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While traveling billions of lightyears, photons lose their energy. During my physics degree, the question of where that energy goes was never answered. I always thought it must be losing it to climbing in some gravitational potential, but which potential?? Sparked by a Veritasium video, I started pondering again. Maybe a part of its energy is lost to black-body radation?

Let's do a thought experiment. Imagine a (e.g. dust) particule in a spherical cavity of radius $R$, outside of which is a thermal bath, all in thermal equilibrium at constant temperature. If the cavity grows (slowly), the density of black-body radiation remains constant, and thus the particule's temperature as well. My reasoning is that even though the boundary of the cavity is further away, there is more of it that radiates into the cavity. Those cancel out exactly because had we started at a slightly larger radius, that would also be in thermal equilibrium. So when the radius grows to that size from a smaller radius to a bigger, we arrive in the state where it is still in equilibrium (assuming slow growth).

I would say our dust particule floating through space for billions of years does not enjoy this equilibrium. Imagine that $R$ is cosmological in scale, that the thermal bath is many many many galaxies, of the same temperature. When the galaxies recede, not more of them radiate to our particule (like our thermal bath's wall did). From our particule's perspective, a growing area of the sky darkens. Also less radiation from each galaxy hits the particule because of the nverse-square law. It's a crude model of cosmic expansion, but I hope it drives the point. Because the equilibrium is not maintained, I would conclude that our particule's temperature will decrease?

I chose a dust particule to delay dealing with the ill-defined temperature of a single fundamental particle, but let's consider a photon now instead:

  • assuming that also a photon emits black-body radiation (here is a paper that seems to suggest so?)
  • and assuming that if the photon emits more black-body radiation than it receives, it loses energy and thus redshifts
  • and assuming that my simple model of receding distant galaxies sufficiently approximates cosmic expansion

Does a photon's redshift over the billions of years come from the cosmic expansion in part through this mechanism of black-body radiation being out of equilibrium?

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    $\begingroup$ The starting assumption is wrong: photons don’t emit radiation. $\endgroup$ Commented Apr 15 at 22:57

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the question of where that energy goes was never answered.

Why should energy be conserved at all? Noether’s theorem says it should if time-translation invariance is obeyed, i.e. if the Lagrangian is invariant under the flow of time, but it’s not because the Universe is expanding. In FLRW spacetime (the standard cosmological model of the expanding Universe, although with new data it might be updated) you end up with a Ricci scalar of

$$R=\frac{6}{c^2}\Bigg(\frac{1}{a}\frac{\text d^2a}{\text dt^2}+\frac{1}{a^2}\bigg(\frac{\text da}{\text dt}\bigg)^2\Bigg)$$

and since the Einstein-Hilbert action is

$$S=\frac{c^4}{16\pi G}\int R\sqrt{g}\text d^4x,$$

the action and Lagrangian are obviously not time-invariant. As such on cosmological timescales you should expect energy to appear/disappear from nowhere just because there’s no reason to expect it should. On reasonable human timescales this effect becomes insignificant, but if a photon has been travelling long enough, it will become apparent. The energy doesn’t go anywhere, it disappears completely.

Knowing all this, postulating about photons emitting radiation can be forgone.

Yes, this is in violation of the first law of thermodynamics. Yes, it is also possible to violate the second law, because the second law is just a statistical statement about what happens at the macroscale based on observations, not an actual hard-and-fast law. The laws are both convenient for learning most physics but they have their edge cases.

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  • $\begingroup$ Thank you. I didn't mention or aim for energy conservation, I was just asking which part of energy loss is due to blackbody radiation. You're saying that any would be a new postulate, agreed, a redundant one at that, also agreed, and so if I understand you correctly you're saying that there is zero redshift due to blackbody radiation? $\endgroup$ Commented Apr 16 at 7:50

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