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l6 - Relativistic momentum Concluding relativity Particle...

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Concluding relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativistic momentum We found E 2 = E 2 0 + ( pc ) 2 We found ( E / c ) 2 p 2 x p 2 y p 2 z = 0 is frame invariant This led to Lorentz transforms for momentum and energy: p x , 2 = γ parenleftBig p x , 1 v ( E 1 c 2 ) parenrightBig and p y , 2 = p y , 1 and p z , 2 = p z , 1 (1) E 2 = γ ( E 1 vp x , 1 ) . Energy and momentum are intertwined when comparing relativistic reference frames!
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Concluding relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativisic collision I A particle A with rest mass m 0 and velocity v A = 0 . 80 c in the ˆ x direction collides with an initially-stationary particle B with rest mass 2 m 0 . Note that for β = 4 / 5 we can find γ = 5 3 . In the frame S 1 , we then use p 1 , y = γ m 0 v 2 , y and E = γ m 0 c 2 to find p x , A , 1 = γ m 0 v x , A = 5 3 m 0 4 5 c = 4 3 m 0 c . p y , A , 1 = p z , A , 1 = 0 E A , 1 = γ m 0 c 2 = 5 3 m 0 c 2 for particle A , and p x , B , 1 = γ m 0 v x , B = 0 p y , B , 1 = p z , B , 1 = 0 E B , 1 = γ m 0 c 2 = 1 ( 2 m 0 ) c 2 for particle B .
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Concluding relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativistic collision II The total energy in frame S 1 is then E 1 = E A , 1 + E B , 1 = parenleftbigg 5 3 + 2 parenrightbigg m 0 c 2 = 11 3 m 0 c 2 , or E 1 = 3 . 67 m 0 c 2 . Now let’s find the center-of-momentum frame. Using p x , 2 = γ ( p x , 1 v ( E / c 2 ) ) from Eqs. 2, we find p x , A , 2 + p x , B , 2 = 0 = γ bracketleftbigg ( p x , A , 1 v ( E A , 1 / c 2 ) ) + ( p x , B , 1 v ( E A , 1 / c 2 ) ) bracketrightbigg = ( 4 3 m 0 c 5 3 m 0 v ) + ( 0 2 m 0 v ) radicalbig 1 v 2 / c 2 Thus we want ( 4 3 c 5 3 v 2 v ) m 0 = 0 which gives v = ( 4 / 11 ) c .
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Concluding relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativistic collision III We found that v = ( 4 / 11 ) c lets us move from frame S 1 where particle B was at rest, to frame S 2 where the two particles come in with equal and opposite momentum. The total energies of the individual particles in S 2 can be found using E 2 = γ ( E 1 vp x , 1 ) to be E A , 2 = E A , 1 vp x , A , 1 radicalbig 1 v 2 / c 2 = 5 3 m 0 c 2 ( 4 11 c )( 4 3 m 0 c ) radicalBig 1 ( 4 11 ) 2 = 1 . 27 m 0 c 2 E B , 2 = E B , 1 vp x , B , 1 radicalbig 1 v 2 / c 2 = 2 m 0 c 2 0 radicalBig 1 ( 4 11 ) 2 = 2 . 15 m 0 c 2 , or E 2 = 3 . 24 m 0 c 2 . As promised, energy is not conserved!
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Concluding relativity Particle accelerators Blackbody Cavity modes Density of states Thermodynamic occupancy Blackbody redeaux Blackbody measurements Max Planck Planck’s fix Relativity and particle accelerators We compared the same two-particle collision in two different frames.
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