Can the Bullet's Gamma Factor Exceed This Limit in Relativistic Collisions?

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Homework Help Overview

The discussion revolves around a relativistic collision problem involving a bullet and a stationary target, focusing on the post-collision gamma factor of the bullet. Participants are exploring the implications of conservation of momentum and the conditions under which the gamma factor can be bounded.

Discussion Character

  • Conceptual clarification, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants are examining the relationship between the four-momentum of the bullet and the target, questioning how to prove that the square of the momentum difference is non-negative in the center of mass frame. There is also discussion about the implications of using the center of mass system and the conditions for the gamma factor.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the mathematical properties of four-vectors and the implications of the hint provided. Some participants express confusion about the proof and the relevance of the center of mass frame, while others are attempting to connect physical arguments to the mathematical inequalities involved.

Contextual Notes

Participants are under time constraints due to an upcoming exam, which may influence the urgency and depth of their inquiries. There is also a noted complexity in the notation and understanding of four-momentum, which is contributing to the confusion in the discussion.

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consider a head on elastic collision of a bullet of rest mass M with a stationary target of rest mass m.
prove that the post-collision gamma factor of the bullet cannot exceed (m^2+M^2)/(2mM)

im given as a hint to prove that if P,P' are the pre and post 4 momentum of the bullet and Q,Q' are of the target, then in the CM frame (P'-Q)^2>=0
well i understand how from the hint i get the answer, but how to arrive at this, i mean the momentum is conserved, i.e: P'+Q'=P+Q
so by interchanging we get:
P'-Q=P-Q'
now in the cm, we must have: (P'-Q)^2-(P-Q')^2=(M+m)^2c^2
but i don't quite see why this is correct, i mean
P=(E/c,p)
Q=(mc,0)
P'=(E'/c,p')
Q'=(E''/c,p'')
where in the cm we have: p'=-p''
so (cause the 3 momentum there is zero) (P'-Q)^2=(E'/c)^2+(mc)^2-2*E'*m
(P-Q')^2=(E/c)^2+(E''/c)^2-2EE''/c^2
here's where I am stuck, can anyone help?
 
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Are you asking why the square of a real number must be non-negative? :confused: Maybe I'm misunderstanding...
 
your'e correct, i thought about it myself, and there's not a lot to prove here, but then why is the hint to use the cm system to prove that (P'-Q)^2>=0?
 
Gokul43201 said:
Are you asking why the square of a real number must be non-negative? :confused: Maybe I'm misunderstanding...

P and P' are four-vectors, so I assume that (P - P')^2 is actually
(P - P&#039;)_0^2 - (P - P&#039;)_1^2 - (P - P&#039;)_2^2 - (P - P&#039;)_3^2[/tex]<br /> where the P_i are real numbers, in which case it is not trivial at all that the &quot;square&quot; is positive.
 
so how would you go around proving it?
my exam is tomorrow... (-:
 
CompuChip said:
P and P' are four-vectors, so I assume that (P - P')^2 is actually
(P - P&#039;)_0^2 - (P - P&#039;)_1^2 - (P - P&#039;)_2^2 - (P - P&#039;)_3^2[/tex]<br /> where the P_i are real numbers, in which case it is not trivial at all that the &quot;square&quot; is positive.
Next time, I&#039;ll actually read the question properly before I shoot my mouth... err, keyboard off!
 
I'm a bit confused by all the primes, I think you still need that
(E/c)^2+(E''/c)^2-2EE''/c^2 >= 0.
Can't you use some estimate like E > E'', which would give
(E/c)^2+(E''/c)^2-2EE''/c^2 >= (E/c)^2+(E/c)^2-2E^2/c^2 = 0
where the estimate is justified by a physical argument?
 

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