Many particle system - problem with cross product

In summary, in a many particle system, the center of mass is denoted by R and the position vector of the ith particle with respect to the center of mass is r_i. Therefore, the position vector measured from the origin is given by R_i = R + r_i. The reason why R times the sum of the mass of all particles (m_i) times the rate of change of their position vectors (r_i) equals 0 is because by definition of the center of mass, the sum of m_i times r_i must equal 0. This is because the center of mass is a point where the total mass of the system can be assumed to be concentrated, and therefore, the sum of the products of mass and position
  • #1
lavster
217
0
in a many particle system we have a center of mass R and position vector of the ith particle with respect to centre of mass is r_i. hence the position vector measured from the origin is R_i=R+r_i.

why does [tex]R\times\sum (m_i \dot{r}_i) =0[/tex], where [tex]\dot{r}_i[/tex] denotes the rate of change wrt time and m_i is the mass of the ith particle?

thanks
 
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  • #2
hi lavster! :smile:

(have a sigma: ∑ and try using the X2 tag just above the Reply box :wink:)

because by definition of centre of mass,

∑ miri' = (∑ miri)' = 0' :wink:
 

1. What is a many particle system?

A many particle system refers to a group of particles or objects that interact with each other through various forces, such as gravity or electromagnetism. This system is often studied in the field of physics and can be used to model and understand complex phenomena such as fluid dynamics or molecular interactions.

2. What is the problem with using cross product in a many particle system?

The cross product, which is a mathematical operation between two vectors, is often used to determine the direction of a resulting force in a many particle system. However, this approach can become problematic when dealing with a large number of particles, as it becomes computationally expensive and may not accurately reflect the complex interactions between all the particles.

3. How can the problem with cross product be addressed in a many particle system?

One way to address this issue is by using numerical methods, such as the N-body simulation, to accurately calculate the forces between particles in a many particle system. This approach takes into account the interactions between all particles and can provide more accurate results compared to using the cross product.

4. Are there any alternative methods to using cross product in a many particle system?

Yes, there are alternative methods such as using the Barnes-Hut algorithm, which approximates the forces between particles by dividing the system into smaller subgroups. This method is more efficient and can accurately model the interactions between particles in a many particle system.

5. Can the problem with cross product be completely eliminated in a many particle system?

No, the problem with cross product cannot be completely eliminated, but it can be minimized by using more sophisticated and accurate numerical methods. However, the cross product can still be a useful tool for understanding the overall behavior of a many particle system, as long as its limitations are taken into consideration.

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