chapter 2 special theory of relativity 300 fall 18... · 2020. 3. 17. · fall 2018 prof. sergio b....

Special Theory of Relativity

Fall 2018 Prof. Sergio B. Mendes 1

CHAPTER 2

Topics


• 2.0 Inertial Frames of Reference• 2.1 Conceptual and Experimental Inconsistencies• 2.2 The Michelson-Morley Experiment• 2.3 Einstein’s Postulates• 2.4 The Lorentz Transformation• 2.5 Time Dilation and Length Contraction• 2.6 Addition of Velocities• 2.7 Experimental Verification• 2.8 Twin Paradox• 2.9 Space-Time• 2.10 Doppler Effect• 2.11 Relativistic Momentum• 2.12 Relativistic Energy• 2.13 Computations in Modern Physics• 2.14 Electromagnetism and Relativity

How to describe an event?


The precise description of an event must be

characterized by its location in space 𝒓𝒓 and its

time 𝑡𝑡.

𝑧𝑧

The set of (calibrated) rulers and (synchronized) clocks form a frame of reference that can be used to characterize

events.

What is an Inertial Frame of Reference ?


Such a frame of reference is established when a body subjected to a null net external force, 𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟎𝟎,

is observed to move with constant velocity (either rectilinear motion at constant speed or at rest).

𝒂𝒂

𝑚𝑚𝒈𝒈 + 𝑻𝑻 = 𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = 𝑚𝑚 𝒂𝒂

𝒗𝒗 = 𝟎𝟎 𝒗𝒗 = 𝒄𝒄𝒄𝒄𝒏𝒏𝒄𝒄𝒏𝒏𝒂𝒂𝒏𝒏𝒏𝒏𝒂𝒂 = 𝟎𝟎 𝒂𝒂 = 𝟎𝟎

One in which Newton’s laws are valid !!

𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟎𝟎 𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟎𝟎

A non-inertial observer reaches conclusions that don’t agree with

Newton’s laws


𝒂𝒂

𝒂𝒂′ = −𝒂𝒂

𝒂𝒂′ = 0

𝒂𝒂′ = −𝒂𝒂

𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 ≠ 𝟎𝟎

𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟎𝟎 𝑭𝑭𝒏𝒏𝒏𝒏𝒏𝒏 = 𝟎𝟎

𝒂𝒂′ 𝒂𝒂′

Therefore, we must use an inertial frame of reference to describe the laws of Mechanics



Once we have found one inertial frame of reference (K), then any frame of reference moving at constant velocity 𝒗𝒗𝒄𝒄 with respect to K is also an inertial frame of reference.

Q: Well, how many inertial frames of reference are out there?

A: An infinite number.

Proof


• Consider that K is at rest.

𝑥𝑥′ = 𝑥𝑥 − 𝑣𝑣𝑜𝑜,𝑥𝑥 𝑡𝑡

𝑦𝑦′ = 𝑦𝑦 − 𝑣𝑣𝑜𝑜,𝑦𝑦 𝑡𝑡

𝑧𝑧′ = 𝑧𝑧 − 𝑣𝑣𝑜𝑜,𝑥𝑥 𝑡𝑡

𝒓𝒓′ = 𝒓𝒓 − 𝒗𝒗𝒄𝒄 𝑡𝑡

𝒓𝒓′𝒓𝒓

𝒗𝒗𝒄𝒄 𝑡𝑡

𝒂𝒂′ = 𝒂𝒂

𝒗𝒗′ = 𝒗𝒗 − 𝒗𝒗𝒄𝒄

𝑡𝑡′ = 𝑡𝑡

K

K’

• Consider that the two frames of reference coincide at 𝑡𝑡 = 0.

• Consider that K’ is moving with constant velocity𝒗𝒗𝒄𝒄 with respect to K.


• If Newton’s laws are valid in one frame of reference (K), then they are also valid in another frame of reference (K’) moving at a uniform velocity relative to the first system.

𝒂𝒂′ = 𝒂𝒂 𝒎𝒎′ 𝒂𝒂′ = 𝒎𝒎′ 𝒂𝒂 𝑭𝑭′ = 𝑭𝑭𝒎𝒎′ = 𝒎𝒎

• So, both are inertial frames of reference.

Principle of (Classical) Relativity


𝒓𝒓′ = 𝒓𝒓 − 𝒗𝒗𝒄𝒄 𝑡𝑡

• The laws of Mechanics (Newton’s laws) are the same in all inertial frames of reference

Galilean Transformation

Inertial frames of reference are related by:

• All inertial frames of reference are equivalent.

𝑡𝑡′ = 𝑡𝑡

2.1 Conceptual and Experimental Inconsistencies


Conceptual Inconsistencies:


Although Newton’s laws of motion had the same form under the Galilean transformation, Maxwell’s equations did not.

𝑭𝑭 = 𝑞𝑞 𝑬𝑬 + 𝑞𝑞 𝒗𝒗 × 𝑩𝑩

�𝑆𝑆𝑬𝑬 𝒓𝒓 .𝒅𝒅𝑨𝑨 =

𝑄𝑄𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝑖𝜖𝜖0

�𝑆𝑆𝑩𝑩 𝒓𝒓 .𝒅𝒅𝑨𝑨 = 0

�𝐶𝐶𝑬𝑬 𝒓𝒓 .𝒅𝒅𝒄𝒄 = −

𝑑𝑑Φ𝐵𝐵

𝑑𝑑𝑡𝑡

�𝐶𝐶𝑩𝑩 𝒓𝒓 .𝒅𝒅𝒄𝒄 = 𝜇𝜇𝑜𝑜 𝐼𝐼 + 𝜇𝜇𝑜𝑜 𝜖𝜖𝑜𝑜

𝑑𝑑Φ𝐸𝐸

𝑑𝑑𝑡𝑡

Experimental Inconsistencies:


