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Definitions of Momentum and Work

  1. Nov 17, 2011 #1
    In classical mechanics, if we consider the motion of a particle of mass [itex]m[/itex], then

    [tex]m=constant[/tex][tex]\vec{v}=d\vec{r}/dt[/tex][tex]\vec{a}=d\vec{v}/dt[/tex][tex]\vec{j}=d\vec{a}/dt[/tex][tex]\ldots[/tex]
    Definition of Momentum [itex](\vec{M})[/itex]

    [tex]\vec{M} \; = \int_a^b m \, \vec{a} \, dt \; = \int_a^b m \,d\vec{v} \; = \Delta \; m \, \vec{v}[/tex]
    [tex]If \quad \vec{a} = 0 \: \; \rightarrow \; \: \int_a^b m \, \vec{a} \, dt = 0[/tex]
    [tex]\rightarrow \; \: \Delta \; m \, \vec{v} = 0[/tex]
    [tex]\rightarrow \; \: m \, \vec{v} = constant[/tex]
    [tex]\rightarrow \; \: \vec{P} = constant[/tex]

    Definition of Momentum 2 [itex](\vec{M}_2)[/itex]

    [tex]\vec{M}_2 \; = \int_a^b m \, \vec{j} \, dt \; = \int_a^b m \,d\vec{a} \; = \Delta \; m \, \vec{a}[/tex]
    [tex]If \quad \vec{j} = 0 \: \; \rightarrow \; \: \int_a^b m \, \vec{j} \, dt = 0[/tex]
    [tex]\rightarrow \; \: \Delta \; m \, \vec{a} = 0[/tex]
    [tex]\rightarrow \; \: m \, \vec{a} = constant[/tex]
    [tex]\rightarrow \; \: \vec{P}_2 = constant[/tex]

    Definition of Work [itex](W)[/itex]

    [tex]W \; = \int_a^b m \, \vec{a} \cdot d\vec{r} \; = \int_a^b m \,\frac{d\vec{v}}{dt} \cdot \vec{v} \, dt \; = \Delta \; {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2}[/tex]
    [tex]If \quad \vec{a} = constant \: \; \rightarrow \; \: \int_a^b m \, \vec{a} \cdot d\vec{r} = \Delta \; m \, \vec{a} \cdot \vec{r}[/tex]
    [tex]\rightarrow \; \: \Delta \; {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2} + \Delta \; \left( - \; m \, \vec{a} \cdot \vec{r} \right) = 0[/tex]
    [tex]\rightarrow \; \: {\textstyle \frac{1}{2}} \, m \, \vec{v}^{\: 2} + \left( - \; m \, \vec{a} \cdot \vec{r} \right) = constant[/tex]
    [tex]\rightarrow \; \: T + V = constant[/tex]

    Definition of Work 2 [itex](W_2)[/itex]

    [tex]W_2 \; = \int_a^b m \, \vec{j} \cdot d\vec{v} \; = \int_a^b m \,\frac{d\vec{a}}{dt} \cdot \vec{a} \, dt \; = \Delta \; {\textstyle \frac{1}{2}} \, m \, \vec{a}^{\: 2}[/tex]
    [tex]If \quad \vec{j} = constant \: \; \rightarrow \; \: \int_a^b m \, \vec{j} \cdot d\vec{v} = \Delta \; m \, \vec{j} \cdot \vec{v}[/tex]
    [tex]\rightarrow \; \: \Delta \; {\textstyle \frac{1}{2}} \, m \, \vec{a}^{\: 2} + \Delta \; \left( - \; m \, \vec{j} \cdot \vec{v} \right) = 0[/tex]
    [tex]\rightarrow \; \: {\textstyle \frac{1}{2}} \, m \, \vec{a}^{\: 2} + \left( - \; m \, \vec{j} \cdot \vec{v} \right) = constant[/tex]
    [tex]\rightarrow \; \: T_2 + V_2 = constant[/tex]

