part two: relativistic mechanics phys 141: principles of mechanics

PART TWO: RELATIVISTICMECHANICS

PHYS 141: PHYS 141: Principles of MechanicsPrinciples of Mechanics

I. Basic Principles A. Spacetime1. Graphical Depiction: Before we look at how

space and time are connected through special relativity, let’s establish how we can describe basic motion: spacetime diagramsspacetime diagrams.

x

t1-D motion in time asmeasuredfrom a ParticularReferenceFrame



x

t

“PAST”

“FUTURE”

“PRESENT”

1-D motion in time asmeasuredfrom a ParticularReferenceFrame



x

t

“SOMEWHEREAHEAD”“HERE”

“SOMEWHEREBEHIND”


I. Basic Principles A. Spacetimea. Event: a particular location and time.

x

t

“event”(x,t)

“OVER THERE,THEN”1-D motion

in time asmeasuredfrom a ParticularReferenceFrame

I. Basic Principles A. Spacetimea. Event: a particular location and time.

x

t

“event 1”(x1,t1)



I. Basic Principles A. Spacetimeb. World Lines are Spacetime trajectories: How

events are ordered.

x

t



“Stay at x = x1

for t2-t1 seconds”


I. Basic Principles A. Spacetimeb. World Lines: How events are ordered.

x

t



“Instantaneously moveFrom

x1 to x2.” Note: this is impossible.


I. Basic Principles A. Spacetimec.c. Normalized Spacetime diagramsNormalized Spacetime diagrams. Let w = ct.

Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light.

x

w = ct

“event 1”(x1,w1)

“event 2”

(x2,w2)

Speed from E1 to E2 is

dw/dx = c/v =1,so

v/c = 1.

1-D motion in time asmeasuredfrom a ParticularReferenceFrame A

I. Basic Principles A. Spacetimec.c. Normalized Spacetime diagramsNormalized Spacetime diagrams. Let w = ct.

Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light.

x

w = ct1-D motion in time asmeasuredfrom a ParticularReferenceFrame A

I. Basic Principles A. SpacetimeNormalized Spacetime diagrams. Let w = ct. Then the slope of a trajectory in spacetime is dw/dx = cdt/dx = c/v, with c = speed of light.

x

w=ct

v/c = 1

dw/dx = c/v >1, sov/c <1

dw/dx = c/v <1, sov/c >1


I. Basic Principles A. Spacetimed. Lightcones:Lightcones: Regions of spacetime in which

events are connected by paths with speed v/c ≤ 1.

x

w

“Futurelight cone”

for A

“Past light cone”

for A

v/c = 1

v/c = 1


I. Basic Principles A. Spacetimed. Lightcones:Lightcones: Regions of spacetime in which

events are connected by paths with speed v/c ≤ 1.

x

wv/c =

1. “Futurelight cone”

for A

“Past light cone”

for A

Inaccessible to A

according torelativity

Inaccessible to A

according torelativity

v/c = 1.

Constant speed(x’,y’,z’,t’)

I. Basic Principles B. Galilean Transformations

For inertial systems, the Galilean TransformationGalilean Transformation allows us to translate between frames: all we have to do is (basically) subtract out the motion of primed frame (the train in this case).


For inertial systems, the Galilean TransformationGalilean Transformation allows us to translate between frames: all we have to do is (basically) subtract out the motion of primed frame (the train in this case).

1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction)

x

y

a) Galilean transformations:coordinates for a point P in the

movingprimed frame as seen from the

(unmoving) unprimed frame (and vice versa).x = x’ + vt’. (I.B.1-4)

y = y’.z = z’.t = t’.


x

y x = x’ + vt’.y = y’.z = z’.t = t’.

x’

y’v

P

x

x’


1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction) a) Galilean transformations:

coordinates for a point P in the moving

primed frame as seen from the (unmoving)

unprimed frame (and vice versa).

x

y x = x’ + vt’.y = y’.z = z’.t = t’.

x’

y’v

P

x

vt’ x’


1. Coordinate Transformations: transformations between two reference frames. Consider a primed reference in motion relative to an unprimed one (let v be along the positive x-direction) a) Galilean transformations:

coordinates for a point P in the moving

primed frame as seen from the (unmoving)

unprimed frame (and vice versa).