In Maxwell’s theory, the speed of light in terms of the permeability and permittivity of free space was given by:

𝜕𝜕2𝐸𝐸𝜕𝜕𝑥𝑥2

= 𝜖𝜖𝑜𝑜 𝜇𝜇𝑜𝑜𝜕𝜕2𝐸𝐸𝜕𝜕𝑡𝑡2

𝜕𝜕2𝐵𝐵𝜕𝜕𝑥𝑥2

= 𝜖𝜖𝑜𝑜 𝜇𝜇𝑜𝑜𝜕𝜕2𝐵𝐵𝜕𝜕𝑡𝑡2

𝑐𝑐 =1𝜖𝜖𝑜𝑜 𝜇𝜇𝑜𝑜

The carrier medium for light, ether:


Following the tradition of the time (that every wave has a medium to carry its propagation) the luminiferous ether was considered as the carrier medium for light propagation, where 𝑐𝑐 = 1

𝜖𝜖𝑜𝑜 𝜇𝜇𝑜𝑜

It also had to have an elasticity to support the high velocity of light waves

Ether had to have such a low density that the planets could move through it without loss of energy

An Absolute Inertial Frame of Reference


• Ether was proposed as an absolute reference system in which the speed of light was this constant and from which other measurements could be made.

• The Michelson-Morley experiment was an attempt to show the existence of ether.

2.2 The Michelson-Morley Experiment


Albert Michelson (1852–1931)was the first U.S. citizen toreceive the Nobel Prize forPhysics (1907). He built anextremely precise devicecalled an interferometer tomeasure the minute phasedifference between two lightwaves traveling in mutuallyorthogonal directions.

The Michelson Interferometer


How does it work ?


1. AC is parallel to the motion of the Earth inducing an “ether wind”

2. Light from source S is split by mirror A and travels to mirrors C and D in mutually perpendicular directions

3. After reflection the beams recombine at A slightly out of phase due to the “ether wind” as viewed by telescope E.

A Typical Interference Pattern


The Analysis


Time 𝑡𝑡1 from A to C and back:

Time 𝑡𝑡2 from A to D and back:

So that the change in time is:

assuming the Galilean Transformation !!

𝑡𝑡1 =𝑙𝑙1

𝑐𝑐 + 𝑣𝑣+

𝑙𝑙1𝑐𝑐 − 𝑣𝑣

𝑡𝑡2 =𝑙𝑙2

𝑐𝑐2 − 𝑣𝑣2+

𝑙𝑙2𝑐𝑐2 − 𝑣𝑣2

∆𝑡𝑡 = 𝑡𝑡2 − 𝑡𝑡1 =2 𝑙𝑙2𝑐𝑐

1

1 − 𝑣𝑣2𝑐𝑐2

−2 𝑙𝑙1𝑐𝑐

1


=2 𝑙𝑙1𝑐𝑐

1


=2 𝑐𝑐 𝑙𝑙1𝑐𝑐2 − 𝑣𝑣2

=2 𝑙𝑙2𝑐𝑐

1


The Analysis (continued)


Upon rotating the apparatus by 90º, the optical path lengths 𝑙𝑙1 and 𝑙𝑙2 are interchanged producing a different change in time:

∆𝑡𝑡′ = 𝑡𝑡′2 − 𝑡𝑡′1 =2 𝑙𝑙2𝑐𝑐

1



1


∆𝑡𝑡 = 𝑡𝑡2 − 𝑡𝑡1 =2 𝑙𝑙2𝑐𝑐

1



1


Difference in times upon rotation:


∆𝑡𝑡′ − ∆𝑡𝑡 =2 𝑙𝑙1 + 𝑙𝑙2

𝑐𝑐1


−1


≅𝑙𝑙1 + 𝑙𝑙2 𝑣𝑣2

𝑐𝑐3

Increasing the pathlength:


Crunching the numbers:


𝑓𝑓𝑓𝑓𝑓𝑓𝑐𝑐𝑡𝑡𝑓𝑓𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓 𝑓𝑓𝑓𝑓𝑡𝑡𝑖𝑖𝑓𝑓𝑓𝑓𝑖𝑖𝑓𝑓𝑖𝑖𝑓𝑓𝑐𝑐𝑖𝑖 𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑖𝑖 =7 × 10−16 𝑠𝑠2 × 10−15 𝑠𝑠

≅ 0.4

𝑓𝑓𝑓𝑓𝑠𝑠𝑡𝑡𝑓𝑓𝑖𝑖𝑚𝑚𝑖𝑖𝑓𝑓𝑡𝑡 𝑓𝑓𝑖𝑖𝑠𝑠𝑓𝑓𝑙𝑙𝑖𝑖𝑡𝑡𝑓𝑓𝑓𝑓𝑓𝑓 ≅ 0.01

𝑣𝑣 = 3 × 104 𝑚𝑚/𝑠𝑠

𝑐𝑐 = 3 × 108 𝑚𝑚/𝑠𝑠

𝑙𝑙1 ≈ 𝑙𝑙2 = 11 𝑚𝑚

𝑇𝑇 =𝜆𝜆𝑐𝑐

∆𝑡𝑡′ − ∆𝑡𝑡 =𝑙𝑙1 + 𝑙𝑙2 𝑣𝑣2

𝑐𝑐3= 7 × 10−16 𝑠𝑠

𝜆𝜆 = 589 × 10−9 𝑚𝑚

𝑐𝑐 = 3 × 108 𝑚𝑚/𝑠𝑠=

589 × 10−9 𝑚𝑚3 × 108 𝑚𝑚/𝑠𝑠

= 2 × 10−15 𝑠𝑠


“The Experiments on the relative motion of the earth and ether

have been completed and the result decidedly negative. The

expected deviation of the interference fringes from the zero

should have been 0.40 of a fringe – the maximum displacement

was 0.02 and the average much less than 0.01 – and then not in

the right place. As displacement is proportional to squares of the

relative velocities it follows that if the ether does slip past the

relative velocity is less than one sixth of the earth’s velocity.”

