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Gaitskell

PH0008Quantum Mechanics and Special Relativity

Lecture 6 (Special Relativity)

Minkowski Space

Use of Lorentz-Einstein TransformationMinkowski Space

Prof Rick Gaitskell

Department of PhysicsBrown University

Main source at Brown Course Publisher

background material may also be available at http://gaitskell.brown.edu

PH0008 Gaitskell Class Spring2002 Rick Gaitskell

Section: Special Relativity Week 3

• Homework (none due for M 3/4)• (see”Assignments” on web pages)

• [Please start on next homework)

• Reading (Prepare for 2/4)

o SpecRel (also by French)• Ch3 Einstein & Lorentz Transforms

• Ch4 Realtivity: Measurement of Length andTime Inetrvals

• Lecture 5 (M 3/4)o Lorentz Transformation

• Worked Example: Rod and Single Clock— Time Dil.,

— Lorentz Cont.,

— Relativity of Simultaneity

o Minkowski Space

• Lecture 6 (W 3/6)o Minkowski Space

• More Worked Example: Two Rods

— Time Dil.,

— Lorentz Cont.,

— Relativity of Simultaneity

• Lecture 7 (F 3/8)o Review with Further Worked Example

• Reading (Prepare for 3/11)

o SpecRel (also by French)• Ch5 RelativisticKinematics

• Ch6 Relativistic Dynamics: Collisions andConservation Laws

• (Review)• Ch3 Einstein & Lorentz Transforms

• Ch4 Realtivity: Measurement of Lengthand Time Inetrvals

• Homework #7 (M 3/11)o Start early - tough

(see web “Assignments”)


Homework

• I have moved last question to week after …o See web site

• Please pick up your HW #1-3 from outside my office B&H 516


Question SectionQuestion Section


Question SpecRel L06-Q1

•Where will the Dow Jones (~10,000 today) be whenyou graduate?

o(1) >+60%

o(2) 40_60%

o(3) 20_40%

o(4) 0_20%

o(5) -20_0%

o(6) -40_-20%

o(7)-60_-40%

o(8) <-60%



•Where will the Dow Jones (~10,000 today) be whenyou graduate?

o(1) >+60%

o(2) 40_60%

o(3) 20_40%

o(4) 0_20%

o(5) -20_0%

o(6) -40_-20%

o(7)-60_-40%

o(8) <-60%

Jan‘99

Oct‘87

1965 ~ 1,000

1,000

10,000

Oct‘71‘83



•How do we view these events? (see demo)o(1) A and B simultaneous

o(2) A before B

o(3) B before A

o(4) None of above


CanCan’’t determine relative time of ant determine relative time of anevent without specifying positionevent without specifying position


Use of LorentzUse of LorentzTransformationTransformation

•to study rod and single clock events


Single Disk and Rod (3)

• Use each Lorentz Transformation in turn

†

(x1,t1)

†

( ¢ x 1, ¢ t 1)

†

(x2,t2)

†

( ¢ x 2, ¢ t 2)

Event #1

Event #2

†

(1) ¢ x 2 = g x2 - bct2( ) fi gDx0 = gbcDtDx0 = vDt

(2) c ¢ t 2 = g ct2 - b x2( ) fi cD ¢ t 0 = g cDt - bDx0( )= g cDt - bbcDt( )

D ¢ t 0 = gDt 1- b 2( )=

1g

Dt

Dt = gD ¢ t 0

(3) x2 = g ¢ x 2 + bc ¢ t 2( ) fi Dx0 = gbcD ¢ t 0= gvD ¢ t 0= gD ¢ x

D ¢ x =1g

Dx0

(4) ct2 = g c ¢ t 2 + b ¢ x 2( ) fi cDt = gcD ¢ t 0



• What do they meano (1) The velocity of disk is v in S rod frame

• The time interval between events in rod frame issimply L/v

• This must be the case…

o (2) Clock tick of disk when observed in rod frameis slower

• Moving clocks appear slower

o (3) Apparent length of rod measured in diskframe is shorter

• Moving lengths appear shorter

o (4) We already knew this…

†

(x1,t1)

†

( ¢ x 1, ¢ t 1)

†

(x2,t2)

†

( ¢ x 2, ¢ t 2)

Event #1

Event #2

†

(1) Dx0 = vDt(2) Dt = gD ¢ t 0(3) D ¢ x =

1g

Dx0

(4) Dt = gD ¢ t 0



• Consider Event #3o The right hand end of the rod when Event #1occurs in rod frame S

†

(x1,t1)

†

( ¢ x 1, ¢ t 1)

Event #3

†

In rod frame Sx3 = Dx0 = x2 t3 = t1 = 0

In disk frame ¢ S

¢ t 3 =?

¢ t 1 = 0†

(x3,t3)

†

( ¢ x 3, ¢ t 3)

• No !!! - don’t use “common” senseo Use Lorentz transforms

†

(5) ¢ x 3 = g x3 - bct3( ) fi ¢ x 3 = g Dx0( )

(6) c ¢ t 3 = g ct3 - b x3( ) fi c ¢ t 3 = g -bDx0( )

¢ t 3 = -gvc 2 Dx0

†

v is velocity of frame ¢ S measured in S¢ x = g x - bct( ) x = g ¢ x + bc ¢ t ( )¢ y = y y = ¢ y ¢ z = z z = ¢ z

c ¢ t = g ct - b x( ) ct = g c ¢ t + b ¢ x ( )b = v c g = 1- b 2( )

- 12



• Consider Event #3o At right hand end of rod, an eventsimultaneous with Event #1 when in the rodframe, S

†

(x1,t1)

†

( ¢ x 1, ¢ t 1)