II. Simultaneity A. Set Up1. Consider two observers, A&B, stationary with

respect to each other and reference frame (w,x). How do we calibrate their identical clocks?

x

w

A B

II. Simultaneity A. Set Up1. Consider two observers, A&B, stationary with

respect to each other and reference frame (w,x). How do we calibrate their identical clocks?

x

w

A B

wB

Eventsare simultaneous(clocks calibrated)By backtracking

To the equidistantSpacetime position

Between light paths.

Events WA and WB

are simultaneous in this frame.

wA

Note:dotted lineindicates

same TIMEfor unprimed

frame

II. Simultaneity A. Set Up2. What happens to the axes for a Galilean

Transformation?

x

w

(x,t)


Transformation? Now consider a frame moving with speed v wrt the original frame.

x

t

(x,t)

x=(x’ + vt’)



x

t

(x,t)

x=(x’ + vt’)

x=(x’ + v(2t’))



x

t

(x,t)

x=(x’ + v(3t’))

x=(x’ + vt’)

x=(x’ + v(2t’))



x

t

(x,t)

x=(x’ + v(3t’))

x=(x’ + vt’)

x=(x’ + v(2t’))

x’



x

t

(x,t)

x’



x

t

(x,t)

x’

t’=t



x

t

x’

II. Simultaneity A. Set Up3. Why? Newtonian addition of speeds.

x

tx’

t’



x

t

+=

I. Basic Principles C. The ‘Principle of Relativity’

“The Laws of Mechanics are the same in every inertial frame, and the Galilean Transformation is valid.”

Problem: Electromagnetism (1850)

I. Basic Principles D. Postulates of Special Relativity

1. The speed of light in vacuum is a constant, independent of the motion of the source, the observer, or both.

2. The Laws of Physics are everywhere the same for inertial frames, and the connection between frames is the Lorentz Transformation.

3. When v/c is small, then the LT reduces to the GT: x’ = (x - vt) & t’ = t.

x’x

v

P

x’ = (x - vt)/{1 - (v/c)2}1/2.y’ = y.z’ = z.

t’ = t - vx/c2)/{1 - (v/c)2}1/2.(I.D.1-4)

II. Simultaneity A. Set Up2. What happens to the axes for a Lorentz

Transformation?

x

w

(x,w)

II. Simultaneity A. Set Up

x

w

(x’,w’)

II. Simultaneity A. Set Up

x

w

II. Simultaneity B. Result1. Now what is measured by another observer moving

with respect to the original frame with speed (v’)?

x

w

A B

w’

1. Now what is measured by another observer moving with respect to the original frame with speed (v’)?

x

w

A B

w’

x’

II. Simultaneity B. Result

II. Simultaneity B. Results1. Now what is measured by another observer moving


x

w

A B

w’

x’

wB’



x

w

A B

w’

wA’

SimultaneityIs frame

dependent

wB’

Events wA’ and wB’are simultaneous in

the red frame.

However, now the events are NOT

simultaneous in theunprimed frame.

