THE GEOMETRY OFSPECIAL RELATIVITY

Tevian Dray

Oregon State University

I: Circle Geometry

II: Hyperbola Geometry

III: Special Relativity

IV: What Next?

CIRCLE GEOMETRY

.1 (cos , sin )

r = arclength

4

53

cos =4

5= tan =

3

4

WHICH GEOMETRY?

Euclidean

ds2 = dx2 + dy2

y

B

A

y

x

x

4

53

tan = 34

B = (cos , sin )

A = ( sin , cos )

Trigonometry!

Return

MEASUREMENTS

Width:

1

1

1

cos

1

cos

Apparent width > 1

Slope:y

x

y

x

y

x

m 6= m1 + m2

HYPERBOLA GEOMETRY

.1

(cosh , sinh )

x2 y2 = 1

r = arclength

ds2 = |dx2 dy2|

cosh =1

2

(

e + e)

sinh =1

2

(

e e)

HYPERBOLIC TRIANGLE TRIG

d cosh

d sinhd

4

5

3

tanh = 3/5

5

4

3

tanh = 4/5

RIGHT TRIANGLES

B

y y

A

x

x

right angles are not angles!

WHICH GEOMETRY?

signature

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

ds2 = c2 dt2 + dx2ct

A

B

ct

x

x

4

5

3

tanh = 35

B = (cosh , sinh )

Special Relativity!

Compare

SPECIAL RELATIVITY

ct

A

B

ct

x

x

v

c= tanh

tanh( + ) =tanh + tanh

1 + tanh tanh=

uc +

vc

1 + uvc2

Einstein addition formula!

Compare

LENGTH CONTRACTION

x

ctct

x

= cosh

x

ctct

x

=

cosh

TIME DILATION

x

ctct

x

COSMIC RAYS

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

Consider -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 -mesonsdecay into other particles is 1.5 microseconds (1.5 106 s). Assuming it doesnt decay, how long would it take a -meson to reach the

surface of the earth?60 km

3 108 ms= 200 s

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

200 s3

2s per half-life

=400

3half-lives

In actual fact, roughly 18

of the mesons would reach the earth! How fast arethey going?

COSMIC RAYS

400

3half-lives

3 half-lives=

400

9

9400

v

c= tanh =

4002 92

400 .99974684

COSMIC RAYS

(60 km)(1000 mkm

)

(4.5 106 s)(3 108 ms )=

400

9

400

9

v

c= tanh =

4004002 + 92

.99974697

TWIN PARADOX

One twin travels 24 light-years to star X at speed 2425c; 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?

24

25 7

cosh =25

7

7

q

q =7

cosh =

49

2549/25

7

24

25

Straight path takes longest!

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.math.oregonstate.edu/~tevian/geometry

WHICH GEOMETRY?

signature flat curved

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

ds2 = r2(d2 + sin2 d2)

Tidal forces!

WHICH GEOMETRY?

signature flat curved

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

General Relativity!

ds2 = dt2 + a(t) dx2Cosmology!

(c = 1)ds2 =

(

1 2mr)

dt2 +dr2

(

1 2mr)

+ r2(

d2 + sin2 d2)

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THE GEOMETRY OFSPECIAL RELATIVITY

Tevian Dray

Oregon State University

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