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Could be detected by infrared radiation?

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    $\begingroup$ I've deleted a number of inappropriate comments and/or responses to them. $\endgroup$ – David Z May 18 at 9:23
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    $\begingroup$ These are 2 different questions. Which one are you looking for an answer for? $\endgroup$ – Dancrumb May 18 at 15:52
  • $\begingroup$ The clarifying question I am interesting in prerequisites the titled quest. $\endgroup$ – elias2010 May 18 at 21:08
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    $\begingroup$ @Dancrumb "the title and the body are two very different questions, so it's not unreasonable for an answer to address just one" (or both, preferably). $\endgroup$ – Rob Jeffries May 19 at 14:13
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    $\begingroup$ They might be different questions but they're obviously quite closely related. Thermal radiation scales with $T^4$, so if it did emit thermal radiation it would emit a lot more (and hence be much easier to detect) if it was hot than if it was cold. They are also related in that if you were able to detect its thermal radiation, you would then be able to infer its temperature. $\endgroup$ – Nathaniel May 19 at 18:19
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Dark matter might be in the form of thermal relics (e.g. WIMPs) or they may be produced non-thermally in phase transitions (e.g. axions). In the latter case, it does not make any sense to talk about a temperature at any stage.

For "thermal relics", the dark matter particles would thermally decouple from the rest of the universe as it cooled. Roughly speaking this occurs when the temperature in the universe falls below $\sim m_x c^2/k_B$ when other particles are not energetic enough to produce the dark matter particles. The density of dark matter particles would then drop to zero as they annihilated, but because the universe expands and becomes less dense, a state is attained where the dark matter particle interaction rate is too slow and they "freeze out". A detailed treatment shows that this happens at about $T_d\sim m_xc^2/20 k_B$.

The above would mean that the particles were non-relativistic when they decoupled, with an approximate Maxwell Boltzmann distribution at the decoupling temperature $\propto \exp(-p^2/2m_x k_BT_d)$.

From there, the distribution is fixed, but the particle momentum (with respect to the co-moving frame) decreases as the inverse of the scale factor, because it's de Broglie wavelength expands in the same way as that of light, and $p = hc/\lambda$. Thus the effective temperature of the dark matter goes as the inverse square of the scale factor.

This is quite different to the behaviour of particles that decouple while relativistic. For them, their momentum distribution is fixed e.g. $$ n(p) \propto (\exp [pc/k_BT_d] \pm 1)^{-1},$$ with the sign in the bracket depending on whether they are fermions or bosons. Here, you can see that the temperature of this distribution only decreases as the inverse of the scale factor as the universe expands.

The bottom line is that if dark matter was non-relativistic when it decoupled (e.g. WIMPS), then it is stone-cold now compared with, for example, the temperature of the photons (relativistic bosons) in the cosmic microwave background. The exact value would depend on the mass of the particle, but is not relevant, since they would behave like a pressure-free fluid under the influence of gravity.

Even neutrinos (with mass) as dark matter will be cold (now). That is because massless neutrinos would of course be relativistic and would have a similar temperature as the CMB, except that electron-positron annihilation after neutrino decoupling raises the photon temperature. However, neutrinos with mass decouple while still relativistic (for sensible neutrino masses), but become non-relativistic as the universe expands (when $p < m_\nu c$). As a result their temperature then declines more rapidly than the CMB.

The answer to your second question is that dark matter is called that because it does not have electromagnetic interactions. So it does not absorb or emit light of any wavelength.

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  • $\begingroup$ Thank you, very nice answer ! A curiosity, what do you mean with "produced non-thermally in phase transitions"? $\endgroup$ – Quillo May 18 at 10:36
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    $\begingroup$ @mrcntn I mean it isn't produced by any equilibrium process and doesn't have a momentum/energy distribution that can be characterised by a temperature. For example if the dark matter is produced by the decay of another particle that is already out of thermal equilibrium. $\endgroup$ – Rob Jeffries May 18 at 11:55
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No. Because it does not take part in the electromagnetic interaction, dark matter neither absorbs nor emits electromagnetic waves. This means that if dark matter particles do have a "temperature" then we do not have any direct means of measuring it.

