What E= mc^2 actually does

We are bounded in a nutshell of Infinite space: Blog Post #21, Worksheet # 7.1, Problem #5: What \(E= mc^2 \) actually does

5. Gamma rays from the radioactive decay of nickel into iron drive most of the optical luminosity of a Type-Ia supernova. The process is given by: \[^{56}Ni \to ~ ^{56}Co + \gamma \to ~ ^{56}Fe + \gamma \] , where \(\gamma\) represents photons. 

The atomic weights of \(^{56} Ni\) and \(^{56} Fe\) are 55.942135 and 55.934941 amu, respectively. Let’s calculate the total energy radiated in the optical wavelengths during the event, given that the characteristic times for the two decay processes are 8.8 days and 111 days, respectively.

(a) Let’s balance the decay process from \(^{56} Ni\) to \(^{56} Fe\) for a single atom, ignoring the intermediate step. According to the first law of thermodynamics, energy cannot be created or destroyed. Use the fact that \(E= mc^2 \) to balance the equation.

(b) How many nickel atoms are there in the white dwarf? Use this number to estimate the total energy emitted in photons.

(c) Now combine the characteristic times for the two processes to find a total characteristic time. Divide the energy you found in part (b) by this time scale to find a characteristic luminosity.

(a) For this problem, a bit of chemistry could be necessary, but more important than that, its understanding the nature of what we are describing. A White Dwarf is a star that has been compressed and become composed of mostly heavy materials, such as nickel, iron, and other, heavier atoms. Because of the density, a White Dwarf, although small, is incredibly massive, and can thus release enormous amounts of energy in spectacular fashions. White dwarfs are also important because they are believed to be the star that initiates a type Ia Supernova explosion, a major part of the cosmic distance ladder we have explored in previous posts.

Now, we’ll see just how luminous these explosions are, which is when a White Dwarf starts initiating a reaction ending in a dazzling flash that outshines the galaxy it inhabits. This process begins, as the problem describes, with \[^{56}Ni \to ~ ^{56}Co + \gamma \to ~ ^{56}Fe + \gamma , \] a chemical process which releases the mass of the atoms in the form of pure energy, as photons. So as Nickel turns into Cobalt and then into Iron, the difference in the mass between the last and first step is: \[M(^{56}Ni) - M(^{56}Fe) = 55.942135 amu - 55.934941 amu , \] \[M(^{56}Ni) - M(^{56}Fe) = 0.007194 amu , \] where amu is atomic mass unit, and this mass represents the mass of the atoms turned into pure energy in the form of part of the electromagnetic spectrum in photons.

Taking this number and changing it into a cgs standard measurement, we use the mass of a proton per amu and multiply it by the change in mass: \[M_{Proton} = 1.6726219 \times 10^{-24} \frac{g}{amu} , \] \[\Delta M \cdot M_{Proton} = 0.007194 amu \cdot 1.6726219 \times 10^{-24} \frac{g}{amu} ,\] \[\Delta M \cdot M_{Proton} = .012032842 \times 10^{-24} g  ~.\]

Now we use the famous equation, \(E= mc^2 \), and plug in the value of the mass changed into energy per atom and constant of the universe, c, the speed of light. \[E = mc^2 , \] \[E_{Photon} = .012032842 \times 10^{-24} g   \cdot \left(3 \times 10^{10} \frac{cm}{s} \right)^2 ,\] \[E_{Photon}  = 1.083 \times 10^{-5} ergs . \]

(b) From earlier problems, we know the maximum mass a White Dwarf can have is \( M_{WD} = 1.4 M_\odot \) and if we were to divide this total mass by the mass of a single atom of 56 Nickel, we would find the total amount of Nickel atoms in a White Dwarf. So, knowing the mass of the sun is: \[M_\odot = 2 \times 10^{33} g ,\] \[ M_{WD} = 1.4 \times \times  2 \times 10^{33} g ,\] \[ M_{WD} = 2.8 \times 10^{33} g.\] And now finding the mass of an atom of nickel in grams as we found the mass turned  into photons: \[ M(^{56}Ni )= 55.942135 amu,\] \[M_{Proton} = 1.6726219 \times 10^{-24} \frac{g}{amu} , \] \[ M(^{56}Ni) = 55.942135 amu \times 1.6726219 \times 10^{-24} \frac{g}{amu} , \] \[ M(^{56}Ni )= 9.35700040134 \times 10^{-23}  g/atom .\] And now we divide the total mass of the star by the mass of an atom: \[\frac{M_{WD}}{ M(^{56}Ni )} = \frac {2.8 \times 10^{33} g }{9.35700040134 \times 10^{-23}  g/atom} \] \[^{56} +Ni~atoms~in~a~WD = 2.9424 \times 10^{55} atoms .\]

Since we have already found the energy produced by a single atom of 56 Nickel, we can take this photon energy, multiply it by the amount of atoms, and understand the energy the White Dwarf releases: \[Energy~Emitted= ~^{56}Ni~atoms~in~a~WD \cdot E_{Photon}  ,\] \[ Energy~Emitted = 2.9424 \times 10^{55} atoms \cdot 1.083 \times 10^{-5} \frac{ergs}{atom} ,\] \[ Energy~Emitted = 3.2406 \times 10^{50} ergs .\]

(c) Now that we know the energy that is produced in total by the explosion of the White Dwarf in photons, we can use this to find its Luminosity, a measure of energy over time. From the problem, we know the time period for the entire atomic process to occur is 8.8 days + 111 days, a total of 119.8 days. And turning this timescale into seconds, we have: \[ 119.8 days \times 24 \frac{hours}{day} \times 60 \frac{min}{hour} \times 60 \frac{seconds}{min} = 10,350,720 ~seconds\] Take these two values, and dividing them, we have the characteristic luminosity of the White Dwarf/Supernova Explosion: \[L = \frac{Energy~Emitted}{Period} ,\] \[L = \frac{3.2406 \times 10^{50} ergs }{1.0350720 \times 10^7 s},\] \[L= 3.1308 \times 10^{43} \frac{ergs}{s} .\]

One Theory for how White Dwarfs become Ia Supernovae

And another one

References:

https://astrofauna.files.wordpress.com/2012/04/merger_animation_rd255.gif

http://hetdex.org/images/dark_energy/supernova_explosion_34452.jpg