Center of Momentum Energy problem

E=c*Sqrt((Ma^2)c^2+Pa^2)=c*Sqrt((Ma^2)c^2+(Mb^2)c^2)=c*Sqrt(((Ma^2)c^2+(Mb^2)c^2)/2)=c*Sqrt((Ma^2+Mb^2)c^2)/2=S*Sqrt((Ma^2+Mb^2)c^2)/2. This is the desired outcome for the center of momentum energy of particle A.
  • #1
bweinert89
1
0

Homework Statement


Hi, I'm a bit confused about Center of Momentum Energy.

It has particles A and B collide to make C and D. It then asks to show that the CM energy of A is ((S+Ma^2-Mb^2)c^2)/(2Sqrt(S))


Homework Equations


The question gives the Mandelstam variables S=((Pa+Pb)^2)/c^2, t=((Pa-Pc)^2)/c^2, u=((Pa-Pd)^2)/c^2.


The Attempt at a Solution


I'm not quite sure what to do. I tried setting up the equation Ea+Eb = Ec+Ed where c and d only have resting energy. I square both sides and separate try to get c*Sqrt((Ma^2)c^2+Pa^2) by itself since that is what the regular Ea equals but i don't get the desired outcome on the otherside.

-Ben
 
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  • #2
jamin

Dear Benjamin,

The center of momentum energy is a concept used in particle physics to describe the energy of a system of particles in their collective center of mass frame. In this case, we have particles A and B colliding to form particles C and D. The total energy of the system can be expressed as the sum of the individual energies of each particle in their respective rest frames. However, in this case, we are interested in the energy in the center of mass frame, which is the frame in which the total momentum of the system is zero.

To determine the center of momentum energy of particle A, we need to calculate the energy of particle A in the center of mass frame. This can be done using the Mandelstam variable S, which is defined as S=((Pa+Pb)^2)/c^2, where Pa and Pb are the momenta of particles A and B, respectively. In the center of mass frame, the total momentum of the system is zero, so we have Pa+Pb=0. This means that S=-2Pa*Pb/c^2.

To calculate the energy of particle A in the center of mass frame, we can use the formula for the energy of a particle with mass m and momentum p: E^2=m^2c^4+p^2c^2. In this case, the momentum of particle A in the center of mass frame is given by Pa=-Pb, so we have E^2=m^2c^4+(-Pb)^2c^2. Substituting this into the equation for S, we get S=-2Pa*Pb/c^2=2Pb^2/c^2. Solving for Pb, we get Pb=c*Sqrt(S/2). Substituting this into the equation for E, we get E=c*Sqrt((Ma^2)c^2+S/2). This is the energy of particle A in the center of mass frame.

To get the desired outcome, we need to simplify this expression. First, we can use the fact that S=((Pa+Pb)^2)/c^2 to rewrite the equation as E=c*Sqrt((Ma^2)c^2+Pa^2). We can then use the fact that Pa=c*Sqrt(S/2) to rewrite the equation as E=c*Sqrt((Ma^2)c^2+c^2*S/2). Finally,
 

1. What is the Center of Momentum Energy problem?

The Center of Momentum Energy problem is a theoretical physics problem that involves finding the center of mass of a system and calculating the total energy of the system based on the mass and velocity of its individual components.

2. Why is the Center of Momentum Energy problem important?

This problem is important because it allows scientists to understand and predict the behavior of a system of objects in motion. It is also a key concept in fields such as astrophysics, where the movement and interactions of celestial bodies are studied.

3. How is the Center of Momentum Energy problem solved?

The problem is solved using the principles of conservation of momentum and conservation of energy. These principles state that the total momentum and energy of a system remains constant unless acted upon by an external force.

4. Can the Center of Momentum Energy problem be applied to any system?

Yes, the principles used to solve the Center of Momentum Energy problem can be applied to any system, as long as the system is isolated and not affected by external forces.

5. How does the Center of Momentum Energy problem relate to other physics concepts?

The Center of Momentum Energy problem is closely related to other physics concepts such as Newton's laws of motion, work and energy, and torque. Understanding this problem can also provide insights into more complex systems, such as collisions and oscillations.

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