Special Relativity

Read P98 to 105

The principle of special relativity: The laws of nature look exactly the same for all observers in inertial reference frames, regardless of their state of relative velocity.

(Note: accelerated reference frames are not inertial since the Newtonʼs law of inertia are not satisfied.)

Position in ′S compared to position in S.′x = x − vt ′y = y ′t = t

Galilean Transformation (Rohlf P99)

v

u

xʼ

x

S Sʼ

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

Velocity

′ux =d ′xdt

= dxdt

− v ′uy =d ′ydt

= dydt

=uy ′uz =d ′zdt

= dzdt

= uz d ′t = dt

′ux =ux − v ′uy =uy ′uz =uz d ′t = dt

Acceleration

′ax =d ′uxdt

= duxdt

= ax ′ay =d ′uydt

=duydt

= ay ′az =d ′uzdt

=duzdt

= az

′ax = ax ′ay = ay ′az = az

Galilean Transformation

Newtonʼs second law is invariant under Galilean Transformation

F = ma ⇒ Fi = mai

ai → ′ai = ai

∴Fi = mai → Fi = mai′ = Fi = mai

Perform Galilean Transformation

Invariance of Maxwellʼs Equations (Rohf p104-105)

∇2 E − µ0ε0∂2E

∂t 2= 0. ⇒ 3 equations:∇2Ei − µ0ε0

∂2Ei∂t 2

= 0.

∂2Ei∂x2

+∂2Ei∂y2

+∂2Ei∂z2

− µ0ε0∂2Ei∂t 2

= 0

∂∂x

∂Ei∂x

+ ∂∂y

∂Ei∂y

+ ∂∂z

∂Ei∂z

− µ0ε0∂∂t

∂Ei∂t

= 0

The electromagnetic wave equation is not invariant under Galilean Transformation?

The wave equation for an electric field in a vacuum.

In Class Exercise Show that this is not invariant with respect to a Galilean transformation.

∂2Ei∂ ′x 2 − µ0ε0

∂2Ei∂ ′t 2 ≠

∂2Ei∂x2

− µ0ε0∂2Ei∂t 2

≠ 0

Orient the axes so that the wave travels along the x axis

Then show:

Evaluate the first derivatives using general chain rule.

First derivatives: space dependent part.

∂Ei∂x

= ∂Ei∂ ′x

∂ ′x∂x

+ ∂Ei∂ ′t

∂ ′t∂x

′x = x− vt ′t = t

Show: ∂Ei∂x

= ∂Ei∂ ′x

Evaluate the first derivatives using general chain rule.

First derivatives: space dependent part.

∂Ei∂x

= ∂Ei∂ ′x

∂ ′x∂x

+ ∂Ei∂ ′t

∂ ′t∂x

′x = x− vt ′t = t

∂ ′x∂x

= 1 ∂ ′t∂x

= 0

∂Ei∂x

= ∂Ei∂ ′x

′x = x− vt ′t = t 

Show:        ∂Ei∂t

= ∂Ei∂ ′t

− v ∂Ei∂ ′x

First derivatives: time dependent part.

′x = x− vt ′t = t ∂Eidt = ∂Ei

∂ ′x∂ ′x∂t + ∂Ei

∂ ′t∂ ′t∂t

∂ ′x∂t = −v ∂ ′t

∂t = 1,

∂Ei∂t = ∂Ei

∂ ′t− v ∂Ei

∂ ′x

Finally,

∂Ei∂x = ∂Ei

∂ ′x          ∂Ei

∂t = ∂Ei∂ ′t

− v ∂Ei∂ ′x

First derivatives: time dependent part.

∂Ei

∂x=∂Ei

∂ ′x               

∂Ei

∂t=∂Ei

∂ ′t− v

∂Ei

∂ ′x

Second derivatives: space dependent part.

Summary of first derivatives

Show: ∂2Ei∂x2 = ∂2Ei

∂ ′x 2

∂Ei∂x = ∂Ei

∂ ′x               ∂Ei

∂t = ∂Ei∂ ′t

− v ∂Ei∂ ′x

Second derivatives: space dependent part.

Summary of first derivatives

∂∂x

∂Ei

∂x= ∂2Ei

∂ ′x 2∂ ′x∂x

+ ∂2Ei

∂ ′t ∂ ′x∂ ′t∂x

∂2Ei

∂x2= ∂2Ei

∂ ′x 2 (1)+∂2Ei

∂ ′t ∂ ′x(0) = ∂2Ei

∂ ′x 2

∂2Ei

∂x2= ∂2Ei

∂ ′x 2

Show: ∂2Ei

∂t 2 =∂2Ei

∂ ′t 2 + v2 ∂2Ei

∂ ′x 2 − 2v∂2Ei

∂ ′x ∂ ′t

Second derivative: time dependent part.