Note:dotted lineindicates

same TIMEfor primed

frame


with respect to the original frame with speed (v)?

x

w

t3t2t1



x

w w’

x’

t3’



x

w w’

x’

t3’

Example:x = 1 lys, t=0s, v/c = 0.5

x’= (1 lys - 0.5c(0s))/{1-0.52}1/2

= 1.15*(5.5) = 1.15 lys

t’ = (0 s - 0.5c(1 ly-s)/c2)/{1-0.52}1/2

=1.15*(-.5) = -0.6 second.



x

w w’

x’

t3’

Example:x = 5 lys, t=0s, v/c = 0.5

x’= (5 lys – 0.5c(0s))/{1-0.52}1/2

= 1.15*(5) = 5.8 lys

t’ = (0 s – 0.5c(5 lys)/c2)/{1-0.52}1/2

=1.15*(-2.5) = -2.9 second.



x

w w’

x’



x

ww’

x’

2 2 2 2( ) .c t x y z

(II.C.1)

Invariant: independent of frame that is

measured

What is the space time interval on a lightcone? 2

2

2

0

0

0

s timelike

s lightlike

s spacelike

II. Simultaneity C. Spacetime Interval

Consider a light pulse that travels in a sphere.

2 2 2 2( ') ' ' ' .c t x y z

2 2 2 2 2( ).s c t x y z

x

w=ct

ds2 = 0,lightlike

ds2>0,timelike,Causally

connected

ds2<0,Spacelike,Causally

unconnected

A

1. What is measured by another observer moving with respect to the original frame with speed (v’)?

x

w w’

w*

We know thatw = 3w* = 2

both measuredfrom the black

Reference frame clock.

What is w’ measured from red

Reference Frame clock?

III.Time Dilation A. Definition


x

w w’

w*

w’ = w*w=w’, and so

w = w*.



x

w

w’w*

w’ = w*w=w’, and so

w = w*.

w’ = w/w’ < w

Time divisions in theprimed frame are differentthan in the unprimed frames


1. Clocks moving relative to an observer are measured by that observer to run slow compared to a clock at rest.

2. Another thought experiment: Albert on a train w/speed v.

t’ = 0: light

leaves

Note: only one clock is needed to measure emission and return

of light.

III.Time Dilation B. Derivation

Light travelsa total distance

equal to 2d.

Light requirest’ = 2d/c seconds to reach the floor.

2. And now Albert’s view from the platform. Now we’ll need two synchronized clocks to measure both events:

Clock 1 Clock 2

3. Mathematical formulation (pseudo-classical):

L = vt.

d = ct’.D = (d2 + L2)1/2.

c = D/t = (d2 + L2)1/2 /t;c2 = (d2 + L2)/t2.

t2 = (d/c)2 + (vt/c)2;t2 (1- (v/c)2) = (d/c)2.

However, d = ct’, and sot = t’/{1- (v/c)2}1/2.

t’ = t{1- (v/c)2}1/2.Measured from platformMeasured on train

t = t’/{1- (v/c)2}1/2. (III.B.1) t’ = t{1- (v/c)2}1/2. (III.B.2)

Note: a) t is the time in the train as measured by the

platform observer.b) t’ is the time in the train as measured by the

train observer.c) t > t’: moving clocks run slow as

measured in the platform frame.

III.Time Dilation

III. Time Dilation C. Implications

Let = 1/{1- (v/c)2}1/2

(III.B.3) t = t’. (III.B.4)

t’ = t/. (III.B.5)

1. Verifications & Implicationsa) Muon lifetime

Particle decay time longer as measured in lab (rest) frame

b) Atomic clock experimentsc) The “Twin Paradox”

III. Time Dilation C. Implications

III. Time DilationTwin “Paradox”

x

w

w’

w*

Rocket and EarthClocks synchronizedAt w’ and w*


x

w**

w’

w*

On return,Rocket and EarthClocks synchronizedAt w’ and w**


x

w’

x’


x

w’

2. Example: An astronaut travels to a star 100 lys away with a speed equal to 0.99c. How much time elapses on the ship as measured by someone in the ship?

The time on the ship is given by B.2

t’ = t{1- (v/c)2}1/2.

Note: t, the time elapsed on the Earth is {100 ly/0.99 ly/y} = 101 y, since v = 0.99c. Thus,

t0 =t’ = t{1- (v/c)2}1/2 ~ (101 yrs)(1 - .992)1/2 = 14 yrs.“proper time”

III. Time Dilation D. Implications

1. Definition: the length of an object L is measured to be shorter when it is moving relative to the observer than when it is at rest.

