Size of a nuclear blast

Detailed calculation of how to find the radius of a blast wave using dimensional analysis and the work of G.I. Taylor to model the first nuclear explosion.
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G.I. Taylor's paper open access royalsocietypu...
My paper with Sam Rigby: link.springer....
Tom Scott visits Sam Rigby's lab • No-one knows how explo...
Series on Nuclear Weapons Physics • Physics of Nuclear Wea...
Credits:
Shot Easy, Operation Buster-Jangle footage, Los Alamos National Laboratory
Rudolf Peierls and John von Neumann portraits courtesy of AIP Emilio Segrè Visual Archives
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Klaus Fuchs' Los Alamos identity badge, Los Alamos National Laboratory
G.I. Taylor's paper open access by The Royal Society Publishing
Shot Grable, Operation Upshot-Knothole image and footage, Los Alamos National Laboratory
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[@jkzero]

To try everything Brilliant has to offer-free-for a full 30 days, visit brilliant.org/JKzero/ The first 200 of you will get 20% off Brilliant’s annual premium subscription.

[@jeremyocassan]

Another assumption I believe must be added is that all the energy suddenly released at a point goes into a shock wave. The initial energy release of a nuclear explosion is largely put into x-rays as well as high ionized plasma. A ball of x-rays expands as it eats surrounding air. Energy transfer inside is mostly radiative. Absorption by air contains the radiant energy so what is called the isothermal sphere expands at a rate dependent on how fast the swarm of x-rays can progress through air and convert it into x-ray producing plasma. The pressure is the same everywhere inside this sphere and there is no shock front. Once the isothermal sphere has cooled and the x-rays become UV does a shock front develop with a characteristic shock peak.

[@jkzero]

thanks fro this, you have a very good point. In the follow-up video I was more careful and called E0 the blast energy instead of the yield of the bomb

[@steveinmidtown]

no matter how many times I see the Trinity explosion, it never ceases to be terrifying.

[@jkzero]

I have been obsessed with the Trinity test since I was a kid, for many years this made me feel some gilt for liking something so horrible until one day I came up with the right wording for its description: terrifying beauty

[@Markoul11]

@@jkzero S=c/(speed of light on air)=1.0003

[@goldengoat1737]

@@jkzero I have been obsessed with the nuclear explosions since I was a kid too… I always wondered why I was so interested in them

[@jkzero]

@@goldengoat1737 same here

[@goldengoat1737]

@@jkzero I mean they are objectively really cool looking!😂 I think it is more than that though

[@marcmarc172]

Thanks for continuing to upload quality content.

[@jkzero]

thanks for the comment and support, creating this type of content is a pleasure when there is a community that appreciates the effort and gets engaged

[@RicksterX-92fs]

Please don’t waste your money on Brilliant. It certainly isn’t. It is expensive and poorly designed - no refunds when unsatisfied and you won’t be satisfied.

[@jkzero]

Sorry to read about your experience with Brilliant. I have used the app since last year and I still find it valuable.

[@vbfilipp]

Ha! I studied at the department of the Faculty of Mechanics and Mathematics, where Sedov was the head of the department. Unfortunately, I entered in 1999, the year when he passed. Still heartwarming to see him being mentioned, and knowing that this was one of his most known contributions to the field

[@jkzero]

Thanks for sharing, I only have two connections to the bomb: I got my doctorate at the physics department where Emil Konopinski taught, he is the person who calculated and confirmed that the Trinity test would not ignite the atmosphere; the other is that Rudolf Peierls, the first person to correctly calculate the critical mass for an atomic bomb, what the PhD advisor of the PhD advisor of my PhD advisor :D

[@michaelcheverie7579]

I just now found your video on Taylor's use of dimensional analysis. Ironically, when you were making this video three months ago I was discussing this very idea with a group of my high-school calculus students. I will be sharing your video with those students tomorrow. Very cool! Thank you!

