4. Relativity and its use in real life with applications

Physics topic relativity for engineering subject physics-1

1. 1,000,000 ms
-1
1,000,000
ms
-1
■HowfastisSpaceshipAapproaching
SpaceshipB?
■BothSpaceshipsseetheotherapproachingat
2,000,000ms
-1
.
■ThisisClassicalRelativity.

2. Einstein’s Special Relativity
1,000,000 ms
-1
0 ms
-1
300,000,000
ms
-1
Both spacemen measure the speed of the approaching ray of light.
How fast do they measure the speed of light to be?
•Stationary man 300,000,000 ms
-1
•Man travelling at 1,000,000 ms
-1
–301,000,000 ms
-1
?
–Wrong!
TheSpeedofLightis
thesameforallobservers

3. Observer in car
Observer on earth
Here we can see the Effect of Reference frame on the
Observations taken by the observer.

4. SPECIAL THEORY OF
RELATIVITY

5. “……..Undoubtedly mechanics was a
snapshot of slow and real motions, while
new physics is a snapshot of fabulously
swift and real motions……..”
V.I.Lenin

6. Fundamentals of Physics~ R.Resnick,
D.Halliday,
Fifth Edition~ John Wiley & Sons,
Inc.~ Singapore.
Fundamentals of Modern Physics~
Arthur Beiser ~ Fifth Edition ~1995

7. Special Relativity
•Special Relativity
–Galilean transformation
–Basic Postulates.
–Lorentz transformation
•Length contraction
•Time dilation
•Rest mass is least, relativistic mass.
•Energy mass equivalence.

8. •SpecialRelativity:Allmotionisrelative;thespeed
oflightinfreespaceisthesameforanobserver.
(Einstein1905)
•Framesofreferences;SandS’
–Inertialframeofreference;Newton’slaw’sofmotion
hold.
–Nouniversalframeofreference.“absolutemotion”
Generalrelativity:Describestherelationshipbetween
gravityandthegeometricalstructure.(notinsyllabus)

9. What is an inertial frame of reference?
A frame of reference is a system of
co-ordinates. An inertialframe is one
that’s not accelerating.

10. Frames of Reference

11. 2 Simple Postulates
•“Thelawsofphysicsarethesamein
everyinertialframeofreference”
–ThePrincipleofRelativity
•“Thespeedoflightinvacuumisthe
sameinallinertialframesofreference,
andisindependentofthemotionofthe
source”
–Invarianceofthespeedoflight

12. First Postulate
“The laws of physics are the same in
every inertial frame of reference”

13. Spot the Difference?

14. Newtonian Principle of Relativity
•IfNewton’slawsarevalidinonereferenceframe,then
theyarealsovalidinanotherreferenceframemovingata
uniformvelocityrelativetothefirstsystem.
•ThisisreferredtoastheNewtonianprincipleof
relativityorGalileaninvariance.
x
z
y
If the axes are also parallel, these frames are said to be Inertial
Coordinate Systems
O
S

15. Galilean-Newtonian Relativity

16. Experiment at rest
Experiment in moving frame
Same result. Ball rises and ends up in the thrower’s hand. Ball in
the air the same length of time.
Experiment looks different from ground observer (parabolic
trajectory, speed as a function of time) and observer on the truck.
However, they both agree on the validity of Newton’s laws.

18. Second Postulate
“The speed of light in vacuum is the same
in all inertial frames of reference, and is
independent of the motion of the source”
(Speed of light c= 3 x 10
8
metres/second
= 670,616,629 mph)

19. v

20. v
uu-v

21. v
c

22. Galilean transformation:
•Theequationrelatingthecoordinatesofa
particleintwoinertialframes(whose
relativevelocityisnegligibleincomparison
tothespeedoflight)arecalled
GALILEANTRANSFORMATION .

