EM radiation has a relativistic mass (see for instance, Does a photon exert a gravitational pull?), and therefore exerts a gravitational pull.

Intuitively it makes sense to include EM radiation itself in the galactic mass used to calculate rotation curves, but I've never actually seen that done before...

So: if we were to sum up all the electromagnetic radiation present in a galaxy, what fraction of the dark matter would it account for?


I found it surprisingly hard to find an authoritative statement of the density of the CMB. According to this article it's about $5 \times 10^{-34}\mathrm{g\ cm}^{-3}$, and since the critical density is somewhere in the range $10^{-30}$ to $10^{-29}\mathrm{g\ cm}^{-3}$ photons don't make a significant contribution.

Photons wouldn't be dark of course. If there were enough photons to make a significant contribution to the mass/energy of the universe we'd see them, just as we can see the CMB.

Response to comment: oops yes, I didn't read your question properly - sorry!

Anyhow, my comment that photons aren't dark matter still applies, but it's easy to make an estimate of the gravitational contribution of the EM radiation in e.g. the Solar System. The Sun converts about $4 \times 10^9$ kg of matter to energy every second. Since it weighs about $2 \times 10^{30}$ kg every second it loses about $2 \times 10^{-19}$% of it's mass every second.

If you're prepared to assume the photon density in the Solar System is dominated by the Sun's output (which seems plausible) and take the size of the Solar System to be Neptune's orbit, i.e. $1.5 \times 10^4$ light seconds then the mass/energy of photons in the Solar System is $3 \times 10^{-15}$% of the Sun's mass. So it's utterly insignificant.

The reason photons make a much lower contribution to the Solar System than to the universe as a whole is because mass is much more concentrated in the Solar System than in the universe as a whole.

  • $\begingroup$ Granted, but this is for the CMB. But I was not asking about EM contribution on a cosmological scale, but on a galactic scale. So, freely rephrased: what is the effective gravitating mass of a star system? I suspect the radius of a star's total gravitating mass would be substantially larger than just the mass of the star itself, while it is the star itself that we're observing (through interactions w/ other objects) $\endgroup$ – Rody Oldenhuis Aug 19 '12 at 11:54
  • $\begingroup$ The Solar mass $M_S = 1.989\times 10^{30}\ \mathrm{kg}$, and its expected lifetime is around $T_S \approx 10^9\ \mathrm{y} = 3.16\times 10^{17}\ \mathrm{s}$. Then, the total gravitational mass emitted as EM radiation would be $4\times 10^9 T_S/M_S \approx 10^{-3}M_S$. So the Sun would have converted roughly one thousandth of its mass to EM. Granted, this is all rather simplistic, nevertheless the Sun alone contributes 0.1% of its mass to the gravitating "goo" in between the stars. Suppose this number was 10-folded by its lifetime alpha/beta/neutrino emissions...that is not insignificant. $\endgroup$ – Rody Oldenhuis Aug 20 '12 at 9:01
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    $\begingroup$ The thing is that the Milky way, which is a fairly large galaxy, is only about $10^5$ light years across, so only the last $10^5$ years worth of solar radiation is still in the galaxy. The total radiation emitted by the Sun during it's life would be spread over a sphere about 10 billion light years in diameter. That makes it gravitationally insignificant compared to the mass of the Sun, unless of course you're looking at length scales of over a billion light years. $\endgroup$ – John Rennie Aug 20 '12 at 19:24
  • $\begingroup$ This is of course true. I hadn't expected it to be much, and indeed, it would seem that the contribution is pretty tiny. I consider this case closed. $\endgroup$ – Rody Oldenhuis Aug 20 '12 at 19:29
  • $\begingroup$ @JohnRennie But of course you need to multiply that by the numbers of stars in the Milky Way. Or if you take the mass the sun loses every second and multiply by the time light takes to travel the Milky Way, which is roughly $3 \times 10^{14}$ seconds, that comes to about $6 \times 10^{-5}$% of the galaxy mass as photons in transit. This is of course neglecting light from other galaxies, which have lower average luminosity than local sources. (Actually Rob's answer has a better estimate using the luminosity of our galaxy) $\endgroup$ – Real Sep 22 '16 at 14:26

The luminosity of the Galaxy is currently estimated to be around $5\times10^{36}$ W and thus an integrated "mass loss" in the form of radiation of of order $10^{-3} M_{\odot}$/yr. But how much radiation is present in the Galaxy? An order of magnitude estimate could be that the Galaxy (including the dark matter) is of order 100,000 light years in radius and so contains about 100,000 years worth of mass in the form of radiation - i.e. about $100M_{\odot}$.

If the CMB has a "mass" density of $5\times10^{-34}$ g/cm$^{3}$, the equivalent mass of CMB photons in the same volume is a few hundred $M_{\odot}$.

These numbers are uncannily similar and of course both are completely negligible in a gravitational sense of order 1 part in $10^{10}$.


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