[@jkzero]

Nothing makes happier than teacher finding the content of my videos suitable for their students. Yes, please, as long as the credit is given to the source, I am happy that my content is shared. Viewers have also made use of other videos in classrooms and I find it a great act of appreciation. I hope they like it and if there are questions please do not hesitate to post them here. I am curious to know what brings viewers to the channel, were you searching for something in particular or did the 'mighty algorithm' find you?

[@michaelcheverie7579]

I think I had watched your video about the math trick that led to quantum mechanics, and then this video about G.I. Taylor's analysis popped up in my feed. I was delighted, particularly so because of my discussion with my students. I've since subscribed to your channel, and I forward the link to your video to my students. By the way, I took a B.S. in physics from UCLA in 1979. I've never lost my interest in physics. Thank you for your videos!

[@duaalok]

Great work and thanks for sharing your hard work! Would be happy to see more of your work in the field of blast engineering. For a user of FE methods to simulate blast wave propagation, what do you feel about the efficacy of blast wave simulation using LS DYNA specifically for near-field explosions? Are the underlying equations relevant?

[@jkzero]

Thanks for watching and the positive feedback. I cannot guarantee to be able to fulfill all the requests but I am collecting suggestions, thanks. I am not an expert in blast engineering myself but I do know some who are so maybe in the future I could have him as a guess and have a Q&A session, what do you think?

[@duaalok]

That would be great.

[@aaronnorman9755]

Your videos are addictive!

[@jkzero]

thanks, I hope that means that you are binge-watching the channel's content. You can support me by liking, sharing, and subcribing

[@aaronnorman9755]

@@jkzero will do, perfect ratio of application to the nitty gritty detail, just wish you had more haha

[@jkzero]

@@aaronnorman9755 thanks again for your kind feedback, I am glad that people find the content of value. More coming soon.

[@fightingforcatalonia]

eres un fenómeno. Gracias por tu labor!

[@jkzero]

gracias por el comentario y apoyo, un gusto crear este tipo de contenido cuando es apreciado. El apoyo viene en las discusiones así como fomentando más vistas para que el algoritmo haga su magia. Un saludo y una pregunta: me interesa saber ¿qué te trajo por acá? ¿En busca de algún tema en especial?

[@fightingforcatalonia]

Soy físico y el tema detonaciones me lo debió recomendar YT. También sigo 3blue1brown por ejemplo. Sería un placer tomar algo contigo algún día!@@jkzero

[@alexandervoytov4966]

I’d like to mention discussed assumptions provided main estimations of a blast physics till advanced sensors became used to measure blast development. Simple theoretical models couldn’t accommodate numbers from experiments and numerical models of blast were created. Numerical models required new assumptions like an assumption on EOS, equation of state, for detonation products, assumptions on theoretical models of chemical reactions for chemical explosions etc. A topic for different video, IMHO.

[@jaimesiles8563]

Why does an upward air stream form when a fireball is very high temperature but it doesn’t happen when it’s just a warm air cloud?

[@jkzero]

It is due to the extremely high temperature of the fireball. A so-called Rayleigh-Taylor instability forms and the cooler air around pushes the hot fireball upwards.

[@t850]

...if I got this right, answer boils down to the simple phyisical law that: Work (Energy) = Force x Distance where: Force = Force whit what the air (atmosphere) is resisting to the aceleration = Mass (of the air witnin the "fireball" that is being "pushed out") x Acceration (of that same air) Distance = Radious of the "fireball" Since: Acceleration = Distance / Time^2 Mass (of the air) = Volume (of the "fireball") x Density equation can be expressed as follows: Energy = [Volume (of the "fireball") x Density] x [Distance / Time^2] x Distance Now we express volume and distance in the terms of Radious of the "fireball" (volume of a sphere) and revrite the whole equation as follows: Energy = [4 / 3 x Pi x Radious^3 x Density] x [Radious / Time^2] x Radious or Energy = 4/3 x Pi x Density x Radious^5 / Time^2 Since Pi is close to 3 the constant at the front can be approximated to 4 so the term takes the following form: Energy = 4 x Density x Radious^5 / Time^2 Now we express the Radious in the terms of Energy, Density and time as follows: Radious = 4 x Energy^(1/5) x Density^(-1/5) x Time ^(2/5) ...:)

