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Home , Galilean invariance

Newtonian Relativity ¾A reference frame in which Newton’s laws are valid is called an inertial frame ¾Newtonian or Galilean invariance

„ If Newton’s laws are valid in one reference frame, then they are also valid in a reference frame moving at uniform velocity relative to the first system

„ Thus this moving frame is also an inertial frame

1 Newtonian Relativity ¾Consider

2 Newtonian Relativity ¾

¾Note time is the same in both systems

3 Newtonian Relativity ¾Note Newton’s laws are valid in both frames

„ The and acceleration are the same in both frames

„ There is no way to detect which frame is moving and which is at rest

d 2 y′ d 2 y F′ = ma′ = m = m = ma = F y y dt′2 dt2 y y d 2 x′ d 2 (x − vt) d 2 x F′ = ma′ = m = m = m = F x x dt′2 dt2 dt2 x

4 ¾We immediately see the Galilean transformation is inconsistent with Einstein’s postulates „ If the velocity of light = c in frame K, the velocity of light = c – V in frame K’ ¾The Lorentz transformation satisfies Einstein’s postulates and also reduces to the Galilean transformation at low velocities ¾A derivation is given in Thornton and Rex p30-31

5 Lorentz Transformation x′ = γ (x − vt) x = γ (x′ + vt′) y′ = y y = y′ z′ = z z = z′ t′ = γ (t − vx / c2 ) t = γ (t′ + vx′/ c2 )

γ 1 1 where = = 1− β 2 v2 1− c2 6 Lorentz Transformation ¾ revisited ’ ’ ’ „ Let Δt = t2 –t1 be the interval ’ ’ measured by a clock fixed at x0 in K γ x′V x′V ′ 2 ′ 0 t2 = (t2 + 2 γ) = (t2 + 2 ) γ c c x′V x′V t = (t′ + 1 ) = (t′ + 0 ) 1 1 c2 γ 1 c2

Δt = t2 − t1 = γΔt′

¾The clocks in S read a time longer than the proper time. The moving clock in S’ runs slow. 7 x Lorentz Transformation ¾ revisited ’ „ Consider a measuring rod with Δx = ’ ’ 2 -x1 . The interval Δx as viewed in S must have the positions measured at the same time t0 in S. x′ = γ (x −Vt ) = γ (x −Vt ) 2 γ 2 2 2 0 x′ = (x −Vt ) = (x −Vt ) 1 1 1 γ 1 0 Δx = Δx′/γ

¾The length of the moving object as measured in S is shorter than the proper length

8 Lorentz Transformation ¾Clock synchronization revisited

„ Consider two clocks synchronized in S’. Clock B’ at ’ ’ x2 and clock A’ at x1 . What times do they read at time t0 in S? γ ⎛ x2V ⎞ tB′ = ⎜t0 − ⎟ ⎝ c2 ⎠ γ ⎛ x1V ⎞ t′A = ⎜t0 − ⎟ ⎝ c2 ⎠ V V L V t′ − t′ = −γ ()x − x = −()x′ − x′ = − 0 B A 2 1 c2 2 1 c2 c2

„ Agrees with results from the homework 9 Beta and Gamma

β =

10 Invariants ¾Invariant quantities have the same value in all inertial frames

„ In the next homework, you’ll show s2 = x2 + y2 + z2 − (ct)2 s′2 = x′2 + y′2 + z′2 − (ct′)2 s2 = s′2 2 „ s is the same for all inertial frames

11 Invariants ¾Consider two events 1 and 2 ¾We define the interval as 2 2 2 „ Δs = Δx -(cΔt) ¾Three cases 2 „ Lightlike (Δs =0) Š The two events can be connected only by a light signal 2 „ Spacelike (Δs >0) Š The two events are not causally connected. We can find an inertial frame where the events occur at the same time but at different positions in space 2 „ Timelike (Δs <0) Š The two events are causally connected. We can find an inertial frame where the events occur at the same position in space but at different times

