Vern J. OstdiekDonald J. Bord

Chapter 12Special Relativity

http://xkcd.com/

V=1%c V=0.01c Y=1.00005V=10%c V=0.1c Y=1.005V=50% V=0.5c Y=1.155V=60% V=0.6c Y=1.250V=80% V=0.8c Y=1.667V=90% V=0.9c Y=2.294V=99% V=0.99c Y=7.088

It is easy to find the Gamma factor.

Find the ratio between v and c v/c.So if V= 60% the speed of light, that means V/c = 0.6

2

1

1

−

=

cv

γ

( )25.1

6.01

12

=−

=γ

4

Rest energy● Our definitions of kinetic energy and

momentum no longer work in the realm of special relativity.

● In order for conservation of kinetic energy and momentum to hold, we must include a rest energy, E0, in our calculations.

● The rest energy is 20E mc=

5

Rest energy, cont’d

● The total relativistic energy of a particle has two terms. the rest energy the kinetic energy

22

rel rel 2 21mcE KE mcv c

= + =−E=Y mc2

At high speed the energy and therefore the mass of an object increases.We can't never reach the speed of light because the mass increasesAnd so is the energy required to accelerate.

Eo=mc2 is the rest energy . Mass is energy if we increase the energy of a system,Its mass increases.

6

Rest energy, cont’d

● The relativistic kinetic energy is then

This reduces to our usual formula for very low speeds:

22

rel 2 21mcKE mcv c

= −−

212KE mv=

7

Example

In an x-ray tube, an electron with mass m = 9.1×10-31 kg is accelerated to a speed of 1.8×108 m/s (60%c). How much energy does the electron possess? Give the answer in joules and MeVs (million electron-Volts).

(energy = Y mc2 and 1 eV=1.6 10-19J)

8

Example

ANSWER:The relativistic energy is then

( )( )( )

231 8

rel 2

14

13

9.1 10 kg 3 10 m/s

1 0.60

8.2 10 J0.80

1.0 10 J

E-

-

-

´ ´=

-

´=

= ´

9

Example

ANSWER:To express this in MeV

( )

13rel

1319

1.0 10 J1eV1.0 10 J

1.6 10 J640,000 eV0.640 MeV

E -

-

= ´

= ´ ´´

==

FIND THE REST ENERGY of the electron Eo=mc2

10

Example

DISCUSSION:Notice that the rest energy of an electron

is

( )( )2

0231 8

14

9.1 10 kg 3 10 m/s

8.2 10 J512,000 eV 0.512 MeV

E mc-

-

=

= ´ ´

= ´= =

11

So, so far:

Erel

= KErel

+ Eo with E

o=mc2

You found : Erel=0.640 MeV Eo=0.512 MeV

Find the relativistic kinetic energy KErel

(energy of motion of the electron)

12

Find the classical kinetic energy:KE = 0.5 m V2

m=9.1 10-31 kg and V=0.6c c =3108m/s

Find the difference between the classicalAnd relativistic kinetic energy.Find the %difference

KErel=0.640-0.512=.128 instead of 0.092That's %error = 28%

13

Find the classical kinetic energy:The classical result for the energy is

KErel=0.640-0.512=.128 instead of 0.092That's %error = 28%The increase in energy = increase in mass.

( )( )21

2231 81

2

14

9.1 10 kg 1.8 10 m/s

1.5 10 J0.092 MeV.

KE mv-

-

=

= ´ ´

= ´=

14

Rest energy (system not moving in frame of reference) is Eo=mc2

c2 is a conversion factor. It changes mass to energy.

If you increase the energy of the system, you increase its mass.In a fusion reaction, energy is produced, the mass of the productIs less than the mass of the reactants. Same for a fission reaction.

In a chemical reaction if energy is produced (like when O2 burns

Fat) the mass also decreases but is is a lot less than for nuclear Reaction. Try:A Bunsen burner adds 1000J of heat energy to a beaker ofWater. What is the increase in the mass of the water ?

(hint: ΔE=Δmc2) Is it a lot ?.

15

1)Suppose that an object has a mass-energy of 200 joules when it is atRest (Eo due to its mass).

A) What is its total energy when it is moving with a velocity of 0.9c ?(hint: E = Y Eo , use table to find Y)

Note: So the object has an increase of energy due to its speed.The increase is KErel

B) What is the kinetic energy of the particle at this speed?(E=KE + Eo)

2)If a 1kg mass is completely converted into energyu, how much energy , inJoules, would be released ?

3) A particle at rest energy 140MeV (Eo) moves at a sufficiently high speedThat its total relativistic energy is 280 MeV (E). How fast is it traveling ?E=KE + Eo

4) p. 353 7,8,10,11,12 (you need to google the mass of a protonAnd remember 1 eV = 1.6 10-19 J and 1 mega = 106)