^{2}\vec{i}[/tex]
    Hence, the tube must be held fast, in order not to be swept along in the direction of the velocity of the incoming jet.
     
    Last edited: Jan 28, 2005
  6. Jan 28, 2005 #5
    To be honest, whenever someone speaks of the force on an object whose mass is not constant my 'caution' radar goes up. E.g. a cooling object is losing mass and as such the object's momentum is changing yet the velocity of the object may remain constant. Thus we have an object whose time rate of change of momentum is non-zero and yet its not accelerating. One does not ordinarily think of this has having a force act upon it. So I can't say that I disagree with your assertion.

    Interesting topic! Thanks. :smile:

    More later when I have more time to think about these comments.

    Pete
     
  7. Jan 30, 2005 #6

    arildno

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    The transport theorem&Galilean invariance:

    pmpphy:
    I am not quite sure about your example; are you talking about the RELATIVISTIC mass decrease a material system/particle experiences as a result of cooling?

    In the classical approximation, unless constituent particles left the object of study, its mass would remain the same. The velocity loss of the constituent molecules would be of random, thermal motion, and hence would not necessarily influence the macroscopically observed object velocity.
    Perhaps you could clarify for me what you referred to?

    MATERIAL VS. GEOMETRICAL SYSTEMS: Reynolds' transport theorem
    In classical&relativistic physics, the fundamental reference object is the material particle (or system). Material particles are subject to, and sources of, forces, and the particle's momentum is related to the external forces acting upon it through the laws of motion.
    (Note: the force/particle conception is a rather outdated view of nature, in light of QM (from what I know); however, I'll leave that aside in the following).

    Classical material particles, were also found to obey the mass conservation law (this is not true in a reltivistic treatment).
    That is, if we have a material system which occupies a region in space [tex]V_{m}(t)[/tex] the material, classical system obeys the following two important laws:
    [tex]\frac{d}{dt}\int_{V_{m}(t)}\rho{dV}=0[/tex]
    and:
    [tex]\vec{F}=\frac{d}{dt}\int_{V_{m}(t)}\rho\vec{v}dV=\frac{d\vec{P}_{m}}{dt}[/tex]
    the first being the mass conservation law ([tex]\rho[/tex] being the density of a constituent material particle),
    the second, Newton's 2.law of motion ([tex]\vec{v}[/tex] the velocity of a constituent material particle, [tex]\vec{P}_{m}[/tex] the momentum of the said system).

    Now, Newton's 2.law in this form is certainly a Galilean invariant:
    Set [tex]\vec{v}=\hat{\vec{v}}+\vec{v}_{0}[/tex]
    where [tex]\vec{v}_{0}[/tex] is a constant vector.
    We have (since "t" is a Galilean invariant):
    [tex]\frac{d}{dt}\int_{V_{m}(t)}\rho\vec{v}dV=\frac{d}{dt}\int_{V_{m}(t)}\rho\hat{\vec{v}}dV+\vec{v}_{0}\frac{d}{dt}\int_{V_{m}(t)}\rho{dV}=\frac{d}{dt}\int_{V_{m}(t)}\rho\hat{\vec{v}}dV[/tex]
    due to the mass conservation law.
    Hence, the forces in either reference frame must be equal, in accordance with the demand of Galilean invariance.

    So far, so good!
    However, it is not always profitable to have as our object of study a system which consist of the same material particles over time (i.e, a material system)!
    Fluid mechanics is rife with such situations:
    In pipe flows, for example, we couldn't care less about calculating the trajectories of individual fluid particles; we are much more interested in questions like:
    How does the pressure at some location vary of time?
    What is the velocity at a given location?
    The last should be given the meaning: what is the velocity of whatever fluid particle happens to drop by at the location where we've got the measurement instrument?
    Over time, it will be DIFFERENT fluid particles dropping by; we are primarily interested in the field quantities, not the particle quantities.


    But this situation leads to an immediate philosophical problem:
    We want to study the dynamics in a region which does not consist of the same material particles over time; yet forces are intimately related to those PARTICLES, not to the arbitrarily chosen geometric region we're looking at!

