7.consequences of the special theory of...

PHGN300/310: Consequences of the special theory of relativityFred Sarazin ([email protected])Physics Department, Colorado School of Mines

Consequences of the special theory of relativity


Question: you are “riding” a light beam, while another light beam comes towards you. What speed do you measure for the approaching beam ?

1. v = 02. v = c3. v = -c4. v = 2c5 v = -2c6. I cannot ride a beam of light (although it would be fun)

Relativistic velocity transformation


• Two reference frames K and K’, with K’ moving away from K in the x positive direction at constant velocity v. From the Lorentz transformation equations, we get:

Velocity addition with Lorentz transformation (K’ à K)

𝑑𝑥 = 𝛾 𝑑𝑥% + 𝑣𝑑𝑡%

𝑑𝑦 = 𝑑𝑦%

𝑑𝑧 = 𝑑𝑧%

𝑑𝑡 = 𝛾 𝑑𝑡% +𝛽𝑑𝑥%

𝑐

• Defining velocities like 𝑢. =/./0

,

𝑢1 =/1/0

… and 𝑢.% =/.%/0% , 𝑢1% =

/1%/0%

…, we get*:

𝑢. =𝑑𝑥𝑑𝑡 =

𝛾 𝑑𝑥% + 𝑣𝑑𝑡′

𝛾 𝑑𝑡% + 𝛽𝑑𝑥%

𝑐=

𝑢.% + 𝑣

1 + 𝛽𝑢.%𝑐

𝑢6 =𝑑𝑧𝑑𝑡 =

𝑢6%

𝛾 1 + 𝛽𝑢.%𝑐

𝑢1 =𝑑𝑦𝑑𝑡 =

𝑢1%

𝛾 1 + 𝛽𝑢.%𝑐

(*see derivation)Note: uy ≠ uy’ and uz ≠ uz’ because dt ≠ dt’ !


• K is moving away from K’ with a velocity –v, hence:

Velocity addition with Lorentz transformation (K à K’)

𝑢.% =𝑑𝑥′𝑑𝑡′ =

𝛾 𝑑𝑥 − 𝑣𝑑𝑡

𝛾 𝑑𝑡 − 𝛽𝑑𝑡𝑐=

𝑢. − 𝑣

1 − 𝛽𝑢.𝑐

𝑢6% =𝑑𝑧′𝑑𝑡′ =

𝑢6

𝛾 1 − 𝛽𝑢.𝑐

𝑢1% =𝑑𝑦′𝑑𝑡′ =

𝑢1

𝛾 1 − 𝛽𝑢.𝑐


So…

Question: you are “riding” a light beam, while another light beam comes towards you. What speed do you measure for the approaching beam ?

YOU

v=c

At rest in (K’)ux=-c

We know the relative velocity between K’ and K (v) and the velocity ux in K (for example), so we want to find ux’.

𝑢.% =𝑢. − 𝑣

1 − 𝛽𝑢.𝑐=−𝑐 − (𝑐)

1 − 𝑐(−𝑐)𝑐:=−2𝑐2 = −𝑐ANSWER:


The space station is threatened by an asteroid!A space ship is sent to intercept the asteroid. It is moving away from the space station at a speed of 0.8c. It fires a proton beam parallel to its direction of motion towards the asteroid. The beam moves at a speed of 0.6c relative to the ship. What is the speed of the missile for the observer on the space station ?

The spaceship is moving at 0.8c with respect to the space station: v = 0.8c. The missile is moving at 0.6c with respect to the spaceship: ux’=0.6c. One needs the speed of the missile observed by an observer in (K): ux

Exercise: terror in space!

𝑢. =𝑑𝑥𝑑𝑡 =

𝛾 𝑑𝑥% + 𝑣𝑑𝑡′

𝛾 𝑑𝑡% + 𝛽𝑑𝑥%

𝑐=

𝑢.% + 𝑣

1 + 𝛽𝑢.%𝑐=

0.6𝑐 + 0.8𝑐

1 + 0.8𝑐 0.6𝑐𝑐:

= 0.946𝑐ANSWER:


The asteroid is approaching the space station at 0.2c, while the proton beam is racing towards it at a velocity of <answer from the previous slide>. If there is a life form on the asteroid, at which speed would it measure the velocity of the proton beam ? And so on, and so on…

Exercise: terror in space (II)!


