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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 .
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!
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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