    Hence, if we want to switch our perspective to the study of a GEOMETRICAL system, we must take care to get the appropriate physical laws!!
    (Those were originally not formulated for geometric systems, but for material systems).

    The solution lies in Reynolds' transport theorem:
    Consider some material quantity [tex]\vec{B}_{m}=\int_{V_{m}(t)}\rho\vec{b}dV[/tex]
    (at places where there aren't any material particles, the density should be given the value zero)
    Now, consider a region [tex]V_{gcv}(t)[/tex] which at a particular instant t* happens to coincide with [tex]V_{m}[/tex]
    That is, we have:
    [tex]\vec{B}_{gcv}(t*)=\int_{V_{gcv}(t*)}\rho\vec{b}dV=\int_{V_{m}(t*)}\rho\vec{b}dV=\vec{B}_{m}(t*)[/tex]

    Reynolds' transport theorem relates the derivatives of [tex]\vec{B}_{m},\vec{B}_{gcv}[/tex] at time t* as follows (I'll suppress the arbitrary t* in the following):
    [tex]\frac{d\vec{B}_{m}}{dt}=\frac{d\vec{B}_{gcv}}{dt}+\int_{S_{gcv}}\rho\vec{b}\vec{v}_{rel}\cdot\vec{n}dS[/tex]
    Here, [tex]S_{gcv}[/tex] is the surrounding surface at t*,
    [tex]\vec{v}_{rel}=\vec{v}-\vec{v}_{gcv}[/tex] is the relative velocity of a material particle with velocity [tex]\vec{v}[/tex] to the velocity of the coincident geometrical surface point [tex]\vec{v}_{gcv}[/tex]
    [tex]\vec{v}\cdot\vec{n}[/tex] is therefore the local normal velocity of the material surface, whereas [tex]\vec{v}_{gcv}\cdot\vec{n}[/tex] is the local normal velocity of the geometrical control surface.
    The difference, [tex]\vec{v}_{rel}\cdot\vec{n}[/tex] measures therefore the rate by which a material particle LEAVES the geometrical control surface.

    EDIT: Due to some strange grumpiness of LATEX, this continues in the next post.
     
    Last edited: Jan 30, 2005
  8. Jan 30, 2005 #7

    arildno

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    When we let [tex]\vec{b}[/tex] equal the scalar 1, we find the appropriate mass conservation law:
    [tex]\frac{d}{dt}\int_{V_{gcv}}\rho{dV}+\int_{S_{gcv}}\rho\vec{v}_{rel}\cdot\vec{n}dS=0[/tex]
    and, with [tex]\vec{b}=\vec{v}[/tex] the appropriate version of Newton's 2.law:
    [tex]\vec{F}=\frac{d}{dt}\int_{V_{gcv}}\rho\vec{v}dV+\int_{S_{gcv}}\rho\vec{v}\vec{v}_{rel}\cdot\vec{n}dS[/tex]

    It is easy to show the Galilean invariance of the last expression.
    I'll post a few examples later on.
     
    Last edited: Jan 30, 2005
  9. Jan 30, 2005 #8

    arildno

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    An example:

    First, we rewrite the mass conservation law as:
    [tex]\dot{m}=-m_{f}[/tex]
    [tex]\dot{m}=\frac{d}{dt}\int_{V_{gcv}}\rho{dV},m_{f}=\int_{S_{gcv}}\rho\vec{v}_{rel}\cdot\vec{n}dS[/tex]

    That is, the only way in which the mass of a classical system can change (measured by [tex]\dot{m}[/tex]), is that material particles chooses to leave (or enter) our geometrical system (measured by the outward mass flux [tex]m_{f}[/tex])

    Furthermore, if the particle velocity at the boundary of the control surface is constant [tex]\vec{v}_{b}[/tex], Newton's 2.law may be written as:
    [tex]\vec{F}=\frac{d\vec{P}}{dt}+m_{f}\vec{v}_{b}[/tex]