∂∂tf = ∂f

∂ ′x∂ ′x∂t

+ ∂f∂ ′t

∂ ′t∂t

∂∂t

∂Ei

∂t= ∂∂ ′x

∂Ei

∂t⎛⎝⎜

⎞⎠⎟∂ ′x∂t

+ ∂∂ ′t

∂Ei

∂t⎛⎝⎜

⎞⎠⎟∂ ′t∂t

substitute into above∂Ei

∂t=∂Ei

∂ ′t− v

∂Ei

∂ ′x, ∂ ′t

∂t= 1

Second derivative: time dependent part.

∂∂t

∂Ei

∂t= ∂∂ ′x

∂Ei

∂t⎛⎝⎜

⎞⎠⎟∂ ′x∂t

+ ∂∂ ′t

∂Ei

∂t⎛⎝⎜

⎞⎠⎟∂ ′t∂t

substitute into above∂Ei

∂t=∂Ei

∂ ′t− v

∂Ei

∂ ′x, ∂ ′t

∂t= 1

∂∂t

∂Ei

∂t= ∂∂ ′x

∂Ei

∂ ′t− v

∂Ei

∂ ′x⎛⎝⎜

⎞⎠⎟

(−v) + ∂∂ ′t

∂Ei

∂ ′t− v

∂Ei

∂ ′x⎛⎝⎜

⎞⎠⎟

(1)

∂2Ei

∂t 2 =∂2Ei

∂ ′x ∂ ′t− v

∂2Ei

∂ ′x 2

⎛⎝⎜

⎞⎠⎟

(−v) +∂2Ei

∂ ′t 2 − v∂2Eii

∂ ′t ∂ ′x⎛⎝⎜

⎞⎠⎟

Second derivative: time dependent part.

Second derivative: time dependent part.

∂∂tf = ∂f

∂ ′x∂ ′x∂t

+ ∂f∂ ′t

∂ ′t∂t

f =∂Ei

∂t⎛⎝⎜

⎞⎠⎟

∂∂t

∂Ei

∂t= ∂∂ ′x

∂Ei

∂t⎛⎝⎜

⎞⎠⎟∂ ′x∂t

+ ∂∂ ′t

∂Ei

∂t⎛⎝⎜

⎞⎠⎟∂ ′t∂t

substitute into above∂Ei

∂t=∂Ei

∂ ′t− v

∂Ei

∂ ′x, ∂ ′t

∂t= 1

∂∂t

∂Ei

∂t= ∂∂ ′x

∂Ei

∂ ′t− v

∂Ei

∂ ′x⎛⎝⎜

⎞⎠⎟

(−v) + ∂∂ ′t

∂Ei

∂ ′t− v

∂Ei

∂ ′x⎛⎝⎜

⎞⎠⎟

(1)

∂2Ei

∂t 2 =∂2Ei

∂ ′x ∂ ′t− v

∂2Ei

∂ ′x 2

⎛⎝⎜

⎞⎠⎟

(−v) +∂2Ei

∂ ′t 2 − v∂2Eii

∂ ′t ∂ ′x⎛⎝⎜

⎞⎠⎟

∂2Ei

∂t 2 =∂2Ei

∂ ′t 2 + v2 ∂2Ei

∂ ′x 2 − 2v∂2Ei

∂ ′x ∂ ′t

∂2Ei

∂x2− µ0ε0

∂2Ei

∂t 2= 0

∂2Ei

∂x2=∂2Ei

∂ ′x 2∂2Ei

∂t 2=∂2Ei

∂ ′t 2 − v2∂2Ei

∂ ′x 2 − 2v ∂2Ei

∂ ′x ∂ ′t

∂2Ei

∂ ′x 2 − µ0ε0∂2Ei

∂ ′t 2 + v2∂2Ei

∂ ′x 2 − 2v ∂2Ei

∂ ′x ∂ ′t⎛⎝⎜

⎞⎠⎟= 0

The Full Equation:

Maxwellʼs equation for an electromagnetic wave is not invariant under a Galilean Transformation. The problem is with the time dependence.

Either Maxwellʼs equations are not laws of physics or Galilean transformations are not the correct transformations for which the laws of physics must be invariant.

Luminiferous Ether Rohlf P100-101

•  Medium of propagation-luminiferous ether

•  Ether may have mechanical properties which may be inferred from the velocity of light relative to the etherʼs rest frame.