    If [itex]\vec{a}[/itex], [itex]\vec{j}[/itex], [itex]\ldots[/itex] are not constant but [itex]\vec{a}[/itex], [itex]\vec{j}[/itex], [itex]\ldots[/itex] are functions of [itex]\vec{r}[/itex], [itex]\vec{v}[/itex], [itex]\ldots[/itex] respectively, then the same final result is obtained; even if Newton's second law were not valid (even if [itex]\sum \vec{F} \ne m\;\vec{a}[/itex])
     
  2. jcsd
  3. Nov 18, 2011 #2
    General Definition of Momentum [itex]\vec{c}[/itex] [itex](\vec{M}_{\vec{c}})[/itex]

    [tex]\vec{M}_{\vec{c}} \; = \int_a^b m \, ( d\vec{c}/{dt} ) \; dt \; = \Delta \; m \, \vec{c}[/tex]
    where [itex]\vec{c}[/itex] can be: [itex]\ldots, \vec{r}, \vec{v}, \vec{a}, \vec{j}, \ldots[/itex]


    [tex]If \quad ( d\vec{c}/{dt} ) = 0[/tex]
    [tex]\rightarrow \; \; \int_a^b m \, ( d\vec{c}/{dt} ) \; dt \; = 0[/tex]
    [tex]\rightarrow \; \; \Delta \; m \, \vec{c} = 0[/tex]
    [tex]\rightarrow \; \; m \, \vec{c} = constant[/tex]
    [tex]\rightarrow \; \; \vec{P}_{\vec{c}} = constant[/tex]
     
  4. Nov 18, 2011 #3
    General Definition of Work [itex]\vec{c}[/itex] [itex](W_{\vec{c}})[/itex]

    [tex]W_{\vec{c}} \; = \int_a^b m \, ( d^2 \, \vec{c}/{dt^2} ) \cdot d\vec{c} \; = \Delta \; {\textstyle \frac{1}{2}} \, m \, ( d\vec{c}/{dt} )^2[/tex]
    where [itex]\vec{c}[/itex] can be: [itex]\ldots, \vec{r}, \vec{v}, \vec{a}, \vec{j}, \ldots[/itex]


    [tex]If \quad ( d^2 \, \vec{c}/{dt^2} ) = constant[/tex]
    [tex]\rightarrow \; \; \int_a^b m \, ( d^2 \, \vec{c}/{dt^2} ) \cdot d\vec{c} = \Delta \; m \, ( d^2 \, \vec{c}/{dt^2} ) \cdot \vec{c}[/tex]
    [tex]\rightarrow \; \; \Delta \; {\textstyle \frac{1}{2}} \, m \, ( d\vec{c}/{dt} )^2 - \Delta \; m \, ( d^2 \, \vec{c}/{dt^2} ) \cdot \vec{c} = 0[/tex]
    [tex]\rightarrow \; \; {\textstyle \frac{1}{2}} \, m \, ( d\vec{c}/{dt} )^2 - \; m \, ( d^2 \, \vec{c}/{dt^2} ) \cdot \vec{c} = constant[/tex]
    [tex]\rightarrow \; \; T_{\vec{c}} + V_{\vec{c}} = constant[/tex]
     
  5. Nov 18, 2011 #4
    Are you actually trying to make a point?
    Of course the basic calculus expressions will work but your secondary definitions for energy and momentum do not express the same physical quantities(energy or momentum)In all your secondary definitions energy E and momentum p are not conserved.For example your second situation says that in the absence of a jerk the force is constant which is obvious but of course momentum and energy are of course not conserved.

    Newton second law is a definition F=dp/dt it does not make much sense to ask what if it wasn't. If we called d^2/dp^2 force we would still have to find a nice name for dp/dt.
     
  6. Nov 19, 2011 #5
    bp_psy:

    "For example your second situation says that in the absence of a jerk the force is constant which is obvious but of course momentum and energy are of course not conserved."

