The 2015 AQA GCSE Physics specification (4.6.3.1) asks that students understand that:

In my experience, students find it natural to accept that bodies absorb electromagnetic radiation — but surely only extremely hot objects (like the filament of a light bulb) emit electromagnetic waves?

This is a consequence of the fact that our eyes can only detect the tiny slice of the vast electromagnetic spectrum.

That tiny sliver known as ‘visible light’ looks insultingly small even on a diagram with a logarithmic scale as shown above. On a linear scale, it’s even worse: if we represented the em spectrum by a line stretching from London to New York, then the range of wavelengths that human eyes can detect would be a strip two centimetres wide.

This calls to my mind some lines quoted many years ago by Arthur C. Clark in his wonderful essay ‘Things We Cannot See’: A being who hears me tapping / The five-sensed cane of mind / Amid such greater glories / That I am worse than blind.

Seeing the unseeable

A Leslie’s cube is a cuboid with black and silver coloured faces that can filled with hot water.

In visible light, there is no difference between its appearance when at room temperature (say 15 degrees Celsius) and when filled with hot water (say 70 degrees Celsius).

However, seen through an infrared camera, things look very different: the hot sides glow brightly, emitting huge amounts of infrared em waves.

There is another effect: the black coloured side throws out more infrared than the silver side. Why? Because any object which is good at absorbing em radiation is also good at emitting radiation.

As the AQA GCSE spec puts it:

By this definition, the Sun is a good approximation of a black body since it absorbs nearly all of the radiation falling on it (from other stars! — as well as the odd photon bounced back from minuscule specks like the Earth) as well being highly effective at emitting em radiation.

Black body radiation curves

4.6.3.2 of the AQA GCSE Physics spec says:

One of the ways to cover this is to look at the radiation curves of two black bodies at different temperatures (pdf here). Both of these objects are at relatively low temperatures, so they emit most of their energy in the infrared part of the em spectrum. The visible light range is shown by the coloured bar just to the right of the y-axis.

Because it is a perfect emitter — as well as absorber — of radiation, the intensity (power per unit area) of emitted radiation from a black body depends only on the temperature of the black body.

Estimating the total power emitted per unit area

We can estimate the total power emitted per unit area by approximating the area underneath the curve. We’re going to count any square which is larger than a half square as one whole square and ignore any part squares which are smaller than a half square.

The area under the blue curve is 325 W/m^2.

The intensity of radiation emitted by the hot object (red curve) is larger than the intensity emitted by the cold object (blue curve).

You can download an unannotated pdf copy of the graph by clicking on the link below.

We will look at the distribution of wavelengths in a later post.

Whoa, black body (bam-ba-lam)

Physics can occasionally, go better with a song.

Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
Black body e.m. waves (Bam-ba-lam)
Come out with peak shapes (Bam-ba-lam)
Planck said, “I’m worryin’ outta mind (Bam-ba-lam)
UV won’t align!” (Bam-ba-lam)
He said, oh black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)

Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
Planck set out to Quantise (Bam-ba-lam)
UV don’t catastrophize! (Bam-ba-lam)
Theory rock steady (Bam-ba-lam)
“No prob now,” said he (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)

Get it!

Whoa, black body (Bam-ba-lam)
Whoa, black body (Bam-ba-lam)
More heat, peak shifts left
Waves out with more zest
Wavelengths out not alike
Some hues power spike!
Whoa, black body (Bam-ba-lam)
Whoa, black body 
Bam-ba-laaam, yeah yeah

(with apologies to Huddie Ledbetter)