THE GEOMETRY OFSPECIAL RELATIVITY

Tevian Dray

Oregon State University

I: Circle Geometry

II: Hyperbola Geometry

III: Special Relativity(a) Basics

(b) Mechanics

(c) Electromagnetism

IV: What Next?

CIRCLE GEOMETRY

.1 θθ(cos , sin )

θ

rθ = arclength

4

53

θ

cos θ =4

5=⇒ tan θ =

3

4

WHICH GEOMETRY?

Euclidean

ds2 = dx2 + dy2

y

B

θ

θ

A

y’

x’

x

4

53

θ

tan θ = 3

4

B = (cos θ, sin θ)

A = (− sin θ, cos θ)

Trigonometry!

Return

MEASUREMENTS

Width:

θ

1

θ

1

1

cos θ

1

cos θ

Apparent width > 1

Slope:y

xφ

y’

x’

y

x

φθ

m 6= m1 + m2tan(θ + φ) = tan θ+tan φ1−tan θ tan φ = m1+m2

1−m1m2

HYPERBOLA GEOMETRY

.1

ββ(cosh , sinh )

β

x2 − y2 = 1

rβ = arclength

ds2 = |dx2 − dy2|

cosh β =1

2

(

eβ + e−β)

sinh β =1

2

(

eβ − e−β)

HYPERBOLIC TRIANGLE TRIG

βd cosh

βd sinhd

β

4

5

3

β

tanh β = 3/5

5

4

3

α tanh α = 4/5

RIGHT TRIANGLES

β

βB

y y’

A

x’

x

• • •

“right angles” are not angles!

WHICH GEOMETRY?

signature

(+ + ... +) Euclidean(− + ... +) Minkowskian

ds2 = −c2 dt2 + dx2

ct

A

B

ct’

x’

xβ

β4

5

3

β

tanh β = 3

5

B = (cosh β, sinh β)

Special Relativity!

Compare

SPECIAL RELATIVITY

ct

A

B

ct’

x’

xβ

β

v

c= tanh β

tanh(α + β) =tanh α + tanh β

1 + tanh α tanh β=

uc + v

c

1 + uvc2

Einstein addition formula!

Compare

LENGTH CONTRACTION

x’

ct’ct

x

β

ℓ

ℓ ′

•ℓ = ℓ ′ cosh β

x’

ct’ct

x

β

ℓ

ℓ ′

•

ℓ ′ =ℓ

cosh β

TIME DILATION

x’

ct’ct

x

COSMIC RAYS

(Taylor & Wheeler, 1st edition, Ex. 42, p. 89.)

Consider µ-mesons produced by the collision of cosmic rays with gas nuclei inthe atmosphere 60 kilometers above the surface of the earth, which then movevertically downward at nearly the speed of light. The half-life before µ-mesonsdecay into other particles is 1.5 microseconds (1.5 × 10−6 s).

• Assuming it doesn’t decay, how long would it take a µ-meson to reach thesurface of the earth?

60 km

3 × 108 ms

= 200 µs

• Assuming there were no time dilation, about what fraction of the mesonsreaches the earth?

200 µs3

2µs per half-life

=400

3half-lives

• In actual fact, roughly 1

8of the mesons reach the earth!

How fast are they going?

COSMIC RAYS

400

3half-lives

3 half-lives=

400

9

9400

α

v

c= tanh α =

√4002 − 92

400≈ .99974684

COSMIC RAYS

(60 km)(1000 mkm

)

(4.5 × 10−6 s)(3 × 108 ms )

=400

9

400

9

α

v

c= tanh α =

400√4002 + 92

≈ .99974697

TWIN PARADOX

One twin travels 24 light-years to star X at speed 24

25c; her twin brother stays

home. When the traveling twin gets to star X, she immediately turns around,

and returns at the same speed. How long does each twin think the trip took?

β

•24

25 7

cosh β =25

7

•

7

q

q =7

cosh β=

49

2549/25

7

24

25

β

Straight path takes longest!

CLASSICAL CONSERVATION LAWS

Momentum:∑

mivi =∑

m̂jv̂jLab Frame

∑

mi(v′i + v) =

∑

m̂j(v̂′j + v)Moving Frame

∑

miv′i =

∑

m̂jv̂′j ⇐⇒

∑

mi =∑

m̂j

Mass must be conserved!