[@jkzero]

in terms of physical dimensions, what you did is correct; however, conceptually it is not because you are using formulas valid for point particles to describe a complex system (fluid under an explosive shock). The only way to describe this properly is in the form of a differential equation that has to be integrated over the fluid. In fact your analysis makes no reference to an explosion. You could ask: what is the distance: is the the radius of the object, is the distance that it moves? what is the density, is it air density or the density of the moving object? what is the energy E, is it the work performed to deform the object? is it its kinetic energy? With no context for using each equation there is no way to know what physical system it is describing, despite the fact that the physical units match.

[@cewkins721]

Great video as usual, keep it up!

[@jkzero]

thanks, follow-up coming soon

[@brownruns]

I have a question about assumption number 2: for higher-yield bombs (Megaton region), would this assumption hold true? Glasstone and Dolan in The Effects of Nuclear Weapons talk about power over time and how there is a time where there is peak power output, but in higher-yield bombs energy is not all released at once. Thanks so much for your hard work!

[@jkzero]

this is a good question, the assumption of instantaneous release is closely linked to the relative size of the explosive device with respect to the yield. This is what makes chemical explosions fail the assumption #2 during the early stages (first microseconds) because for chemical explosives the relation between size and yield is much larger than for nuclear devices. In other words, assumption #2 is equivalent to "the ratio between explosive before the explosion and yield is very small," which is true for most nuclear and thermonuclear explosions. You can check the paper that I wrote (see Fig. 4 in link.springer.com/article/10.1007/s00193-022-01089-z), there you will see how the blast-wave evolution deviates from the Sedov-Taylor-von Neumann solution in early times but it works fine for all nuclear tests. For late times all explosions, chemical and nuclear, deviate when the blast wave decays into an acoustic wave, this is the breakdown of assumption #3 mentioned in the video.

[@domenicobarillari2046]

The nuclear energy release even in a 3-stage thermonuclear weapon takes not much longer than about 1 microsecond, and possibly up to an order of magnitude more or less. This can be estimated in detail with some information about bomb component sizes, but is characterized by average collision times for fast neutrons and for the travel time of X-rays to the secondary and its subsequent compression, so you can imagine why these are the right times scales. Now, you may be reading, in S&G about the propagation of energy out of the bomb material plasma - T ~ 1 giga Kelvin - where there are indeed distinct, new stages of energy transfer to the surrounding air. Largely, it is the somewhat prolonged and complex stage of X-ray ionization of air to fairly deep depths, with the blast originating in the expansion of layers of the latter (much like lightning). The simplest pattern is a "classic" two pulse phenomenon noted atmospheric tests. regards DKB

[@SuperYTPmaster]

Here before 100k followers (currently 2.56k)

[@jkzero]

thanks for the optimistic forecast, I hope to get there at some point so sharing, liking, and subscribing are highly appreciated so that the almighty algorithm can do its magic

[@driesvanoosten4417]

I have done (experimental) work on plume formation during femtosecond laser ablation. It processes similar to one-dimensional ST blasts...

[@jkzero]

that's interesting, you might want to check Sec. 4.4 in the paper that I mentioned in the video, we defined a dimensionless and yield-normalized system of time and space coordinates so that you can visualize a variety of shock waves on the same plot; there we included laser-induced shock waves and from nuclear tests in the same plot (see Fig. 4) link.springer.com/article/10.1007/s00193-022-01089-z

[@Grak70]

So in Taylor’s dimensional analysis, the radius goes as the ambient pressure to -1/5 power. What assumption did he make that causes the estimate to fail for an explosion in a near perfect vacuum where rho0~0?