Albert Abraham Michelson, 1887

2.3 Einstein’s Two Postulates


1. The principle of relativity: The laws of physics are the same in all inertial frames of reference. There is no way to detect absolute motion and no preferred inertial system exists.

2. The constancy of the speed of light:Observers in all inertial frames of reference measure the same value for the speed of light when propagating in vacuum.

Consequences:


Source and Observer at rest:

Observer in motion with respect to the Source:

c

Source in motion with respect to the Observer:c

c

v

-v

Consequences


In Newtonian physics, we previously assumed that 𝒏𝒏 = 𝒏𝒏’.

Einstein realized that events considered simultaneous in K may not be in K’.

• Therefore K and K’ would always agree if two events happen at the same time (simultaneous) or not.

Because speed of light is absolutethen simultaneity is relative


K’

K K

K’


• Two events that are simultaneous in one reference frame (K) are not necessarily simultaneous in another reference frame (K’) moving with respect to the first frame.

• This suggests that each coordinate system must have its own set of observers with their own set of synchronized clocks.

2.4 Lorentz Transformations


The special set of linear transformations that preserve the constancy of the speed of light between inertial observers.

Two inertial frames of reference


𝓞𝓞′

As seen by 𝓚𝓚

𝓞𝓞𝒗𝒗𝑜𝑜

𝓚𝓚 and 𝓚𝓚’1. The axes along (𝑥𝑥,𝑦𝑦, 𝑧𝑧) are parallel to the corresponding

axes along (𝑥𝑥𝑥,𝑦𝑦𝑥, 𝑧𝑧𝑥).

𝓞𝓞′𝓞𝓞−𝒗𝒗𝑜𝑜

As seen by 𝓚𝓚’

2. The relative motion between the two inertial frames of reference is along the x-axis (and x’-axis).

3. Consider that the origins 𝓞𝓞 and 𝓞𝓞′ of the two systems coincide at 𝑡𝑡 = 𝑡𝑡𝑥 = 0

As seen by 𝓚𝓚 ∶


𝓚𝓚: (𝑥𝑥, 𝑡𝑡)

𝒗𝒗𝑜𝑜

𝓚𝓚′: (𝑥𝑥′, 𝑡𝑡′)

𝒗𝒗𝑜𝑜

𝒗𝒗𝑜𝑜

𝒗𝒗𝑜𝑜

𝒗𝒗𝑜𝑜

As seen by 𝓚𝓚′:


𝓚𝓚′: (𝑥𝑥′, 𝑡𝑡′)

𝓚𝓚: (𝑥𝑥, 𝑡𝑡)

− 𝒗𝒗𝑜𝑜





As seen by 𝓚𝓚:


• A flashbulb goes off at the origins when 𝑡𝑡 = 𝑡𝑡𝑥 = 0.

𝒗𝒗𝑜𝑜

As seen by 𝓚𝓚′:


• A flashbulb goes off at the origins when 𝒏𝒏 = 𝒏𝒏𝑥 = 𝟎𝟎.

−𝒗𝒗𝑜𝑜


• According to Postulate 2, the speed of light will be c in both systems !!

𝓞𝓞′𝓞𝓞

𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 = 𝑐𝑐2 𝑡𝑡 2 𝑥𝑥′ 2 + 𝑦𝑦′ 2 + 𝑧𝑧′ 2 = 𝑐𝑐2 𝑡𝑡′ 2

𝓚𝓚𝓚𝓚′

• The wavefronts observed in both systems must be spherical with respect to their own coordinates and time.


𝑦𝑦 = 𝑦𝑦′

𝑧𝑧 = 𝑧𝑧′

as in Galilean transformation

𝑥𝑥 2 + 𝑦𝑦 2 + 𝑧𝑧 2 − 𝑐𝑐2 𝑡𝑡 2 = 0

0 = 𝑥𝑥′ 2 + 𝑦𝑦′ 2 + 𝑧𝑧′ 2 − 𝑐𝑐2 𝑡𝑡′ 2

𝑥𝑥 2 − 𝑐𝑐2 𝑡𝑡 2 = 𝑥𝑥′ 2 − 𝑐𝑐2 𝑡𝑡′ 2


𝑥𝑥′ = 𝑥𝑥 − 𝑣𝑣𝑜𝑜 𝑡𝑡

𝑥𝑥 = 𝑥𝑥′ + 𝑣𝑣𝑜𝑜 𝑡𝑡′

𝛾𝛾

𝛾𝛾due to symmetry

linear modification

= 𝛾𝛾 𝛾𝛾 𝑥𝑥 − 𝑣𝑣𝑜𝑜 𝑡𝑡 + 𝑣𝑣𝑜𝑜 𝑡𝑡′

𝑡𝑡′ = 1𝛾𝛾 𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 𝑥𝑥 + 𝛾𝛾 𝑡𝑡solve for:

in Galilean transformation


𝑥𝑥 2 − 𝑐𝑐2 𝑡𝑡 2 = 𝑥𝑥′ 2 − 𝑐𝑐2 𝑡𝑡′ 2

𝑥𝑥′ = 𝛾𝛾 𝑥𝑥 − 𝑣𝑣𝑜𝑜 𝑡𝑡 𝑡𝑡′ = 1𝛾𝛾 𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 𝑥𝑥 + 𝛾𝛾 𝑡𝑡


