Relativistic dynamic

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Spiral Physics

Modern Physics

The Special Theory of Relativity:

Dynamics

2

Copyright © 2003

Paul D’Alessandris

Spiral Physics

Rochester, NY 14623

All rights reserved. No part of this book may be reproduced or transmitted in any

form or by any means, electronic or mechanical, including photocopying, recording,

or any information storage and retrieval system without permission in writing from

the author.

This project was supported, in part, by the National Science Foundation. Opinions expressed are those of

the author and not necessarily those of the Foundation.

3

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The Special Theory of Relativity: Dynamics

Relativistic Momentum, Force and Energy

Once Einstein revolutionized our understanding of space and time, physicists were faced with

a monumental task. All of physics, before Einstein, was based on the idea of absolute space

and time. Once these concepts were found to be erroneous, all of classical physics had to be

re-examined in this light. In this section, we will “re-examine” our understanding of

momentum, force, and energy.

Momentum In classical physics, momentum is defined as

mvp

However, using this definition of momentum results in a quantity that is not conserved in all

frames of reference during collisions. However, if momentum is re-defined as

mvp

it is conserved during particle collisions. Therefore, experimentally, momentum is defined by

the above equation.

Force Once nature tells us the proper formula to use for calculating momentum, mathematics tells us

how to measure force and energy. Force is defined as the time derivative of momentum

dt

dpF

(In classical physics, where p = mv, this reduces to F = ma.) Substituting the correct

relationship for momentum yields

mac

v

dt

dmvF

dt

dvmmv

dt

dF

dt

mvdF

])1[(

)()(

)(

2/1

2

2

4

Use the chain rule to evaluate the derivative of ,

maac

v

c

vmvF

madt

dv

c

v

c

vmvF

])1([

])2

()1(2

1[

2/3

2

2

2

2

2/3

2

2

Factor out the common factor ma,

]1

1

[

2

2

2

2

c

v

c

v

maF

Find a common denominator and simplify,

maF

c

vmaF

c

v

c

v

c

v

maF

3

2

2

2

2

2

2

2

2

]

1

1[

]

1

1

[

This is the relativistically correct form of Newton’s Second Law.

Energy The kinetic energy of an object is defined to be the work done on the object in accelerating it

from rest to speed v.

v

FdxKE0

Using our result for force derived above yields

v

madxKE0

3

5

The variable of integration is x, yet the integrand is expressed in terms of a and v (v is hidden

inside ). To solve this problem,

v

v

v

v

c

v

mvdvKE

mvdvKE

dvdt

dxmKE

dxdt

dvmKE

0 2/3

2

2

0

3

0

3

0

3

)1(

)(

)(

This integral can be done by a simple u-substitution,

22

2

2

2

0

2

2

2

2/12

0

2/3

2

2

2

2

]1

1

1[

]

1

1[

]2[2

2

2

1

mcmcKE

c

vmcKE

c

vmcKE

umc

KE

u

dumcKE

dvc

vdu

c

vu

v

v

Rearranging this yields,

RestTotal

22

EKEE

mcKEmc

Einstein identified the term mc2 as the total energy of the particle. Thus, the total energy is

the sum of the kinetic energy and a completely new form of energy, the rest energy. Particles

have rest energy just by virtue of having mass. In fact, mass is simply a form of energy.

6

Momentum and Energy

An electron is accelerated through a potential difference of 80 kilovolts. Find

the kinetic energy, total energy, momentum and velocity of the electron.

The following collection of equations express the relationships between momentum, energy,

and velocity in special relativity. (Momentum is often easier expressed as “pc” rather than “p”

as you will see once you begin working problems.)

2222

2

2

2

2

)()(

)1(

)(

mcpcE

mcKE

mcKEE

mcE

c

vmcpc

mvp

total

total

total

The last equation is particularly useful in that it allows a direct relationship between energy

and momentum without the need to calculate the velocity. The proof of this relationship is left

as an exercise.

From electrodynamics, the kinetic energy of a charge accelerated through a potential

difference V is simply the product of the charge and the potential difference,

qVKE

Rather than substituting the numerical value of the charge on an electron (-e = -1.6 x 10-19

C)

into this expression (and obtaining the kinetic energy in joules), we will leave “e” in the

equation and use “eV” as a unit of energy. Note 1.0 eV = 1.6 x 10-19

J.

keVKE

VxeKE

qVKE

80

)1080( 3

Thus, the kinetic energy of the electron is 80 keV.

The total energy of the electron is then

keVE

keVkeVE

mcKEE

total

total

total

591

51180

2

7

The momentum is

keVpc

pc

mcEpc

mcpcE

total

total

297

)511(591

)(

)()(

22

222

2222

(Again, momentum is often easier expressed as “pc” rather than “p”)

The speed of the electron is

cv

c

v

c

vmcpc

503.0

)(591297

)(2

Collisions and Decays I

A neutral pion (rest energy 135 MeV) moving at 0.7c decays into a pair of

photons. The photons each travel at the same angle from the initial pion

velocity. Find this angle and the energy of each photon.