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    $\begingroup$ That's fine for the electromagnetic emission part, but it doesn't mean dark matter doesn't have kinetic energy. We can use EM emission to measure the temperature of normal matter, but temperature exists independently of that. Of course, it's quite possible it's not going to be particularly useful, depending on how dark matter interacts. $\endgroup$ – Luaan May 18 at 9:21
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    $\begingroup$ Just as an example of how we would be able to measure the temperature of Dark Matter without blackbody radiation: If we knew the interaction between dark matter and gravity (its mass), then we could the spatial distribution around galaxies to calculate the temperature: The temperature determines how spread out it would be from the potential minima. $\endgroup$ – Rasmus Damgaard Nielsen May 18 at 11:56
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    $\begingroup$ @RobJeffries the title and the body are two very different questions, so it's not unreasonable for an answer to address just one. $\endgroup$ – Dancrumb May 18 at 15:52
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These are actually two questions.

  1. The temperature of the dark matter is an important value of many cosmological models. There are "hot" dark matter, "warm" dark matter and so on, models, proposing different energies of the proposed dark matter particles. This temperature (=kinetic energy of the particles) is an important factor for the dark matter behaviour even without the electromagnetic indication. Depending on their temperature, these particles, for example, may or may not have the escape velocity to get out of a galaxy, orbit the galaxy or just clump into the central black hole.

  2. As for the infrared: whatever they are, the dark matter doesn't have much of a coupling to photons, so there is no electromagnetic emission of the dark matter that we are aware of. We can imagine, however, that if the dark matter decoupled early enough, its thermodynamic temperature is probably less than the microwave background. So even if there is a method of emission, it's probably not infrared either. It will be radio waves.

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    $\begingroup$ This answer would lead someone to think that"hot" and "cold" refer to present day temperatures of the particles, which they don't. $\endgroup$ – Rob Jeffries May 18 at 5:48
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    $\begingroup$ Hot and cold dark matter is related to present day temperatures, it is just pretty much relative in a sense that hot dark matter is hotter than the cold one. Even the hottest (modelled) DM is still dead cold by our (electromagnetic-dominated heat exchange) standards. $\endgroup$ – fraxinus May 18 at 8:50
  • $\begingroup$ If DM decoupled early, shouldn't it be warmer than today's microwave background since it lack of interactions prevents it from equillibrating with the rest of the matter in the universe? $\endgroup$ – proton May 19 at 6:19
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    $\begingroup$ The mechanism of cooling is not some interaction, but the universe expanding. $\endgroup$ – fraxinus May 19 at 13:25
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    $\begingroup$ @Michael you may also think about it like adiabatic expansion. $\endgroup$ – fraxinus May 20 at 8:29
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As others already mentioned, the evidence is that dark matter neither absorbs nor emits electromagnetic radiation---hence its name. To be a little more precise, if it does interact with electromagnetic radiation then it does so only very weakly, and so glimmers much too dimly to be detected by our instruments. However, there are indirect ways to get at the possible motions of dark matter, and these suggest that it is not so hot as to have extreme thermal motion, at speeds of the order of the speed of light. If it were then it would not have clumped together in the way it has. This is why it is called 'cold' dark matter in the scenario which is our best working model. So that word 'cold' here signifies merely that the temperature is not huge.

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It would not be dark if it emitted (significant) amounts of electromagnetic radiation.

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In our Milky Way specifically, dark matter is well approximated by an "isothermal halo", i.e. a huge cloud of dark matter particles with the same temperature everywhere. This is equivalent to saying that dark matter in the Milky Way (and other galaxies) roughly follows a Maxwell-Boltzmann distribution. This distribution has a maximum at some speed. The more massive the galaxy, the larger that speed; for our Milky Way, it's about 200km/s. If you knew or assume a mass of the dark matter particles you could convert that speed into a temperature in units of Kelvin.

No, dark matter does not interact with light and therefore also does not emit blackbody radiation.

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