𝛾𝛾 =1

1 − 𝛽𝛽2𝛽𝛽 ≡

𝑣𝑣𝑜𝑜𝑐𝑐

𝑥𝑥 2: 1 = γ2 −𝑐𝑐2

𝛾𝛾2 𝑣𝑣𝑜𝑜2 1 − 𝛾𝛾2 2 1

1 − 𝛾𝛾2= −

𝑐𝑐2

𝛾𝛾2 𝑣𝑣𝑜𝑜2

1𝛾𝛾2

− 1 = −𝑣𝑣𝑜𝑜2

𝑐𝑐2

𝑡𝑡 2: −𝑐𝑐2 = 𝛾𝛾2 𝑣𝑣𝑜𝑜2 − 𝑐𝑐2 𝛾𝛾2

1𝛾𝛾2

= 1 −𝑣𝑣𝑜𝑜2

𝑐𝑐2

2 𝑥𝑥 𝑡𝑡: 0 = −𝛾𝛾2 𝑣𝑣𝑜𝑜 − 𝑐𝑐21𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 𝛾𝛾2 𝑣𝑣𝑜𝑜 = −𝑐𝑐21𝑣𝑣𝑜𝑜

1 − 𝛾𝛾2 −𝑣𝑣𝑜𝑜2

𝑐𝑐2 =1𝛾𝛾2

− 1

Lorentz Transformations:


𝑥𝑥′ = 𝛾𝛾 𝑥𝑥 − 𝛾𝛾 𝑣𝑣𝑜𝑜 𝑡𝑡

𝑡𝑡′= −𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐2

𝑥𝑥 + 𝛾𝛾 𝑡𝑡

𝛽𝛽 ≡𝑣𝑣𝑜𝑜𝑐𝑐

𝛾𝛾 =1

1 − 𝛽𝛽2

Inverse Lorentz Transformations:


𝑥𝑥 = 𝛾𝛾 𝑥𝑥′ + 𝛾𝛾 𝑣𝑣𝑜𝑜 𝑡𝑡′

𝑡𝑡 =𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐2

𝑥𝑥′ + 𝛾𝛾 𝑡𝑡′

𝛽𝛽 ≡𝑣𝑣𝑜𝑜𝑐𝑐

𝛾𝛾 =1

1 − 𝛽𝛽2

Relativistic Factor 𝛾𝛾


𝛽𝛽 =

2.5 Time Dilation and Length Contraction



Time Dilation:

𝒗𝒗𝑜𝑜

Time Dilation:


𝑡𝑡′1 = −𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐2

𝑥𝑥1 + 𝛾𝛾 𝑡𝑡1

𝑡𝑡′1

𝑡𝑡′2 = −𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐2

𝑥𝑥2 + 𝛾𝛾 𝑡𝑡2



𝑥𝑥1 = 𝑥𝑥2

Time Dilation & Proper Time:


𝑡𝑡′1 𝑡𝑡′2

𝑡𝑡′2 − 𝑡𝑡′1 = 𝛾𝛾 𝑡𝑡2 − 𝑡𝑡1𝑡𝑡2 − 𝑡𝑡1 = proper time,

time duration measured at same location


Length Contraction:

𝒗𝒗𝑜𝑜

Length Contraction:


𝑡𝑡′2 = 𝑡𝑡′1

𝑥𝑥′2 − 𝑥𝑥′1=1𝛾𝛾

𝑥𝑥2 − 𝑥𝑥1

𝑥𝑥2 − 𝑥𝑥1 = proper length, length measured at rest

𝑥𝑥 = 𝛾𝛾 𝑥𝑥′ + 𝛾𝛾 𝑣𝑣𝑜𝑜 𝑡𝑡′

𝑥𝑥1 = 𝛾𝛾 𝑥𝑥′1 + 𝛾𝛾 𝑣𝑣𝑜𝑜 𝑡𝑡′1 𝑥𝑥2 = 𝛾𝛾 𝑥𝑥′2 + 𝛾𝛾 𝑣𝑣𝑜𝑜 𝑡𝑡′2

2.6 Addition of Velocities


Taking the differentials:





𝑑𝑑𝑥𝑥′ = 𝛾𝛾 𝑑𝑑𝑥𝑥 − 𝛾𝛾 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡

𝑑𝑑𝑡𝑡′= −𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐2

𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑𝑡𝑡

𝑦𝑦′ = 𝑦𝑦

𝑧𝑧′ = 𝑧𝑧

𝑑𝑑𝑦𝑦′ = 𝑑𝑑𝑦𝑦

𝑑𝑑𝑧𝑧′ = 𝑑𝑑𝑧𝑧

Along x-axis





𝑑𝑑𝑥𝑥′

𝑑𝑑𝑡𝑡′=

𝛾𝛾 𝑑𝑑𝑥𝑥 − 𝛾𝛾 𝑣𝑣𝑜𝑜 𝑑𝑑𝑡𝑡

−𝛾𝛾 𝑣𝑣𝑜𝑜𝑐𝑐2 𝑑𝑑𝑥𝑥 + 𝛾𝛾 𝑑𝑑𝑡𝑡

𝑖𝑖′𝑥𝑥 ≡𝑑𝑑𝑥𝑥′


𝑖𝑖𝑥𝑥 − 𝑣𝑣𝑜𝑜1 − 𝑣𝑣𝑜𝑜 𝑖𝑖𝑥𝑥

𝑐𝑐2

Along y-axis





𝑑𝑑𝑦𝑦′


𝑑𝑑𝑦𝑦


𝑖𝑖′𝑦𝑦 ≡𝑑𝑑𝑦𝑦′


𝑖𝑖𝑦𝑦𝛾𝛾 1 − 𝑣𝑣𝑜𝑜 𝑖𝑖𝑥𝑥

𝑐𝑐2

Along z-axis





𝑑𝑑𝑧𝑧′


𝑑𝑑𝑧𝑧


𝑖𝑖′𝑧𝑧 ≡𝑑𝑑𝑧𝑧′


𝑖𝑖𝑧𝑧𝛾𝛾 1 − 𝑣𝑣𝑜𝑜 𝑖𝑖𝑥𝑥

𝑐𝑐2

In Summary, Addition of Velocities:


𝑖𝑖′𝑦𝑦 =𝑖𝑖𝑦𝑦

𝛾𝛾 1 − 𝑣𝑣𝑜𝑜 𝑖𝑖𝑥𝑥𝑐𝑐2

𝑖𝑖′𝑧𝑧 =𝑖𝑖𝑧𝑧

𝛾𝛾 1 − 𝑣𝑣𝑜𝑜 𝑖𝑖𝑥𝑥𝑐𝑐2

𝑖𝑖′𝑥𝑥 =𝑖𝑖𝑥𝑥 − 𝑣𝑣𝑜𝑜

1 − 𝑣𝑣𝑜𝑜 𝑖𝑖𝑥𝑥𝑐𝑐2

Inverted Relations:


𝑖𝑖𝑦𝑦 =𝑖𝑖′𝑦𝑦

𝛾𝛾 1 + 𝑣𝑣𝑜𝑜 𝑖𝑖′𝑥𝑥𝑐𝑐2

𝑖𝑖𝑧𝑧 =𝑖𝑖′𝑧𝑧

𝛾𝛾 1 + 𝑣𝑣𝑜𝑜 𝑖𝑖′𝑥𝑥𝑐𝑐2

𝑖𝑖𝑥𝑥 =𝑖𝑖′𝑥𝑥 + 𝑣𝑣𝑜𝑜

1 + 𝑣𝑣𝑜𝑜 𝑖𝑖′𝑥𝑥𝑐𝑐2

Example


𝒗𝒗𝑜𝑜 = −0.600 𝑐𝑐

𝒖𝒖𝑥𝑥 = +0.990 𝑐𝑐

𝑖𝑖′𝑥𝑥 =𝑖𝑖𝑥𝑥 − 𝑣𝑣𝑜𝑜

1 − 𝑣𝑣𝑜𝑜 𝑖𝑖𝑥𝑥𝑐𝑐2

=0.990 𝑐𝑐 − −0.600 𝑐𝑐

1 − −0.600 𝑐𝑐 0.990 𝑐𝑐𝑐𝑐2

= 0.997 𝑐𝑐


2.7 Experimental Verification of Special Relativity

Cosmic Rays and Muon Decay


𝜏𝜏 = 1.52 𝜇𝜇𝑠𝑠

𝑁𝑁𝑁𝑁𝑜𝑜

= 𝑖𝑖−𝑙𝑙𝑖𝑖 2 𝑡𝑡

𝜏𝜏


(Incorrect) Classical Calculation


= 𝑖𝑖−𝑙𝑙𝑖𝑖 2 𝑡𝑡

𝜏𝜏

𝑡𝑡 =ℎ𝑣𝑣𝑣𝑣 = 0.98 𝑐𝑐

ℎ = 2000 𝑚𝑚= 6.80 𝜇𝜇𝑠𝑠

= 𝑖𝑖−𝑙𝑙𝑖𝑖 2 6.80 𝜇𝜇𝑖𝑖

1.52 𝜇𝜇𝑖𝑖 = 4.5%

Don’t agree with experiment𝜏𝜏 = 1.52 𝜇𝜇𝑠𝑠

Relativistic Calculation


𝑣𝑣 = 0.98 𝑐𝑐 𝛾𝛾 =1

1 − 0.982≅ 5

𝑡𝑡′ =𝑡𝑡𝛾𝛾 =

6.80 𝜇𝜇𝑠𝑠5

≅ 1.36 𝜇𝜇𝑠𝑠


= 𝑖𝑖−𝑙𝑙𝑖𝑖 2 𝑡𝑡′

𝜏𝜏 = 𝑖𝑖−𝑙𝑙𝑖𝑖 2 1.36 𝜇𝜇𝑖𝑖

1.52 𝜇𝜇𝑖𝑖 = 54%Agrees with experiment

Atomic Clock Measurement


Velocity Addition


𝑖𝑖′𝑥𝑥 = 𝑐𝑐 𝑣𝑣𝑥𝑥 = 𝑐𝑐

𝜋𝜋0 → 𝛾𝛾 + 𝛾𝛾

𝑖𝑖𝑥𝑥 =𝑖𝑖′𝑥𝑥 + 𝑣𝑣𝑜𝑜

1 + 𝑣𝑣𝑜𝑜 𝑖𝑖′𝑥𝑥𝑐𝑐2

𝑣𝑣𝑜𝑜 = 0.99975 𝑐𝑐

2.10 Relativistic Doppler effect for light waves in vacuum


Although light velocity in vacuum is always constant c, the frequency will change for a relative motion between source and observer

Higher Frequency Lower Frequency

Source-Observer

approaching

Source-Observer receding

Relativistic Doppler effect for light waves in vacuum


𝑓𝑓 =𝑐𝑐 𝑇𝑇 − 𝑣𝑣 𝑇𝑇

𝜆𝜆

=𝑐𝑐 𝑓𝑓

𝑐𝑐 𝑇𝑇 − 𝑣𝑣 𝑇𝑇

𝑙𝑙𝑖𝑖𝑓𝑓𝑓𝑓𝑡𝑡ℎ 𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑖𝑖 𝑤𝑤𝑓𝑓𝑣𝑣𝑖𝑖 𝑡𝑡𝑓𝑓𝑓𝑓𝑓𝑓𝑓𝑓 = 𝑐𝑐 𝑇𝑇 − 𝑣𝑣 𝑇𝑇

