Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Newton's third law in terms of inertial position vectors for n-body system

  1. Dec 23, 2012 #1

    ato

    User Avatar

    Assuming
    $$\vec{r_{a}}$$ and $$\vec{r_{b}}$$ is calculated from an inertial frame of reference.

    then for any two objects (named a and b) in a system of more than two objects,
    Is this the newton's third law,

    $$\frac{d^{2}}{dt^{2}}m_{a}\vec{r_{a}}=-\frac{d^{2}}{dt^{2}}m_{b}\vec{r_{b}}$$

    i think this cant be right because then this implies

    $$\frac{d^{2}}{dt^{2}}m_{i}\vec{r_{i}}=0$$ for every object in that system.

    so i think i have misunderstood the law, so my question is can anyone state the law in terms of above variables for n-body system ?

    Edit 1 (fix)
    fixed a embarrassing mistake d/dt -> d^2/dt^2

    thank you
     
    Last edited: Dec 23, 2012
  2. jcsd
  3. Dec 23, 2012 #2
  4. Dec 24, 2012 #3

    ato

    User Avatar

    No it does not :(

    i think the problem i am facing is which frame of reference to use, for example

    1. if i use a frame of reference such as its origin and the center of mass of the object from which the position vector is to be calculated, coinsides .so

    $$\vec{R_{AB}}$$ is position vector of object A from a frame of reference such as its origin and center of mass of object B coinsides.
    and
    $$\vec{R_{BA}}$$ is position vector of object B from a frame of reference such as its origin and center of mass of object A coinsides.

    and the law would be

    $$\frac{d^{2}}{dt^{2}}m_{A}\vec{R_{AB}}=-\frac{d^{2}}{dt^{2}}m_{B}\vec{R_{BA}}$$

    but the problem what $$\theta_{AB},\theta_{BA}$$ to choose, certainly the above equation is not true for arbitrary $$\theta_{AB},\theta_{BA}$$ .

    2. may be a inertial frame of reference is that frame and following eqaution is the law,

    $$\frac{d^{2}}{dt^{2}}m_{A}(\vec{r_{A}}-\vec{r_{B}})=-\frac{d^{2}}{dt^{2}}m_{B}(\vec{r_{B}}-\vec{r_{A}})$$

    hence

    $$\frac{d^{2}}{dt^{2}}\left(m_{A}-m_{B}\right)\left(\vec{r_{A}}-\vec{r_{B}}\right)=0$$

    where $$\vec{r_{A}},\vec{r_{B}}$$ is calculated from an inertial frame of reference .

    please tell me which one is correct ? if both not correct please state the law too in terms of $$\vec{r_{A}},\vec{r_{B}}$$ ?

    thank you
     
  5. Dec 24, 2012 #4
    it just says
    F12=-F21,where F12=m2 dv2/dt(acting on second one).similarly for F21.Don't count others position vector into first one or so.
     
  6. Dec 24, 2012 #5

    Doc Al

    User Avatar

    Staff: Mentor

    Newton's 3rd law simply says:
    [tex]\vec{F}_{ab} = - \vec{F}_{ba}[/tex]
    When you drag in the acceleration you are really talking about Newton's 2nd law, which involves the net force. No reason to think that the net force on particle a will be equal and opposite to the net force on particle b if other particles exist.
     
  7. Dec 24, 2012 #6

    ato

    User Avatar

    ok but since $$\vec{F}_{AB}$$ is a vector what is the frame of reference ?

    can $$\vec{F}_{AB}$$ be expressed in terms of any positional vectors or any other variables ?
     
  8. Dec 24, 2012 #7
    ANY inertial reference frame will work.
     
  9. Dec 24, 2012 #8

    ato

    User Avatar

    i think i got it.

    $$\vec{F}_{AB}=\left|\vec{F}_{AB}\right|\hat{F}_{AB}$$
    where $$\left|\vec{F}_{AB}\right|$$ is all those forces like gravitational etc .

    i did not know how to express $$\vec{F}_{AB}$$ . i thought it was same force as defined in 2nd law.

    thanks andrien, Doc Al .
     
  10. Dec 24, 2012 #9

    Doc Al

    User Avatar

    Staff: Mentor

    Well, that would be true for any vector.
    Not sure what you are saying here.

    When I wrote Newton's 2nd law as [tex]\vec{F}_{ab} = - \vec{F}_{ba}[/tex]
    [itex]\vec{F}_{ab}[/itex] stood for the force on a exerted by b and [itex]\vec{F}_{ba}[/itex] stood for the force on b exerted by a.
     
  11. Dec 24, 2012 #10

    ato

    User Avatar

    what i mean is that before tackling the third law, it is necessary that the notion $$\vec{F}_{ab}$$ be understood, means how to calculate or express it , which i did not. the point is $$\vec{F}_{ab}$$ is completely different than $$\vec{F}_{a}$$ defined in 2nd law. actually its confusing that both are referred as force.

    i think the law is saying that when defining (or at least for all defined) inter-body forces (like gravitational,electric or magnetic), you only need (to define)
    $$\vec{F}_{ab-interbody-force}$$
    the third law would automatically define,
    $$\vec{F}_{ba-interbody-force}$$

    i suppose its for individual interbody force , then thats what i am saying . i am just trying to avoid any personification i can. but if its not for individual forces, then i dont know how to get $$\vec{F}_{ab}$$ if there are more than two forces involved.
    my guess would be for two forces gravitational and electric,

    $$\left(\vec{F}_{ab-gravitational}+\vec{F}_{ab-electrical}\right)=-\left(\vec{F}_{ba-gravitational}+\vec{F}_{ba-electrical}\right)$$

    but then i think newton's laws would be insufficient to imply this
    $$\vec{F}_{ab-gravitational}=-\vec{F}_{ba-gravitational}$$
    and
    $$\vec{F}_{ab-electrical}=-\vec{F}_{ba-electrical}$$

    please tell me if i am wrong .
    thanks
     
  12. Dec 24, 2012 #11

    Doc Al

    User Avatar

    Staff: Mentor

    Newton's 3rd law concerns individual forces (interactions between two bodies) not the net force on a body.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Newton's third law in terms of inertial position vectors for n-body system
  1. Newtons Third Law (Replies: 2)

Loading...