Relativistic kinetic energy
By Lisa Jacobson
The Basics
• Mass increases as the speed of an object
increases
– Therefore energy also increases (mass is a form
of energy)
• Einstein tried to find mathematical mass-energy
relationship, but KE=1/2 mv2 doesn’t work at
high speeds .......................................................
But this formula does:
KE = mc2 – moc2
E = mc2
(energy of object with velocity)
Eo = moc2
(rest energy)
E = moc2 + KE (total energy = rest energy
+kinetic energy)
DE = (Dm)(c2)
(changing energy in ANY system will change the
mass in the system)
An electron (mo = 9.11 x 10-31 kg) travels at a speed
v = 0.85c = 2.5 x 108 m/s.
What is the kinetic energy? Compare to a classical
calculation.
An electron (mo = 9.11 x 10-31 kg) travels at a speed
v = 0.85c = 2.5 x 108 m/s.
What is the kinetic energy? Compare to a classical
calculation.
mo
9.11 x 10-31 kg
-30 kg
m=
v2 1/2 =
=
1.73
x
10
(1 – 2 )
(1 – 0.852 )1/2
c
An electron (mo = 9.11 x 10-31 kg) travels at a speed
v = 0.85c = 2.5 x 108 m/s.
What is the kinetic energy? Compare to a classical
calculation.
mo
9.11 x 10-31 kg
-30 kg
m=
v2 1/2 =
=
1.73
x
10
(1 – 2 )
(1 – 0.852 )1/2
c
KE = (m – mo)c2 = (-9.11 x 10-31 kg + 1.73 x 10-30
kg)c2
KE = 7.37 x 10-14 J
An electron (mo = 9.11 x 10-31 kg) travels at a speed
v = 0.85c = 2.5 x 108 m/s.
What is the kinetic energy? Compare to a classical
calculation.
mo
9.11 x 10-31 kg
-30 kg
m=
v2 1/2 =
=
1.73
x
10
(1 – 2 )
(1 – 0.852 )1/2
c
KE = (m – mo)c2 = (-9.11 x 10-31 kg + 1.73 x 10-30
kg)c2
KE = 7.37 x 10-14 J
1
1
2
KE = mov = (9.11 x 10-31 kg)(2.5 x 108)2
2
2
KE = 3.0 x 10-14 J [less then half of correct result]
How much energy would be released if a po meson
(mo = 2.40 x 10-28 kg) decayed completely into
electromagnetic radiation?
2.16 E -11
How much energy would be released if a po meson
(mo = 2.40 x 10-28 kg) decayed completely into
electromagnetic radiation?
Eo = moc2
2.16 E -11
How much energy would be released if a po meson
(mo = 2.40 x 10-28 kg) decayed completely into
electromagnetic radiation?
Eo = moc2 = (2.40 x 10-28 kg)c2 = 2.16 x 10-11 J
or kgm2/s2
Relativistic momentum
m ov
p = mv =
v2 1/2
(1 – 2 )
c
(Relativistic momentum)
E = mc2
E2 = m2 c4 = m2 c2 (c2 + v2 – v2 )
= m2c2v2 + m2c2(c2 – v2)
(since p = mv and m = mo/(1- v2/c2)1/2 )
= p2c2 + m2oc4(1 – v2/c2)/(1 – v2/c2)
E2 = p2c2 + mo2c4
(Energy-momentum relation)
An electron’s mass gets ten times as large as its rest mass (9.11 E -31)
when it is traveling at 2.00 E 8 m/s. What is its momentum?
E2 = p2c2 + m2oc4
(10*9.11E-31 *c2)2 = p2c2 + (9.11E-31)2 c4
= 2.72 E -21 kgm/s
2.72 E -21 kgm/s