---
title: "From absorbed starlight to grain temperature"
description: "Derive the dust heating balance with spectral units, the four-pi cooling factor and a worked gray-grain example."
canonical_url: "https://sandboxphysics.com/stories/interstellar-dust/grain-absorption-and-equilibrium-temperature/"
language: "en"
page_type: "story"
alternate_languages:
  zh-CN: "https://sandboxphysics.com/zh/stories/interstellar-dust/grain-absorption-and-equilibrium-temperature/"
last_modified: "2026-11-04"
---

[Interstellar dust, infrared light and polarization](https://sandboxphysics.com/stories/interstellar-dust/)

Interstellar dust · L1

# From absorbed starlight to grain temperature

Sandbox Physics · 2026-11-04

A grain temperature follows from a power balance. First calculate how much of the incident spectrum the material absorbs per second. Then find the temperature at which the same material emits that power. Keeping the units visible prevents several plausible but incorrect shortcuts.

## Before you begin

Read “Why a dark dust cloud shines in infrared” first. You should know power, area and a definite integral; the Planck function is introduced below.

## An effective collecting area

Imagine a narrow beam carrying $F=1\,\mathrm{W\,m^{-2}}$. An object with absorption cross section $C_{\mathrm{abs}}=10^{-12}\,\mathrm{m^2}$ receives $10^{-12}\,\mathrm W$ of heat from that beam. A cross section is an effective collecting area; it need not equal the geometric area at every wavelength.

For many grains it is convenient to divide cross section by dust mass. The absorption opacity $\kappa_\nu$ then has units $\mathrm{m^2\,kg^{-1}}$. The kilogram here means dust, not hydrogen and not gas plus helium. Changing that convention would change the inferred mass. The Lab chooses a teaching opacity per dust mass; the Astrodust tables use their own explicitly stated per-hydrogen normalizations.

A star supplies a spectrum, so we replace total beam flux by $F_\nu$ in $\mathrm{W\,m^{-2}\,Hz^{-1}}$. Multiplying by $\kappa_\nu$ and integrating over frequency gives a heating rate per dust mass. Absorption belongs in this expression. An extinction opacity would incorrectly heat the grain with light that was merely scattered.

$$
H=\int_0^\infty \kappa_\nu F_\nu\,d\nu,\qquad [H]=\mathrm{W\,kg^{-1}}
$$

This is the isotropic-material, single-beam version. Several incoming directions can be summed; an aligned material must also resolve the two electric-field modes, as in the next chapter.

Reading: Hensley & Draine, arXiv:2208.12365v1 (2022) · Dust absorption, heat and polarization: standalone science

## Why cooling has an angular factor

The Planck function $B_\nu(T)$ specifies thermal radiance: power per projected area, per frequency, per solid angle. It has units $\mathrm{W\,m^{-2}\,Hz^{-1}\,sr^{-1}}$. For an isotropic material, thermal emission by a small dust mass into one direction is $m\kappa_\nu B_\nu$ per solid angle. Integrating all directions supplies $4\pi$.

This statement does not mean that a black surface emits four times its Stefan–Boltzmann flux. A surface and an absorption cross section use different areas. For a spherical gray grain, the beam intercepts its projected disk while emission comes from its entire surface. The two descriptions give the same result when their geometry is kept consistent.

Set outgoing power per mass equal to absorbed power per mass. The cooling integral increases with temperature, so a positive heating rate has a unique root for the material used here. Solving that one-dimensional equation is easier than interpreting a complicated infrared map; the difficulty lies in calculating the correct heating and material response.

$$
H=4\pi\int_0^\infty \kappa_\nu B_\nu(T)\,d\nu
$$

Radiative equilibrium for isotropic equilibrium grains. The aligned producer uses the angular mean of its tensor; its trace is three times this scalar opacity, so the same scalar cooling integral results.

Reading: Dust absorption, heat and polarization: standalone science

## Worked example: a gray spherical grain

Let a spherical grain absorb and emit perfectly at every relevant wavelength. A parallel beam of total flux $F$ supplies $\pi a^2F$, where $a$ is the grain radius. Its surface radiates $4\pi a^2\sigma_{\mathrm{SB}}T^4$. Cancel the same geometric factor before taking the fourth root.

For $F=1\,\mathrm{W\,m^{-2}}$, the result is $T=45.82\,\mathrm K$. Doubling the flux gives $54.49\,\mathrm K$, a factor $2^{1/4}$, not twice the temperature. The grain gets better at losing energy as it warms.

The radius cancels only because this example is gray and has the same absorption and emission efficiency at all wavelengths. Real small grains generally have size-dependent efficiencies. The Lab’s opacity is also not gray: it turns over near a micrometre and becomes approximately quadratic in frequency at long wavelengths. This calculation checks the geometry and units; it does not replace that material model.

$$
T=\left(\frac{F}{4\sigma_{\mathrm{SB}}}\right)^{1/4}
$$

Gray spherical grain under a parallel beam, with no additional radiation bath. The factor four is the surface-to-projected-area ratio.

![Gray-grain temperature rises with beam flux more slowly than linearly.](https://sandboxphysics.com/stories/interstellar-dust/materials/gray-grain-temperature.svg)

Independent analytic spherical-gray-grain balance, recomputed by the supplied figure producer. This is a geometry limit, not the Lab opacity.