23. The Galilean Transformation
•For a point P:
–In one frame S: P = (x, y, z, t)
–In another frame S’:P = (x’, y’, z’, t’)
x
z
yS
P = (x, y, z, t)
P = (x’, y’, z’, t’)

24. Inertial Frames S and S’
•SisatrestandS’ismovingwithvelocity‘v’
•Axesareparallel
•SandS’aresaidtobeINERTIALCOORDINATE
SYSTEMS
24
S
S’

25. The Galilean Transformation
For a point P
In system S: P = (x, y, z, t)
In system S’: P = (x’, y’, z’, t’)
25
x
K
P
K’
x’-axis
x-axis
S
S’
vt X’

26. Conditions of the Galilean
Transformation
•Parallel axes
•K’ has a constant relative velocity in the x-direction
with respect to K
•Time(t) for all observers is a Fundamental invariant,
i.e., the same for all inertial observers
26
S S’

27. The Inverse Relations
Step 1. Replace with .
Step 2. Replace “primed” quantities with
“unprimed” and “unprimed” with “primed.”
27x x vt
S S’

29. Anobserverinthelaboratoryseestwoparticlescollidingat
x=20.5m,y=0,z=0andt=7.2s.Whataretheco
ordinatesofthiseventinaframemovingat30.5m/swith
respecttothelaboratoryframe?
Ans:x’=-199.1m,y’=0,z’=0andt’=7.2s.
x
z
yK
P = (x, y, z, t)
P = (x’, y’, z’, t’)
Numerical

30. The constancy of the speed of light
ConsiderthefixedsystemSandthemovingsystemS’.
Att=0,theoriginsandaxesofbothsystemsare
coincidentwithsystemS’movingtotherightalongthex
axis.
Aflashbulbgoesoffatbothoriginswhent=0.
Accordingtopostulate2,thespeedoflightwillbecinboth
systemsandthewavefrontsobservedinbothsystems
mustbespherical.S

31. The constancy of the speed of light is not
compatible with Galilean transformations.
Spherical wavefronts in S:
Spherical wavefronts in S’:
Note that this cannot occur
in Galilean transformations:
There are a couple
of extra terms (-
2xvt + v
2
t
2
) in the
primed frame.vx x t
yy
zz
tt



 2 2 2 2 2 2 2 2 2 2
( 2 v v )x y z x x t t y z c t          

32. The Lorentz Transformations
The special set of linear transformations that:
preservetheconstancyofthespeedoflight(c)
betweeninertialobservers;withtheassumption
thatthetimeisnotanabsolutequantity
Therefore t t
known as the Lorentz transformation equations
32

33. Derivation
33


Consider a system of two inertial frames of references S and S .
S: is at rest and S : is a moving frame with constant velocity v relative to S
(x,y,z,t) coordinates of event P for an observer O in
     

the stationary frame S
(x ,y ,z ,t ) coordinates of event P for an observer O in the moving frame S
Let x = (x-vt) ....(1) γ : proportionality constant.
According to first
  


postulate of Special Theory of Relativity
Therefore x= (x +vt ) ....(2) where t t
According to second postulate of Special Theory of Relativity
The wavefront along x, x axis must sa




tisfy
x=ct and x =ct
Thus substitute the value of x and x in eq (1)and (2) respectively.
ct = (c-v)t and ct= (c+v)t
Multiply these equations and solve it for 22
1
1 v / c



34. 2
x
x [ (x - vt) vt ] (x - vt) vt
x x x 1
t t (1 )
v v v
x = (x-vt) ....(1) x= (x +vt ) ....(2)
From eq(1) and eq(2)
or
t = or t =
similarly t
  
      


      

2
x1
t (1 )
v
=

  
 Derivation2
2
2
xv
t
c
t
v
1
c


 2
2
2
xv
t
c
t
v
1
c





35. Properties of γ22
1
1 v / c


Lorentz Transformation Equations

36. 22
v
1 v /
xt
x
c


 2
22
v/
1 v /
t x c
t
c


 yy zz The complete Lorentz
Transformation
If v << c, i.e., v/c≈0 and ≈ 1, yielding the familiar Galilean transformation.
Space and time are now linked, and the frame velocity cannot exceed c.22
v
1 v /
xt
x
c


 2
22
v/
1 v /
t x c
t
c


 yy zz
Length
contraction
Simultaneity
problems
Time
dilation

37. Consequence of Relativity
The Major Consequences To This Theory are:-
Length Contraction
Time Dilation
Mass Expansion