2. Mathematical formulation:

L = L0{1- (v/c)2}1/2. (IV.D.1)

3. Example: in the previous example, if the ship has a length of 100m as measured in its rest frame, what is its length as seen from the Earth?

L = L0{1- (v/c)2}1/2 = (100m)(1 - .992)1/2 = 14 m.

IV. Length Contraction A. Formulation

1. Relative velocity is not computed by simply adding (or subtracting velocity vectors.

2. Mathematical Formulation:

u = (v ± u’)/(1 ± vu’/c2).(V.A.1)

v u’

V. Velocity Addition A. Formulation

3. Example: two particles collide with individual speeds of .9c and .5c, respectively as measured in the lab. What is the relative speed of the second particle as seen by the first particle?

Classically, we would just add these speeds and get 1.4c, which is clearly wrong if SR is valid.

u = (v + u’)/(1 + vu’/c2) = (1.4c)/(1 + .45c/c) = 0.97c.

v u’

V. Velocity Addition A. Formulation

V. Velocity Addition B. Spacetime1. How does velocity addition appear in ST diagrams?

a) Galilean addition

x

w=ctv/c =

1

= arctan(.5) = 27o

Let a trainmove atv/c = 0.5 wrtthe platform.Let’s considerAn object at reston the trainas it speeds by the platform.

x’

v’/c = 0.5


a) Galilean addition

x

w=ctv/c =

1

= arctan(.5) = 27o


Now consider someone runningDown the train at u’/c = 0.5

x’

v’/c = 0.5

u’/c = 0.5

Galileo: combined

velocity = c. D’oh!


a) Lorentz addition

x

w=ctv/c =

1

= arctan(.5) = 27o


Now consider someone runningdown the train at u’/c = 0.5

x’

v’/c = 0.5

u’/c = 0.5

Einstein: combined

velocity = .8c.

A. Momentum

p = mv/{1 - 2}1/2,= mv = mc. (VI.A.1)

Equation A.1 => v for a massive object can never be ≥ c.

B. Energy1. Relativistic Kinetic Energy:

K = mc2 - mc2 = ( - 1)mc2, or (VI.B.1)K = mc2( 1/{1 - 2}1/2 - 1).

Note: for << 1, then 1/{1 - 2}1/2 ~ 1 + (1/2)2, and

K~ mc2(1 + (1/2)2 - 1) = (1/2)mv2, (VI.B.2)

VI. Momentum & Energy

2. Total Energy

E = mc2 = mc2 + K, where (B.3)

The quantity mc2 is the rest energy of the object. Using the

momentum equation (A.1), we can also write:

E2 = p2c2 + (mc2)2. (B.4)

The consequences of (B.3, 4) are that mass and energy are equivalent, and that they can be converted into each other.


3. Example: Suppose our astronaut has a mass of 70kg. Compare her classical and relativistic KE.

K(class.) = 1/2mv2 = 3.2 x 1018 J.K(rel.) = mc2 - mc2 = 3.8 x 1019 J.

4. Example: How much energy would be released if our astronaut were converted completely into energy?

E(rest) = mc2 = 6.3 x 1018 J.E(tot) = E(rest) + K ~ 4 x 1019 J ~ 10,000 Megatons of TNT.


5. Another example: An electron has a total energy of

10.0 “Mega-electron volts” (MeV) as measured in a particular laboratory frame. What is the value of its K, v?

Note: Eelectron(rest) = melectronc2 = 8.2 x 10-14 J = 0.511 MeV.

Thus, K = E – mc2 = 9.5 MeV.E = mc2; = E/mc2 = 19.6;

2 = {1 – 2}-1 = 384 => 2 = 1 - 1/384 = 0.997.

= 0.999


part two: relativistic mechanics phys 141: principles of mechanics

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