Lect15_[Compatibility_Mode]

Lect15_[Compatibility_Mode] - Physics 344 Foundations of...

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Physics 344 Foundations of 21 st Century Physics: Relativity, Quantum Mechanics and heir Applications Their Applications Instructor: Dr. Mark Haugan Office: PHYS 282 haugan@purdue.edu TA: Dan Hartzler Office: PHYS 7 dhartzle@purdue.edu Grader: Shuo Liu Office: PHYS 283 liu305@purdue.edu Office Hours: If you have questions, just email us to make an ppointment. e enjoy talking about physics! appointment. We enjoy talking about physics! Notices: Help session Tuesday 1:30 to 3:30 in room 160. Exam I will be at 8:00pm Thursday, October 7 in MTHW 210
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After introducing the 4-momentum vector of a particle last lecture we looked back at the recitation problem involving a particle of mass m = 3 kg moving at speed v = 4/5 in the y direction relative to a Home-frame coordinate system. The problem asked what energy an observer in an Other frame oving in the x direction at speed 3/5 relative to the Home frame would moving in the x direction at speed V = 3/5 relative to the Home frame would measure the particle to have. We considered this highly relativistic situation to illustrate the way in which orking with 4- ectors can be used to make physical predictions and to solve working with 4 vectors can be used to make physical predictions and to solve problems. In the process, we obtained an expression for the Other-frame energy of the article that was correct for any speeds nd I meant to use this to show particle that was correct for any speeds v and V . I meant to use this to show the result reduces to the expected Newtonian result in the limit that both speeds are small compared to c , but forgot to do this.
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We saw that the Home-frame components of the particle’s 4-momentum 5/ 3 m m γ ⎡⎤ vector were related to its relativistic energy and momentum components measured in that frame and that in this particular case they were 0 4/ 3 p px py mv P m v ⎢⎥ = ± where 22 11 5 3 ( 4 / 5 ) p v == = −− 0 pz ⎣⎦ We determined the Home-frame components of the Other observer’s 4-velocity nd used its relationship to the dimensionless unit vector parallel to the t’ axis and used its relationship to the dimensionless unit vector parallel to the t axis of the Other coordinate system to solve the problem. 5/4 3/4 00 0 obs V u = ± where 5 4 ( 3 / 5 ) V =
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Specifically, this particle’s energy measured in the Other frame is 0 ˆ p m V E Pt m γγ γ ⎡⎤ ⎢⎥ = ⋅= ±i 0 00 p p mv
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Lect15_[Compatibility_Mode] - Physics 344 Foundations of...

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