Event #3

†

In rod frame Sx3 = Dx0 = x2t3 = t1 = 0

In disk frame ¢ S ¢ x 3 = gDx0

c ¢ t 3 = -gbDx0

†

(x3,t3)

†

( ¢ x 3, ¢ t 3)• In the disk frame S’ Event #3

o Occurs before t’=0 (Event #1)• i.e. before Event #1

o It is a distance >Dx0 from Event #1• Not “shortened”, further away

• But remember it does not occur at sametime as t1‘

Let’s introduce a 2nd diskseparated by rigid bar to helpvisualise what is going on


Two Disks , a Rod, and an “Excuse Me?” (7)

• Consider Event #3o Event #1 & #3 simultaneous in rod frame

†

In rod frame Sx3 = Dx0 = x2t3 = t1 = 0

In disk frame ¢ S ¢ x 3 = gDx0

c ¢ t 3 = -gbDx0• In the disk frame S’…o Event # 3 occurs before Event #1

• t3‘<0

o Event #3 is a distance >Dx0 from Event #1• The disks are further apart than Dx0

• But remember it does not occur at sametime as t1‘

†

(x1,t1)Event #3

†

(x3,t3)Event #1

Viewed in rod frame

†

( ¢ x 1, ¢ t 1) †

( ¢ x 3, ¢ t 3)

Viewed in (two) disk frame

Event #3

Event #1


Space-Time DiagramsSpace-Time Diagrams•Help visualize consequences of Lorentz Transforms


Simple 1-D (space) world (Minkowski, 1908)

• Add time as 2nd dimension

x

ct


Simple 2-D (space) world (Video)

• Time as extra dimension

x

ct

y


Space and Time Become Mixed …

• Note that variable in S’ (x’,t’) are formed from both (x,t) and vise versa

†

¢ x = g x - bct( ) x = g ¢ x + bc ¢ t ( )¢ y = y¢ z = z

c ¢ t = g ct - b x( ) ct = g c ¢ t + b ¢ x ( )

b = v c ,v is velocity of frame ¢ S measured in S)

g =1

1- v 2 c 2=

11- b 2

Note the use of (ct) rather than t which accentuates the symmetry of the transforms


Minkowski

• Path is described by unique locus in (x,t)

x

ct


Minkowski (Equally valid)

• Even though axis are not orthoganal, locus is still unique

x’

ct’


Minkowski: Trajectories

• Consider particles with different velocities in frame S

x

ctLight-Ray

Prohibited trajectory

Allowed trajectoryAllowed(constant v)


Minkowski: Stationary

• Stationary point in frame S

x

ct Path of x=constant (i.e. point stationary in S)


Minkowski: 2nd frame

• Consider 2nd frame S’ of reference with constant velocity v

x

ct Path of x’=0 (i.e. point stationary in S’)

†

¢ x = g x - bct( )If ¢ x = 0fi x = bct = vt

This could be Galilean?



•The axis shown could be Galilean?o(1) Yes

o(2) No


Minkowski: 2nd frame

• Consider 2nd frame S’ of reference with constant velocity v

x


†

If ¢ x = 0¢ x = g x - bct( ) fi x = bct = vt

Consider also ¢ t = 0

c ¢ t = g ct - b x( ) fi x =1b

ctx’

ct’

• Note symmetrical arrangement of x’ & ct’


Minkowski: -b

• Consider 2nd frame S’ of reference with constant velocity -v

x


†

If ¢ x = 0¢ x = g x + bct( ) fi x = -bct = -vt

Consider also ¢ t = 0

c ¢ t = g ct + b x( ) fi x = -1b

ct

x’

ct’


Minkowski: Light Path

• Light traces same velocity in either frame !

x

ct

†

x = ct¢ x = c ¢ t

x’

ct’

Light-Ray


Video

• Video 00:30:30 -> 00:37:50


Minkowski: Calibrating Axes

• Calibrating axeso If we define x=1, where is x’=1?

x

ct

x’

ct’

Light-Ray

†

Consider "invariant"x[ ]2

- ct[ ]2= g ¢ x + bc ¢ t ( )[ ]2

- g c ¢ t + b ¢ x ( )[ ]2

= g 2 ¢ x [ ]2+ 2 ¢ x bc ¢ t [ ] + bc ¢ t [ ]2 ...- c ¢ t [ ]2

- 2 ¢ x bc ¢ t [ ] - b ¢ x [ ]2

È

Î Í Í

˘

˚ ˙ ˙

= g 2 1- b 2( ) ¢ x [ ]2- c ¢ t [ ]2( )[ ]

= ¢ x [ ]2- c ¢ t [ ]2

†

Draw hyperbolax[ ]2

- ct[ ]2=1

Sincex[ ]2

- ct[ ]2= ¢ x [ ]2

- c ¢ t [ ]2=1

So point where it intersects ¢ x - axisc ¢ t = 0 fi ¢ x [ ]2

=1This is true generally for any ¢ S

x=1x’=1


Lorentz Transformation

• Two frames S and S’ moving at relative velocity v

x’

ct’

x

ctLight-Ray

Video†

¢ x = g x - bct( ) x = g ¢ x + bc ¢ t ( )¢ y = y¢ z = z

c ¢ t = g ct - b x( ) ct = g c ¢ t + b ¢ x ( )

b = v c ,v is velocity of frame ¢ S measured in S)

g =1

1- v 2 c 2=

11- b 2

Note the use of (ct) rather than t which accentuates the symmetry of the transforms


Consider a stationary rod in S

• Stationary Rod in S

x’

ct’

x

ct


Consider a stationary rod in S’

• Stationary Rod in S’

x’

ct’

x

ct


Two Watches

• Discuss

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