40. Lorentz Contraction
A fast-
moving
plane at
different
speeds.
v = 10% c
v = 80% c
v = 99.9% c
v = 99% c

41. Length contraction
Z
X
Y
Z’
Y’
X’
S
S’v2
o 2 1
1
0
22
22
22
oo 22
xx L
L
vv
11
cc
vv
or L L 1 , 1 1, So L L
cc







    

A rod is lying along the x axis of the moving frame S
In S frame, real length
L =x -x
Using Larentz Transformation
L is observed lenthe gth
Motion produces elastic froces which in turn
produce contra
(apparent le
ction in ato
of the rod
in S
mic constitution
ngth)
of m
fra
at
me.
ter.

42. 2 2 2 2 2 2 2 2 2 2
y z c t x y z c t
Imp: Prove that
x
under Lorentz Transformation
          22
o
1 v / c
o
Therefore L=L
The length (L ) of an object in motion with respect
to an observer (in S Frame) appears to be shorter,
this phenomenon is knownLorentz FitzGerald
contraction or Length
as
Cont
o
.
Case I : if v=0, L= ?
Case II: if v=c, L=?
Exercise : Find L when L =1m
and v
ract
=0
ion
.9c. Length contraction

43. •Time between ‘ticks’ = distance / speed of light
•Light in the moving clock covers more
distance…
–…but the speed of light is constant…
–…so the clock ticks slower!
•Moving clocks run more slowly!
V
Time Travel!

44. Time Dilation
2L
o
L
o

45. The time interval required for the
pulse to travel from O ' to mirror
back is
2distance traveled
speed
  
o
p
L
t
c Time DilationThe proper timeintervalisthetime interval
betweentwoeventsmeasuredbyanobserver
who seestheeventsoccuratthesame point
inspace.
L
o
2L
o

46. Time Dilation
2L
o

47. 22
2
2
2
2
2
1
1;
2
2
2
p
p
2
c t v t
22
22
Solve for t=
c
1
c
UsingEquation Where
1
t
tt
1
cc







o
oo
L
LL
t
v v
c
vv
 
   

   
   






 According to the second postulate of the special theory of relativity, both
observers must measure c for the speed of light. Because the light travels
farther according to O, the time interval measured by O is longer than
the time interval measured by O'. To obtain a relationship between these
two time intervals, we use this figure p
thetimeintervalΔtmeasuredby
anobservermovingwithrespecttoaclockislongerthantheti
Be
meintervalΔt
measuredbyanob
causeγ isalwaysgreaterthanun
serveratrestwithrespecttoth
ity
ecl
,thus
ock. Time Dilation
L
o

48. Time Dilation

49. Let us see the effect of Time dilationon a Radio active mass.
When the Disintegrating Mass is at Restw.r.t. the Observer.
Here Each Dot in the radioactive atom represents an atom
OOOO……..
So in 10 seconds 2out of 10
atoms are left.

50. When the Disintegrating Mass is at Movingw.r.t. the Observer.
Here Each Dot in the radioactive atom represents an atom
Impossible!!!!
Howinthesametimei.e.
10s,5atomsareleftoutof
10atoms.
Here the active mass is movingat an velocity comparable to c.

51. Aparticlewithaproperlifetimeof1µsmoves
at0.9c.Whatisitslifetimeasmeasuredby
observersinthelaboratory?
Ans:2.294μs.
Numerical

52. OnEarth,mostnaturallyoccurringmuonsarecreatedbycosmicrays,
whichconsistmostlyofprotons,manyarrivingfromdeepspaceatvery
highenergy
.
About10,000muonsreacheverysquaremeteroftheearth'ssurfacea
minute;thesechargedparticlesformasby-productsofcosmicrays
collidingwithmoleculesintheupperatmosphere.
Travelingatrelativisticspeeds,muonscanpenetratetensofmetersinto
rocksandothermatterbeforeattenuatingasaresultofabsorptionor
deflectionbyotheratoms.
What is muons?