    Let us consider a case which is typically handled by the fallacious equation F=dP/dt:
    Let a conveyor belt move with constant velocity [tex]V\vec{i}[/tex] and let sand rain down upon the conveyor belt with velocity [tex]\vec{v}_{s}[/tex]; the rate of mass increase being [tex]\dot{m}=\alpha>0[/tex]
    We assume that the sand particles are instantaneously accelerated to obtain velocity [tex]V\vec{i}[/tex]
    We enclose the conveyor belt+attached particles within a fixed geometrical control volume, and analyze the situation with that in mind
    We therefore have:
    [tex]\vec{F}=\frac{d\vec{P}}{dt}+m_{f}\vec{v}_{s}=\alpha{V}\vec{i}-\alpha\vec{v}_{s}=\alpha(V\vec{i}-\vec{v}_{s})[/tex]
    since the conveyor belt and attached particles does not experience any acceleration.
    In the particular case where the horizontal velocity of incoming particles is zero, we see that the fallacious treatment gives the correct expression for the additional HORIZONTAL force acting upon the conveyor belt; the treatment is still dead wrong.
    The above, correct, treatment, is evidently Galilean invariant, something the fallacious treatment is not.
     
    Last edited: Jan 30, 2005
  10. Jan 30, 2005 #9
    In cases with time varying mass I always use:

    [tex]F=\frac{d(mv)}{dt}=m\frac{dv}{dt}+\frac{dm}{dt}v[/tex]

    Where this last term is the mass flux. What's the problem?
     
  11. Jan 30, 2005 #10

    arildno

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    da willem:
    Your right-hand side is not a Galilean invariant.
    That's one error.
     
  12. Jan 30, 2005 #11
    Sorry for that. I guess that was pretty unclear. I was talking about radiative cooling. I.e. place an object in deep space where there is a near perfect vacuum. Let the temperature initially be 75F. The body will cool by radiative cooling whereby the body emits (blackbody) radiation. The decrease in the rest energy of the body due to the loss of energy will result in a decrease in rest mass.

    Pete
     
  13. Jan 30, 2005 #12

    arildno

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    I thought it might be; it is an interesting system, but I wouldn't dare to undertake the proper relativistic treatment of the situation..(I'll leave that to persons like yourself who evidently master such subjects. I don't..)
     
  14. Jan 30, 2005 #13
    I recommend studying this particular system. Doing so will give you a good idea of the mass-energy relationship. I've elaborated on that here

    http://www.geocities.com/physics_world/sr/mass_energy_equiv.htm

    In that page I derive E = mc2 by defining m through p = mv. This provides the relationship E0 = m0c2 too.

    Pete
     
  15. Jan 31, 2005 #14

    arildno

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    The rocket equation

    First off: Thanks, Pete, your link looks very interesting indeed.

    In order to derive the rocket equation, it is simplest to choose a geometrical control volume surrounding and following the ship+remaining fuel system throughout time.
    The outlet of ejected gas is at the boundary of our control volume.

    In addition, we look at this in the NON-INERTIAL reference frame in which the ship is at rest.
    Hence, every particle of mass [tex]\delta{m}[/tex] included in our control volume at a given time "t" experiences a pseudo-force [tex]-\delta{m}\vec{a}[/tex] where [tex]\vec{a}[/tex] is the acceleration of the ship, measured in an inertial reference frame.
    These pseudo-forces add up to [tex]-m(t)\vec{a}[/tex] where m(t) is the enclosed mass at time "t".
    The exhaust velocity of the gas is [tex]\vec{v}_{e}[/tex]

    We therefore have:
    [tex]-m(t)\vec{a}=\frac{d\vec{p}}{dt}+m_{f}\vec{v}_{e}[/tex]
    Now, the momentum enclosed within our control volume is always zero, and we also have [tex]m_{f}=-\dot{m}=-\frac{dm}{dt}[/tex]
    Hence, we gain:
    [tex]m\vec{a}=\frac{dm}{dt}\vec{v}_{e}[/tex]
    which is the rocket equation.
     