[@jkzero]

one of the assumptions is that the explosion occurs in a medium, in free air to be more specific. His formula has to be cleverly corrected for explosions under water and under ground for this reason.

[@Grak70]

@@jkzero now that I think about it, isn’t that just a consequence of the choice of variables? The blast pressure isn’t a convenient variable because it isn’t constant or easily known, so eliminating P0 would make the problem difficult to simplify.

[@Grak70]

@@jkzero I still don’t see why the analysis doesn’t work in a vacuum. In the video you said this is not a shock front situation, so why does it matter that there’s a rarified medium at all vs a dense one? A thin/thick atmosphere would still give results, but as a predictor of radius vs time, this is going to diverge as pressure goes to zero. Where’s the breakdown? Does the value of S actually depend on the ambient pressure?

[@jkzero]

When you solve the system of partial differential equations you need boundary conditions, for a blast wave these are the Rankine-Hugoniot relations, which characterize the discontinuity of physical variables at the shock front produced by the compressed medium. In vacuum there would be no medium. You have to check the equations that describe the system instead of the final result; otherwise, the assumptions get hidden away. The details are presented in the paper below that I wrote a couple of years ago: set ho_0=0 and all the boundary conditions vanish eprints.whiterose.ac.uk/189567/1/D%C3%ADaz-Rigby2022_Article_BlastWaveKinematicsTheoryExper.pdf

[@Grak70]

@@jkzero thanks!

[@Martini_GP]

Thanks!

[@jkzero]

Your continuous support is much appreciated

[@KevinBalch-dt8ot]

Was this used by Fermi to calculate the yield with his confetti drop experiment?

[@jkzero]

this is not related to the legendary yield estimate by Fermi and his pieces of paper. I have searched for a clear explanation on how he did it but Fermi simply wrote that from the 2.5 m that his papers were displaced he inferred the yield to be 10 kt. I have found some articles about what Fermi did but nothing really satisfactory.

[@petro6049]

Très bon vidéo! Merci

[@jkzero]

thanks for watching

[@hypercomms2001]

Fantastic!

[@jkzero]

Thanks, glad you like it!

[@goldengoat1737]

Oh ok that equation totally makes sense 😂😂😂

[@bernardosaenz83]

Muy bueno jorge!

[@jkzero]

gracias, me alegra que a la audiencia le guste. Me interesa saber ¿qué te trajo por acá? ¿En busca de algún tema en especial?

[@R-C-R]

Love your content please upload h bomb working video

[@jkzero]

I am glad you find the content of interest, I cannot guarantee to be able to fulfill all the requests but I am collecting suggestions, thanks

[@Markoul11]

S=c/(speed of light on air)=1.0003

[@jkzero]

this is the index of refraction of air, it is completely unrelated to the function S(\gamma) shown in the video. It is also the wrong value, S(air)=1.033

[@Markoul11]

@@jkzero I see. Thank you for clarifying this.

[@driesvanoosten4417]

It never ceases to amaze me that people just spread nonsense (like Taylor's supposedly assuming S=1). It just shows they didn't read past the first page, if they read the paper at all. If I remember correctly, he says somewhere that S is a constant expected to be of order unity, which is ofcourse something very different from S=1...

[@jkzero]

totally agree, I didn't say it the video because I didn't want to sound patronizing, but for all the people/videos/blogs that simple say "Taylor used S=1" I want to say: dude, read the f-ing paper!" The whole paper is about calculating S!

[@benquinneyiii7941]

Typhoon George Washington

[@satanofficial3902]

The heavens declare the glory of the Bomb, and the firmament showeth Its handiwork. Glory be to the Bomb, and to the Holy Fallout. As it was in the beginning, is now, and ever shall be. World without end. May the Blessings of the Bomb Almighty, and the Fellowship of the Holy Fallout, descend upon us all. This day and forever more. Amen.

[@jkzero]

whoa.