The classical law gives every star one colour

A star’s colour gives its surface temperature from two brightness measurements and nothing else, which is the cheapest useful measurement in astronomy. It works for a reason that had to be discovered, and the discovery is usually told as a story about the ultraviolet rather than about thermometers.

Count the standing electromagnetic waves that fit in a cavity and there are $8\pi\lambda^{-4}$ of them per unit volume per unit wavelength. Give each the $kT$ that classical equipartition allots to every degree of freedom and the spectrum is

$$B_\lambda = \frac{2ckT}{\lambda^4},$$

which rises without limit towards short wavelengths and has a divergent integral. That failure is famous.

The failure that matters here is quieter and is in the same expression. The temperature appears once, as a multiplier, and the wavelength appears alone — so the ratio of the law at two wavelengths contains no temperature whatsoever.

The cancellation, in one line #

A colour index is a ratio of fluxes through two passbands, converted to magnitudes:

$$B - V = -2.5\log_{10}\frac{\int B_\lambda S_B(\lambda)\,d\lambda}{\int B_\lambda S_V(\lambda)\,d\lambda} + \text{const}.$$

Substitute the classical law. The $2ckT$ is a constant with respect to $\lambda$, so it comes out of both integrals and cancels in the ratio, leaving

$$B - V = -2.5\log_{10}\frac{\int \lambda^{-4} S_B\,d\lambda}{\int \lambda^{-4} S_V\,d\lambda} + \text{const},$$

which depends only on the shapes of the two filters. Every star in a classical universe has the same colour index, whatever its temperature, and a photometrist measuring two magnitudes learns the filters’ effective wavelengths and nothing about the star.

It is worth dwelling on how complete the cancellation is. It does not depend on the filters being narrow, or on their shapes, or on their separation; it does not depend on the temperature being in any particular range; and it does not depend on the star being a blackbody in any sense beyond having the classical spectrum. Every classical thermal source in the universe, at any temperature, has one colour index per pair of filters.

So colour photometry is a purely quantum measurement. Not in the sense that its interpretation requires quantum mechanics — in the sense that the quantity it measures is identically zero in a classical world.

That statement can be sharpened. What breaks the cancellation is the $e^{hc/\lambda kT} - 1$ in Planck’s denominator, and the temperature enters only through the combination $hc/\lambda kT$. So the colour index measures $hc/kT$ against the filters’ wavelengths, and a colour temperature is a measurement of Planck’s constant divided by Boltzmann’s, in units of the passband separation.

Where the two laws part company #

The ratio of the two expressions is instructive because it is a function of one variable.

$$\frac{B_\lambda^{\text{RJ}}}{B_\lambda^{\text{Planck}}} = \frac{e^x - 1}{x}, \qquad x = \frac{hc}{\lambda kT}.$$

At small $x$ — long wavelength, hot source — the ratio approaches one and the classical law is correct. At large $x$ it grows exponentially.

For the Sun, $x$ is 5.60 in the B band and 4.52 in V. The classical law therefore overstates $B_\lambda$ by a factor of 48 at B and 20 at V, and the ratio of those is 2.38 — which is 0.94 magnitudes. Classical physics gets the Sun’s colour wrong by almost a magnitude, and gets every other star’s wrong by the same amount, which is why the error is invisible in a ratio and fatal to a thermometer.

The number $x$ is the whole of the argument. It is the ratio of a photon’s energy at that wavelength to the thermal energy available, and it says how many photons of that energy a mode of the field can afford. Where $x \ll 1$ the mode holds many quanta and behaves classically; where $x \gg 1$ it holds essentially none and the classical allotment of $kT$ is a fiction.

Visible light from a star has $x$ between 4 and 6, which is squarely in the quantum regime. A star’s colour is the part of its spectrum where the classical theory is worst, and that is not a coincidence — it is because the peak of the spectrum is at $x = 4.965$ by construction.

How steeply the thermometer reads #

The sensitivity of a colour index to temperature is worth computing, because it decides what the measurement is good for and where it stops working.

Differentiating the colour index with respect to $\log T$ gives, for two narrow bands at $\lambda_1$ and $\lambda_2$,

$$\frac{d(m_1 - m_2)}{d\ln T} = -1.086\left(\frac{x_1}{1 - e^{-x_1}} - \frac{x_2}{1 - e^{-x_2}}\right),$$

which goes to zero as both $x$ go to zero — the classical limit again, arrived at by differentiation instead of by cancellation — and grows as the bands move into the quantum regime.

For B and V at solar temperature the derivative is about 1.6 magnitudes per e-folding of temperature, so a colour measured to 0.01 magnitudes gives a temperature to about 0.6 per cent, which is 35 K for the Sun.

