Lecture 18: Lorentz Transformation - Zhejiang Universityzimp.zju.edu.cn/~xinwan/courses/physI17/handouts/lecture18.pdf · An observer in system K' moving with uniform speed v along

General Physics IGeneral Physics I

Lecture 18: Lorentz Lecture 18: Lorentz Transformation Transformation

Prof. WAN, Xin （万歆）

[email protected]://zimp.zju.edu.cn/~xinwan/

OutlineOutline

● Experimental verification of the special theory● Lorentz transformation● The invariant interval

Muon DecayMuon Decay

● Cosmic rays interact with particles in the earth's outer atmosphere and create particles called muons. Muons are unstable with half-life 1.52 ms in their own rest frame, i.e.

● Detector on top of a 2000-m mountain counts 1000 muons traveling at v = 0.98c over a given period of time.

● Now at the sea level, how many muons do you expect to detect at v = 0.98c over the same period of time?

See B. Rossi and D. B. Hall, Phys. Rev. 59, 223 (1941).

N = N02−t /t1/2

Nonrelativistic ViewNonrelativistic View

● Classically, muons travel 2000 m in 6.8 ms. Expect only 45 muons to survive.

See B. Rossi and D. B. Hall, Phys. Rev. 59, 223 (1941).

Experimental measurement indicates 542 muons survive, a factor of 12 more. Why?

1000 muons

2000 m

45 muons

6.8 ms

1000×2−6.8/1.52 = 45

2000m

0.98⋅3×108m / s= 6.8 μ s

Relativistic, Earth FrameRelativistic, Earth Frame

● At the high speed v = 0.98c of the muons relative to the experimenters fixed on the earth, we conclude that they observe the muons' clock to be running slower by a slowing-down factor s = 0.2.

● In the earth frame, the halflife of a muon is not 1.52 ms, but 7.6 ms.

● The radioactive decay law predicts that 538 muons survive. Hence the measurement (542 muons) verified the special theory of relativity within experimental uncertainties.

1000×2−6.8 /7.6 = 538

Relativistic, Muon FrameRelativistic, Muon Frame

● In the muons' rest frame, the mountain is moving at the high speed v = 0.98c relative to the muons. We conclude that muons observe the mountain’s height shrinks by a shrinking factor 0.2. So the height of the mountain is not 2000 m, but 400 m.

● The flight time over 400 m is 1.36 ms. Again, we obtain

1000×2−1.36/1.52 = 538

ComparisonComparison

Relativistic,Muon Frame

Relativistic,Earth Frame

Nonrelativistic

Distance 400 m 2000 m 2000 m

Time 1.36 ms 6.8 ms 6.8 ms

Halflives 0.895 0.895 4.47

Surviving 538 538 45

The relativistic approach yields agreement with experiment and is greatly different from the non-relativistic result. Note that the muon and ground frames do not agree on the distance and time, but they agree on the final result. One observer sees time dilation, the other sees length contraction, but neither sees both.

Atomic Clock Time MeasurementAtomic Clock Time Measurement

● Cerium-133 atom has a well-defined transition at a frequency of 9,192,631,770 Hz and can be used as an accurate measurement of a time period.

J.C. Hafele and R. E. Keating, Science 177, 166 (1972)

In October 1971, Joseph C. Hafele and Richard E. Keating took four cesium-beam atomic clocks aboard commercial airliners. They flew twice around the world, first eastward, then westward, and compared the clocks against others that remained at the US Naval Observatory.

Atomic Clock Time MeasurementAtomic Clock Time Measurement

● General relativity predicts an increase in gravitational potential due to altitude further speeds the clocks up. That is, clocks at higher altitude tick faster than clocks on Earth's surface.

J.C. Hafele and R. E. Keating, Science 177, 166 (1972)

Galilean Transformation Galilean Transformation

The two inertial observers agree on measurements of acceleration.

Lorentz TransformationLorentz Transformation

x

y

zx '

y '

z '

v

Frank Mary

Expect to find a linear map

where a, b, c, and d depend on v.

How to determine a, b, c, and d?

t = t ' x = x ' + vt '

The Galilean transformation:

t = at ' + bx'

x = ct ' + dx'


x

y

zx '

y '

z '

v

Frank Mary



1. Relative speed v: For the origin in Frank's frame,

x = 0 x ' = −vt '

c /d = v

x = d(x ' + vt ')

t = at ' + bx'

x = ct ' + dx'

The Tale of Two FramesThe Tale of Two Frames

LF

0 -TM

v

0 0

DF

TM 0

v

TF

TF

DB

LF

0 0

0 TF

DM

LM

Train Frame (Mary's)Track Frame (Frank's)

v

T F = DFv /c2

TM = LMv /c2By the principle of relativity,

0

TF


x

y

zx '

y '

z '

v

Frank Mary



2. Rear clock ahead

t = 0 t ' = −x 'v /c2

b /a = v /c2

t = a(t ' + x 'v /c2)

t = at ' + bx'

x = ct ' + dx'

Length ContractionLength Contraction

LF

0 -TM

v

0 0

DF

TM 0

v

TF

TF

DF

LF

0 0

0 TF

DM

LM

Train Frame (Mary's)Track Frame (Frank's)

v

LM = DM = sDF

s = √1 − v2/c2where


x

y

zx '

y '

z '

v

Frank Mary



3. Length contraction

t ' = 0 x ' = sx

d = 1 /s

t = at ' + bx'

x = ct ' + dx'

Time DilationTime Dilation

L0

v

L

T' v

sL

0

v

sL

T'

v

Stick Frame Clock Frame

T ' = s(L /v) (time dilation)In the clock frame,

T = L /v

In the stick frame (at rest),

for clock to arrive at the right end.

