Register to reply 
Newton's third law in terms of inertial position vectors for nbody system 
Share this thread: 
#1
Dec2312, 03:43 AM

P: 30

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 nbody system ? Edit 1 (fix) fixed a embarrassing mistake d/dt > d^2/dt^2 thank you 


#2
Dec2312, 04:01 AM

P: 1,020

see if this helps
https://docs.google.com/viewer?a=v&q...dUaHm8yd0n31kg 


#3
Dec2412, 05:00 AM

P: 30

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 


#4
Dec2412, 06:19 AM

P: 1,020

Newton's third law in terms of inertial position vectors for nbody system
it just says
F_{12}=F_{21},where F_{12}=m_{2} dv_{2}/dt(acting on second one).similarly for F_{21}.Don't count others position vector into first one or so. 


#5
Dec2412, 06:20 AM

Mentor
P: 41,323

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. 


#6
Dec2412, 06:28 AM

P: 30

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 ? 


#7
Dec2412, 06:32 AM

P: 1,020

ANY inertial reference frame will work.



#8
Dec2412, 08:54 AM

P: 30

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 . 


#9
Dec2412, 10:20 AM

Mentor
P: 41,323

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. 


#10
Dec2412, 11:28 AM

P: 30

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) interbody forces (like gravitational,electric or magnetic), you only need (to define) $$\vec{F}_{abinterbodyforce}$$ the third law would automatically define, $$\vec{F}_{bainterbodyforce}$$ my guess would be for two forces gravitational and electric, $$\left(\vec{F}_{abgravitational}+\vec{F}_{abelectrical}\right)=\left(\vec{F}_{bagravitational}+\vec{F}_{baelectrical}\right)$$ but then i think newton's laws would be insufficient to imply this $$\vec{F}_{abgravitational}=\vec{F}_{bagravitational}$$ and $$\vec{F}_{abelectrical}=\vec{F}_{baelectrical}$$ please tell me if i am wrong . thanks 


Register to reply 
Related Discussions  
Expressing cartesian unit vectors in terms of spherical unit vectors  General Math  9  
Vectors, velocity in terms of unit vectors  Calculus & Beyond Homework  3  
Acceleration in terms of position  Introductory Physics Homework  4  
Position Vectors, Velocity Vectors, and Acceleration Vectors  Introductory Physics Homework  3  
Relativity: position/acceleration 4vectors to newton's laws  Advanced Physics Homework  0 