That is remarkable precision from two brightness measurements, and it degrades rapidly upward. At 20,000 K both $x$ are near unity, the derivative has fallen by a factor of four, and the same photometric precision gives a temperature good to only a few per cent. At 40,000 K the optical colours are almost useless and the measurement has to move to the ultraviolet, where $x$ is large again.

The practical rule is that a colour index works as a thermometer when the passbands straddle or lie blueward of the spectrum’s peak, and fails when both lie on the Rayleigh–Jeans tail. For a hot star the whole optical is on that tail.

What Planck actually did #

The received account is that the catastrophe forced the quantum, and the order of events was the other way round.

Planck was working on the thermodynamics of cavity radiation and had two limiting expressions. Wien’s law fitted the short-wavelength data and failed at long wavelengths; the Rayleigh–Jeans form, derived later but implicit in the classical argument, fitted the long-wavelength data and failed at short ones. In October 1900 he found an interpolation between the two entropy expressions that fitted the whole measured curve, and presented it as a formula without a derivation.

The derivation came eight weeks later and required that the energy of an oscillator be quantised in units of $h\nu$. He described the step as an act of desperation and spent years trying to avoid it.

The catastrophe was named in 1911, eleven years after the formula it is supposed to have forced. The measurements that drove Planck were the long-wavelength ones, where Wien’s law failed and the classical law worked — not the ultraviolet, where nobody had good data.

Wien’s own law, $B_\lambda \propto \lambda^{-5}e^{-hc/\lambda kT}$, is the short-wavelength limit of Planck’s and does have a temperature-dependent colour: dropping the $-1$ from the denominator leaves the exponential, and the exponential does not cancel. So the colour thermometer would have worked in a Wien universe as well as in a Planck one, and would have given slightly wrong temperatures.

That is the honest form of the argument. It is the $-1$ that is classical and the exponential that is quantum, and the cancellation this essay is about happens only in the regime where the exponential has linearised away entirely.

That convergence is the reason the same photometric system behaves so differently across the stellar temperature range. For an M dwarf at 3,000 K the optical sits deep in the exponential and the colours change rapidly with temperature; for an O star they sit on the tail and change by less than a magnitude across a factor of five in temperature. A colour–magnitude diagram of a cluster is therefore stretched at the bottom and compressed at the top, and the compression is not a property of stars but of where the passbands fall relative to $hc/kT$. The diagram that sorted the stars has that distortion built into its horizontal axis, and every feature read off the hot end of it is being read off a compressed scale.

The one place a stellar spectrum is classical #

The Rayleigh–Jeans law is not merely a historical error; it is used every day, and it is used on stars.

At radio wavelengths $x$ is minuscule — at 1 cm and 5,772 K it is $2.5\times10^{-4}$ — so the exponential linearises and Planck’s law becomes the classical one exactly. Radio astronomers therefore quote intensities as brightness temperatures: the temperature a blackbody would need to produce the observed intensity at that frequency, using the classical relation.

That convention works because the relation is linear in $T$, so a brightness temperature is a direct proxy for intensity and adds and subtracts the way intensity does. It would be useless in the optical, where the relation is exponential and a sum of two brightness temperatures means nothing.

It also means that a radio measurement of a star carries no colour information at all, in exactly the sense this essay has been about. Two radio bands give two intensities whose ratio is $\lambda^{-2}$ — the Rayleigh–Jeans form in frequency — regardless of temperature. A radio spectral index that is not $-2$ is a statement that the source is not thermal, and that is the main thing radio spectral indices are used for.

The measurement that is genuinely classical #

There is one astronomical colour measurement that survives in a classical universe, and it is worth naming because it shows what the cancellation does and does not remove.

If two stars have the same temperature and different amounts of dust in front of them, their colours differ — because dust removes blue light preferentially and the removal depends on wavelength through the grains’ properties rather than through Planck’s law. That difference is a measurement of the reddening, and it is there whatever the source spectrum is.

So a classical universe would still have reddening, still have a colour excess, and still be able to map interstellar dust by photometry. What it would not have is any way to tell an intrinsically red star from a reddened blue one, because in that universe there is no such thing as an intrinsically red star.

Every star would have the same colour, and any departure from it would be dust. Which would make the dust measurement easier and everything else impossible, and is a reasonable summary of what the quantum bought: it made the intrinsic colour a variable, at the cost of making it degenerate with reddening. Separating the two has been a standing difficulty ever since.

Defined constants, idealised filters, and one exact cancellation #

Both laws are evaluated exactly, with the 2019 SI values of $h$, $c$ and $k$ — all three of which are now defined rather than measured.

The passbands are idealised as Gaussians at the published effective wavelengths and widths of the Johnson–Cousins system. A real filter is not a Gaussian and is not symmetric, and the widths used here are right while the shapes are not. That matters for the third decimal of a colour index and not for the argument being made, with one exception: a Gaussian has no cut-off, so a fraction of each band’s flux is drawn from wavelengths a real filter passes nothing at. For the classical law, which diverges towards the blue, that tail is not negligible, and the quoted classical index is therefore a number for these idealised filters rather than for real ones.