Velocity of light is very large and difficult to measure Doppler effect. Interference effects on phases change with respect to motion relative to the medium.

Albert Michelson

First American Nobel Prize -1907

Michelson-Morley experiment Rohlf P100-101

Measured differences in detected speed of two rays of light coming from the same source, as measured by an observer traveling on the earth moving toward the source of light relative to the ether, the other perpendicular to the source of light, which is not moving relative to the ether.

Found no difference, which contradicts expectations of lumeniferous ether

ve

taca = L′cac

+ L′cca

= Lc + ve

+ Lc − ve

=

= Lc 1+

vec + ve

2

c2+ ⋅ ⋅ ⋅1−

vec + ve

2

c2⎛

⎝⎜⎞

⎠⎟≈ 2Lc 1+

ve2

c2⎛⎝⎜

⎞⎠⎟

taba = L′cab

+ L′cba

= Lc2 − ve2

+ Lc2 − ve2

= 2Lc1

1− ve2 / c2

= 2Lc 1−ve2

2c2+ ⋅ ⋅ ⋅

⎛

⎝⎜⎞

⎠⎟≈ 2Lc 1−

ve2

2c2⎛⎝⎜

⎞⎠⎟

Δt = taca − taba =Lve2

c3

Δt = Lve2

c3 ΔX = cΔt = Lve2

c2

Δφ = ΔXλ

= Lve2

λc2

Rotate apparatus clockwise 90°

Δφ = − ΔXλ

= − Lve2

λc2 Total Δφ = 2Lve2

λc2

L = 10m ve = 4 ×104 m/s λ = 500 nm = 5 ×10−7  m⇒ Δφ = 0.4

ve

Expect shift in interference pattern by 0.4 fringes. Actually, no shift observed.

Lorentz (1892) and independently Fitzgerald:

Bodies moving in the luminiferous ether are shorter in the direction of motion due to their interactions with the ether.

Lx = γ (Lx0 − vt) Ly = L0y , Lz = L0z

β ≡ vc

γ ≡ 11− β 2

Larmor (1900)

Considered also time, and wrote down the “Lorentz Transformation”

′x = γ (x − vt) ′y = y, ′z = z ′t = γ (t − vc2x)

Lorentz (1904)

Maxwellʼs equations in vacuum are invariant under the Lorentz transformation. But, he didnʼt get the charge transformation right. The two systems are not exactly equivalent.

Poincare (1906) was the first to coin the phrase “Lorentz Transformation”

Precursers to Einstein

What did Einstein do?

.

.

.

The most notable result of the the principle of special relativity is that time intervals are not the same from one observer to another. That is, the passage of time is relative.

A simple example of Einsteinʼs postulate about the speed of light.

D

d s

L = vΔ ′t

′D

′Sv

d s

D

S

Δt = 2Dc

′D

Δ ′t =2 ′D

c

Since ′D > D Δ ′t >Δt

Show: Δ ′t = Δt1− β 2

= γΔt

Δ ′t =2 L2

4+ D2

c= L2 + 4D2

c= v2Δ ′t 2 + 4D2

c

c2Δ ′t 2 = v2Δ ′t 2 + 4D2 , Δ ′t 2c2 (1− v2

c2) = 4D2 , Δt '2 (1− v

2

c2) = 4D

2

c2= Δt 2

Δ ′t = Δt

1− v2

c2

, γ ≡ 11− β 2

  ,      β ≡ vc, Δ ′t = Δt

1− β 2= γΔt

D

D2 + (L / 2)2

d s

L = vΔ ′t

D2 + (L / 2)2

′Sv

d s

D

S

Δt = 2Dc

Textbook, Foundations of Special Relativity: P130, Problems 2, 3

Homework 4 DueTues. Sept. 15.

Derivation of the relativistic transformations.

