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

[ato]

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 can't 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

 

[andrien]

see if this helps
https://docs.google.com/viewer?a=v&q=cache:Ox-lcMhv_m8J:www.physics.usu.edu/torre/3550_Fall_2012/Lectures/03.pdf+&hl=en&gl=in&pid=bl&srcid=ADGEESij72jFit_PAfkzuI0qeYbsxYsbaQX_VkPE6utWGEOKpshxDJ4dk8GlJhJ3bK7hwO6B09ngIEHrC0OgUfPo6DdOgLiC6H6p38ZVVw5LxeTLhrI-vQxwqISfkGdVzfzsa4M14kjh&sig=AHIEtbTfhVkqv5ZxlAFXdUaHm8yd0n31kg

 

[ato]

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

 

[andrien]

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.

 

[Doc Al]

Newton's 3rd law simply says:
$$\vec{F}_{ab} = - \vec{F}_{ba}$$
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.

 

[ato]

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 ?

 

[andrien]

ANY inertial reference frame will work.

 

[ato]

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 .

 

[Doc Al]

i think i got it.

$$\vec{F}_{AB}=\left|\vec{F}_{AB}\right|\hat{F}_{AB}$$

Well, that would be true for any vector.

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

Not sure what you are saying here.

When I wrote Newton's 2nd law as $$\vec{F}_{ab} = - \vec{F}_{ba}$$
$\vec{F}_{ab}$ stood for the force on a exerted by b and $\vec{F}_{ba}$ stood for the force on b exerted by a.

 

[ato]

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,
$$\vec{F}_{ba-interbody-force}$$

$$\vec{F}_{ab} = - \vec{F}_{ba}$$
$\vec{F}_{ab}$ stood for the force on a exerted by b and $\vec{F}_{ba}$ stood for the force on b exerted by a.

i suppose its for individual interbody force , then that's what i am saying . i am just trying to avoid any personification i can. but if its not for individual forces, then i don't 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

 

[Doc Al]

Newton's 3rd law concerns individual forces (interactions between two bodies) not the net force on a body.