We now conclude our introduction to Einstein's theory of special relativity with what many regard as its most fundamental conclusion: the inertia of energy, or light. This is embodied in Einstein's famous equation:
E = m c^2
which is more accurately posed as:
E = (delta m) c^2
where delta m is the "mass defect" or difference, say in a nuclear reaction, and c is the velocity of light.
Before looking at examples, it's useful to consider the relativistic mass of a particle, in terms of its rest mass m(o). The rest mass, as the term implies is the mass of the object at rest or:
m(o) = m [(1 - v^2/c^2)^½ ]
Thus, if v = 0 (particle at rest) then we have:
m(o) = m(1)^½ = m
so the mass and rest mass are identical.
Now, the relativistic mass is then:
m = m(o) / [(1 - v^2/c^2)^½ ]
and again, if v = 0 then m(o) = m
But what if v = c? (Object moving at the speed of light?)
Then:
m = m(o) / [(1 - c^^2/c^2)^½ ] = m(o)/ [1 -1]^½ = m(o)/0 = oo
In other words, m would be infinite! This is another way of saying that to try to achieve the velocity of light one would have to overcome infinite inertia! In other words, it can't be done...not for a material object.
From this, we can also see the relativistic momentum must be:
p = mu = m(o) u / [(1 - c^^2/c^2)^½ ]
again, this approaches the classical value (p = mu) as u -> 0
Newton's 2nd law in the relativistic format is simply:
F = ma = m (du/dt) = d/dt[m(o) u / [(1 - v^^2/c^2)^½ ]
The relativistic energy is found by taking the integral of
(dp/du) u du
from 0 to u and obtaining:
W = mc^2/ [(1 - u^^2/c^2)^½ ] - mc^2
And by the work -energy theorem:
W = K(f) - K(i)
where K(i) is just the initial rest energy, or m(o)c^2
Then W = m(o)c^2/ [(1 - u^2/c^2)^½ ] - m(o) c^2 = (total energy - rest energy)
Example Problem:
Let's apply the basic mass-energy equation, E = (delta m) c^2, to the case of nuclear fusion.
Consider:
1H2 + 1H2 -> 2He3 + 2He3 + o n 1
which actually occurs in the Sun.
We now compile the masses (in atomic mass units) on each side:
2.015 u + 2.015 u -> 3.017 u + 1.009 u
or:
4.030 u -> 4.026 u
Now, the right side is less than the left by an amount equal to the mass defect or:
delta m = 4.030 u - 4. 026 u = 0.004 u
To get the energy E:
E = (0.004 u)(931 MeV/u) = 3.7 MeV
where 931 MeV/u is the conversion factor incorporating c^2
To transfer to Joules:
3.7 MeV = 3.7 MeV x (1.6 x 10^-13 J/MeV)= 6.0 x 10^-13 J
Example Problem (2):
Determine the energy required to accelerate an electron from 0.50c to 0.90c.
By the work -energy theorem:
W = K(f) - K(i)
K(i) = m(o)c^2/ [(1 - u1^2/c^2)^½ ]
u1 = 0.50 c
K(f) = m(o)c^2/ [(1 - u2^2/c^2)^½ ]
u2 = 0.90c
where: m(o) = 9.1 x 10^-31 kg
K(f) - K(i) = m(o)c^2/ [(1 - (0.90c)^2/c^2)^½ ]
- m(o)c^2/ [(1 - (0.50c)^2/c^2)^½ ]
K(f) - K(i) = m(o)c^2/[(1 - 0.81]^½ -m(o)c^2/[(1 - 0.25]^½
K(f) - K(i) = 2.294m(o)c^2 - 1.155 m(o)c^2 = 1.134m(o)c^2
Or:
K(f) - K(i) = 9.32 x 10^-14 J = 0.583 MeV
Other Problems:
(1) A spaceship of mass 10^8 kg is to be accelerated to 0.6c using a matter-antimatter mix engine.
(a) How much energy does this require?
(b) How many kilograms of matter and antimatter will it take to provide this much energy?
(2) Consider the decay:
24 Cr 55 -> 25 Mn55 + e
The Cr 55 nucleus has a mass of 54.9279 u and the Mn 55 nucleus has a mass of 54.9244u.
(a) Calculate the mass difference between the two nuclei.
(b) What is the maximum kinetic energy of the emitted electrons?
(3) Find the energy required to remove a simngle proton from 19 K 41.
(4)Find the speed and mass of an electron whose kinetic energy is 50 MeV.
(5) A rocket ship is to be accelerated to a speed of 0.5c. If propulsion is to be by using nuclear fuel, what fraction of the intial rest mass of the ship would have to be converted into kinetic energy to attain the desired speed?
What time dilation results if the speed is v = 0.5c?
Would this be sufficient to allow one generation of humans to reach the star Proxima Centauri (4.2 light years distant)?
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