Using dimensional analysis to estimate the energy released by an atomic bomb

Legend has it that in the early 1950s, British physicist G. I. Taylor was visited by some very serious men from the military authorities. His crime? He had apparently secured unauthorised access to worryingly accurate and top secret information about the energy released by the first atom bomb.

Sir G. I. Taylor (1896-1965)

Taylor explained that, actually, he hadn’t: he had estimated the energy yield from a series of photographs of the first atomic test explosion published by Life magazine. Taylor had used the standard physics technique known as dimensional analysis.

Part of the sequence of photographs of the Trinity atomic weapon test (16/7/45) published by Life magazine in 1950

The published pictures had helpfully included a scale to indicate the size of the atomic fireball in each photograph and Taylor had been able to complete a back-of-the-envelope calculation which gave a surprisingly accurate value for what was then the still highly classified energy yield of an atomic weapon.

This story was shared by the excellent David Cotton (@NewmanPhysics) on Twitter, and included a link to a useful summary which forms the basis of what follows. (NB Any errors or omissions are my own.)

It is presented here for A-level Physics teachers to consider using as an example of the power of dimensional analysis beyond the usual “predicting the form of the equation for the period of a simple pendulum”(!)

Taylor’s method: step one

Taylor began by assuming that the radius R of the fireball would depend on:

  • The energy E released by the bomb. The larger the energy released then the larger the fireball.
  • The density of the air ρ. The greater the density of the air then the smaller the fireball since more work would have to be done to push the air out of the path of the fireball.
  • The time elapsed t from the explosion. The longer the time then the larger the size of the fireball (until the moment when it began to collapse).

These three factors can be combined into a single relationship:

k is an unknown arbitrary constant. Note that we would expect the exponent y to be negative since R is expected to decrease as ρ increases. We would, however, expect x and z to be positive.

Taylor’s method: step two

Next we think of the dimensions of each of the values in terms of the basic dimensions or measurements of length [L], mass [M] and time [T].

  • R has the dimension of length, so R = [L].
  • E is in joules or newton metres (since work done = force x distance). From F=ma we can conclude that the dimensions of newtons are [M] [L] [T]-2. This makes the dimensions of energy [M] [L]2 [T]-2.
  • ρ is in kilograms per cubic metre so it has the dimensions [M] [L]-3.
  • t has the dimension of time [T].

Taylor’s method: step three

Next we write equation 1 in terms of the dimensions of each of the quantities. We can ignore k as we assume that this is a purely numerical value with no units. This gives us:

Simplifying this expression, we get:

Taylor’s method: step four

Next, let’s look at the exponents of [M], [L] and [T].

Firstly, we can see that x + y = 0 since there is no [M] term on the left hand side.

Secondly, we can see that 2x – 3y = 1 since there is an [L] term on the left hand side.

Thirdly, we can see that z – 2x = 0 since there is no [T] term on the left hand side.

Taylor’s method: step five

We now have a system of three equations detailing three unknowns.

We can solve for x, y and z using simultaneous equations. This gives us x=(1/5), y=(-1/5) and z=(2/5).

Taylor’s method: step six

Let’s rewrite equation 1 using these values. This gives us:

Rearranging for E gives us:

Taylor’s method: step seven

Next we read off the value of t=0.006 s and estimate R=75 m from the photograph. The density of air ρ at normal atmospheric pressure is ρ=1.2 kg/m3.

If we substitute these values into equation 6 (assuming that k=1) we get E= 7.9 x 1013 joules.

Conclusion

Modern sources estimate the yield of the Trinity test as being equivalent to between 18-20 kilotons of TNT. Let’s take the mean value of 19 kilotons. One kiloton is equivalent to 4.184 terajoules. This means that, according to declassified sources that were not available to Taylor, the energy released by the Trinity test was 7.9 x 1013 joules.

As you can see, Taylor’s “guesstimated” value using the dimensional analysis technique was remarkably close to the actual value. No wonder that the military authorities were concerned about this apparent “leak” of classified information.

Through a glass, lightly

Any sufficiently advanced technology is indistinguishable from magic.

Profiles of the Future (1962)

So wrote Arthur C. Clarke, science fiction author and the man who invented the geosynchronous communications satellite.

Clarke later joked of his regret at the billions of dollars he had lost by not patenting the idea, but one gets the impression that what really rankled him was another (and, to be fair, uncharacteristic) failure of imagination: he did not foresee how small and powerful solid state electronics would become. He had pictured swarms of astronauts crewing vast orbital structures, having their work cut out as they strove to maintain and replace the thousands of thermionic valves burning out under the weight of the radio traffic from Earth . . .

A communications nexus that could fit into a volume the size of a minibus, then (as the technology developed) a suitcase and then a pocket seemed implausible to him — and to pretty much everyone else as well.

And yet, here we are, living in the world that microelectronics has wrought. We are truly in an age of Clarkeian magic: our technology has become so powerful, so reliable and so ubiquitous — and so few of us have a full understanding of how it actually works — that it is very nearly indistinguishable from magic.