Reading: Dust absorption, heat and polarization: standalone science

## A different material changes the temperature scaling

Suppose, only over the frequencies carrying most of the cooling, $\kappa_\nu\propto\nu^\beta$. Substitute $x=h\nu/(k_{\mathrm B}T)$ into the Planck integral. The frequency powers and differential together give cooling proportional to $T^{4+\beta}$. Thus $\beta=2$ gives a sixth-power cooling law.

If the shape of the illuminating spectrum stays fixed and only its strength changes, heating is proportional to that strength. In the long-wavelength approximation, temperature then changes as its $1/(4+\beta)$ power. This is a useful local expectation. It is not an exact law for moving a cloud whose internal spectrum also reddens.

In the producer, each cell receives the spectrum that actually survived its upstream column. The program integrates that changed spectrum to obtain the local heating. A single scaled template misses this change of spectral shape, so shielded cells require the full local spectrum.

Reading: Hensley & Draine, arXiv:2208.12365v1 (2022) · Dust absorption, heat and polarization: standalone science

## Frequency and wavelength must agree

A spectrum per hertz is not numerically the same as a spectrum per metre. The same finite power must occupy corresponding bins, so $F_\lambda=F_\nu c/\lambda^2$ at $\nu=c/\lambda$. The Planck functions obey the same Jacobian. The opacity is a cross section per mass evaluated at the corresponding frequency; it does not acquire another spectral-density Jacobian.

A useful independent check integrates the same heating once in frequency and once in wavelength. Agreement tests both the coordinate conversion and the integration limits. Merely relabeling a frequency axis as wavelength would fail this test.

A finite spectral grid also misses its two tails. The scientific replay bounds the omitted source and cooling powers separately. A tiny root residual on a truncated integral would otherwise certify only that truncated equation, even if physically important heating had been left out.

$$
\int \kappa_\nu F_\nu\,d\nu=\int \kappa_{c/\lambda}F_\lambda\,d\lambda
$$

The limits on the right run in increasing wavelength and correspond to the reversed frequency limits; the Jacobian is already contained in the positive spectral density.

Reading: Dust absorption, heat and polarization: standalone science

## A solved root is not the final physical check

An optically thick infrared cloud can absorb its own thermal emission. Then the direct starlight balance alone is incomplete: neighboring grains heat one another. The present producer accepts only cases whose bolometric upper bound on neglected infrared absorption is below one part in a thousand. It also bounds neglected heating by the cosmic microwave background and accepts only its stated warm range.

These checks answer different questions. Deposited power matching emitted power tests the local cooling balance. A cloud-mass check tests geometric quadrature. Image refinement tests the representation of the same cells through a finite point-spread function. None can stand in for the others.

Try increasing dust mass until the infrared-escape condition fails. The last valid calculated cloud remains visible with a refusal notice. Treat that as information about the approximation: the requested case needs more radiative transfer, even if a numerical temperature root could still be found.

Reading: Dust absorption, heat and polarization: standalone science

## Exercises and explanations

1. A gray grain receives four times the beam flux. What is its temperature ratio? What ratio would the approximate $\beta=2$ cooling law give if heating changed by the same factor?

   Read the explanation

   The gray ratio is $4^{1/4}=1.414$. The quadratic-opacity approximation gives $4^{1/6}=1.260$. Both rise less than the heating ratio because cooling increases steeply with temperature. The latter requires the stated power-law cooling domain and unchanged incident spectral shape.

2. Why can a code report excellent absorbed–emitted equality while still giving an incomplete physical temperature?

   Read the explanation

   It may solve an incomplete equation accurately: omitted incident ultraviolet power, infrared reabsorption or an external radiation bath can all change the heating. Separate tail and omitted-physics bounds are needed in addition to the root residual.

## Try the prediction

### [Follow the same dust from absorbed light to thermal emission](https://sandboxphysics.com/labs/interstellar-dust-polarization/)

1. Save the illuminated-cloud record. In its exported cells, distinguish deposited power in watts from temperature in kelvins and mass in kilograms.
2. Change luminosity with the cloud position and mass fixed. Compare the temperature response with a sixth-power cooling expectation, explaining why the exact material integral need not match it perfectly.
3. Open the numerical bounds. Compare the cooling residual with the infrared-escape upper bound, and name the distinct physical question each answers.

[Previous: Why a dark dust cloud shines in infrared](https://sandboxphysics.com/stories/interstellar-dust/why-dark-dust-shines-infrared/) [Next: Why transmitted and thermal polarization turn opposite ways](https://sandboxphysics.com/stories/interstellar-dust/aligned-grains-and-two-polarizations/)

## Sources and calculation scope

1. [Hensley & Draine, arXiv:2208.12365v1 (2022)](https://arxiv.org/abs/2208.12365v1)

   Fixed first manuscript: sections 2.1–2.3, 4–6; equations 1–11/27 and Figures 6/11/12/18. Material cross sections, temperature distributions and prescribed size-dependent alignment are distinct from the teaching tensor. Its 2022 forecasts are not current-status claims.

2. [Dust absorption, heat and polarization: standalone science](https://sandboxphysics.com/stories/interstellar-dust/materials/dust-transfer-science-v1.zip)

   MIT scientific code plus separately licensed CC0 native inputs. Ten genuinely recomputed offline steps, retained coarse-grid failures, independent volume/Planck references and optional first-network real exercise. The site supplement records a later successful real first acquisition; the frozen archive remains unchanged.