E: EmitterR: Receiver

E R

L0

• Flash of light emitted by E and received by R

• For the observer attached to the system: Δ𝑡C =:DEF

• The system is moving at a velocity v with respect to the laboratory frame. The outside observer sees:

E R E R E R

LL

vDt

v v v

Relativity of time


• For the outside observer:

– The system has moved 𝑑 = 𝑣Δ𝑡, while the light has traveled Δ𝑡 = :DF

– Pythagoras theorem: 𝐿 = 𝐿C: +H∆0:

:�

• With 𝐿C =F∆0E:

and a bit of calculation*:

Dt0: Proper time, e.g. time measured by the clock attached to the systemDt >Dt0: The clock attached to the system appears to run slower from the outside

observer point-of-view.

Time dilation

∆𝑡 = 𝛾Δ𝑡C TIME DILATION

(*see derivation)


A simplified version of an experiment, which actually took place to measure the prediction of the special theory of relativity: a cesium atomic clock (a highly precise timing device) is flown around the earth at 1000 km/h. How much time does it lose relative to a similar clock that remains at the origin ? (Circ. Earth = 40,000 km) [SIMPLIFICATION: fixed earth frame: non-rotating, non-orbiting, flight at constant speed (acceleration and deceleration neglected)]

v = 1,000 km/h = 106m/h = 278 m/sT = 40,000/1,000 = 40h = 40 x 3600 s

Taylor expansion:g = 1 + 0.5(v/c)2 = 1 + 0.5 (278 / 3.108)2 = 1 + 4.3×10-13

Each tick of the moving clock is a bit longer than the tick of the clock at rest (by4.3×10-13s every 1s). After 40h, the moving clock lost: 4.3×10-13×40×3600 = 62ns

Example


E: EmitterR: Receiver

ER

L0

ER

vE

R

ER

v

L+vDt1L-vDt2

Relativity of length

Reference frame of the instrument.

Outside (laboratory) reference frame


• For the outside observer:– It takes Dt1 for the light to do the first part of the trip and Dt2 for the second

part.cDt1 = L+vDt1 Dt1 = L/(c-v)cDt2 = L-vDt2 Dt2 = L/(c+v)

– After considering time dilation: ∆𝑡 = 𝛾Δ𝑡C, we get*:

L0: rest length or proper length ; objects appear contracted in the direction of motion

NOTE: the length contraction is experienced in both reference frames! An Earth observer may see a spaceship contracted in the direction of motion, but a person on board the spaceship will also see the Earth contracted in the direction of motion!

Length contraction

∆𝑡 = ∆𝑡K + ∆𝑡: = 2𝐿𝑐

1

1 − 𝑣:

𝑐:

𝐿 = 𝛾LK𝐿C LENGTH CONTRACTION

(*see derivation)


What length will observers measure for a spaceship moving relative to them at 0.99c if its occupants determine the length of the ship to be 50m ?

Exercise

𝐿 = 𝛾LK𝐿C = 1 −0.99𝑐 :

𝑐:�

×50 = 7.1𝑚ANSWER:


• The two descriptions should be equivalent [see Muon decay]

• Both observers see the time dilation and length contraction affecting the other observer [see Twin Paradox]

• In both cases, if v<<c à L»L0 and Dt »Dt0 (we are back to classical mechanics)

A word on time dilation and length contraction


• Muons originates from cosmic-rays

• Experiment: 1000 muons traveling at 0.98c detected at 2000m, ~540 muons “survived” (did not decay) at see level

• Radioactive decay law: 𝑁 𝑡 = 𝑁C𝑒L STUVW/U

0,

with 𝑇K/: = 1.52𝜇𝑠, the half-life of the muon.

• It takes :CCCC.\]F

= 6.8×10L^𝑠(6.8𝜇𝑠) for the muons to travel 2000m. Hence, according to classical mechanics (and the radioactive decay law), only 𝑁 6.8𝜇𝑠 ~45 muons should be left at see level!

• What is the problem?

Muon (µ) decay


• For the observer on earth, time appears dilated ∆𝑡 = 𝛾Δ𝑡C , the muons appear to live longer than their rest (or proper) half-life 𝑇K/:: 𝑇 ab0c = 7.64𝜇𝑠 (see calculation).