12 Addition of Velocities ¾Recall the Galilean transformation between two frames K and K’ where K’ moves with velocity v with respect to K

„ Consider an object moving with velocity u in K and u’ in K’ x′ = x − vt dx′ d(x − vt) dx = = − v dt′ dt dt u′ = u − v

13 Galilean Transformation

x′ = x − vt x = x′ + vt y′ = y y = y′ z′ = z z = z′ t′ = t t = t′

¾Note time is the same in both systems

14 Addition of Velocities ¾We know the Lorentz transformation should be used instead so

dx γ (dx′ + vdt′) u′x + v ux = = = dt ⎡ ⎛ v ⎞ ⎤ ⎛ v ⎞ γ dt′ + dx′ 1+ u′ ⎢ ⎜ 2 ⎟ ⎥ ⎜ 2 ⎟ x ⎣ ⎝ c ⎠ ⎦ ⎝ c ⎠ similarly

u′y uy = ⎡ ⎛ v ⎞ ⎤ γ 1+ u′ ⎢ ⎜ 2 ⎟ x ⎥ ⎣ ⎝ c ⎠ ⎦

u′z uz = ⎡ ⎛ v ⎞ ⎤ γ 1+ u′ ⎢ ⎜ 2 ⎟ x ⎥ ⎝ c ⎠ ⎣ ⎦ 15 Lorentz Transformation x′ = γ (x − vt) x = γ (x′ + vt′) y′ = y y = y′ z′ = z z = z′ t′ = γ (t − vx / c2 ) t = γ (t′ + vx′/ c2 )

γ 1 1 where = = 1− β 2 v2 1− c2 16 Addition of Velocities ¾Swapping primed and unprimed variables and letting v go to –v u − v u′ = x x ⎛ v ⎞ 1− ⎜ ⎟ux ⎝ c2 ⎠

uy u′y = ⎡ ⎛ v ⎞ ⎤ γ 1− u ⎢ ⎜ 2 ⎟ x ⎥ ⎣ ⎝ c ⎠ ⎦

uz u′z = ⎡ ⎛ v ⎞ ⎤ γ 1− ⎜ ⎟u ⎢ c2 x ⎥ ⎣ ⎝ ⎠ ⎦ 17 Addition of Velocities

¾Example - let u’x=c, u’y=0, u’z=0 (c +V ) u = = c, u = 0, u = 0 x ⎛ cV ⎞ y z ⎜1+ ⎟ ⎝ c2 ⎠

¾Example - let u’x=0, u’y=c, u’z=0 c u = V , u = , u = 0 x y γ z θ u γV tan = x = uy c 18 Addition of Velocities ¾A rocket blasts off from the earth at v=0.90c ¾A second rocket follows in the same direction at velocity 0.98c ¾What is the relative velocity of the rockets using a Galilean transformation ¾What is the relative velocity of the rockets using a Lorentz transformation?

19 Lorentz Transformation ¾Last time we argued that y = y′ and z = z′ γ ¾The most general linearα transformation for x=f(x’,t’) is γ ⎛ α ⎞ x = x′ + t′ = ⎜ x′ + t′⎟ ⎝ γ ⎠ ¾At low velocities, γ→1 and α/γ→V x = γ (x′ +Vt′) ¾The inverse transformation is the same except for the sign of relative motion x′ = γ (x −Vt)

20 Lorentz Transformation ¾For a light pulse in S we have x=ct ¾For a light pulse in S’ we have x’=ct’ ¾Then x = ct = γ (ct′ +Vt′)= γ (c +V )t′ x′ = c()()t′ = ct −Vt = c −V t γ () ct′ = c −V ()cγ+V t′ γ γ c 1 γ = V 2 1− c2 21 Lorentz Transformation ¾For the t transformation x′ = γ (x −Vt) γx′γ= []()x′ +Vt′ −Vt γ ⎛ 1 ⎞ x′ t = t′ +γ ⎜ − ⎟ ⎝ ⎠ V γ ⎛ Vx′⎞ t = γ ⎜t′ + ⎟ ⎝ c2 ⎠

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