53. Example of Time Dilation
6.610
2
m
Muon is created
Muon decays
Thelifetimeofμmesonsis2.2μsandtheir
speed0.998c,sothattheycancoveronlya
distanceof0.998cx2.2μsor0.66kmin
theirlifetime,andyettheyarefoundin
profusionatsealevel,i.e.,atadepthof10.4
kmfromtheupperatmospherewherethey
areproduced.Howmaythisbeexplainedon
thebasisof(i)lengthcontraction(ii)time
dilation?

54. Example of Time Dilation
Muon is created
Muon decays22
(i) Length contraction:
t 2.2 , L =2.2 0.998 0.66
Distance on the earth,
L L 1 / L 10.42 10
p
oo
s s c km
v c km km
  
     22
(ii) Time Dilation
t
t= 34.8
1/
distance travelled on the earth
34.8 0.998 10.42 10
p
ts
vc
s c km km





   

56. Twin paradox

57. Twin paradoxQuestion: Twins A and B are 20 years each. Twin A travels towards a star 30 light years
away at a speed 0.8c. He thenreturns home. Twin B remains at the home. What are the
two observations different? A. In B's frame of reference, The time spent in journey is 75
o
2l
t= years
0.8c
 75 0.6 45


p
t
The time spent by A t years   
 Hence in B's frame of reference, Age of A=20+45=65 years
Age of B=20+75=95 years B. From A's point of view, In his moving frame of reference, length appears to be contracted 
2
2
30 1 0.8 18
o 2
v
L L 1 light years
c
     18
22.5
0.8
L
Thustimetakeninthe outward trip is year
v
 Similarly time taken in the return trip is 22.5 years. 
p
Therefore time lapsed in A frame of reference t =45 years. 45 0.6 75
p
For B on the ground t= t years    Hence in A's frame of reference, Age of A=20+45=65 years
Age of B=20+75=95 years

58. Consider two lights which can be switched on at the
same time.
IfDobsonisexactlythesamedistancefrombulbA
asbulbB,andisnotmoving,hewillseeboth
lightsgoonatthesametime.
The eventof the lights turning on appears to
Dobson to be simultaneous.
NowsupposeDobsonismovingtotheright.Atthe
instantheisatthecenter,thelightsare
switchedon.WhatwillDobsonsee?
A B
A B
The relativity of simultaneity:
"Events which are simultaneous in one reference
frame may not be simultaneous in another."
Relativity of Simultaneity

59. Lorentz Consequence of Simultaneity of Events Consider two inertial frames S and S'. S' moving with velocity v relative to S
along positive direction of x-axis. If t is the time of an event at position x in
frame S, time t' in frame S' is gi
2
'
2
2
1
ven by
vx
t
c
t
v
c


 12
1 2 1 2 1 2
Suppose two events occurs simultaneously at instant (t = t ) at different positions
x and x (x x ) in reference frame S. The corresponding times t ' and t '
in frame S' are given by
 12
12 22
12
22
22
' & ' ;
11
vx vx
tt
cc
tt
vv
cc


 Will the two events occurring
simultaneously in S appea
Since space and time are relative, the question arises;
r simultaneously in S' or vi

The answece-ve r isrsa? No. Therefore   
2 1 1 2 2
21
2
2
''
1
  


v
t t x x
c
tt
v
c

60.  
 
   
122
1 2 1 2 2 1 2 1
2
2
1 2 1 2
& ' ' 0; ' '
1
v
xx
c
Butt t x x t t t t
v
c
That is two events simultaneously t t at different positions x x
in frame S are, in general, not simultaneously in another reference frame
Thus t
S'.
h

       


e two events do not appear simultaneous in S'.
Or
Simultaneity is not absolute. Thereisnothinglike“absolutetime”whichissameforallobserver.
Timeisrelativeandisdifferentforobserversinrelativemotion.

