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


by ato
Tags: inertial, nbody, newton, position, terms, vectors
ato
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#1
Dec23-12, 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 n-body system ?

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

thank you
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andrien
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#2
Dec23-12, 04:01 AM
P: 977
see if this helps
https://docs.google.com/viewer?a=v&q...dUaHm8yd0n31kg
ato
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#3
Dec24-12, 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

andrien
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#4
Dec24-12, 06:19 AM
P: 977

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


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
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#5
Dec24-12, 06:20 AM
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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.
ato
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#6
Dec24-12, 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 ?
andrien
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#7
Dec24-12, 06:32 AM
P: 977
ANY inertial reference frame will work.
ato
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#8
Dec24-12, 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 .
Doc Al
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#9
Dec24-12, 10:20 AM
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Quote Quote by ato View Post
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 [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.
ato
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#10
Dec24-12, 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) inter-body forces (like gravitational,electric or magnetic), you only need (to define)
$$\vec{F}_{ab-interbody-force}$$
the third law would automatically define,
$$\vec{F}_{ba-interbody-force}$$

Quote Quote by Doc Al View Post
[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.
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
Doc Al
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#11
Dec24-12, 01:11 PM
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Newton's 3rd law concerns individual forces (interactions between two bodies) not the net force on a body.


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