    For example: in the absence of a jerk [itex]\vec{j} = 0[/itex] the "acceleration" is constant, therefore


    [itex]m \, \vec{a} = constant[/itex]

    [itex]\vec{P}_2 = constant[/itex]

    [itex]{\textstyle \frac{1}{2}} \, m \, \vec{a}^{\: 2} + \left( - \; m \, \vec{j} \cdot \vec{v} \right) = constant \quad \quad \vec{j} = 0 \rightarrow \; \: {\textstyle \frac{1}{2}} \, m \, \vec{a}^{\: 2} = constant[/itex]

    [itex]T_2 + V_2 = constant \quad \quad \vec{j} = 0 \rightarrow \; \: T_2 = constante[/itex]


    I think that there are impulses, momentums, works, energies, ... of different classes:


    [itex]\ldots, \quad \vec{P}_0 = m \, \vec{r}, \quad \vec{P}_1 = m \, \vec{v}, \quad \vec{P}_2 = m \, \vec{a}, \quad \vec{P}_3 = m \, \vec{j}, \quad \ldots[/itex]

    [itex]\ldots, \quad T_0 = {\textstyle \frac{1}{2}} \, m \, \vec{r}{\: ^2}, \quad T_1 = {\textstyle \frac{1}{2}} \, m \, \vec{v}{\: ^2}, \quad T_2 = {\textstyle \frac{1}{2}} \, m \, \vec{a}{\: ^2}, \quad T_3 = {\textstyle \frac{1}{2}} \, m \, \vec{j}{\: ^2}, \quad \ldots[/itex]


    [itex]\ldots \ne \vec{I}_0 \ne \vec{I}_1 \ne \vec{I}_2 \ne \vec{I}_3 \ne \ldots[/itex]

    [itex]\ldots \ne \vec{P}_0 \ne \vec{P}_1 \ne \vec{P}_2 \ne \vec{P}_3 \ne \ldots[/itex]

    [itex]\ldots \ne W_0 \ne W_1 \ne W_2 \ne W_3 \ne \ldots[/itex]

    [itex]\ldots \ne T_0 \ne T_1 \ne T_2 \ne T_3 \ne \ldots[/itex]

    [itex]\ldots \ne V_0 \ne V_1 \ne V_2 \ne V_3 \ne \ldots[/itex]

    [itex]\ldots \ne E_0 \ne E_1 \ne E_2 \ne E_3 \ne \ldots[/itex]


    On the other hand, "Newton second law is a definition F=dp/dt"

    I think that Newton's second law is not a definition, is an empirical principle.
     
  7. Nov 19, 2011 #6
    The definitions are best presented in the new post:

    Physics Forums > Physics > Classical Physics > General Definitions of Impulse and Work
     
  8. Nov 20, 2011 #7
    What you have there are some trivial results from calculus .Any function that has its nth derivative in some region equal to 0 will have its (n-1)th derivative constant in that region.Calling all time derivatives of m*r momenta isn't done because it isn't really useful and can be confusing.

    In order for Newton second Law it to be an empirical principle we would have to have a definition of force that does not depend on dp/dt. We would have to be able to measure force independently of dp/dt and then show that they are the same. This can't be done in classical mechanics.We can only measure dp/dt and by convention we call it force.All three laws are in fact a single definition of force in a inertial frame of reference.
     
  9. Nov 20, 2011 #8

    Matterwave

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    You can go on to define a plethora of objects...an infinity of objects. But if those objects don't help you calculate anything, why bother?
     
  10. Nov 20, 2011 #9

    Therefore, the Newtonian dynamics (all three laws) is a tautology ?

    [itex]\sum \vec{F} = d\vec{p}/dt[/itex]. Therefore, in classical mechanics: [itex]\sum \vec{F} = m \, \vec{a}[/itex]

    where:

    [itex]\sum \vec{F} = \sum \vec{F}_g + \sum \vec{F}_e + \ldots[/itex]

    where:

    [itex]\sum \vec{F}_g[/itex] sum of the gravitational forces, [itex]\sum \vec{F}_e[/itex] sum of the electrical forces, [itex]\ldots[/itex]


    I think that the definitions are simple and they can be useful in theoretical physics.
     
  11. Nov 20, 2011 #10

    "I think that the definitions are simple and they can be useful in theoretical physics."
     
