How Would a Neutron Star Actually Appear?

  • Having seen many pictures produced by artists of neutron stars and planets that orbit some of them, I was wondering how a pulsar would appear to a human being, in visible light (assuming the intense radiation etc. doesn't kill us in the process).

    As I understand, the pulsar's beam is projected from the star's magnetic poles rather than rotational poles, which are not necessarily in line with each other. Given that pulsars rotate extremely quickly and the beam could be visible across vast distances - such as if it were shining through the pulsar's nebula - would it appear as a straight line, curved line or perhaps a cone? This is assuming the beam can be seen in visible light.

    Given the incredible density of neutron stars and their small physical sizes, would the night sky be visibly distorted to the point where (for example) just after sunset on a hypothetical planet, one could possibly observe other planets near or behind the star that would otherwise be blocked by it?

    Given their small surface areas, would a neutron star still appear as luminous as say, the Sun, at a similar distance? How close would you have to get to a neutron star for its apparent magnitude to match the Sun's from Earth?

    Not related to your question, but what things would look like on the surface of a Neutron Star is much more interesting. Because of the way light bends, the sky when standing on the surface of a Neutron Star would be squeezed into a tiny circle and the planet would visibly appear to rise up around you, taking up most of what you can see.

    @userLTK It's a fascinating link, and a negatively curved horizon would be amazing to see to say the least!

    Does anyone know if such "ultracompact" neutron stars actually form?

  • ProfRob

    ProfRob Correct answer

    7 years ago

    Your question is too general, you need to get to specific examples.

    First, very few neutron stars are pulsars. Pulsars are either a brief phase during a pulsar's spin-down at the start of a neutron star's life, or they are the product of the spin-up of a neutron star in a binary system. Most neutron stars fall in neither of these categories.

    A standard neutron star will look like any other star at a similar temperature. Most of them will be very hot indeed - 100,000 K or more, though the cooling histories of neutron stars are still uncertain and depend on some exotic physics. Such an object is "white hot" - it emits black body radiation at all frequencies visible to the eye (as well as lots more at UV wavelengths).

    How close would you have to get for it's apparent luminosity/magnitude to match the Sun? Well that depends on the size and temperature of the neutron star. Most are thought to have a diameter of 20 km. The way you would do the calculation is equate the blackbody radiative flux per unit area at a given distance to the solar radiation constant of about 1300 W per square metre. However, there are two wrinkles for a neutron star: First, the radiation is gravitationally redshifted, so the temperature we measure is lower than the temperature at the surface. Second, General Relativity tells us that we can see more than just a hemisphere of the neutron star - i.e. we can see around the back - and this increases the flux we observe. These are roughly factor of two effects, so just to get an order of magnitude estimate, ignore GR and assume a 10 km radius NS with $T=10^{5}$ K.

    Using Stefan's law for a blackbody, then at a distance $d$, we have that
    $$\frac{4 \pi r^2}{4\pi d^2} \sigma T^4 = 1300\ W\ m^{-2},$$
    where $\sigma$ is the Stefan-Boltzmann constant.

    For $r=10$ km, then $d=7 \times 10^{8}$ m, which is coincidentally about a solar radius. Of course this distance depends on the square of the temperature, so a younger NS with $T=10^6$ K, then $d \sim 1$ au.

    These are the distances where the total flux at all wavelengths would be similar to that from the Sun. To do the calculation just for the visible range we need to account for the bolometric correction, which converts a visual magnitude to a bolometric magnitude. The bolometric correction for the Sun is $\sim 0$, whereas the bolometric correction for a very hot star could be -5 mag. This means that only 1% as much flux from the hot neutron star emerges in the visible band compared with sunlight. This means that the distances calculated above, if we require the visual brightness of the neutron star be similar to the Sun, must be reduced by a factor of 10.

    To turn to pulsars. Note that the pulsed radiation does have an optical component and pulsed optical radiation has been seen from a number of pulsars. Optical synchrotron emission would just appear to be a periodic, intense brightening of the pulsar, as the beam sweeps across the line of sight. If you were not in the line of sight, then you would not see the pulsed optical emission. If you could observe the beam passing through nebulosity or some other medium around the pulsar then yes there may well be some effects you could see in terms of ionisation or scattered light coming from along the beam path.

    Lastly, the gravitational lensing effect. Yes, this should be strong close to a neutron star. The deflection angle (in radians) is given by
    $$ \alpha = \frac{4GM}{c^2 b},$$
    where $b$ is how close the light passes to the neutron star and $M$ is the neutron star mass. Expressing $b$ in terms of the 10km radius of the neutron star:
    $$ \alpha \simeq 0.83 \left(\frac{M}{1.4M_{\odot}}\right) \left(\frac{b}{10 km}\right)^{-1},$$
    where strictly speaking this formula is only valid for $\alpha \ll 1$.

    So consider a planet directly behind the neutron star at a distance of 1 au. The light from this would only need to be bent through an angle of $\sim 2 \times 10\ km/1\ au \sim 10^{-7}$ radians in order to be seen from a planet diametrically opposite at a distance of 1 au. So this is easily possible. However, the image would likely be highly distorted, especially if the neutron star was spinning. It would not look dissimilar to this simulated black hole image, but with a bright neutron star in the middle rather than a black disc.

    distorted images

    A very interesting answer. I had imagined a neutron star's luminosity would be higher than what would be calculated due to light emitted from its 'far side' being bent towards an observer, but I did not realise it would also be redshifted in such a way as to make the star appear cooler.

    *Does* the lensing increase the observed flux in this case? Thinking in terms of light rays emitted from the surface, some emitted nonradially from the back hemisphere will be seen, but this also means that some emitted from the front hemisphere that "would have been" observed *won't* be, because they will bend to miss the observer. ... For a hypothetical nonrotating neutron star, spherical symmetry implies only the redshift matters due to energy conservation. For a more realistic one, it would depend on relative orientation.

    @StanLiou that does sound correct. It can't be brighter in all directions.

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Content dated before 7/24/2021 11:53 AM