  16. Jan 31, 2005 #15
    Well giving it another thought, you're absolutely right, systems with varying mass do not obey F=dp/dt. It's not helping that [itex]v dm/dt[/itex], with v the velocity of the mass leaving, is indeed the force that accelerates a rocket though...
     
  17. Jan 31, 2005 #16

    arildno

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    Glad you agree, da willem!
    Basically, the problem which has been swept under the carpet, is that calculating the change in momentum contained within an arbitrarily chosen region, is not the same procedure as summing together the momentum changes of a fixed set of material particles.

    When your region is not defined in terms of a fixed set of particles, momentum changes may occur within that region as a result that some particles originally inside choose to leave (carrying their own momentum with them).
    Similarly, the momentum contained within the chosen region may change when "new" particles enter.

    The momentum flux term is crucial in order to tie together the momentum change of a MATERIAL system (which is related to forces through Newton/Einstein) and the change of momentum in our arbitrarily chosen geometrical system (which, at a particular INSTANT, is coincident with some material system).
     
    Last edited: Jan 31, 2005
  18. Jan 31, 2005 #17

    arildno

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    The "coincident" material system&velocity jumps at the boundary

    The material system (which determines the forces) "coinciding" with the geometric system at a particular instant, can be determined more precisely, in such a manner that we may accept velocity jumps occurring at the boundary (this has already been implivitly used).

    Consider a given geometric contol volume [tex]V_{gcv}(t)[/tex] at a time "t".

    Consider the material system [tex]V_{m}[/tex] whose constituent particles comes in 3 categories:
    1. Constituent particles included in [tex]V_{gcv}[/tex] and in [tex]V_{gcv}(t+\delta{t})[/tex]
    2. Exiting particles which are included in [tex]V_{gcv}[/tex] but NOT in [tex]V_{gcv}(t+\delta{t})[/tex]
    3.Entering particles which are NOT included in [tex]V_{gcv}[/tex] but included in [tex]V_{gcv}(t+\delta{t})[/tex]

    The exiting particles may be thought to sit at the inner boundary of [tex]V_{gcv}[/tex] at time "t", whereas the entering particles sit at the outer boundary of [tex]v_{gcv}[/tex] at time "t"
    We let [tex]\delta{m}_{in},\delta{m}_{out}[/tex] be the (infinitesemally small) masses of entering and exiting particles respectively.

    The velocity of an entering particle at time "t" is [tex]\vec{v}^{-}_{in}[/tex]
    whereas the velocity of an exiting particle at time [tex]t+\delta{t}[/tex] is [tex]\vec{v}^{+}_{out}[/tex]

    We have then the following equalities between the momenta at "t" and "[tex]t+\delta{t}[/tex]" :
    [tex]\vec{p}_{gcv}(t)=\vec{p}_{m}(t)-\delta{m}_{in}\vec{v}^{-}_{in}[/tex]
    [tex]\vec{p}_{gcv}(t+\delta{t})=\vec{p}_{m}(t+\delta{t})-\delta{m}_{out}\vec{v}^{+}_{out}[/tex]
    Or, subtracting the first equation from the last, rearranging and taking the limit, we get:
    [tex]\frac{d\vec{p}_{gcv}}{dt}+\frac{\delta{m}_{out}}{\delta{t}}\vec{v}^{+}_{out}-\frac{\delta{m}_{in}}{\delta{t}}\vec{v}^{-}_{in}=\frac{d\vec{p}_{m}}{dt}=\vec{F}[/tex]
    Clearly, the rate of change of mass inside the control volume fulfills:
    [tex]\dot{m}=\frac{\delta{m}_{in}}{\delta{t}}-\frac{\delta{m}_{out}}{\delta{t}}[/tex]
    It is these last form which basically has been used in the conveyor belt example and the rocket equation, rather than the "continuous" version given in Reynolds' transport theorem.
    In the conveyor belt example, there are no exiting particles
    ([tex]\delta{m}_{out}\equiv0[/tex]),
    whereas there are no entering particles in the rocket equation
    ([tex]\delta{m}_{in}\equiv0[/tex])
     
    Last edited: Jan 31, 2005
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