Energy:

1

2

∑

miv2

i =1

2

∑

m̂jv̂2

jLab Frame

1

2

∑

mi(v′i + v)2 =

1

2

∑

m̂j(v̂′j + v)2Moving Frame

Automatically conserved!

RELATIVISTIC MOMENTUM

u =dx

dt= c tanh α

p = mdx

dτ= mc sinh α

α

dx

c dt c dτ

RELATIVISTIC CONSERVATION LAWS

Momentum: (p = mdxdτ = mc sinh α)∑

mic sinh αi =∑

m̂jc sinh α̂jLab Frame

∑

mic sinh(α′i + β) =

∑

m̂jc sinh(α̂′j + β)Moving Frame

sinh(α + β) = sinh α cosh β + cosh α sinh β =⇒∑

mic sinh α′i =

∑

m̂jc sinh α̂′j

⇐⇒∑

mic cosh α′i =

∑

m̂jc cosh α̂′j

⇐⇒∑

mic cosh αi =∑

m̂jc cosh α̂j

What is mc cosh α?

RELATIVISTIC CONSERVATION LAWS

mc cosh α = mc√

1 + sinh2 α

≈ mc

(

1 +1

2sinh2 α

)

Energy: E = mc2 cosh α ≈ mc2 +p2

2m

≈ mc2 +1

2mv2

Momentum conserved ⇐⇒ Energy conserved!

Mass need not be conserved!

RELATIVISTIC MOMENTUM

p = mdx

dτ= mc sinh α

E = mc2 cosh α = mc2dt

dτ

α

dx

c dt c dτ

(

Ecp

)

= m ddτ

(

ctx

)

α

p

Ec mc

ELECTROMAGNETISM

Capacitor at Rest: (charge density ±σ0)

~E0 =σ0

ǫ0̂

~B0 = ~0

Moving Capacitor:

σ = σ0 cosh α

~κ = σ ~u = σc tanh α ı̂~E =

σ

ǫ0̂ = K cosh α ̂

~B = −µ0~κ × ̂ = −1c K sinh α k̂

ELECTROMAGNETISM

Lab Frame:

~E = K cosh α ̂~B = −1

c K sinh α k̂

Moving Frame:

~E′ = K cosh(α + β) ̂

~B′ = −1

c K sinh(α + β) k̂

cosh(α + β) = cosh α cosh β + sinh α sinh β = Ey

K cosh β − cBz

K α sinh β

sinh(α + β) = sinh α cosh β + cosh α sinh β = −cBz

K cosh β + Ey

K sinh β

cosh(α + β) = cosh α cosh β + sinh α sinh β = Ey

K cosh β − cBz

K α sinh β

sinh(α + β) = sinh α cosh β + cosh α sinh β = −cBz

K cosh β + Ey

K sinh β

=⇒ E′y = Ey cosh β − cBz sinh βB′z = Bz cosh β − 1

c Ey sinh β

LORENTZ TRANSFORMATIONS

Λ =

cosh β − sinh β 0 0

− sinh β cosh β 0 0

0 0 1 0

0 0 0 1

v =

ct

x

y

z

F =

0 1

c Ex 1

c Ey 1

c Ez

−1

c Ex 0 Bz −By

−1

c Ey −Bz 0 Bx

−1

c Ez By −Bx 0

v′ = Λv

F′ = ΛFΛT

SUMMARY

• Lorentz transformations are hyperbolic rotations

• Beautifully treated in Taylor & Wheeler, 1st ed

• Removed from Taylor & Wheeler, 2nd edition

• Not currently covered in existing texts

http://www.physics.oregonstate.edu/coursewikis/GSR

http://www.math.oregonstate.edu/~tevian/geometry

WHICH GEOMETRY?

signature flat curved

(+ + ... +) Euclidean Riemannian(− + ... +) Minkowskian

ds2 = r2(dθ2 + sin2θ dφ2)

Tidal forces!

WHICH GEOMETRY?

signature flat curved

(+ + ... +) Euclidean Riemannian(− + ... +) Minkowskian Lorentzian

General Relativity!

ds2 = −dt2 + a(t) dx2

Cosmology!

(c = 1)ds2 = −

(

1 − 2mr

)

dt2 +dr2

(

1 − 2mr

)

+ r2(

dθ2 + sin2θ dφ2)

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THE GEOMETRY OFSPECIAL RELATIVITY

Tevian Dray

http://www.physics.oregonstate.edu/coursewikis/GSR

http://www.math.oregonstate.edu/~tevian/geometry