61. Velocity Addition
Letusconsider,somethingismovingalongXaxis
relativetobothS(rest)andS’(movingwithvelocityv).
AnobserverinSmeasuresthreecomponentofvelocityto
be
ToanobserverinS’theyare
BydifferentiatingtheLorentztransformequationsforx’,
y’,z’andt’,weobtain   
2
2 2 2 2
dx-vdt dt-vdx/c
dx = , dy =dy, dz =dz, dt =
1-v /c 1-v /c x y z
dx dy dz
u = , u = , u =
dt dt dt   
  
  
x y z
dx dy dz
u = , u = , u =
dt dt dt ˆˆˆ
x y z
u=iu +ju +ku ˆˆˆ' ' '
x y z
u'=iu +ju +ku

62.  
x
x 2 2 2
x
Now we can write
u -vdx dx-vdt dx/dt-v
u = = = =
dt dt-vdx/c 1-v/c dx/dt 1-vu /c
This is relativistic velocity transformation equation.
Its Inverse transformation equation v = -v is


 Velocity Addition

x
x 2
x
u +v
u=
1+vu /c

63. Velocity Addition



yz
22
y 2
22
2
22
y
y 2
x
By applying the same technique we can
obtain transformation for u and u as
dy 1-v /cdy
u = =
dt 1-vdx/c
dy/dt 1-v /c
=
1-v/c dx/dt
u 1-v /c
u=
1-vu /c 
22
z
z 2
x
Similarly,
u 1-v /c
u=
1-vu /c

64. Example
Try more numerical
1.Twophotonsapproacheachother.Whatistheirrelative
velocity?
2.Onephotonandoneelectron(0.9c)approacheachother.
3.Twoelectrons(0.7c)approacheachother.x
x
x 2
x
uc
uv cv
u
1 vu / c 1 v
Let , means if a ray of light is emitted in the moving reference
frame S’ in the opposite direction of motion relative to S, an observer
in frame S will measure the velocity

 


2
c(c v)
c
c / c (c v)




65. Relativity of Mass
AccordingtoNewtonianmechanicsthe
massofabodyisunaffectedwithchangein
velocity.
Butspaceandtimechange……..
Therefore“mass”ofabodyisnolongerbe
unaffected

66. Herewewillseehowmasschangeswithvelocity.
ConsidertwoframesSandS’.
Consideranelasticcollisionbetweentwoperfectlyelastic
sphericalballsAandBinS’andviewitfromframeS.
Relativity of Mass

67. Relativity of Mass
Lettwoballshavingmassm
1andm
2aretravelingwithvelocitiesu’
and–u’paralleltox-axisinsystemS’.
Supposetwobodiescollideandcollapseintoone
AftercollisiontheyareatrestinsystemS’(asthevelocityoftwo
wasequalandopposite).

68. Relativity of Mass
LetusseethiscollisionfromframeS.
Asperlawofadditionofvelocities,thevelocitiesu
1andu
2inframe
Scorrespondingtou’and–u’inS’are12
22
u ' v u ' v
u u ...(2)
u 'v u 'v
11
cc
...(1) and
  


Applyingtheprincipleofconservationof
momentum:
m
1u
1+ m
2u
2= (m
1+m
2) v

69. Aftercollisionthetwobodiescollapseintooneandmovewith
velocityvasS’ismovingwithvelocityvw.r.t.S.(Thisisobserved
fromS) 
1 1 2 2 1 2
12
1 2 1 2 22
12 22
2
1
2
2
m u m u (m m )v (3)
u ' v u ' v
m m (m m )v
1 (u ' v / c ) 1 (u ' v / c )
u ' vu ' v
m v m v
1 (u ' v / c ) 1 (u ' v / c )
m1 (u ' v / c )
m 1 (u ' v / c )
  
  
  


  
 



....
substituting the value of u and u

2
2
1 2
....(4)
u ' v
1 (u ' v / c )




Now squaring eq(1)
u
Relativity of Mass

70.   
 