  12. Nov 20, 2011 #11
    A tautology is special type of formula in propositional logic ,It must contain a set of premises, some logical relations and a conclusion that is true no matter the value of the premisses.Newtons laws do not have this structure .There are no premises ,logical relations or conclusions.The laws just say that there is such a thing as a force and what a force does.They are not meant as a derivation or a proof of anything.
    You can use Lagrangian/Hamiltonian mechanics to avoid just making some definitions but you still have some definitions and axioms at the base of the theory.

    The thing that we can measure is dp/dt even for the gravitational and electromagnetic forces we can then use dp/dt to determine the form of the force. F=dp/dt is the definition ,we measure dp/dt for a system we do the math and we find that dp/dt=kMm/r^2.We then say that the gravitational force between two objects is F=GMm/r^2.In the case F=GMm/r^2 we have an empirically derived law.In the case of F=dp/dt we only have a definition since you can't measure both side of the equality independently and you can't derive F from dp/dt.In fact force is just a label for dp/dt.

    Where?
     
  13. Nov 20, 2011 #12

    ok, I change the question:
    Therefore, the Newtonian dynamics (all three laws) is not an empirical theory ?


    The thing that we can also measure is [itex]\sum \vec{F}[/itex]. For example, if we consider a single gravitational force acting on a particle, then [itex]\sum \vec{F} = G M m / r^2[/itex]


    I still do not know.
     
  14. Nov 20, 2011 #13
    Newtonian mechanics is empirical because the predictions it makes coincide with what we observe in a experiment that is in the region where classical mechanics is valid.The three laws are not directly derivable from observation.Nature didn't force us pick the definitions for force momentum and energy, we picked them because they are convenient.You could define momentum as pi*sqrt(m)*v and still do classical mechamics.

    What you can measure is usually position at a given time from that you you get velocity ,acceleration,momentum as functions of time.Once you have those you can derive a force law such as F = G M m / r^2 and make predictions on the future behavior of the systems.This is what Newton did.He had some observations,he derived a force law and made some predictions based on that law.The measurements were still positions of planets in the sky not measurements of G M m / r^2 directly.
     
  15. Nov 20, 2011 #14

    Therefore, the Newton's third law is a definition ? But there are forces that do not follow the Newton's third law.

    In addition, the Newton's laws of motion should be called the Newton's definitions of motion (?)

    In the previous example, we can measure [itex]GMm/r^2[/itex] directly.


    The End
     
  16. Nov 20, 2011 #15

    Matterwave

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    Newton's first 2 laws are more or less definitions, I prefer to think of the first law as a definition of inertial frame of reference, and the second law as a definition of force. Definitions are not tautologies. 1+1=2 could be construed as a definition of the symbols "1" and "2" (assuming you already defined "+" and "=" elsewhere), but this is not a tautology. It is not trying to draw a logical conclusion.

    Newton's 3rd law is a statement of conservation of momentum of particles. This is not always true because momentum can be carried by fields. In this sense, Newton's 3rd law is an empirical statement.

    Where Newtonian mechanics really comes in is the postulating of the forms of different forces. F=GMm/r^2 is an empirical hypothesis which says that the gravitational force is proportional to the product of the masses and the inverse square of the distance between them. This was verified by Cavendish and many others. F=-kx is another empirical hypothesis on the force provided by a spring. F=-mu*N is another for friction, and on and on we go.

    Newton's (first 2) laws define the framework in which we want to work. The empirical hypotheses come later.

    You can choose not to work in Newton's framework (indeed, for many problems it's simpler to work in Lagrangian or Hamiltonian mechanics), but that's just a choice.
     
  17. Nov 20, 2011 #16
    "A statement is tautologous if it is logically true, that is, if it is logically impossible for the statement to be false"

    However,

    If a definition is a tautologous statement or if a definition is not a tautologous statement, it is not the purpose of this post.

    If the Newton's laws of motion are definitions or if the Newton's laws of motion are not definitions, it is not the purpose of this post.
     
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