𝑓𝑓 =𝑐𝑐𝜆𝜆

𝑣𝑣 𝑇𝑇𝑣𝑣𝑐𝑐 𝑇𝑇𝑐𝑐


𝑓𝑓 =𝑝𝑝𝑓𝑓𝑓𝑓𝑝𝑝𝑖𝑖𝑓𝑓 𝑡𝑡𝑓𝑓𝑚𝑚𝑖𝑖 𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑖𝑖 𝑠𝑠𝑓𝑓𝑖𝑖𝑓𝑓𝑐𝑐𝑖𝑖𝑡𝑡𝑓𝑓𝑚𝑚𝑖𝑖 𝑝𝑝𝑖𝑖𝑓𝑓𝑓𝑓𝑓𝑓𝑑𝑑 𝑓𝑓𝑓𝑓 𝑡𝑡ℎ𝑖𝑖 𝑠𝑠𝑓𝑓𝑖𝑖𝑓𝑓𝑐𝑐𝑖𝑖 =

𝑇𝑇′𝑜𝑜�1𝑓𝑓′𝑜𝑜

𝑇𝑇 = 𝛾𝛾 𝑇𝑇′𝑜𝑜 𝑓𝑓 =𝑇𝑇 𝑓𝑓′𝑜𝑜𝛾𝛾

𝑓𝑓 =𝑐𝑐 𝑓𝑓

𝑐𝑐 𝑇𝑇 − 𝑣𝑣 𝑇𝑇 =1 + 𝛽𝛽1 − 𝛽𝛽

𝑓𝑓′𝑜𝑜=𝑐𝑐

𝛾𝛾 𝑐𝑐 − 𝑣𝑣𝑓𝑓′𝑜𝑜

Relativistic Doppler effect for light propagating in vacuum


𝑓𝑓 =1 + 𝛽𝛽1 − 𝛽𝛽

𝑓𝑓′𝑜𝑜 𝑓𝑓 =1 − 𝛽𝛽1 + 𝛽𝛽

𝑓𝑓′𝑜𝑜

redshiftedblueshifted

Source-Observer

approaching

Source-Observer receding

Rotation of Venus


Laser Cooling


Laser Radar Technology


2.11 Relativistic Linear Momentum


Classical Expressions from Galileo and Newton:


Newton’s law

𝑝𝑝 ≡ 𝑚𝑚 𝑖𝑖 �⃗�𝐹 =𝑑𝑑𝑝𝑝𝑑𝑑𝑡𝑡

Linear Momentum

Newton’s law is invariant under Galilean transformations, but notunder Lorentz transformations.

Relativistic Linear Momentum


𝑝𝑝 = 𝛤𝛤 𝑚𝑚 𝑖𝑖

Invariance of Newton’s law under Lorentz transformations will be shown

to lead to:

𝛤𝛤 ≡1

1 − 𝑖𝑖𝑐𝑐

2

Conservation of Linear Momentum in an Elastic Collision:


