The plasma of the ITER reactor is planned to be at 150 million K. Using the Stefan-Boltzmann law, setting the surface area as $1000\,\mathrm{m}^2$ (the plasma volume is $840\,\mathrm{m}^{3}$ so this is being generous), and the emissivity as $0.00001$ (emissivity is empirical so I just plugged in an extremely low value) yields a power of $2.87\times 10^{23}\,\mathrm{W}$. It would require somewhere on the order of $10^{35}$ fusion reactions per second just to break even, which clearly is not happening.

How can fusion researchers confine plasmas for several minutes if the blackbody radiation is this extreme? It seems like that with this level of heat, the plasma would just cool down within a few nanoseconds, and everyone in the vicinity would be torn to shreds by gamma rays, but evidently this does not happen. How?

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    $\begingroup$ Basically the fallacy here is: “$.00001$ is an extremely low value”. $\endgroup$ Commented Jul 4, 2018 at 10:02

3 Answers 3


The plasma in a fusion reactor is typically "optically thin"; the radiation isn't really in equilibrium with itself and the plasma particles.

Generally, instead of just modeling the plasma as a black body, people look at specific radiation processes. Kenneth Gentle (UT) has a nice set of slides that works through that.

Hot plasmas are almost transparent -- optically thin, except for a few special cases and stellar interiors.

• A picture of a hot plasma in the visible is empty except for possible bright spots at the edge where incoming neutrals are ionized.

• The principal, and most important, exception is at the electron cyclotron frequency. (For n>10^19/m3 and Te>100 eV, the plasma is a black body radiating at Te, at least in certain directions and polarizations.


Blackbody radiation at 100 eV (1 million degrees; 200 times the surface of the sun) is quite significant. Were it not confined to a narrow band of frequencies near Ωe, it would be a very large power.

So that's not the wide-bandwidth process of true black-body radiation. And other process have their own limitations:


Since hot plasma “collisionless”, intensity far less than blackbody -- detectable only because it extends over a broad frequency range. The radiation is broad-band, but peaks at hν~Te, like blackbody radiation, because there are fewer electrons at higher energies capable of emitting such photons, and the density of states decreases with decreasing frequency.

Line radiation from surfaces and impurities can also be strong, but (1) it's not the black-body you're asking about and (2) they work really hard to reduce the impurities.

  • $\begingroup$ So, basically the plasma's density is so low that collisional effects are negligible, and thus energy is not being constantly released in the form of blackbody radiation. Whereas in an environment such as the solar core, where the plasma is extremely dense, collisional effects are relevant and so blackbody radiation is released. $\endgroup$ Commented Jul 4, 2018 at 11:49
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    $\begingroup$ @NikhilMurali: Even a rarefied plasma emits thermal radiation, just not a lot of it per unit of volume. The Sun's photosphere has roughly the same density as the vacuum in low earth orbit -- but it is a lot deeper than any laboratory plasma, so the total thermal emission from all of the layers nevertheless add up to a honest-to-god blackbody spectrum of equilibrium intensity. $\endgroup$ Commented Jul 5, 2018 at 0:47
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    $\begingroup$ @NikhilMurali Yes, but you don’t need mechanical collisions, just enough depth that the probability of photons interacting inside gets toward 1. A solid black body is solid: the entire surface is atoms. You can’t see “through” it. A gas blackbody has to be thick and/or dense for that to be true. $\endgroup$ Commented Jul 5, 2018 at 12:51

In ITER the average electron and ion temperatures are about 8 keV and the electron density is about $10^{20}$ m$^{-3}$.

At such a low density, the plasma is optically thin (see for example How large should an optically large fusion reactor be?) and therefore does not emit blackbody radiation.

The primary means of emission will be (optically thin) thermal bremsstrahlung. The link gives an approximate formula to calculate the total power emitted per unit volume. $$P \simeq 1.7 \times 10^{-38} Z^2 n_i n_e \left(\frac{T}{{\rm eV}}\right)^{1/2}\ {\rm W/m}^3$$

For the parameters above and assuming $Z=1$ and $n_i=n_e$, then $P = 1.5\times 10^{5}$ W/m$^3$. So for 840 m$^3$, the power lost in bremsstrahlung is 12.7 MW; a small fraction of the intended power output of the fusion reactor.

The mean free path for photons due to Thomson scattering by electrons (which will be the primary opacity at those temperatures; or possibly a bit higher for Compton scattering) is $(n_e \sigma_T)^{-1} \sim 1.5\times 10^{8}$ m, or about $10^7$ times the linear dimension of the reactor.


Fully stripped atoms can't radiate by having electrons jump between energy levels anymore, because there are no bound electrons. So, that removes the biggest radiation channel unless impurities with high atomic numbers are introduced (such as tungsten). Heavy elements don't get fully stripped of electrons and then the radiation losses are just as bad as you're suggesting and can cause complete disruption and termination of the plasma discharge.

Fusion plasmas with mostly hydrogen isotopes and only lightweight impurities like carbon will typically only radiate strongly from the cold edge layers, where radiation actually has the beneficial (as long as it STAYS at the edge) effect of distributing heat exhaust.


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