  
  
2 2 2 2
2
1
22
2
2 2 2 2
2
22
1
2 2 2 2
2
22
2
1
2
1 v / c 1 u / c
c
1 (u ' v / c )
1 v / c 1 u / c
1 (u ' v / c )
1 (u / c )
1 v / c 1 u / c
1- (u ' v / c )
1 (u / c )
m1 (u ' v / c
m











u
1-
or
Similarly,
Substituting these values in eq (4), we get
222
2
2 2 2
1
2
2o
o
1
2
1
1 (u / c ))
1- (u ' v / c ) 1 (u / c )
0, m
m
m
1 (u / c






If the body of mass m is at rest or moving with zero velocity, in
stationary frame S that is u before collision , the rest mass
of the body ,
2
) Relativity of Mass

71. This is the relative formula for variation of mass with
velocity.
m
0is called the rest mass and m is the effective mass.   
1 2 1
2 2 2 2
In commonly used notation m , m and u
1
or m
1 / 1 /
o
o
o
m m v
mm
m
v c v c
  


Relativity of Mass

72. Whenvissmallascomparedtoc,v
2
/c
2
<<<1
m=m
0i.e.atvelocitiesmuchsmallerthancthe
massofthemovingobjectissameasthemassat
rest.
Whenviscomparabletoc,the1-v
2
/c
2
<1,m>m
0
massofthemovingobjectisgreaterthanatrest.
Whenvc,v
2
/c
2
1,som=orimaginary.Thisisnon
senseconcept. 
22
m
1/
o
m
vc


Relativity of Mass

73. KineticEnergy=TotalEnergy-Rest massEnergy Relation between Mass and Energy
Limiting Case: When v << c2
2 2 2 0
00
22
1/
mc
K mc m c m c
vc
   
 2
0
1
Show that .
2
K E m v
Limiting Case: When v c
K.E. tends to infinity. to accelerate the particle to the speed
of light infinite amount of work would be needed to be done .

74. 0 0 0
()
()
Integrating by parts ( )
s s mv
o
o
d m v
KE Fds ds vd m v
dt
xdy xy ydx

  

  
 o
o
o
o
o
o
o o o

75. Relativistic mass
Mass at rest.
Note:
1.Length contraction,
2.Time Dilation
3.Rest mass is least
Numerical:Astationarybodyexplodesintotwofragments
eachofmass1kgthatmoveapartatspeedsof0.6crelative
totheoriginalbody.Findthemassoftheoriginalbody.
Ans:2.5kg
Proof: do it yourself 
22
m
1/
o
m
vc

 2
2 0
22
2
0
TotalEnergy
1/
Rest mass energy or Potential energy
mc
E mc
vc
mc
  


Relativity of Mass

76. (1)PairProduction:whenaphotonofenergyequaltoorgreater
than1.02MeVpassesclosetoanatomicnucleus,itdisappears
andapairofelectronandpositroniscreatedi.e.
(GammaPhoton)=e
-
(electron)+e
+
(positron)
Positronisaparticleofsamemassaselectronbutequaland
oppositecharge.
(2)PairAnnihilation:whenanelectronandpositroncomeclose
together,theyannihilateeachotherandequalamountofenergy
isproducedintheformofapairof-rayphotons.
e
-
(electron)+e
+
(positron)=+
Examples of mass-energy equivalence

77. (1)PairProduction:whenaphotonofenergyequaltoorgreater
than1.02MeVpassesclosetoanatomicnucleus,itdisappears
andapairofelectronandpositroniscreatedi.e.
(GammaPhoton)=e
-
(electron)+e
+
(positron)
Positronisaparticleofsamemassaselectronbutequaland
oppositecharge.
(2)PairAnnihilation:whenanelectronandpositroncomeclose
together,theyannihilateeachotherandequalamountofenergy
isproducedintheformofaapirof-rayphotons.
e
-
(electron)+e
+
(positron)=+
Examples of mass-energy equivalence

78. Relation between Total energy (E) and momentum(p)  
 
 
2 2 4
22 00
2222
2 2 2
2200
2222
2 4 2 2 2
2 2 2 00
2 2 2 2
2 4 2 2
2 4 2 2 2
02 2 2 00
2 2 2 2
2 2 2 2 4
0
Now we have
1/1/

1/1/
(1 / ) (1 / )
1/
(1 / ) (1 / )
m c m c
E mc E
vcvc
m v m c v
and p mv p c
vcvc
m c m v c
E p c
v c v c
m c v cm c m v c
E p c
v c v c
E p c m c
   

   

   


  

