Image 1 Caption: Blackbody spectra (in an intensity plot) for Kelvin temperatures 3000 K, 4000 K, and 5000 K. The classical Rayleigh-Jeans law for T = 5000 K is also shown in the image. The Rayleigh-Jeans law is only asymptotically correct as λ → ∞.

The horizontal axis is wavelength of electromagnetic radiation in microns (μm). The vertical axis is radiant flux (AKA flux) (energy per unit steradian (sr) per unit area perpendicular to the beam path per wavelength) in appropriate units (i.e., kW/sr/m**2/nm).

Below we explicate thermal radiation and its very important special case is called blackbody radiation.

  1. Thermal Radiation:

    1. All matter emits/absorbs electromagnetic radiation (EMR) from/into its own heat energy pool by various microscopic processes---which are transitions and charge accelerations brought about by thermal interactions. That matter always does this is due to a combination of physical laws, most prominently quantum mechanics and the 2nd law of thermodynamics.

      This kind of EMR is thermal radiation.

    2. By "from/into its own heat energy pool", we mean that the EMR is NOT reflected EMR.

      Reflection is a process in which the EMR does NOT strongly interact with a body or change its state. In reflection, EMR just "bounces off" the body surface changing the surface state only temporarily and briefly in an ideal case.

      Of course, bodies usually emit thermal radiation AND reflect EMR at the same time.

    3. To qualify slight the above statement 1, to emit thermal radiation a body must have temperature above absolute zero (i.e., 0 kelvins (K)). However, this case is virtually always the case.

    4. Dense materials (i.e., solids, liquids, and dense gases) have a continuum (insofar as we know) of emission and absorption channels that extend effectively over all wavelengths that are practically accessible: i.e., over all the electromagnetic spectrum that is practically accessible. So they radiate and absorb continuous spectra.

      Recall continuum means there are NO "missing points". The real numbers are a continuum of numbers. Integers and rational numbers are NOT continuums.

      Actually crystalline solids like rocks can have broad emission and absorption bands (HI-92), and so do NOT necessarily radiate a continuum. We will NOT go into that detail here.

    5. Dilute gases have emission line spectra in emission and absorption line spectra in absorption. Actually, they always have weak (sometimes very weak) continuum emission and absorption too. We will NOT discuss line spectrum further here.

  2. Blackbody Radiation:

    1. A dense body at a SINGLE temperature will emit---NOT counting any reflection---blackbody radiation with a blackbody spectrum.

      The blackbody spectra for 3 temperatures are illustrated in Image 1.

      The blackbody spectra only depend on the temperature of the emitting body, NOT on any other properties: e.g., density (as long as sufficiently high), chemical composition, and shape. This remarkable fact is a result of thermodynamics.

    2. The formula for blackbody spectra is Planck's law
                 2hc**2      1
        B_λ(T) = ------ ----------  ,
                 λ**5   [exp(x)-1]   

      where the Planck constant h = 6.62607015*10**(-34) J·s (exactly), the vacuum light speed c = 2.99792458*10**8 m/s (exactly) ≅ 3*10**8 m/s = 3*10**5 km/s ≅ 1 ft/ns, and the x = hc/(kTλ), where the Boltzmann contant k = 1.380649*10**(-23) J/K (exactly). For reference for the fundamental physical constants, see also NIST: Fundamental Physical Constants.

      Note that Planck's law can be regarded as a function of wavelength λ with temperature T as a parameter or as a function of both wavelength λ and temperature T.

    3. Blackbody radiation is the simple limiting case of thermal radiation.

      Actually, many examples of thermal radiation approximate blackbody radiation so closely that they are just considered blackbody radiation. In fact, thermal radiation can often be considered as consisting of a mixture blackbody radiations of different temperatures.

      Also actually, thermal radiation and blackbody radiation are often used loosely as synonyms. Thermal radiation that is NOT blackbody radiation just being thought of as non-ideal blackbody radiation.

      log-log plot

    4. Image 2 Caption: Blackbody spectra for temperatures between 100 K and 10**4 K plotted on a log-log plot.

      The blackbody spectrum for T = 5777 K approximates the solar spectrum as seen above the Earth's atmosphere. The solar photosphere effective temperature = 5777 K. Effective temperature of a body (e.g., star or planet) is the temperature a perfect blackbody radiator (with a radius equal to the defined radius of the body) that radiates exactly as much radiant flux as the body.

      Effective temperatures for stars are usually good average temperatures for their stellar photospheres.

      The blackbody spectrum for T = 300 K approximates those for the Earth and the human body.

    5. Note that the maximum of the blackbody spectrum on the log-log plot in Image 2 shifts with wavelength with temperature in a systematic way. The wavelength of the maximum, in fact, obeys Wien's law which is explicated in detail file wien_law.html (see also Wikipedia: Wien's displacement law: Derivation from Planck's law). However, to give the short explication, for a blackbody spectrum, Wien's law states that the maximizing wavelength λ_max = hc/(x_max*kT) = (2897.771955185172661 ... μm-K)/T, and so is inversely proportional to temperature T: i.e., λ_max ∝ 1/T. Thus, temperature as a function of maximizing wavelength obeys the inverse Wien's law T = hc/(x_max*λ_max*k)=(2897.771955185172661 ... μm-K)/λ_max. Therefore if we constrain Planck's law by inverse Wien's law, we get a curve which is the maximum of blackbody spectrum as just a function of maximizing wavelength: i.e.,
                       2hc**2      1
        B_λ[T(λ)]max = ------ --------------  ∝ λ**(-5) ,
                       λ**5  [exp(x_max)-1]   

      where one constrains x = x_max = hc/(kTλ) = (4.965114231744276303 ...) which is just a constant. Note that B_λ[T(λ)]max is power law with power p = -5, and so should be a straight line on the log-log plot in Image 2 with slope = -5---which is, in fact, the case. A rough calculation from the log-log plot values gives

        slope = (-1-9)/(1.5-0.5) = -5 
      using the fact that log(0.3) ≅ -1/2 and log(30) ≅ 3/2. We get the exact result -5 fortuitously.

