here - Department of Physics, HKU

More documents

Recommendations

Info

CHAPTER 3. LORENTZ TRANSFORMATIONS 22 Hence, simultaneity depends on the observers. Recall that in Newtonian mechanics, all observers would agree that either one event happened earlier or the two events happened at the same time. (If the two events are light-like related, then all observers will find that they are light-like, and we usually do not discuss their chronological order.) 3.3 Addition of Velocities If a bullet is moving with speed u relative to a train, and the train is moving with speed v relative to the ground in the same direction, then in Newtonian mechanics, the speed of the bullet relative to the ground is u + v. This is not correct when u or v or both are near c. Let us consider the case that the bullet is moving in general direction relative to the moving frame. Then, its coordinates are x ′ = u ′ x t′ (3.29) y ′ = u ′ y t ′ (3.30) z ′ = u ′ z t′ , (3.31) where u ′ ≡ (u ′ x , u′ y , u′ z ) is the velocity of the bullet in the moving frame. By Eq. (3.15) to Eq. (3.18), the coordinates of the bullet in the stationary frame are ⎧ x = γ v (u ′ x ⎪⎨ t′ + vt ′ ) = γ v (u ′ x + v) t′ y = u ′ y t′ z = u ⎪⎩ ′ z t′ t = γ v (t ′ + vu ′ x t ′ /c 2 ) = γ v (1 + vu ′ x/c 2 ) t ′ ⎧ x = u′ x + v 1 + vu ⎪⎨ ′ x/c t 2 u ′ y y = γ v (1 + vu ′ x /c2 ) t (3.32) u ′ z ⎪⎩ z = γ v (1 + vu ′ x/c 2 ) t √ where γ v = 1/ 1 − v 2 /c 2 . Hence, the resulting velocity is u ≡ ( u ′ x + v u ′ y 1 + vu ′ x /c2, γ v (1 + vu ′ x /c2 ) , u ′ ) z γ v (1 + vu ′ x /c2 ) . (3.33) This is called the addition of velocities. One can easily check that if v = c or the speed of the bullet √ u ′ x 2 + u ′ y 2 + u ′ z 2 = c, the resulting speed is still c. This is consistent with the fact that speed of light is constant in any inertial frame.
CHAPTER 3. LORENTZ TRANSFORMATIONS 23 v v −v 1 2 y’ x’ v 1 2 −v v i 1 2 y x 1 2 v f Figure 3.4: Collisions of two identical particles in two frames. 3.4 Energy and Momentum We will examine the conservation of momentum and energy in this section. Consider the collision of two identical particles of rest mass m in two frames. The upper half of Fig. 3.4 is the center of mass frame. Suppose before the collision, the speeds of the two particles are v and −v along the x ′ -axis, as indicated in the left of the figure. After the collision, suppose the velocities of the particles are along the y ′ -axis, as indicated in the right. The lower half of the figure is the laboratory frame. Before the collision, particle 2 is at rest, and let the speed of particle 1 be v i . After the collision, let the velocity of particle 1 be v f . Then, by Eq. (3.33), we have 2v v i = (3.34) 1 + v 2 /c 2 v fx = v (3.35) v fy = v γ v (3.36) where v fx and v fy are the components of v f . (Substitute (u ′ x , u′ y , u′ z ) = (v, 0, 0) to get the first equation, and (u ′ x, u ′ y, u ′ z) = (0, v, 0) to get the last two.) If we defined the momentum by the classical way, p ≡ mv, then the
Page 1 and 2: PHYS 1303 - Special Relativity I Le
Page 3 and 4: Chapter 1 Pre-Special Relativity Ph
Page 5 and 6: CHAPTER 1. PRE-SPECIAL RELATIVITY P
Page 11 and 12: CHAPTER 2. SPECIAL RELATIVITY 9 l 2
Page 13 and 14: CHAPTER 2. SPECIAL RELATIVITY 11 v
Page 15 and 16: CHAPTER 2. SPECIAL RELATIVITY 13 h
Page 17 and 18: CHAPTER 2. SPECIAL RELATIVITY 15 Fr
Page 19 and 20: CHAPTER 3. LORENTZ TRANSFORMATIONS
Page 23: CHAPTER 3. LORENTZ TRANSFORMATIONS
Page 29 and 30: Chapter 4 Relativistic Kinematics W
Page 31 and 32: CHAPTER 4. RELATIVISTIC KINEMATICS
Page 33 and 34: CHAPTER 4. RELATIVISTIC KINEMATICS
Page 35 and 36: Chapter 5 Paradoxes In this last ch
Page 37 and 38: CHAPTER 5. PARADOXES 35 front door
Page 39 and 40: CHAPTER 5. PARADOXES 37 gas cloud t

here - Department of Physics, HKU

You also want an ePaper? Increase the reach of your titles

Delete template?

Save as template?