0

0

TT


x

y

zx '

y '

z '

v

Frank Mary



4. Time dilation

x ' = 0 t ' = st

a = 1 /s

t = at ' + bx'

x = ct ' + dx'


x

y

zx '

y '

z '

v

Frank Mary



c /d = v

b /a = v /c2

d = 1 /s

a = 1 /s

Relative speed v:

Rear clock ahead:

Length contraction:

Time dilation:

t =t ' + vx ' /c2

s

x =vt ' + x'

s

t = at ' + bx'

x = ct ' + dx'


x

y

z

x '

y '

z '

v

Frank Mary

t ' = t

x ' = x − vt

t =t ' + vx ' /c2

√1 − v2 /c2

x =x ' + vt '

√1 − v2/c2

v≪ct ' =t − vx /c2

√1 − v2 /c2

x ' =x − vt

√1 − v2 /c2

When

recover the Galilean transformation.


x

y

zx '

y '

z '

v

Frank Mary

The complete transformation

t ' = γ(t − vx /c2)

x ' = γ(x − vt )

y ' = y

z ' = z

We went all the way from the postulates to the fundamental effects, and finally to the Lorentz transformation. One may also start from the postulates to derive the Lorentz transformation first, then to the relativistic effects. See, e.g., Classical Dynamics of Particles and Systems by S. T. Thornton and J. B. Marion.

1)

2)

3)

4)

γ =1

√1 − v2 /c2where

Example: Length ContractionExample: Length Contraction

● Consider a rod of length l lying along the x-axis of an inertial frame K. An observer in system K' moving with uniform speed v along the x-axis measures the length of the rod in the observer's own coordinate system by determining at a given instant of time t' the difference in the coordinates of the ends of the rod, x'(2) – x'(1). According to Lorentz transformation (Eq. 2)

where x(2) – x(1) = l. Because t'(2) = t'(1), Eq. 1 leads to

x '(2) − x '(1) =[x (2) − x (1)] − v [t (2) − t (1)]

√1 − v2/c2

t (2) − t (1) = [x (2) − x (1)]v /c2

Example: Length ContractionExample: Length Contraction

● The length l' as measured in the K' system is therefore

where

● Therefore, to the observer in K', objects in K appear contracted.

● Similarly, to a stationary observer in K, objects in K' also appear contracted.

● Put together, to an observer in motion relative to an object, the dimensions of objects are contracted by a factor s.

l ' = x '(2) − x '(1) = [x (2) − x (1)]√1 − v2/c2

= sl

s = √1 − v2/c2

The Invariant IntervalThe Invariant Interval

● Under Galilean transformation − the distance between two events is invariant, and− Newton's laws are invariant.

● Under Lorentz transformation− The interval between events is invariant, and − We will consider dynamics in the coming lecture.

(Δ s)2 = (Δ x)2

+ (Δ y )2

+ (Δ z)2

(Δ s)2 = c2(Δ t)2 − (Δ x)2

− (Δ y)2

− (Δ z)2

Proof of the Invariant IntervalProof of the Invariant Interval

● For simplicity we drop D and ignore y and z directions. According to the (inverse) Lorentz transformation

c2t2 − x2 =c2(t ' + vx '/c2)2

1 − v2 /c2−

(x ' + vt ')2

1 − v2 /c2

=t '2(c2 − v2

) − x '2(1 − v2/c2)

1 − v2 /c2

= c2t '2 − x '2

≡ s2

Timelike, Spacelike & LightlikeTimelike, Spacelike & Lightlike

● Timelike separation: s2 > 0− It is possible to find a frame in which the two events happen at

the same place. s/c is called the proper time. − It is possible for a particle to travel from one event to the other.

● Spacelike separation: s2 < 0− It is possible to find a frame in which the two events happen at

the same time. |s| is called the proper distance, or proper length.

● Lightlike separation: s2 = 0− In every frame a photon emitted at one of the events will arrive

at the other.

Ex: Time DilationEx: Time Dilation

● Let frame K' move at speed v with respect to frame K. Consider two events at the origin of K', separated by time t'. The separation between the events is

− in K': (x', t') = (0, t'),− in K: (x, t) = (vt, t).

● The invariant interval implies

c2t '2 − 0 = c2t2 − v2t2 t =t '

√1 − v2 /c2

Note the assumption x' = 0.

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Lecture 18: Lorentz Transformation - Zhejiang Universityzimp.zju.edu.cn/~xinwan/courses/physI17/handouts/lecture18.pdf · An observer in system K' moving with uniform speed v along