• Muons surviving at sea level:

Time dilation!

𝑁 𝑡 = 7.64𝜇𝑠 = 1000𝑒Lde:f.^g ^.] ~540!


• One should be able to get the same results by solving the problem with length contraction.

• You’re a muon! You are to travel at 0.98c a distance of L0=2000m measured by an earth observer, how much distance will you measure according to your own meter stick?

• The muons travel the distance in: ∆𝑡 = i\]C.\]F

= 1.35×10L^𝑠(1.35𝜇𝑠). Lets now apply the radioactive decay law (in the frame of the muon),

Note: Don’t forget to use 𝑇K/: = 1.52𝜇𝑠, the half-life of the muon at rest, since we are in the referential of the muon!

Length contraction!

𝐿 = 𝛾LK𝐿C = 1 − 0.98𝑐 :� ×2000 = 398𝑚LENGTH CONTRACTION

𝑁 𝑡 = 1.35𝜇𝑠 = 1000𝑒Lde:K.k: K.ik ~540! SAME RESULT!


• Two twins: one travels, one stays at home. The traveler ages slower than its twin, who stayed on earth. But, relatively speaking, who is moving ?

• The paradox: both are moving with respect to the other, therefore they should age the same way !

• Answer: The one that is really “moving” (e.g. feeling the effect of time dilation) is the one accelerating and decelerating. [See textbook]

The twin paradox


• Determine the decrease in length of a space shuttle (L0=50m) traveling 17,000 mph relative to the earth observers.

v = 17000 mi/h = (17000 x 1.61 x 103) / 3600 = 7.6 x 103 m/s

Length contraction: L = g-1 L0 = L0 √( 1 - (7.6 x 103 )2/(3.0 x 108)2 )= L0 √( 1 - 6.4 x 10-10 )

Contraction = L0 – L = 1.6 x 10-8 m !!!!

Exercise


• Nothing can travel faster than the speed of light. Therefore, there are regions of space and time that cannot be reached (CAUSALITY).

PHGN310: space – time diagrams


Lightning at position x3:(Left) x1,x2,x3 stationary(Right) x1,x2,x3 moving

PHGN310: simultaneity


MavisA BStanleyA Bx

ct

cc

Stan

ley

(in h

is

own

ref f

ram

e)

x

ct

cc

v

• Mavis: The two events “lightning strike hits (A’)” and “lightning strike hits (B’)” are NOT simultaneous.

• Stanley: The two events “lightning strike hits (A)” and “lightning strike hits (B)” are simultaneous.

PHGN310: the ”Stanley / Mavis” problem (I)

Mav

is (i

n St

anle

y’s

ref f

ram

e)


StanleyA Bx

ct

cc

Stan

ley

(in h

is

own

ref f

ram

e)

• Mavis: The two events “lightning strike hits (A’)” and “lightning strike hits (B’)” are NOT simultaneous.

• Stanley: The two events “lightning strike hits (A)” and “lightning strike hits (B)” are simultaneous.

PHGN310: the ”Stanley / Mavis” problem (II)

Mav

is (i

n he

r ow

n re

f fra

me)

MavisA’ B’x’

ct’

cc


• Length contraction / Time dilation– Length and time are not the same in two inertial frames moving with

respect to one another at a sizeable fraction of c

• Is there an invariant quantity that measures the same in both reference frame? Yes.

Ds2 = (Dx2+Dy2+Dz2) - (cDt)2 ; Ds2 (in K) = Ds’2 (in K’)

• Using only 1-D of space (x): two events described by (x1,t1) and (x2,t2) can be expressed by:

Ds2 = (x1-x2)2 - c2(t1-t2)2 (SPACETIME INTERVAL)

⦁ [interval]2 = [separation in space]2 - [separation in time]2

PHGN310: space – time interval


• Lightlike interval:– Ds2 = 0 (Dx2 = c2Dt2): the two

events can only be connected by a light signal

• Spacelike interval:– Ds2 > 0 (Dx2 > c2Dt2): the two

events cannot be causally connected.

• Timelike interval:– Ds2 < 0 (Dx2 < c2Dt2): the two

events can be causally connected.

PHGN310: space – time interval

CAUSALITY

7.consequences of the special theory of...

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