not in Rohlf

optional material in lecture notes

′x = x − vt ′y = y, ′z = z ′t = t

′x = Ax + Bt ′y = y, ′z = z ′t = Cx + Dt

Galilean Transformation

More General Linear Transformation

Find A,B,C,D

′x = 0x = vt

S Sʼ ′x = Ax + Bt0 = Ax + Btx = vt0 = Avt + BtB = −Av

Case I: S is at rest, S˙moves with velocity v

′x = −v ′tx = 0

Sʼ

′t = Cx + Dt′t = 0 + Dtv ′t = vDt, v ′t = −Bt,∴B = −vDFromCase I B = −vA∴D = A

   ′x = Ax + Bt−v ′t = 0  + Btv ′t = −Bt

Case II: S˙ is at rest, S moves with velocity -v

So far: B = −Av

Case II x=0 Viewed from Sʼ

S

Case I ′x =0Viewed fromS

′x = 0x = vt

S Sʼ

′x = −v ′tx = 0

Sʼ

   ′x = Ax + Bt−v ′t = 0  + Btv ′t = −Bt

Have B and D in terms of A Need to get C in terms of A

′x = Ax + Bt0 = Ax + Btx = vt0 = Avt + BtB = −Av

′t = Cx + Dt′t = 0 + Dtv ′t = vDt, v ′t = −Bt,∴B = −vDFromCase I B = −vA∴D = A

Case I Case II

Case III: light viewed fromS and ′S

′x =Ax+ Bt′x = c ′t , x= ct

1) c ′t =Act + Bt

′t =Cx+Dt′t =Cct +Dt

2) c ′t =Cc2t +Dct w

B = −vA D = A C = − vc2A

′x = A(x − vt), ′t = A(t − vc2x)

In summary, we have

S Sʼ

′x = c ′tx = ct

Equate1)and2) using D = Aand B= -AvAct + Bt =Cc2t +DctAct − vAt =Cc2t + cAtAc− vA=Cc2 + cA−vA=Cc2

C =− vc2 A

B = −vA D = A C = − vc2A

′x = A(x − vt), ′t = A(t − vc2x)

In summary, we have

B = −vA D = A C = − vc2A

′x = A(x − vt) ′t = A(t − vc2x)

In summary, we have

′x = Ax + Bt ′y = y, ′z = z ′t = Cx + Dt

Need to get A

The inverse transformation must be obtained by letting v-v

x = A( ′x + v ′t )        t = A( ′t + vc2

′x ) eq. 2

Solve for x and t in terms of xʼ and t’ !

x = 1

A 1- v2

c2⎛⎝⎜

⎞⎠⎟

′x +v ′t( ) t = 1

A 1- v2

c2⎛⎝⎜

⎞⎠⎟

′t + vc2

′x⎛⎝⎜

⎞⎠⎟

eq.1

Equate the coefficients of eq. 1) and eq. 2) to obtain

1

A 1− v2

c2⎛⎝⎜

⎞⎠⎟

= A,     A = 1

1− v2

c2

= 11− β2

= γ

′x = A(x − vt), ′t = A(t − vc2x) Need to obtain A

Lorentz Transformation

′x = γ (x − vt) ′y = y ′z = z ′t = γ t − vc2x⎛

⎝⎜⎞⎠⎟

x = γ ( ′x + v ′t ) y = ′y z = ′z t = γ ′t + vc2

′x⎛⎝⎜

⎞⎠⎟

′x = x − vt ′y = y ′z = z ′t = t

x = ′x + v ′t y = ′y z = ′z t = ′t

Galilean Transformation

Homework 5. Due. Tues. Sept 18.

Textbook: Space-time. Lorentz Trans: P130-131, Problems 6,9,11,12

Show that Maxwellʼs wave equation is invariant under a Lorentz transformation along the x axis for a wave moving in the x direction .

Use: ∂Ei

∂x=∂Ei

∂ ′x∂ ′x∂x

+∂Ei

∂ ′t∂ ′t∂x        

∂Ei

∂t=∂Ei

∂ ′x∂ ′x∂t

+∂Ei

∂ ′t∂ ′t∂t

′x = γ (x − vt)       ′t = γ t − vc2x⎛

⎝⎜⎞⎠⎟

∂ ′x∂x

= γ ∂ ′t∂x

= −γ vc2

∂ ′x∂t

= −γ v ∂ ′t∂t

= γ

Some Consequences of Special Relativity

Exercise

An event occurs in coordinate system S at x=0, and t=0. A second event occurs at x=0 and time t=2 sec.

An observer in Sʼ moving with velocity β = 0.9 relative to x synchronizes her clock so that she also sees the first event at xʼ=0 and tʼ=0.

1.  What will be the time tʼ for the second event, as measured in Sʼ?

2.  What will be position of the second event as seen in Sʼ?

An event occurs in coordinate system S at x=0, and t=0. A second event occurs at x=0 and time t=2 sec.

An observer in Sʼ moving with velocity β = 0.9 relative to x synchronizes her clock so that she also sees the first event at xʼ=0 and tʼ=0.

What will be the time tʼ for the second event, as measured in Sʼ?

′t = γ t − vc2x⎛

⎝⎜⎞⎠⎟

γ = 11− β 2

= 11− 0.92

= 2.3

′t = 2.3 2.0 − 0.9cc2

× 0⎛⎝⎜

⎞⎠⎟= 4.6 sec

What will be position of the second event as seen in Sʼ?