I want to outline a fictional vignette on the same theme which Clarke wrote in 1961 which has stayed with me since I read it. Please bear with me while I sketch out the story — it will lend some perspective to what follows.

In the novel A Fall of Moondust (1961) set in the year 2100, the lunar surface transport Selene has become trapped under fifteen metres of moondust. Rescue teams on the surface have managed to drill down to the stricken vessel with a metal pipe to supply the unfortunate passengers and crew with oxygen from the rescue ship. Communication would seem to be impossible as the Selene’s radio has been destroyed, but luckily Chief Engineer (Earthside) Lawrence has a plan:

They would hear his probe, but there was no way in which they could communicate with him. But of course there was. The easiest and most primitive means of all, which could be so readily overlooked after a century and a half of electronics.

A few hours later, the rescue team’s pipe drills through the roof of the sunken vessel.

The brief rush of air gave everyone a moment of unnecessary panic as the pressure equalised. Then the pipe was open to the upper world, and twenty two anxious men and women waited for the first breath of oxygen to come gushing down it.

Instead, the tube spoke.

Out of the open orifice came a voice, hollow and sepulchral, but perfectly clear. It was so loud, and so utterly unexpected, that a gasp of surprise came from the company. Probably not more than half a dozen of these men and women had ever heard of a ‘speaking tube’; they had grown up in the belief that only through electronics could the voice be sent across space. This antique revival was as much a novelty to them as a telephone would have been to an ancient Greek.

The humble converging lens as an ‘antique revival’

If you hold a magnifying glass (or any converging lens) in front of a white screen, then it will produce a real, inverted image of any bright objects in front of it. This simple act can, believe it or not, draw gasps of surprise from groups of our ‘digital native’ students: they assume that images can only be captured electronically. The fact that a shaped piece of glass can do so is as much a novelty to them as an LCD screen would have been to Galileo.

Think about it for a moment: how often has one of our students seen an image projected by a lens onto a passive screen? The answer is: possibly never.

The cinema? Not necessarily — many cinemas use large electronic screens now; there is no projection room, no projector painting the action on the screen from behind us with ghostly, dancing fingers of light. School? In the past, we had overhead projectors and even interactive whiteboards had lens systems, but these have largely been replaced by LED and LCD screens.

I believe that if you do not take the time to show the phenomenon of a single converging lens projecting a real image on to a passive white screen to your students, they are likely to have no familiar point of reference on which to build their understanding and lens diagrams will remain a puzzling set of lines that have little or no connection to their world.

Teaching Ray Diagrams

Start with a slide that looks something like this:

What represents the lens? The answer is not the blue oval. On ray diagrams, the lens is represented by the vertical dotted line. F1 and F2 are the focal points of this converging lens and they are each a distance f from the centre of the lens, where f is the focal length.

Now what happens to a light ray from the object that passes through the optical centre of the lens?

The answer, of course, is a big fat nothing. Light rays which pass through the optical centre of a thin lens are undeviated.

Now let’s track what happens to a light ray that travels parallel to the principal axis as shown?

Make sure that your students are aware that this light ray hasn’t ‘missed’ the lens. The lens is the vertical dotted line, not the blue oval. What will happen is that it will be deviated so that it passes through F1 (this is because this is a converging lens; if it had been a diverging lens then it would be be bent so that it appeared to come from F2).

The image is formed where the two light rays cross, as shown below.

We can see that the image is inverted and reduced.

The image is formed close to F1 but not precisely at F1. This is because, although the object is distant from the lens (‘distant’ in this case being ‘further than 2f away’) it is not infinitely far away. However, the further we move the object away from the lens, then the closer to F1 the image is formed. The image will be formed a distance f from the screen when the object in very, very, very large distance away — or an ‘infinite distance’ away, if you prefer.

One of my physics teachers liked to say that ‘Infinity starts at the window sill’. In the context of thinking about lenses, I think he was right . . .

Method for finding the focal length of a converging lens (image from https://www.youtube.com/watch?v=AElLVGW9kxQ)

Some free stuff . . .

The PowerPoint that I used to produce the ray diagrams above is here. It is imperfect in a lot of ways but. truth be told, it has served me well over a number of years. It also features some other slides and animations that you may find useful — enjoy!

Postscript: ‘Through a glass, darkly’

The phrase ‘Through a glass, darkly’ comes from the writings of the apostle Paul:

For now we see through a glass, darkly; but then face to face: now I know in part; but then shall I know even as also I am known.

KJV 1 Corinthians 13:12

The New International Version translates the phrase less poetically as ‘Now we see but a poor reflection as in a mirror.’

It has been argued that the ‘glass’ Paul was referring to were pieces of naturally-occurring semi-transparent mineral that were used in the ancient world as lenses or windows. They tended to produce a recognisable but distorted view of the world — hence, ‘darkly’.

Better technology means that there is much less distortion produced by our glasses — hence, ‘through a glass, lightly’.