 2 4 2 2
0
E m c p c

79. Relation between Kinetic energy (K) and momentum (p)2
0
2 4 2 2
2 4 2 2 2
0
1/ 2
22
2
024
But, E=
K=
or, K= 1 1
o
o
o
K E m c
m c p c
m c p c m c
pc
mc
mc


  





If v<<c, that is , p<<m
oc
2
, then
Neglecting higher order terms of binomial expansion, we get 22
2
0 24
K 1 ... 1
2
o
pc
mc
mc

  
 2
2
K As v<<c, then m =m
2
K (Classical Expression)
2
o
o
p
m
p
m


Limiting value of
Relativistic Kinetic Energy

80. Limiting value of Relativistic kinetic energy
Relativistic kinetic energy is 22
0
1/2
2
2
02
K mc m c
v
1 1 m c
c




  

 24
2
024
1 v 3 v
K 1 ............. 1 m c
2 c 8 c

    


Expanding by Binomial theorem
If v<<c i.e. v/c<<1 So2
22
002
1 v 1
K m c m v
2 c 2

This is expression for kinetic energy in non relativistic case.

81. 1) When m
0=o and v<c, E=0, p=0
2) When m
0=o and v=c, E=0/0, p=0/0
indeterminate form, hence must have E and p.
Mass less Particle? 2
2
2
2 0
2
0
2
mc
E mc and
v
1
c
mv
p mv
v
1
c
We have 




82. Mass less Particle? Answer : is Photon.2 4 2 2
0
22
2
(1)
For massless particle , m 0
or p= (2)
Since = / (3)
(2) (3)
i.e. the velocity of the massless particle is same as that of
light
o
E m c p c
E
E p c pc
c
p mv Ev c
From and v c
    

     
     

2
2
in free space.
The energy of the particle ,
Where m is the mass equivalent to energy E, that is , /
, , .:Examplesof mass Photon Nel ues trs par ions
E pc mc
m
Gravti it
Ec
on tes ccl se


A particle which has zero rest mass

83. Units of energy, mass and momentum
•Energy; eV, keV, MeV, GeV
•Mass; MeV/c
2
,
•Momentum: MeV/c
•Rest mass Energy of
–Electron: m
0c
2
=0.51MeV
–Proton: m
0c
2
=938 MeV
–Neutron: m
0c
2
=931 MeV
–Photon: m
0c
2
=0 as massless particle

84. Ex : Find the mass and speed of 2MeV electron.
m=3.55 10
-30
kg
v=2.90 10
8
ms
-1
Ex:Calculatethespeedofanelectronacceleratedthroughapotential
differenceof1.5310
6
volts.
Givenc=310
8
m/sec,m
o=9.110
-31
Kg,ande=1.610
-19
Coulomb
v=0.968c
Ex:Abodywhosespecificheatis0.2kcal/kg-
o
Cisheatedthrough
100
o
C.Findthe%increaseinitsmass.
%increase=9.3310
-11
%

85. Summary
•Special theory of relativity
–Basic Postulates
•Galilean transformation equations: v <<c
•Lorentz transformation equations: v ≈ c
•Length contraction:
•Time dilation:
•Addition of velocities
•Rest mass is least
•Energy –mass relationship
•Mass-less particle.2
2
1LL
0
c
v
 2
2
1
t
t
0
c
v

 2
2
1
m
m
0
c
v

 2242
0 cpcmE 

86. 2
2 2 2 2
1 t xv / c
LTE(v c) x (x vt) y y z z t
1 v / c 1 v / c

         

; ; , ; Summary2
2
2
2
0
0
t
Length Contraction L L 1 Time Dilation t
1
v
c
v
c
    
 GTE (v<<c) : v ; ; ;x x t y y z z t t         


22 22
y zx
x y z 2 2 2
x x x
u 1-v /c u 1-v /cu +v
Velocity addition u = u = u =
1+vu /c 1-vu /c 1-vu /c 2
2
0
m
•Rest mass is least: m
1
v
c

 Postulates of STR : Newton’s laws hold good, Invariance of speed of light 2 4 2 2
0
Energy mass relationship:
for mass less particle
E m c p c
E pc
  