′x = γ (x − vt) = γ (0 − βct) = −2.3(0.9)(3×108 )(2.0) = −12.4 ×108 m = −12.4 ×105 km

Note: Distance from earth to moon 4 ×105 km

Exercise

Velocity Transformation

′x = γ (x − vt) ′y = y ′z = z ′t = γ t − vc2x⎛

⎝⎜⎞⎠⎟

d ′x = γ (dx − vdt) d ′y = dy d ′z = dz d ′t = γ dt − vc2dx⎛

⎝⎜⎞⎠⎟

′ux =d ′xd ′t

= γ (dx − vdt)

γ dt − vc2dx⎛

⎝⎜⎞⎠⎟

=

dxdt − v

⎛⎝⎜

⎞⎠⎟

1− vc2dxdt

⎛⎝⎜

⎞⎠⎟

′ux =(ux − v)

1− vc2ux

⎛⎝⎜

⎞⎠⎟

′uy =d ′yd ′t

=uy

γ 1− vc2ux

⎛⎝⎜

⎞⎠⎟

′uz =uz

γ 1− vc2ux

⎛⎝⎜

⎞⎠⎟

Exercise:

Sʼ v

S

u

u = ux = 0.9cv = −0.9c

Find ′ux

u = ux = cv = −0.9c Find ′ux

Sʼ v

S

u

u = ux = 0.9cv = −0.9c

′ux =ux − v

1− vc2ux

⎛⎝⎜

⎞⎠⎟

= 0.9c − (−0.9c)

1− −0.9cc2

0.9c⎛⎝⎜

⎞⎠⎟

=1.8c1+ 0.81( ) =

1.8c1.81

= 0.99c

u = ux = cv = −0.9c

′ux =ux − v

1− vc2ux

⎛⎝⎜

⎞⎠⎟

= c − (−0.9c)

1− −0.9cc2

c⎛⎝⎜

⎞⎠⎟

=1.9c1+ 0.9( ) =

1.9c1.9 = c

Solution:

Simultaneity

Sʼ β

S

1 2

Two events happen simultaneously in S.

t2 = t1 ′t1 = γ t1 −vc2x1

⎛

⎝⎜⎞

⎠⎟                 ′t2 = γ t2 −

vc2x2

⎛

⎝⎜⎞

⎠⎟

′t2 − ′t1 = γ t2 − t1( )  −γ vc2

x2 − x1( ) Δ ′t = γΔt   −γ vc2

Δx

Δ ′t = γ 0  −γ vc2

Δx = −γ vc2

Δx

Exercise

Two events happen simultaneously at x1=0 m and x2=1000 m Suppose Sʼ is moving to the right at 0.9 c. Which event occurs first an by how much?

Event 1 occurs at t=0 µs, and event 2 occurs at t=2 µs. When does event 2 occur in Sʼ?

Solution

Two events happen simultaneously at x1=0 m and x2=1000 m Suppose Sʼ is moving to the right at 0.9 c. Which event occurs first an by how much?

Δ ′t = γΔt   −γ vc2

Δx = −2.30.9cc2

1,000 = −6.9 ×10−6 s = − 6.9 µs

Event 1 occurs at t=0 µs, and event 2 occurs at t=2 µs. When does event 2 occur in Sʼ?

Δ ′t = γ Δt  − vc2

Δx⎛⎝⎜

⎞⎠⎟=2.3 2×10−6 − 0.9c

c21,000⎛

⎝⎜⎞⎠⎟= 2.3 2×10−6 − 3×10−6( ) = −1µs

The order of the events is reversed. Does this violate causality?

No. Events can be causally connected only if they can communicate. The fastest communication is at the speed of light.

For event 2 to be influenced by event 1, a signal from event 1 must travel at a speed at least

It would require a signal between the 2 events to travel faster than the speed of light.

ΔxΔt

= 10002 ×10−6 = 0.5 ×109 m/s≈1.5c

Δ ′t = γ Δt   − vc2

Δx⎛⎝

⎞⎠ > 0 Δt   > v

c2Δx (timelike) Δt   < v

c2Δx (spacelike)

x=-ct x=ct

x

ct

light speed world line

Space - Time

Since space and time are not independent, define a new 4-dimensional space-time, with coordinates:

x0 ≡ ct,   x1 ≡ x,    x2 ≡ y,    x3 ≡ z,           or simply xµ

This is called Minkowski space-time. The evolution of a body is a continuum of points which form a curve, called a world line, in space time.

For 1-space and time dimension:

timelike world line

Minkowski Space-Time: 2 spatial and 1 time dimensions.

World line