Day 5.6
Relativistic Energy
Instructions: In all the previous lessons, all of the equations and concepts of relativity have either
been developed from theory or experiment. This last equation is just given without justification.
There are conceptual approaches, but I feel that they are not worth the effort at this level. One
example can be found in the Perimeter Institute’s Revolution in Science resource.
1) Momentum and energy are related in relativity by E2 = m2c4 + p2c2 = 2m2c4
a) What is the formula for energy when the object is a rest? E = mc2
Instructions: This is the most famous physics equation and it is the approximation that we will be
using here. In the last unit, quarks to cosmos we will be looking at the other aspects of this equation.
b) Atomic masses are usually measured in atomic mass units, u, which is equal to 1.66053886 x 10-27
kg – about the mass of a proton. How much energy can you get from 1 u? (Use c = 2.99792458 x 10 8
m/s)
1.4924179 x 10-10 J
c) The joule is too large a unit so the MeV is preferred. What is the energy of one atomic mass unit in
MeV? (Use e = 1.60217646 x 10-19 C.)
931.1494988 MeV
2) Energy in radioactive decay: Uranium-236 can decay into Th-232 and an alpha particle.
a) How many neutrons and protons were there at the start and at the finish? Is mass conserved?
A periodic table shows that U has 92 protons and 236-92 = 144 neutrons. Th has 90 protons and 23290=142 neutrons. An alpha particle has 2 protons and 2 neutrons, so protons and neutrons are
conserved. are conserved. At this precision, mass is conserved.
b) More precise measurements of the mass of uranium-236, thorium-232 and the alpha particle are
236.045562 u, 232.0380510 u and 4.002602 u. How much mass has been lost? 0.004904 u
c) How much energy is emitted? E = 0.004904 x 931.15 MeV = 4.573 MeV.
Correct use of significant digits is crucial here. All of the digits in mass precision are needed in order
to have a non-zero difference in mass. The difference only has 4 significant digits, so you only need
to multiply it by five digits of the mass to energy conversion. You can use all the digits, but it is
unnecessary work.
d) How much energy is produced per mass of uranium? 4.5727/236.05 = 0.01937 MeV/u
3) Energy in fission: If uranium-235 (m = 235.0439299 u) absorbs a neutron (m = 1.008664916 u) to
become U-236 it will split to form krypton-92 (m = 91.92615613 u) and barium-141(m =
140.9144119 u) and three neutrons.
a) How much energy is released? Mass loss can only be precise to seven decimal places because that
is the limit for the given mass of uranium and barium. The mass difference is 0.1860321 and the
energy released is 0.1860321 x 931.14950 MeV = 173.2237 MeV.
b) How much energy is released per u of reactants? 173.2237/236.0526 = 0.7338352 MeV/u
Notice how fission produces almost 40 times more energy per u of reactant than alpha decay.
4) Energy from fusion: A neutron and proton (m = 1.007276466813 u) can form a deuteron (m =
2.014102 u).
a) How much energy is released? Mass loss is 0.001839 u, E = 1.712 MeV
b) How much energy is released per u of reactants? 1.712/2.016 = 0.8492 MeV/u
5) Energy for pair-creation: Dirac suggested that the negative roots of E2 were anti-matter.
a) What is anti-matter? Does it produce anti-gravity? Anti-matter is has the same mass but otherwise
is like the opposite of matter and can annihilate completely with its anti-matter twin. An
antiparticle has opposite charge and other quantum properties. Neutral particles like neutrons and
neutrinos have antiparticles. The photon is its own antiparticle. It does not produce antigravity. It
does not repel.
b) How much energy is released when a proton-antiproton pair annihilate? 2.01455293363 x
931.1494988 = 1875.849954 MeV.
c) How much energy is released per u of reactants? 931.494988 MeV/u! Doesn’t this look familiar?
Matter-antimatter annihilation involves converting all of the mass to energy. This is the maximum
amount of energy that you can get and is much more than fusion and fission.
d) Because mass and energy can be converted to each other, the mass is often given in units of in
MeV/c2 and often as MeV. In this last system of units, the speed is written in units of c, so the
speed of light is just 1. The electron has a mass of 9.11 x 10-31 kg. What is its mass in MeV?
If we use the conversions at the top 9.11/1.6605 x 108 = 5.48(6) x 108 u is 0.511MeV.
We don’t need to use atomic mass units here. Instead m = (9.11 x 9.00)/1.602 x 10-31+16+19 kg =
51.2 x 104eV = 0.512 MeV.
