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Here is a non rotating neutron star emitting perfect black body radiation from its surface. Supposing that the radiation is redshifted by a factor of 2 for distant observers (which is chosen purely for simplicity because neutron stars in reality can’t be that compact), each photon’s energy should be halved uniformly. As a result, the frequency peak should also be halved. Because the frequency peak is linearly proportional to the temperature, the effective temperature perceived by distant observers should be halved as well. Besides the surface of the neutron star also experiences time dilation by a factor of 2, which means the number of photons emitted from the surface per second is twice the number of photons reaching distant sites. Taken together, the effective luminosity of the neutron star should be a quarter (1/4) of the surface luminosity.

However, according to the Stefan-Boltzmann law, the radiation power is proportional to the 4th power of temperature, which means the effective luminosity should be 1/16, not 1/4 of the surface luminosity. Is it due to the gravity lensing effect which makes the neutron star look bigger than its actual size? Another complicating factor is that once below the photon sphere, the surface of the neutron star no longer obeys the lambert emission law because light rays emitted at low angles are directed back towards the surface, which reduces the effective luminosity. These calculations are beyond my capability, so I come to ask for help.

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The light is gravitationally lensed. According to the distant observer, the emitting surface area is increased by exactly the right amount to deal with the apparent contradiction that you have noted.

The relevant equations are (Haensel 2001): $$ L_\infty = 4\pi R_\infty^2 \sigma T_\infty^4 = L(1 -r_s/R)$$ $$ T_\infty = T(1 - r_s/R)^{1/2}$$ $$ R_\infty = R(1 - r_s/R)^{-1/2}$$ where $r_s$ is the Schwarzschild radius and the $\infty$ subscript means the quantity inferred by a distant obsever.

The last of these equations though can only be applied if $R>1.5r_s$. This point marks the minimum value of $R_{\infty, {\rm min}} = 3\sqrt{3}r_s/2$ at a fixed mass. For lower values of the neutron star radius then photons with an impact parameter larger than $R_{\infty, {\rm min}}$ cannot escape from the neutron star.

i.e. The effective radius will be fixed at its minimum value and the first equation becomes $$ L_\infty = \frac{27L}{4}\left(\frac{r_s}{R}\right)^2\left(1 -\frac{r_s}{R}\right)^2$$ for $R<1.5r_s$ and the luminosity falls more steeply as you suspected.

The extreme situation you hypothesise requires $R= 4r_s/3$. This is still above the Buchdahl limit so might be allowed by the hardest of proposed equations of state. You only need $R < 1.76 r_s$ for a distant observer to see the whole of the surface (e.g., see this answer). The limiting $R_{\infty, {\rm min}}$ corresponds to tangentially emitted light orbiting the neutron star multiple times before exiting to the observer at infinity!

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  • $\begingroup$ Thanks for your equations. There is one problem. As the radius of the star approaches the Schwarzschild radius (again, this is purely hypothetical because no hydrostatic equilibrium exists below 9/8 Rs) the redshift diverges, which means the effective radius should diverge as well. However, black holes still have a finite black shadow according to simulations. Maybe the equation above has a cut off when R<1.5Rs. $\endgroup$
    – 哲煜黄
    Mar 17, 2023 at 9:28
  • $\begingroup$ That’s why I suggested that once the size goes below the photon sphere, the effective radius is fixed (which is equal to the size of the shadow of black hole) and the luminosity drops more rapidly due to photons trapped below the photon sphere. $\endgroup$
    – 哲煜黄
    Mar 17, 2023 at 9:31
  • $\begingroup$ @哲煜黄 I think you have it right. See my edit, there is a minimum in $R_{\infty}/r_s$ and it will not look any larger if $R < 1.5r_s$. $\endgroup$
    – ProfRob
    Mar 17, 2023 at 10:54
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You forgot that the emitting area is reduced by $(\frac 1 2)^2$, so the total factor is $\frac 1 {16}$, per unit emitting surface area.

Regarding the non-Lambertian emission of radiation (after lensing), your argument is reasonable, but phase space volume is conserved, so there is no where else for the radiation to go if you consider $4\pi$ steradian...basically you can see the more that half the neutron star's surface so that has to make up for the deficit.

Though it is not gravitational redshift, the CMB remains a near perfect black body at $Z \approx 1100$. 2.7 K is the recombination temperature of hydrogen, divided by $z$ (or $z+1$, not sure).

That Planck's Law, derived in 1900, before the formulation of special relativity in 1905, is Lorentz covariant is fascinating. I believe this was explained to me when I took Physics 136 from both Thorne and Blanford...derived from grand canonical ensemble considerations, but I cannot remember the details. See: http://www.pmaweb.caltech.edu/Courses/ph136/yr2012/ or buy the book, it is truly brilliant: https://www.amazon.com/Modern-Classical-Physics-Elasticity-Statistical/dp/0691159025

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  • $\begingroup$ Could you elaborate a bit about the reduction of the emitting area? $\endgroup$
    – 哲煜黄
    Mar 17, 2023 at 6:28
  • $\begingroup$ I don’t think the gravity lensing will make up the loss. While some photons pointing to other directions are redirected to your direction, photons pointing to you are also directed away which cancels out exactly. The gravity lensing makes the neutron star bigger but not brighter. $\endgroup$
    – 哲煜黄
    Mar 17, 2023 at 6:33
  • $\begingroup$ If some photons are unable to escape the photon sphere, the gravity lensing won’t compensate for it because gravity lensing doesn’t change the number of photons. $\endgroup$
    – 哲煜黄
    Mar 17, 2023 at 6:34

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