E=mc 2 Physics 100 Chapt 19

Einstein’s hypotheses: 1. The laws of nature are equally valid in every inertial reference frame. 2. The speed of light in empty space is same for all inertial observers, regard- less of their velocity or the velocity of the source of light.

Summary (so far) Moving clocks run slow Moving objects appear shorter Moving object’s mass increases By a factor of 

GPS system 3 m precision requires corrections for relativity

What is the significance of the increase mass of a moving object?? m=  m 0 increase in mass = m – m 0 increase in m=  m 0 -m 0 =(  -1)m 0 Summer 1905

Increase in m = (  -1)m 0 1  1 – v 2 /c 2  v/c  -1 =??

Mathematicians tell us: 1  1 – v 2 /c 2  v/c = 1 + ½ + other stuff v2c2v2c2 1  1 – v 2 /c ½1 + ½ v2c2v2c2 The “other stuff” is pretty huge near v=c What about here?

Below v/c = 0.5, “other stuff” is not so large 1 + ½1 + ½ v2c2v2c2 v/c 1  1 – v 2 /c  1 – v 2 /c 2  = 1 + ½ + other stuff v2c2v2c2 At v=0.5c the “other stuff”is about 10% How about down here?

Below v/c = 0.1, “other stuff” is negligible v=0.1c 1 + ½1 + ½ v2c2v2c2 1  1 – v 2 /c 2 1  1 – v 2 /c 2  = 1 + ½ + other stuff v2c2v2c2

Below v/c = 0.1,  = 1+½  – 1 = ½ v2c2v2c2 v2c2v2c2 This covers all velocities ever encountered by mankind in 1905

Increase in mass = (  -1)m 0 for v<0.1c =  -1 = ½ v2c2v2c2 increase in mass = ½ m 0 (increase in mass)c 2 = ½ m 0 v 2 v2c2v2c2 Kinetic energy Multiply both sides by c 2

For v < 0.1c m = m 0 + increase in m mc 2 = m 0 c 2 + (increase in m)c 2 mc 2 = m 0 c 2 + kinetic energy Multiply both sides by c 2

Einstein’s hypothesis: mc 2 = m 0 c 2 + kinetic energy holds for all velocities (not only for v<0.1c) Kinetic energy = (m - m 0 )c 2 = ½ m 0 v 2 + “other stuff” Since this is only significant for huge speeds, Newton, etc had no way of knowing about it

mc 2 = m 0 c 2 + kinetic energy This must also be a form of energy:  ”rest mass energy” This must be the total energy E = mc 2

Mass is a form of energy E = mc 2 The conversion factor from mass to energy is c 2, a huge number! Energy (Joules) = Mass (kg) x (3x10 8 ) 2 =9x10 16 E (Joules) = 90,000,000,000,000,000 x m (kg)

Mass contains an enormous amount of energy

Electric power usage on Oahu Power =1,300 MegaWatts = 1,300 x 10 6 Watts Energy usage = Power x time Power = 1.3 x 10 9 J/s Energy usage/yr =1.3x10 9 J/s x (3.2 x 10 7 s/yr) =1.3 x 3.2 x J/yr =4.2 x J/yr # of secs in 1 yr

How much mass would be needed to produce 4.2 x J ? E = mc 2  m = Ec2Ec2 4.2x10 16 J 9x10 16 m 2 /s 2 m = kg m = 0.47 kg About the mass of an apple

Nuclear Power Plant Converts about 1kg of mass  energy/yr

Another nuclear power plant Sun’s power output=4x10 26 Watts

Sun converts mass to energy Power = energy generated in one sec: E 1sec = 4x10 26 J = m 1sec C 2 m 1sec = E 1sec C 2 = 4x10 26 J 9x10 16 m 2 /s 2 m 1sec = 4/9x kg= 0.44x10 10 kg m 1sec = 4.4x10 9 kg over 4 million tons each second!!

Another nuclear power plant 4.4x10 9 kg  energy / sec