6) Energy in pair-annihilation: Watch http://www.triumf.ca/videos/life-sciences (4.24 minutes)
a) Where do the positrons and electrons in PET come from? The positrons are emitted by radioactive
material, F-18 attached to a biologically active molecule that the patient swallows. The cyclotron
fires protons at O-18 and produces F-18 and a neutron. The electrons are part of the person’s
body.
b) Why are two photons emitted instead of one? This conserves momentum.
c) How do the photons provide an image? The gamma photon pairs are picked up by a scanner. If
two arrive at about the same time, then they are assumed to come from the same annihilation
event and it occurred somewhere along the line between the two detectors. Computer analysis of
many lines allows locations of intersection points and then a construction of an image. Systems
with better time resolution (~ 3 ns) can also use the time of flight to locate the point of
annihilation within 10 cm.
d) Why is the F-18 attached to different molecules? It is attached to a molecule that is known to be
used near the site of interest. For example neurotransmitters like F-dopa go to the brain and
provide information for Parkinson research and FDG for tumour detection.
e) What is the frequency of the photons emitted? What colour is this light? f = E/h.
E = 5.11 x 1.602 x 10-14 J = 8.186 x 10-14 J. f= (8.186/6.63) x 10-14+34 Hz = 1.23 x 1020 Hz.
Visible light is 1015 Hz. This is gamma radiation.
7) In chemistry, you learn that mass is conserved in a chemical reaction. Is it?
No, mass-energy is conserved. However, the change in mass is very small. On page 581 the textbook
goes through an analysis and shows that when 1 kg of coal is burnt, it releases 32 MJ of energy,
which means that 3.6 x 10-10 kg of mass is converted to energy.
8) This mass-energy conversion occurs in variety of places. Which type keeps the Earth’s core molten
(decay), powers the stars (fusion), was used in WWII (fission), is used in H bombs (fission to
detonate, fusion for the energy), is used for nuclear power generation (fission, possibly fusion in the
future), is used for PET (electron-positron annihilation) scans, is used to hunt for the Higgs boson
(quark anti-quark annihilation and from this energy Higgs-antiHiggs pair creation)?
9) Compare the conversion of mass to energy for decay, fission and fusion. Why is fusion used for the
most powerful bombs but fission is used for energy generation? What has the potential to be the best
fuel for space travel? Fusion and fission produce much more energy per fuel mass than decay. Fusion
is more efficient than fission and the reactants are more plentiful and the products less hazardous.
However, it is technologically much easier to split atoms apart (fission) than to squeeze them together
(fusion). It can be done in a bomb but they are still trying to figure out how to do this in a safe,
controlled way for power generation. Matter-antimatter annihilation is the most efficient. However, it
takes a lot of energy to create anti-matter, it is hard to do this in large quantities and very hard to store
because it tends to annihilate its containers.
10) All the previous examples looked at particles at rest, where E = mc2. This question looks at moving
particles. How much work must be done to get an object of mass m to a speed of 3/5 c?
a) non-relativistically E = ½ mv2 = 9/50 mc2= 0.18 mc2
b) relativistically
It starts out with an energy of mc2 and ends up with mc2. The added energy is
(-1) mc2 = 1/4 mc2= 0.25 mc2. This is much more than the classical calculation.
c) Why can’t a particle with mass go at the speed of light? E = mc2and gamma becomes infinite. It
would take an infinite amount of energy.
d) For small speeds, v << c, it can be shown using the first two terms of the binomial expansion that
~ (1 + ½ v2/c2). What is the approximation for the energy of an object at small speeds?
E = mc2 = (1 + ½ v2/c2) mc2 = mc2 + ½ mv2 .For these speeds, rest mass energy and kinetic energy
can be treated as separate quantities. At high speeds, mass and momentum and energy are connected.
e) Light has no mass. Use the energy equation to show that it has momentum.
E2 = m2c4 + p2c2 and m = 0, so p = E/c. This result for light comes out of Maxwell’s equations
without relativity.
11) Go to Sixty Symbols: E = mc2. http://www.sixtysymbols.com/videos/emc2.htm (1:58-5:05) and
Minute Physics: E = mc2 is incomplete http://www.youtube.com/watch?v=NnMIhxWRGNw
Summarize their arguments about what important concepts are missed by E = mc2?
It misses the negative roots - anti-matter.
Light has momentum and can be bent by gravity.
Objects with mass can never go at the speed of light.
12) How does relativity affect the possibility of space travel? Consider t, d, v, m and E.
 Time is dilated so a trip to a star that is 10 light years away will take ten years for light to
travel. Therefore all communications between the expedition and Earth will have a roundtrip
delay of 20 years.
 However, the trip will feel like it takes less than ten years and the faster you go, the shortere
the distance seems and the less time the trip takes for the travellers.
 If they ever try to come back, they will feel like they have travelled into the future.
 It is much harder to reach these high speeds than classical physics predicts.
 E=mc2 gives a source of energy that will help reach these high speeds.
Textbook: 11.3
p. 583 #1-6