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.


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.

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