Has any simulation been done to produce an accurate visualization of a neutron star, as seen from an observer at distances on the order of 1 AU?

(Edit: I suppose instead of 1 AU I mean "distance such that its angular size is comparable to the Sun's from Earth, or even closer")

At a surface temp of approximately 1 million K, it's "real" color I'd imagine to be dark blue, but with such a high redshift might it actually appear dark red? Right on the verge of being visible at all?

Seems very spooky if so.

Edit: a basic search tells me the higher end redshift at a neutron star's surface is perhaps $z=0.4$, which by the following formula:

$$z+1 = \frac{\lambda_\infty}{\lambda_{surf}}$$

is enough to turn blue light (450 nm) into reddish-orange light (630 nm).

My primary question is, has anyone produced a visualization of one, in the same vein as the black hole visualization featured in Interstellar (not necessarily that high end)?

W can calculate that from a far observer's viewpoint any object falling into a black hole will be redshifted into non-being. My thought was that for a neutron star we might see the same trend with the surface of the star in a static fashion, e.g. if we incrementally added mass until it passed the limit of becoming an event horizon.

UPDATE: I suppose my secret hope is that there could be a version of a neutron star around $1.4r_S$ or so that looks like the following – ghostly red and barely visible, about to recede forever beyond the reach of the outer universe with the addition of a bit more mass.

From ProfRob's answer, it sounds like a neutron star about this size could cool over time until its blackbody spectrum fit the below?

(adapted from Astronomy Magazine

Because of the way the blackbody spectrum shifts, I suppose this cooled neutron star might be indistinguishable from a Red Dwarf star, and a visitor would not realize the difference until it was too late.

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    $\begingroup$ Note that, as the temperature of a blackbody increases, its emissions get brighter at every wavelength. Sunlight is white; to an excellent approximation, hotter blackbodies are brighter white. Note also that a redshifted blackbody spectrum is still a blackbody spectrum. $\endgroup$
    – rob
    Aug 26, 2022 at 4:25
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    $\begingroup$ Yes, but never “dark” blue. Compare white sunlight to the blue-white blackbody emissions from a lightning bolt, whose temperature is higher. As the temperature increases further, increase the brightness at every wavelength. $\endgroup$
    – rob
    Aug 26, 2022 at 14:39
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    $\begingroup$ The biggest redshift you will ever see from a neutron star surface is a factor of 2. It will collapse beyond this. $\endgroup$
    – ProfRob
    Aug 27, 2022 at 1:08
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    $\begingroup$ A hot neutron star will look (bluish) white. A redshifted hot neutron star will still look (bluish) white because the blackbody peak is in the ultraviolet $\endgroup$
    – ProfRob
    Aug 27, 2022 at 1:09
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    $\begingroup$ A neutron star like the one in en.wikipedia.org/wiki/PSR_J0348%2B0432 has a radius of ~13 km. It has an angular diameter of 32' (like the Sun) at ~2800 km. You probably don't want to get that close... $\endgroup$
    – PM 2Ring
    Aug 27, 2022 at 6:45

2 Answers 2


To see a neutron star with a similar angular diameter to the Sun, you would have to get much closer than 1 au. They have a radius of approximately 10 km but their apparent radius to a distant observer is larger, due to "light bending" in General Relativity. An observer can see the back side of the neutron star to some extent and can actually see the whole of the neutron star surface if the radius is below 1.76 times the Schwarzschild radius for its mass, $r_s = 2GM/c^2$. See https://physics.stackexchange.com/a/350814/43351 for details and some attempts to visualise this. e.g.

A visualisation of a neutron star

A neutron star with a radius less than $1.5r_s$ would distort a background star field in a similar way to a black hole, including the photon ring at an apparent radius of $2.7r_s$ caused by unstable circular photon orbits at $1.5r_s$. Larger neutron stars would distort the positions of background stars but would lack this photon ring.

A neutron star surface could be approximated as a blackbody emitter. At a temperature of a million K (in your question) the peak of the spectrum would be in the extreme UV and the neutron star would have a surface brightness about a factor $(10^6/5800)^4$ larger than the surface of the Sun. Thus there would be no prospect of you "looking" at it with your own eyes and the flux of radiation received by the observer would be about 1.4 kW/m$^2$ scaled up by a similar factor - mostly in the UV and EUV wavelength range.

Hypothetically, if you built an electronic detector that could survive this and had a response similar to the human eye (why would you do this?), Then the neutron star would appear bluish-white.

You ask how gravitational redshift affects this? Well, assuming that this hasn't already been taken into account (reported observed temperatures for neutron stars are usually the temperature after redshift), then the blackbody temperature is reduced by a factor of $(1-z)^{1/2}$ where $z = r_s/R$. (Haensel 2001 : Incidentally, the apparent radius is increased by a factor $(1-z)^{-1/2}$). For a typical neutron star this redshift factor is about $(1- 0.5)^{1/2} = 0.7$, which shifts the peak of the blackbody spectrum into the far UV. The neutron star would still appear bluish-white.

But you ask whether we could consider an extremely small neutron star with a much smaller temperature redshift factor? The answer to that is that very small neutron stars cannot be stable. There is a fundamental "Buchdahl limit" that is due to pressure and internal kinetic energy contributing to the gravity of a neutron star in GR (see for example https://physics.stackexchange.com/a/265052/43351). In practice this limit is even more stringent because realistic equations of state are limited by causality such that $p < \rho c^2$, where $\rho$ is the density. What this means is that $z$ is limited to be $<0.7$ and the temperature reduction factor could never be smaller than about 0.5.

As a final point, I note that a surface temperature of a million K is only appropriate for very young neutron stars. These cool rapidly by neutrino emission and then later by radiation from the surface. Their surface temperatures will be well below a million K in only a million years or so. How their temperatures evolve after this is highly uncertain and none have been observed. The problem is that neutron stars have a very small heat capacity. They cool easily, but they can also be reheated easily by ohmic dissipation of their strong magnetic fields or accretion from the interstellar medium. Thus the surface temperature of most neutron stars is likely to be much lower than $10^6$ K, but probably significantly higher than the $10^4$ K they would need to be dominated by a visible spectrum.


It would look like a bluish, very bright star. I'll give a rough, order of magnitude calculation.

Its diameter is about $10^{-5}$ of the Sun, so its solid angle would be about $10^{-10}$ of the Sun. In the long wavelength part of a blackbody curve, intensity is proportional to temperature, so about 100 times the Sun per solid angle. Multiply, get $10^{-8}$. That's 20 astronomical magnitudes. The Sun has apparent magnitude -27, so the neutron star would have magnitude -7. Compare to Sirius at -1 or Venus at at -4.

  • $\begingroup$ That's great info. I had revised my initial question to acknowledge that a viewer would have to be much closer even than Mercury's orbit for the neutron star to have a similar solid angle to the Sun viewed from Earth. $\endgroup$
    – RC_23
    Aug 27, 2022 at 1:45

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