Chapter 26

Special Relativity

The Postulates of Special Relativity

THE POSTULATES OF SPECIAL RELATIVITY 1. The Relativity Postulate. The laws of physics are the same in every inertial reference frame. 2. The Speed of Light Postulate. The speed of light in a vacuum, measured in any inertial reference frame, always has the same value of c, no matter how fast the source of light and the observer are moving relative to one another.

L = L0γ

⇒ L ≤ L0Length contraction

Δt = γΔt0 ⇒ Δt ≥ Δt0Time dilation

Some consequences of Einstein’s two postulates of Special Relativity

γ ≡1

1− v2

c2

≥1where, Δt0 =proper time, ΔL0 = proper length, and

Some consequences of Einstein’s two postulates of Special Relativity -- continued

Mass Increase m = γm0 ⇒ m ≥m0

Relativistic momentum

Total relativistic energy

p =mv = γm0v

E =mc2 = γm0c2

m = relativistic mass, m0 = rest mass

For a moving object:

Relativistic Momentum

Relativistic momentum

p = m0v

1− v2

c2

pclassical =m0vNonrelativistic momentum

Comparison between relativistic and nonrelativistic momentum

The Equivalence of Mass and Energy

RELATIVISTIC ENERGY OF AN OBJECT

Total energy of an object (with PE = 0)

Rest energy of an object (v = 0)

E = γm0c2 =

m0c2

1− v2 c2

Eo =m0c2

Kinetic Energy ≡KE = E −Eo =m0c2 γ −1( )

à Intrinsic energy of an object with rest mass m0

Kinetic energy in the classical limit of v à 0 :

KE =m0c2 γ −1( ) =m0c

2 1

1− v2

c2

−1

"

#

$$$$

%

&

''''

1+ x( )α =1+αx +α α −1( )2

x2 +

1

1− v2

c2

= 1− v2

c2"

#$

%

&'

−12=1+ −

12

"

#$

%

&' −

v2

c2"

#$

%

&'+≈1+ v2

2c2for v→ 0

∴KE ≈ m0c2 1+ v2

2c2−1

"

#$

%

&'=12m0v

2

Binomial theorem

α x

Useful relationship between E and p :

E =mc2 p =mv m = γm0 γ =1

1− v2

c2

p2 = m0v2

1− v2

c2

⇒ p2 − p2 v2

c2=m0

2c2 v2

c2⇒

v2

c2=

p2

m02c2 + p2

1− v2

c2=m02c2 + p2( )− p2m02c2 + p2

=m02c2

m02c2 + p2

E 2 =m02c4

1− v2

c2

=m02c4

m02c2

m02c2 + p2

#

$%

&

'(

= p2c2 +m02c4 ⇒ E = p2c2 +m0

2c4

How much energy does it take to accelerate a mass m0 to c ?

v = c ⇒ E =mc2 = γm0c2 =

m0c2

1− c2

c2

=m0c

2

1−1=m0c

2

0→∞

Therefore, it would take an infinite amount of energy to accelerate any non-zero mass to c à c is the absolute speed limit of the Universe!

Van de Graaff generator used as a particle accelerator. An electron is accelerated vertically starting from rest across a +100,000 V potential difference produced by a Van de Graaff. Find the final electron a) kinetic energy, b) rest mass energy, c) total relativistic energy, d) relativistic mass, e) momentum, and f) speed. Compare the result in part f) with the classical calculation.

VB = +105 V

VA = 0 V

electron

a) ΔKE = KE − 0 = −ΔPE = −qΔV = − −1.60×10−19( ) 105( ) =1.60×10−14 J =100 KeVb) E0 =m0c

2 = 9.11×10−31( ) 3.00×108( )2= 8.20×10−14 J = 512 KeV

c) E = KE +E0 =1.60×10−14 +8.20×10−14

= 9.80×10−14 J = 612 KeV

d) E =mc2 ⇒ m =Ec2=9.80×10−14

3.00×108( )2 =10.9×10

−31kg

q = −1.60×10−19C, melectron = 9.11×10−31kg

vA = 0 m s

e) E 2 = p2c2 +m02c4

⇒ p = E 2 −m02c4

c=

9.80×10−14( )2− 9.11×10−31( )

23.00×108( )

4

3.00×108

=1.79×10−22 kg ms

f ) p =mv ⇒ v = pm=1.79×10−22

10.9×10−31=1.64×108m s = 0.547 c

From classical mechanics:

KE = 12m0v

2 ⇒ v = 2KEm0

=2 1.6×10−14( )9.11×10−31

=1.87×108m s

Classical mechanics gives a speed that is about 14% too high in this case.

The Equivalence of Mass and Energy

Example: The Sun is Losing Mass The sun radiates electromagnetic energy at a rate of 3.92 x 1026 W. What is the change in the sun’s mass during each second that it is radiating energy? What fraction of the sun’s mass is lost during a human lifetime of 75 years (msun = 1.99 x 1030 kg).

The Equivalence of Mass and Energy

( )( )( )

kg1036.4sm103.00

s 0.1sJ1092.3 928

26

2 ×=×

×=

Δ=ΔcEm o

( )( ) 1230

79

sun

100.5kg1099.1

s1016.3skg1036.4 −×=×

××=

Δ

mm

75 years

The Equivalence of Mass and Energy

Conceptual Example: When is a Massless Spring Not Massless? The spring is initially unstrained and assumed to be massless. Suppose that the spring is either stretched or compressed. Is the mass of the spring still zero, or has it changed? If the mass has changed, is the mass change greater for stretching or compressing?

PE = 12kx2

Example. A 8 TeV proton from the Large Hadron Collider. The LHC can accelerate protons to 8 TeV kinetic energy. Find γ for an accelerated proton (mproton = 1.67 x 10-27 kg). What is its speed?

E =mc2 = γm0c2

E0 =m0c2 = 1.67×10−27( ) 3.00×108( )

2=1.50×10−10 J = 0.94GeV

γ =E

m0c2 =

KE +E0m0c

2 =8000GeV + 0.94GeV

0.94GeV= 8500

γ =1

1− v2

c2

⇒ v = c 1− 1γ 2

= c 1− 185002

= 0.999999993 c

Proton speed at 8 TeV kinetic energy:

Very close to c !

Example. Energy needed to accelerate an object to a speed near c. If kinetic energy is supplied to an object equal to its rest mass, what speed will the object have?

E = KE +E0 =mc2 = γm0c

2

If KE = E0 ⇒ E = 2E0 = γm0c2

1γ=m0c

2

2E0=m0c

2

2m0c2 =

12

⇒ γ = 2

1γ 2

= 1− v2

c2#

$%

&

'(=14

⇒ v = 32c = 0.866 c

Time slowed to ½ in object’s rest frame Not far from c

How much energy must be supplied to accelerate a 5 metric ton spacecraft to this speed?

E0 =m0c2 = 5000( ) 3.00×108( )

2= 4.5×1020 J

About 1 millionth of the E&M output of the Sun in 1 second (maybe if you had a large enough solar sail over a 1 year period………)