Any process that occurs in nature must obey energy and momentum conservation. To analyze

this particle decay, apply both conservation laws to the process.

8

First, find the Lorentz factor for the pion.

4.1

)7.0(1

1

1

1

2

2

2

2

c

c

c

v

Applying energy conservation yields:

MeVE

E

Ecm

EEE

EE

photon

photon

photonpion

photonphotonpion

afterdecayybeforedeca

5.94

2)135(4.1

22

21

The two photons must have the same energy since they travel in the same direction relative to

the initial pion velocity. This is the only way that momentum in this perpendicular direction

can be conserved.

Applying momentum conservation (actually conservation of “pc”) along the initial direction

of travel and using the relationship 222

)(mcEpc total yields:

o

photonpion

photonpion

photonphotonpion

afterdecayybeforedeca

mcEmcE

cpcp

cpcpcp

pcpc

6.45

)(cos1893.132

))(cos)0(5.94(2)135())135(4.1(

))(cos)((2)(

)(cos2

)(cos)(cos

2222

222222

21

The photons each travel at 45.60 from the direction of the pions initial path.

9

Collisions and Decays II

A photon of energy 500 keV scatters from an electron at rest. The photon is

redirected to an angle of 350 from its initial direction of travel. Find the energy

of the scattered photon and the angle and energy of the scattered electron.

35°

’

e-

To analyze, apply energy conservation:

electronphoton

electronphoton

electronphotonelectronphoton

EE

EE

EEEE

''

''

''

1011

511500

note that the electron initially has only rest energy.

Apply x-momentum conservation (and use 222

)(mcEpc total ):

cos)511(35cos500

cos)511(35cos)0(0)0(500

cos35cos'

22''

22'22'22

electronphoton

electronphoton

ee

EE

EE

cpcpcppc

Apply y-momentum conservation:

sin)511(35sin

sin)511(35sin)0(00

sin35sin'

22''

22'22'

electronphoton

electronphoton

ee

EE

EE

cpcpcppc

10

This yields three equations with the requested three unknowns (E’photon, E

’electron, and ).

If you enjoy algebra, solve this system of equations by hand. If you have better things to do

with your life, use a solver to find:

1.58

586

425

'

'

keVE

keVE

electron

photon

11

. . . . . . .. . .

The Special Theory of Relativity: Dynamics

Activities

12

Six particles (with rest energy E0) are detected in the collision “debris” at a particle accelerator. Each

particle’s total energy (E) is measured.

E0 (MeV) E (MeV)

A 100 200

B 200 300

C 100 400

D 400 500

E 200 400

F 800 1000

a. Rank these particles on the basis of their mass.

Largest 1. _____ 2. _____ 3. _____ 4. _____ 5. _____ 6. _____ Smallest

_____ The ranking cannot be determined based on the information provided.

Explain the reason for your ranking:

b. Rank these particles on the basis of their speed.




c. Rank these particles on the basis of their kinetic energy.




13

The following six particles (with the rest energy E0 and kinetic energy K indicated) are collided with their

antimatter partners traveling in the opposite direction with the same kinetic energy. The resulting matter-

antimatter annihilation produces a pair of photons (traveling in opposite directions).

E0 (MeV) K (MeV)

A 100 200

B 200 100

C 100 400

D 400 200

E 200 400

F 800 100

a. Rank these particles on the basis of their momentum.




a. Rank these particles on the basis of energy of the photons they create.




b. Rank these particles on the basis of speed of the photons they create.




14

An electron is accelerated through a potential difference of 10 million volts. Find its kinetic energy,

momentum and velocity. Compare its velocity to that predicted by classical physics.

Mathematical Analysis

15

A proton is accelerated through a potential difference of 10 million volts. Find its kinetic energy,

momentum and velocity. Compare its velocity to that predicted by classical physics.


16

a. Calculate the kinetic energy and momentum of a neutral pion (0) traveling at 0.6c.

b. Calculate the velocity and momentum of a neutral pion (0) with kinetic energy 200 MeV.

c. Calculate the velocity and kinetic energy of a neutral pion (0) with momentum 200 MeV/c.


17

a. Calculate the kinetic energy and momentum of a psi-meson () traveling at 0.2c.

b. Calculate the velocity and momentum of a psi-meson () with kinetic energy 200 MeV.

c. Calculate the velocity and kinetic energy of a psi-meson () with momentum 200 MeV/c.


18

Electrons are accelerated to high speed in two stages. The first stage accelerates the electrons from rest to

0.990c. The second stage accelerates them from 0.990c to 0.999c.

a. Find the energy needed for each stage.

b. If the energy needed for the second stage is again applied to the electron, what would be its final

speed?

c. If protons were being accelerated, find the energy needed for the first two stages.


19

Electrical energy can be sold for approximately 10 cents per kilowatt hour. If there was a way to convert

mass energy directly into electrical energy, how much would my corpse be worth?