𝑖𝑖′𝑥𝑥 = 0

𝑖𝑖′𝑦𝑦 = ∓𝑖𝑖𝑜𝑜

𝑖𝑖𝑥𝑥 = 0

𝑖𝑖𝑦𝑦 = ± 𝑖𝑖𝑜𝑜

𝑖𝑖𝑦𝑦 =𝑖𝑖′𝑦𝑦

𝛾𝛾 1 + 𝑣𝑣 𝑖𝑖′𝑥𝑥𝑐𝑐2

𝑖𝑖𝑥𝑥 =𝑖𝑖′𝑥𝑥 + 𝑣𝑣

1 + 𝑣𝑣 𝑖𝑖′𝑥𝑥𝑐𝑐2

𝑖𝑖𝑥𝑥 = 𝑣𝑣

𝑖𝑖𝑦𝑦 =∓ 𝑖𝑖𝑜𝑜𝛾𝛾

∆𝑖𝑖𝑥𝑥 = 0

∆𝑖𝑖𝑥𝑥 = 0

∆𝑖𝑖𝑦𝑦 = −2 𝑖𝑖𝑜𝑜

∆𝑖𝑖𝑦𝑦 =2 𝑖𝑖𝑜𝑜𝛾𝛾

∆𝑝𝑝 = 𝑚𝑚 ∆𝑖𝑖 + ∆𝑖𝑖𝑚𝑚

𝑚𝑚

𝑚𝑚 ≠ 0



𝑖𝑖𝑥𝑥 = 𝑣𝑣

𝑖𝑖𝑦𝑦 =∓ 𝑖𝑖𝑜𝑜𝛾𝛾

𝛤𝛤 =1

1 −𝑣𝑣2 + 𝑖𝑖𝑜𝑜2

𝛾𝛾2𝑐𝑐2

𝑖𝑖𝑥𝑥 = 0

𝑖𝑖𝑦𝑦 = ± 𝑖𝑖𝑜𝑜

𝛤𝛤 ∆𝑖𝑖𝑥𝑥 = 0

𝛤𝛤 ∆𝑖𝑖𝑦𝑦 =−2 𝑖𝑖𝑜𝑜

1 − 𝑖𝑖𝑜𝑜2𝑐𝑐2

𝛤𝛤 =1


𝛤𝛤 ∆𝑖𝑖𝑥𝑥 = 0

𝛤𝛤 ∆𝑖𝑖𝑦𝑦 = 𝛤𝛤2 𝑖𝑖𝑜𝑜𝛾𝛾

=2 𝑖𝑖𝑜𝑜


𝑝𝑝 = 𝑚𝑚 𝛤𝛤 𝑖𝑖 𝛤𝛤 ≡1


2∆𝑝𝑝 = 𝑚𝑚 𝛤𝛤 ∆𝑖𝑖 + ∆𝑖𝑖 = 0



𝑝𝑝 = 𝑚𝑚 𝛤𝛤 𝑖𝑖

𝛤𝛤 ≡1


2

2.12 Relativistic Energy


�⃗�𝐹 =𝑑𝑑�⃗�𝑝𝑑𝑑𝑡𝑡

𝑊𝑊12 = 𝐾𝐾2 − 𝐾𝐾1 = �1

2�⃗�𝐹 � 𝑑𝑑𝑓𝑓 = �

1

2 𝑑𝑑𝑝𝑝𝑑𝑑𝑡𝑡

� 𝑑𝑑𝑓𝑓


𝐾𝐾2 − 𝐾𝐾1 = �1

2𝑑𝑑𝑝𝑝𝑑𝑑𝑡𝑡

� 𝑑𝑑𝑓𝑓

�⃗�𝑝 = 𝑚𝑚 𝛤𝛤 𝑖𝑖

= 𝑚𝑚�1

2𝑑𝑑 𝛤𝛤 𝑖𝑖𝑑𝑑𝑡𝑡

� 𝑖𝑖 𝑑𝑑𝑡𝑡

𝑑𝑑𝑓𝑓 = 𝑖𝑖 𝑑𝑑𝑡𝑡

= 𝑚𝑚�1

2𝑖𝑖 𝑑𝑑 𝛤𝛤 𝑖𝑖

𝑑𝑑�⃗�𝑝 = 𝑚𝑚 𝑑𝑑 𝛤𝛤 𝑖𝑖

𝑖𝑖 =𝑑𝑑𝑓𝑓𝑑𝑑𝑡𝑡

Solving the integral:


𝐾𝐾 = 𝑚𝑚�1

2𝑖𝑖 𝑑𝑑 𝛤𝛤 𝑖𝑖

𝛤𝛤 ≡1


2

= 𝑚𝑚 𝑐𝑐2 1

1− 𝑢𝑢𝑐𝑐

2 − 1

= 𝑚𝑚�0

𝑢𝑢 𝑖𝑖


2 �3 2𝑑𝑑𝑖𝑖

Kinetic Energy


𝐾𝐾 = 𝑚𝑚 𝑐𝑐21


2− 1

𝐾𝐾 ≅𝑚𝑚 𝑖𝑖2

2

𝑖𝑖𝑐𝑐

2≪ 1

1


2≅ 1 +

12𝑖𝑖𝑐𝑐

2

Comparison of Classical and Relativistic Kinetic Energy


Experimental Results


Total Relativistic Energy


𝐾𝐾 = 𝑚𝑚 𝑐𝑐21


2− 1

𝐾𝐾 + 𝑚𝑚 𝑐𝑐2 =𝑚𝑚 𝑐𝑐2


2𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙 =

𝐸𝐸0 = 𝑚𝑚 𝑐𝑐2Rest Energy:

Total Energy: = 𝛤𝛤 𝑚𝑚 𝑐𝑐2

Hydrogen Fusion and Solar Energy


1𝐻𝐻 + 1𝐻𝐻 → 2𝐻𝐻 + 𝑖𝑖+ + 𝜈𝜈𝑖𝑖 + 0.42 MeV

2𝐻𝐻 + 1𝐻𝐻 → 3𝐻𝐻𝑖𝑖 + 𝛾𝛾 + 5.49 MeV

3𝐻𝐻𝑖𝑖 + 3𝐻𝐻𝑖𝑖 → 4𝐻𝐻𝑖𝑖 + 2 1𝐻𝐻 + 𝛾𝛾 + 12.86 MeV

Total Energy and Linear Momentum


𝑝𝑝 =𝑚𝑚 𝑖𝑖


2

𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙 =𝑚𝑚 𝑐𝑐2


2

𝑝𝑝2 𝑐𝑐2 =𝑚𝑚2 𝑖𝑖2 𝑐𝑐2


2

𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙 2 =𝑚𝑚2 𝑐𝑐4


2

𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙 2 − 𝑝𝑝2 𝑐𝑐2 = 𝑚𝑚2 𝑐𝑐4

Massless Particles:



𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙 = 𝑝𝑝 𝑐𝑐

𝑖𝑖 = 𝑐𝑐

𝑚𝑚 = 0

Units of Energy:


𝑊𝑊 = 𝑞𝑞 ∆𝑉𝑉

𝑖𝑖 = 1.6022 × 10−19 𝐶𝐶

∆𝑉𝑉 = 1 𝑉𝑉

𝑊𝑊 = 𝑖𝑖 ∆𝑉𝑉 = 1.6022 × 10−19 𝐽𝐽 ≡ 1 𝑖𝑖𝑉𝑉

Units of Mass:


𝐸𝐸0 = 𝑚𝑚 𝑐𝑐2 𝑚𝑚 =𝐸𝐸0𝑐𝑐2

Particle Mass (MeV) / c2 Mass (u)electron 0.511 0.54858 × 10-3

proton 938.27 1.007276neutron 939.57 1.008665

Higgs boson 125,090 134.3

1 𝑖𝑖 ≡ 112𝑚𝑚𝑓𝑓𝑠𝑠𝑠𝑠 𝑓𝑓𝑓𝑓 𝑓𝑓𝑖𝑖𝑖𝑖𝑡𝑡𝑓𝑓𝑓𝑓𝑙𝑙 12𝐶𝐶

= 931.494 𝑀𝑀𝑖𝑖𝑉𝑉/𝑐𝑐2

= 1.66054 × 10−27 𝑘𝑘𝑓𝑓

1 𝑘𝑘𝑓𝑓 = 8.987 × 1016 𝐽𝐽/𝑐𝑐2

Units of Linear Momentum



𝑙𝑙𝑓𝑓𝑓𝑓𝑖𝑖𝑓𝑓𝑓𝑓 𝑚𝑚𝑓𝑓𝑚𝑚𝑖𝑖𝑓𝑓𝑡𝑡𝑖𝑖𝑚𝑚 =𝑖𝑖𝑓𝑓𝑖𝑖𝑓𝑓𝑓𝑓𝑦𝑦

𝑐𝑐

Binding Energy:


