Lect14_[Compatibility_Mode]

# Lect14_[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 [email protected] TA: Dan Hartzler Office: PHYS 7 [email protected] Grader: Shuo Liu Office: PHYS 283 [email protected] Office Hours: If you have questions, just email us to make an ppointment. e enjoy talking about physics! appointment. We enjoy talking about physics! Reading: Chapter 9 of the text. Notices: Exam I will be at 8:00pm Thursday, October 7 in MTHW 210

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Recitation Recap The first problem gave you an opportunity to use some of the new concepts and methods of calculation we’ve been discussing. These are important because of the powerful new perspective on special relativity they make possible. We will develop that perspective further today. The problem focused on the 4-velocity of an object moving at constant speed t V = 3/5 in the x direction relative to a Home inertial coordinate system. The idea was to focus on a couple of 2 21 ˆ tt Δ 21 s Δ 21 t Δ events on the object’s worldline so that we could use the definition 21 s u Δ = 21 ˆ x x Δ 1 21 τ Δ 21 t Δ where Home-frame measurements give, x 21 x Δ 21 21 21 21 ˆ ˆ 0 0 x st t x x ⎢⎥ Δ Δ= Δ + Δ ±
ince = 3/5 we know the ratio of 1 1 = 3/5, so lets use 1 =3s Since V 3/5 we know the ratio of Δ x 21 / Δ t 21 3/5, so lets use Δ x 21 3 s and Δ t 21 = 5 s so the corresponding proper time interval is 22 2 2 21 21 21 21 21 25 9 4 ss t x s s s τ ⇒⇒ Δ= Δ Δ = 2 2 21 21 21 21 21 21 21 21 1 / 5/4 tt tx x t t γ ΔΔ Δ Δ=− Δ Δ Δ = = /4 ⎤⎡ also So, t 2 21 21 3/4 53 ˆ ˆ 00 44 V s ut x ⎢⎥ Δ ⎥⎢ == + = Δ ⎦⎣ ± 21 ˆ Δ 21 s Δ 21 t Δ The “squared magnitude” of this timelike 4-vector is 2 u u ⎛⎞ = = 21 ˆ x x Δ 1 1 uu u ⋅≡ ⎜⎟ ⎝⎠ which we could have guessed because x 21 x Δ 21 21 21 21 21 21 s s u s Δ Δ Δ = Δ

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Since the object is moving at a constant speed in the x direction relative to the Home frame, there is an Other coordinate system in standard orientation in which it is at rest. Obviously, in that Other frame Δ x’ 21 = 0 s, so, 21 21 21 0 ˆ 0 t st t Δ ⎡⎤ ⎢⎥ ′′ Δ= Δ ± 21 0 ˆ 1 0 t ut Δ ⇒= ± t 2 t 0 ⎣⎦ since the coordinate time interval measured between events 1 and 2 in is frame also the proper time 0 21 s Δ x this frame Δ t 21 is also the proper time interval between them.
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Lect14_[Compatibility_Mode] - Physics 344 Foundations of...

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