In this video Murray Gell-Mann, also a Nobel Prize winner, talks about Fermi's ability to do extraordinary calculations with little effort. [](https://www.youtube.com/watch?v=69ol4eKymJ8) Derek Muller from Veritasium has a very interesting video where he goes to Trinity's ground zero, measures the current levels of radiation and inspects some of the effects of the explosion. [](https://www.youtube.com/watch?v=TRL7o2kPqw0&feature=youtu.be&t=246) Here is some [footage](https://www.youtube.com/watch?v=7dfK9G7UDok) of the actual explosion Enrico Fermi was an Italian physicist and Nobel Prize winner. He emigrated to the United States where he worked on the Manhattan Project during World War II. Fermi led the team that designed and built Chicago Pile-1, which in 1942 demonstrated the first artificial self-sustaining nuclear chain reaction. Fermi was one of the few physicists to excel both theoretically and experimentally and was known for his ability to make good approximate calculations with little or no actual data.  Trinity was the code name of the first detonation of a nuclear weapon, conducted by the United States Army at 5:30 am on July 16, 1945, as part of the Manhattan Project. The exact origin of the code name "Trinity" for the test is unknown, but it is often attributed to the physicist Robert Oppenheimer as a reference to the poetry of John Donne, which in turn references the Christian notion of the Trinity (three-fold nature of God). The test was conducted in the Jornada del Muerto desert near New Mexico less than one month before the bombing of Hiroshima and Nagasaki. In this picture we can see Oppenheimer visiting Trinity's ground zero  This is a very interesting statement. According to Fermi the light emitted during the explosion was such that at a distance of 10 miles it would be more intense than in "full daylight". To check if this statement is really true let's do some back of the envelope calculations in Fermi's fashion.  It's known that the solar luminosity ($L_{sun}$), meaning the total power emitted in the form of photons by the Sun that reaches the Earth is $3.8 \times 10^{33} \text{erg/s}$ (the atmosphere then causes a reduction of the extraterrestrial solar luminosity by about 30% on a very clear day). Luminosity is related to the irradiance of the Sun ($I_{sun}$ - power emitted per unit of area) and the distance between the Sun and the Earth ($d_{sun}$ - $9.3 \times 10^7 $ miles) $$ L_{sun} = 4 \pi d_{sun}^2 I_{sun} $$ According to the official records, the energy of the Trinity explosion was $20$ kilotons of TNT = $9.2 \times 10^{20}$ erg. During a nuclear explosion in the atmosphere, [50% of the energy goes to the blast and the other 50% is thermal radiation](https://en.wikipedia.org/wiki/Effects_of_nuclear_explosions) (light - from infrared to visible and ultraviolet), which means that when calculating the luminosity of the explosion ($L_{trinity}$) we can only take into account half of the energy. It's also safe to assume that most of the energy of the explosion was emitted in less than $0.1$ seconds. We can finally calculate $$ L_{trinity}=\frac{1}{2}\frac{9.2 \times 10^{20}}{0.01} = 4.6\times 10^{21} \text{erg/s} $$ Now to see if the light emitted by the explosion is more intense than in full daylight let's compare the two irradiances $$ \frac{I_{trinity}}{I_{sun}}=\frac{d_{sun}^2}{d_{trinity}^2}\frac{L_{trinity}}{L_{sun}}= \frac{(9.3 \times 10^7)^2}{10^2} \frac{4.6\times 10^{21}}{2.53 \times 10^{33}} \simeq 157 $$ Note that the explosion took place at night, since on July 16th 1945 the sunrise in New Mexico was at 6:07 am and the explosion was at 5:30 am. This means that at the moment of the explosion the sky was dark and for a fraction of a second the energy of the bomb was enough to light it up 157 times more intensely than in full daylight. Fermi was right! (as a reference, in a clear day the irradiance is 10 times more intense than in a cloudy day). The value for the energy of the explosion was actually kept secret for a long time. In 1947, photographs of the Trinity Test were released by the US Army. They generated great interest and appeared in newspapers and magazines all over the world.  Based on these photographs a British physicist named G. I. Taylor was able to estimate the power released by the explosion using dimensional analysis. He started by making 2 assumptions: 1. The energy (E) was released in a small space. 2. The shock wave was spherical. We have the radius of the fire ball (R as a function of t) at several times. The question is how does the radius varies depending on the energy (E), time (t) and density of the medium ($\rho$). If we now do a dimensional analysis of the variables involved in the problem: $[R] = L$, $ [E] = ML^2/T^2$, $[t] = T$ and $[\rho] = M/L^3$. We can say $$ [R] = L = [E]^x[\rho]^y[t]^z $$ If we now substitute the units for energy, time and density: $$ [R] = L = M^{x+y}L^{2x−3y}T^{−2x+z} $$ we end up with the following set of equations $$ x + y = 0 \\ 2x − 3y = 1 \\ −2x + z = 0 $$ which means that $x = 1/5, y = −1/5$ and $z = 2/5$. Finally we have $$ E \propto (R^5 \rho)/t^2 $$ Taylor then used the several values for R and t to calculate the energy. Note that at t = 0.006 seconds the radius of the shock wave was approximately 80 meters. $$ E = (80^5) × 1.2/(0.0062) = 1 × 10^{14} kg ∗ m^2/s^2 = \\25 \text{ kilotons TNT} $$  After the explosion the shock wave propagated through the air via multiple compressions and decompressions of the air molecules. As time went by, its energy spread over an increasingly larger sphere and the amplitude of the oscillations of the air molecules decreased. The energy released by an expanding gas across one direction is $$ E= Fd = PAd = P\Delta V $$  To calculate the volume displacement Fermi used the distance covered by the pieces of paper he threw in the air immediately after the shock wave passed by (2.5m). This means that the volume of the spherical shell of displaced air was $$\Delta V = 4\pi d_{trinity}^2 (2.5 \text{m})= 8\times 10^9 \text{m}^3$$ The pressure of the compression wave is approximately equal to $$ P = v_{sound}\rho_{air} wd_{paper} $$ where $v_{sound}$ is the velocity of the sound in the air, $\rho_{air}$ is the density of the air, $w$ is the angular frequency at which the pulse ramps up and down and $d_{paper}$ is the displacement amplitude (distance the pieces of paper moved). For shock waves in the air $w\simeq 1 s^{-1}$, $v_{sound} = 343 \text{m/s} $ and $\rho_{air}= 1.225 \text{kg}/m^3 $, which means that $$ P \simeq 10^3 N/m^2 $$ Calculating the energy of the shock wave we get $$ E_{wave}= P \Delta V \simeq 10^{13} \text{J} $$ Finally, since the energy of the blast accounts for 50% of the energy of the explosion we get $$ E_{total} \simeq 2 \times 10^{13} \text{J} = \frac{2 \times 10^{13} \text{J}}{4.2 \times 10^{9} \text{J/ton TNT}} = 5 \text{ kilotons TNT} $$ As we can see we get fairly close to Fermi's calculation - 10 kilotons TNT. This is of course a reconstruction of what I think his thought process was and there are a few approximations along the way that can account for the difference between the results. In particular, I suspect Fermi might have used a slightly lower percentage for the fraction of the energy allocated to the blast or a different estimator for the pressure. The idea here is to get an order of magnitude of the energy and not an exact number. A few days after the explosion it was confirmed that the exact energy was 20 kilotons TNT.