2 ×

2 ×

2 𝐸𝐸0, 𝑝𝑝 + 2 𝐸𝐸0, 𝑖𝑖 − 𝐸𝐸0, 𝐻𝐻𝑖𝑖 = 𝐸𝐸𝐵𝐵,𝐻𝐻𝑖𝑖

2 1.007276 𝑖𝑖 + 2 1.008665 𝑖𝑖 − 4.001505 u 𝑐𝑐2 = 0.0304 𝑖𝑖 𝑐𝑐2

= 𝐸𝐸𝐵𝐵,𝐻𝐻𝑖𝑖

= 28.3 𝑀𝑀𝑖𝑖𝑉𝑉

2.9 Spacetime Representation


Conventional Representation: (𝑡𝑡, 𝑥𝑥)


Spacetime Representation (𝑥𝑥, 𝑐𝑐𝑡𝑡)


• When describing events in relativity, it is convenient to represent events on a spacetimediagram.

• In this diagram one spatial coordinate x, to specify position, is used and instead of time t, ct is used as the other coordinate so that both coordinates will have dimensions of length.

• Spacetime diagrams were first used by H. Minkowski in 1908 and are often called Minkowskidiagrams. Paths in Minkowski spacetime are called worldlines.

Worldline



𝜃𝜃

𝑐𝑐 ∆𝑡𝑡

∆𝑥𝑥

tan 𝜃𝜃 =∆𝑥𝑥𝑐𝑐 ∆𝑡𝑡

=𝑖𝑖𝑐𝑐

= 𝛽𝛽

tan 𝜃𝜃 =𝑖𝑖𝑐𝑐

= 𝛽𝛽 ≤ 1𝑖𝑖

𝜃𝜃 ≤ 45°

Light Cone


Spacetime and Inertial Frames of Reference


𝑥𝑥

𝑥𝑥’

𝑐𝑐 𝑡𝑡’𝑐𝑐 𝑡𝑡




𝜃𝜃

𝜃𝜃

tan 𝜃𝜃 =𝑣𝑣𝑜𝑜𝑐𝑐

Back to the Lorentz Transformations:







𝑑𝑑𝑥𝑥 2 + 𝑑𝑑𝑦𝑦 2 + 𝑑𝑑𝑧𝑧 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡 2 = 𝑑𝑑𝑥𝑥′ 2 + 𝑑𝑑𝑦𝑦′ 2 + 𝑑𝑑𝑧𝑧′ 2 − 𝑐𝑐2 𝑑𝑑𝑡𝑡′ 2

Invariant: its value does not change among inertial frames of reference

𝑥𝑥𝑦𝑦𝑧𝑧𝑐𝑐𝑡𝑡

four-vector

Example 2.13


𝐾𝐾 = 2.00 𝐺𝐺𝑖𝑖𝑉𝑉 𝐾𝐾 = 2.00 𝐺𝐺𝑖𝑖𝑉𝑉

𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙 = ? ?

= 𝐾𝐾 + 𝑚𝑚 𝑐𝑐2

= 2.00 𝐺𝐺𝑖𝑖𝑉𝑉 + 0.93827 GeV

= 2.94 𝐺𝐺𝑖𝑖𝑉𝑉

Example 2.13



𝑝𝑝 = ? ?


𝑝𝑝 =1𝑐𝑐

𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙 2 − 𝑚𝑚 𝑐𝑐2 2

=1𝑐𝑐

2.938 𝐺𝐺𝑖𝑖𝑉𝑉 2 − 0.93827𝐺𝐺𝑖𝑖𝑉𝑉 2

=2.78 𝐺𝐺𝑖𝑖𝑉𝑉

𝑐𝑐

Example 2.13



𝑖𝑖 = ? ? 𝛽𝛽 = ? ? 𝛤𝛤 = ? ?

𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙 = 𝐾𝐾 + 𝑚𝑚 𝑐𝑐2 =𝑚𝑚 𝑐𝑐2


2= 𝛤𝛤 𝑚𝑚 𝑐𝑐2

𝛤𝛤 =𝐸𝐸𝑡𝑡𝑜𝑜𝑡𝑡𝑡𝑡𝑙𝑙𝑚𝑚 𝑐𝑐2

=2.938 𝐺𝐺𝑖𝑖𝑉𝑉

0.93827 𝐺𝐺𝑖𝑖𝑉𝑉= 3.13

𝛽𝛽 =𝛤𝛤2 − 1𝛤𝛤2

= 0.948

Topics


• 2.0 Inertial Frames of Reference• 2.1 Conceptual and Experimental Inconsistencies• 2.2 The Michelson-Morley Experiment• 2.3 Einstein’s Postulates• 2.4 The Lorentz Transformation• 2.5 Time Dilation and Length Contraction• 2.6 Addition of Velocities• 2.7 Experimental Verification• 2.8 Twin Paradox• 2.9 Space-Time• 2.10 Doppler Effect• 2.11 Relativistic Momentum• 2.12 Relativistic Energy• 2.13 Computations in Modern Physics• 2.14 Electromagnetism and Relativity

chapter 2 special theory of relativity 300 fall 18... · 2020. 3. 17. · fall 2018 prof. sergio b....

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