What is exact regardless of the filters is the cancellation itself, since it depends only on the $T$ coming out of both integrals.

The numbers for the ratio $e^x - 1$ over $x$ are exact arithmetic at the stated wavelengths and temperatures, and they carry no filter assumption at all.

A quantum measurement made with a filter wheel #

There is a pleasing consequence of all this that is worth stating plainly, because it is not the usual way photometry is described.

The list of astronomical measurements that require quantum mechanics to exist rather than to interpret is short. A spectral line is one — an atom’s energy levels are quantised and a classical atom has none. A colour temperature is another, for the reason this essay has computed. The Saha equation is a third, since it is a statement about the occupation of quantised states.

What distinguishes the colour index from the other two is that the apparatus is trivial. A spectrograph is a precision instrument; an ionisation balance is an inference from one. A colour index is two exposures through two pieces of coloured glass, and the quantity it delivers would be identically constant in a universe governed by classical electrodynamics.

Three hundred years of optics could not have produced it, and it was available within a decade of Planck’s paper to anybody with a telescope and two filters. The first photometric colour temperatures were being published in the 1910s.

A star is not a blackbody #

A star is not a blackbody. Its spectrum has absorption lines, a Balmer jump, and in cool stars molecular bands that eat whole regions of the optical. The Planck curves drawn are what a star would emit if it had no atmosphere above its photosphere, and the difference between that and a real star is the whole subject of stellar spectroscopy.

Neither curve is normalised to anything physical. Both are scaled to the solar Planck peak so that all six can share a canvas; the absolute levels differ by the fourth power of the temperature and would put the coolest curve four decades below the hottest.

And no drawing here shows the ultraviolet catastrophe as an infinity. The plot stops at 80 nm, where the discrepancy is a factor of a thousand; continuing it to arbitrarily short wavelengths would show the classical curve rising without bound, which is a statement about a limit rather than something a finite canvas can display.

Why the cancellation is worth noticing at all #

A reader might reasonably object that the classical law is wrong, so its predictions about colour indices are of no interest. The objection misses what the calculation is for.

Physical laws are usually compared against reality by their agreement with measurements. What this comparison does instead is ask what kind of measurement each law makes possible — and the two laws differ not in the accuracy of a colour temperature but in whether the quantity exists.

That distinction recurs whenever a measurement technique is examined for what it is sensitive to rather than for how precise it is. A velocity amplitude gives only a minimum mass because the inclination cancels out of the observable, and no improvement in precision recovers it. A colour index cannot distinguish two of the three things that shift it, for the same structural reason.

A degeneracy is a fact about a law rather than about an instrument, and identifying which quantities a given law makes degenerate is a more useful exercise than computing how well it fits. Here the classical law makes temperature degenerate with nothing at all, because it removes temperature from the observable entirely.

Two failures of the same kind #

It is worth noticing that the classical law fails twice in the same way, and that both failures are about a ratio.

The divergent integral is a failure of the sum over wavelengths: too much energy at short wavelengths because every mode is given $kT$ and there are infinitely many short-wavelength modes.

The colour degeneracy is a failure of the comparison between wavelengths: the same $kT$ given to every mode means the spectrum’s shape is fixed and only its scale responds to temperature.

Both are consequences of equipartition applied to a system with unboundedly many degrees of freedom, and the quantum fixes both at once by making high-frequency modes expensive. The energy cut-off and the shape’s temperature dependence are the same statement, because a cut-off that moves with temperature is a temperature-dependent shape.

That is the deeper reason a colour is a thermometer. It is not measuring how much energy the star has; it is measuring where the exponential cut-off falls, and the cut-off’s position is set by the ratio of a photon’s energy to $kT$ and by nothing else.

Still open: nothing in the physics, and something in the practice #

Planck’s law is not in dispute and neither is anything above. What remains open is a practical question the argument raises: how well a colour index recovers a temperature for a source that is not a blackbody.

The answer is that it recovers a colour temperature, which is the temperature of the blackbody with the same colour, and that quantity differs from a star’s effective temperature by tens to hundreds of kelvin depending on the star. The difference is a property of the star’s atmosphere and is modelled rather than measured.

From here, once the quantum has fixed both failures #

The quantum fixed both failures at once, and the two most-quoted consequences of Planck’s law — Wien’s displacement and Stefan–Boltzmann — are usually presented as two separate laws about radiation.

They are not two. They are the peak and the integral of one curve, and the fourth power in the second is the fifth power of the peak’s height divided by the first power of its width. Fitting all three exponents off the drawn curves finds that only two of them are independent.