20

To compete with e-mail, the U.S. Post Office has introduced Super Express mail, where a letter is sent to its

destination at a speed of 0.999c using a special Letter Accelerator. Assume that a typical letter has a mass

of about 25 g. What should be the minimum cost of a stamp, assuming the post office only charges for the

energy used to accelerate the letter? Assume the post office powers the accelerator using electricity

purchased at 10 cents per kilowatt hour.


21

Based on the total power output of the sun, calculate the approximate decrease in mass of the sun per year.

In a one billion year period, by approximately what percentage does the sun’s mass decrease?


22

Imagine a process in which a portion of an object’s rest energy is directly converted into kinetic energy.

a. For an object initially at rest, find the speed of the object if one-half of its rest energy is directly

converted into kinetic energy. Ignore momentum conservation for this calculation.

b. For an object initially at rest, find the percentage of its mass that must be directly converted into

kinetic energy in order to achieve a speed of 0.99c. Ignore momentum conservation for this calculation.


23

In classical physics, KE = 1/2 mv

2. Below what velocity is the classical expression accurate to within 5%?


24

For a particle of mass m traveling at very close to the speed of light, E ~ pc. Above what velocity is the

approximation E = pc for a material particle accurate to within 5%?


25

Prove that the relationship 2222

)()( mcpcEtotal

is valid.


26

Imagine a head-on elastic collision in the laboratory between an object of mass 3 kg traveling at 0.8c and

an object of mass 4 kg traveling in the other direction at 0.6c. The objects will rebound with exactly the

same speeds, clearly conserving classical momentum in this frame. However, consider the same collision in

a frame initially moving to the right at 0.8c:

before:

3 kg 0.8c 0.6c 4 kg

4 kg 0.6c 0.8c 3 kg

after:

Laboratory Frame

before:

3 kg _______

_ 4 kg

4 kg _______

__ _______ 3 kg

after:

Moving Frame

a. Using the velocity addition formula, calculate the velocities of the two objects in the frame of reference

moving toward the right at 0.8c.

b. Using the classical formula for momentum, calculate the momentum before the collision in the moving

frame.

c. Using the classical formula for momentum, calculate the momentum after the collision in the moving

frame.

The different values in (b) and (c) indicate that the classical formula for momentum does not result in

momentum conservation being valid in all frames of reference. We must either abandon momentum

conservation or abandon the classical formula.


27

A constant force of 1 N acts on an object of mass 1 kg. According to classical physics, how long will it take

the initially stationary object to accelerate to a speed of:

a. 1000 m/s?

b. 0.5 c?

c. c?

d. Using the relativistically correct form of Newton’s Second Law, determine the actual time needed to

reach each of these speeds. (Hint: Express acceleration as a = dv/dt, separate the velocity and time

variables, and integrate the resulting expression.)


28

A neutral pion with kinetic energy 1.0 GeV decays into a pair of photons.

0 => +

Both photons travel parallel to the initial pion velocity. Find each photons energy.


29

A neutral pion with kinetic energy 1.0 GeV decays into a pair of photons.

0 => +

Both photons travel at the same angle from the initial pion velocity. Find this angle.


30

A stationary charged pion decays into a muon and a neutrino.

- => -

+ The mass of a neutrino is so small it is essentially zero. Find the speed of the emitted muon.


31

A charged kaon with kinetic energy 800 MeV decays into a muon and a neutrino.

- => -

+ The mass of a neutrino is so small it is essentially zero. Both particles travel parallel to the initial kaon

velocity. Find the possible energies of the emitted muon.


32

A positron of kinetic energy 2.5 MeV annihilates with an electron at rest, creating two photons.

e+ + e

- => +

One photon emerges at 900 to the initial positron direction. What is the direction of the other photon?


33

A positron of kinetic energy 20 MeV annihilates with an electron at rest, creating two photons.

e+ + e

- => +

One photon emerges at 500 to the initial positron direction. What is the direction of the other photon?


34

An electron-positron pair can be produced by a gamma ray striking a stationary electron,

+ e- => e

- + e

- + e

+.

If the total energy is divided equally among the three end products, what is the initial gamma ray energy?

Assume the electrons and positron all travel parallel to the initial photon velocity.


35

A neutral kaon moving at 0.7c decays into a pair of charged pions.

0 => +

Both pions travel parallel to the initial kaon velocity. What is the energy of each pion?


36

A neutral kaon moving at 0.7c decays into a pair of charged pions.

0 => +

The pions each travel at the same angle from the initial kaon velocity. Find this angle and the energy of

each pion.


37

A neutral psi-meson with kinetic energy 2.0 GeV decays into an electron and positron.

0 => e + e

The electron and positron each travel at the same angle from the initial psi-meson velocity. Find this angle.


38

An omega-minus with kinetic energy 5.0 GeV decays into a neutral lambda and charged kaon.

- => +

Both particles travel parallel to the initial omega-minus velocity. What are the possible kinetic energies of

each end product?


39

A delta with kinetic energy 3.5 GeV decays into a proton and charged pion.

++ => p +

Both particles travel parallel to the initial delta velocity. What are the possible kinetic energies of each end

product?


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Relativistic dynamic

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