    6. In fact, the name blackbody radiation is NOT very good since the a body emitting it is often NOT black. In fact, to the human eye, it can have a colors and look very bright depending on temperature as we discuss below and how much EMR it reflects.

      If the temperature of the emitting body is very low and there is NO reflection, then yes a blackbody radiator will look black to the human eye

      In any case, tradition (beginning in the 19th century) has stuck us with the name blackbody radiation. Thermodynamic equilibrium thermal radiation would be a more accurately descriptive, if longwinded, name.

    7. Blackbody radiation itself (NOT just its emitting body the blackbody radiators) has the temperature that appears in Planck's law. Usually, this will be the temperature of the emitting body. However, adiabatic expansion/contraction of the space occupied by the blackbody radiation can cause the blackbody radiation temperature to change independent of matter. A very important special case of change is that cosmic background radiation (CBR) cools with the expansion of the universe and now is the cosmic microwave background (CMB) with CMB temperature = 2.72548(57) K (Fixsen 2009).

    8. If blackbody radiation and the matter it is in contact with have the same temperature, then the two are in thermodynamic equilibrium with respect to each other.


    10. A perfect blackbody spectrum is a smooth spectrum that reaches from zero to infinite wavelength.

      A perfect blackbody spectrum is an ideal limiting case, but very nearly perfect ones occur both in nature and in the laboratory and many objects (e.g., stars) approximate blackbody radiators to one degree or another.

    11. The temperature of blackbody radiation is always specified on the Kelvin scale.

      The zero point of the Kelvin scale (i.e., T = 0 K) is absolute zero---which is the coldest possible state where all heat energy that can be removed has been removed.

      If true a blackbody radiator was at absolute zero (where there is no heat energy to radiate) and was NOT reflective, it would be black.

    12. The classical Rayleigh-Jeans law for T = 5000 K is also shown in the image.

      Rayleigh-Jeans law asymptotically agrees with the correct blackbody spectrum in the limit that wavelength λ goes to infinity. But as λ goes to zero, the Rayleigh-Jeans law predicts infinite energy. This is the famous ultraviolet catastrophe which, when it was first noticed, was a clue that classical physics failed outside of what we now call the classical realm.

    13. Note the plot illustrates that as temperature increases/decreases the spectra become bluer/redder (more dominated by blue/red light).

    14. What color does the human eye with its peculiar psychophysical response see for a blackbody spectrum?

      For both photopic vision (i.e., human eye psychophysical response under well lit conditions) and scotopic vision (i.e., human eye psychophysical response under dimly lit conditions), it seems that ∼ 1500--11000 K is white hot (see Temperature of a "White Hot" Object, Carine Fang, 2001, bottom of page); going lower in temperature would lead orange to red hot to just visible red hot at ∼ 750 K; going higher temperature would lead to blue hot (see Wikipedia: Thermal radiation: Subjective color to the eye of a black body thermal radiator; Wikipedia: Red heat).

      However, for viewing stars at night, the colors are modified by the Earth's atmosphere which preferential scatters blue light which is why day sky is blue (see Wikipedia: Diffuse sky radiation).

      For example, reddish stars seen in visual astronomy have photospheric temperature which would make them look white hot if seen by visual astronomy in space. it seems that apparent colors are a bit different: star orange-red: ∼ ≤ 3700 K, star yellow-orange: ∼ 3700--5200 K, star yellow-white: ∼ 5200--6000 K (e.g., the Sun), star white: ∼ 6000--7500 K, star white to blue-white: ∼ 7500--10,000 K, star blue-white: ∼ 10,000--30,000 K, star blue: ∼ ≥ 30,000 K (Star Colors Explained, Brian Ventrudo, 2008 December 23).

      Blue hot is seldom seen in everyday life, but hot young OB stars (T ∼ ≥ 30,000 K: Wikipedia: O-type main-sequence star: Properties) are star blue---though no star in the sky as ever looked blue to yours truly. OB stars give the blue color (∼ 0.450--0.495 μm) to the spiral arms of spiral galaxies that you often see in images of spiral galaxies. I think this blue color (∼ 0.450--0.495 μm) in the images is meant to be true color---no one ever says it isn't.

    15. Is there a green hot? NO. Even when a blackbody spectrum peaks in the green color band (∼ 0.495--0.570 μm), the mixture of colors never looks green to the human eye.

  3. Blackbody Keywords:

    See blackbody keywords below (local link / general link: blackbody_keywords.html).


  4. Blackbody Radiation Videos:

    See the Blackbody radiation videos below (local link / general link: blackbody_videos.html).


  1. Credit/Permission: User:Darth Kule, 2010 / Public domain.
    Image link: Wikipedia: File:Black body.svg.
  2. Credit/Permission: © User:Prog, 2017 / CC BY-SA 4.0.
    Image link: Wikimedia Commons: ile:BlackbodySpectrum loglog en.svg.
Local file: local link: blackbody_spectra.html.
File: Blackbody file: blackbody_spectra.html.