The star whose far side can be seen

The bend Newton got half right found that light passing the Sun is deflected by 1.75 seconds of arc, twice what a falling corpuscle would manage. The lens with no focal length and the fifth image a galaxy hides turned that deflection into an instrument: a mass between an observer and a distant source rearranges the source’s light into rings, arcs and odd numbers of images. In every one of those arguments the light belongs to something far behind the mass, and the mass is only a lens.

The same geometry acts on the light a body sends out itself, and for an ordinary star the effect is too small to matter. For a neutron star it is not. Light leaving its surface is bent on the way out by tens of degrees, and the consequence is one of the plainest violations of everyday visual experience that physics offers: from a single direction, an observer sees most of the star, including a large part of its back.

A hemisphere is a statement about straight lines #

The rule that a ball shows half of itself is so familiar that it hardly seems to need a reason. The reason is that every point an observer sees sends a ray in a straight line to the eye, and a straight line from a point on the far side of the ball has to pass through the ball to get there. The boundary between the seen half and the hidden half is the circle of points from which the straight line to the observer just grazes the surface — the limb. For a distant observer it is a great circle, at exactly 90° from the point that faces the observer.

Nothing in that argument is about the ball. It is about the rays. If rays bend, the boundary moves, and if they bend towards the body, it moves round the back. A ray that leaves a point slightly behind the geometric limb, travelling at first along the surface, can be curved by the body’s gravity just enough to come out heading towards the observer. That point is then visible, although no straight line connects it to the eye.

The question is how far round the back this reaches, and it has an exact answer, because the paths of light outside a spherical body are known exactly. Outside any spherical, non-rotating mass the geometry is Schwarzschild’s, whatever the mass is made of and however it is arranged inside, and in that geometry a ray’s path is fixed by a single number: its impact parameter, the distance by which it would miss the centre if gravity were switched off.

The ray that leaves along the surface #

Measure lengths in units of the Schwarzschild radius $r_s = 2GM/c^2$, and let the star’s radius be $R$. A ray leaves a point on the surface at an angle $\alpha$ to the vertical there, measured by an observer standing on the surface. Its impact parameter is

$$b = \frac{R \sin\alpha}{\sqrt{1 - r_s/R}},$$

larger than the straight-line value $R\sin\alpha$ because the local observer’s rulers and clocks are those of a place where time runs slow and radial distance is stretched. From there the ray’s path is fixed. On its way out to infinity it swings round the centre by an angle

$$\psi(\alpha) = \int_0^{r_s/R} \frac{du}{\sqrt{1/b^2 - u^2(1 - u)}}, \qquad u = \frac{r_s}{r},$$

and $\psi$ is the angle, seen from the centre, between the point the ray left and the direction of the observer. With no gravity the integral gives $\psi = \alpha$ exactly: the ray leaves at $\alpha$ to the vertical and arrives at $\alpha$ to the line from the centre, which is the straight-line statement. With gravity the integrand is larger, and $\psi$ exceeds $\alpha$.

The figure shows the integral evaluated for three stars. The interesting ray is the last one, at $\alpha = 90^\circ$: it leaves the surface horizontally, skimming along it, and for a straight line that would be the ray from the limb itself. Here it arrives from further round. A star of five Schwarzschild radii carries its tangential ray 104° from the line of sight; one of three radii carries it 119°; one of two carries it 153°.

Every point with $\psi$ smaller than that maximum sends some ray to the observer, because $\psi(\alpha)$ rises steadily from zero, so the visible region is a cap of the surface whose edge is $\psi_{\max} = \psi(90^\circ)$. The hemisphere was the special case $\psi_{\max} = 90^\circ$.

Three-quarters of a neutron star at once #

A neutron star of 1.4 solar masses has a Schwarzschild radius of 4.1 kilometres, and its actual radius, by the best present estimates, is between about 11 and 13 kilometres. Take 12. That is 2.9 Schwarzschild radii, close to the middle of the range, and the hero figure at the top of this essay traces its rays exactly.

The dotted vertical line is where a straight-ray observer’s view would stop. The rays that reach the observer leave from well behind it. The outermost, drawn in red, leaves the surface tangentially at 121° from the point facing the observer and curves round to join the other rays heading right. The cap it bounds has an area fraction of $(1 - \cos\psi_{\max})/2$, and with $\psi_{\max} = 121^\circ$ that is 76 per cent.

Nothing about this is a projection effect or an artefact of perspective. A camera far away pointed at such a star would record an image disc whose every point is a different point on the star, and three-quarters of the surface would be represented somewhere on it. The back of the star would appear as a bright ring around the edge of the disc, compressed into the outer part of the image, because rays from far round the back all leave close to tangentially and arrive at large impact parameters.

The image disc is also larger than the star. Its edge is the impact parameter of the tangential ray, $b = R/\sqrt{1 - r_s/R}$, which for $2.9\,r_s$ is 1.24 times the radius. A star drawn on its own image would sit inside it with room to spare, a magnification of the whole object by its own field that has no counterpart in the lens arguments, where the lensing mass is dark.

The fraction against radius is the whole story in one curve. Far from compact it tends to a half, slowly: a star of eight Schwarzschild radii still shows 57 per cent of itself, and the excess above a half falls only as the first power of $r_s/R$. The Sun, at 236,000 Schwarzschild radii, shows half of itself plus about two millionths, which is real and immeasurable. The white dwarf of the redshift that weighs a dead star is at a few thousand Schwarzschild radii and shows half of itself plus a few hundredths of a per cent.

Neutron stars sit in the shaded band, where the curve has turned steeply upward. The whole uncertainty in their radius, ten to fourteen kilometres, moves the visible share between 71 and 83 per cent, which is a lever large enough to measure with — and is, as the last part of this essay describes.

The radius at which nothing is hidden #

As the star is compressed the tangential ray swings further round, and at some radius it comes from the point directly behind the star. That radius is found by solving $\psi_{\max}(R) = 180^\circ$, and the exact integral gives $R = 1.76\,r_s$. A star that compact would show its whole surface from any direction. The point directly behind it would appear as a ring around the edge of the image — an Einstein ring of the star’s own back.

Below $1.76\,r_s$ there is more than everything. Points near the back send rays round both sides, and are seen twice: once in the outer ring and once, mirror-reversed, in a thinner ring inside it. Further compression brings the star toward the circle light cannot leave, the photon sphere at $1.5\,r_s$, where a tangential ray no longer escapes at all but circles indefinitely, and rays from every surface point can wind round the star any number of times before leaving. The image would become an infinite nest of rings.

Whether any star can reach those radii is a separate question, answered by the matter inside rather than the geometry outside. The pressure that weighs down what it holds up found that general relativity forbids any static star smaller than $9/8$ of its Schwarzschild radius, whatever it is made of, because the pressure needed to support it becomes infinite at the centre. Real matter fails long before that. Requiring the speed of sound inside to stay below that of light, the most compact neutron star any equation of state can make sits near $1.4\,r_s$, and the stars actually observed, with their measured masses and their inferred radii, lie between about 2 and 3.5. So the whole-surface view is not something any known star provides. What known stars provide is the steep part of the curve, where a few kilometres of radius is worth ten per cent of the surface.

How bright the back looks #

Seeing a point and seeing it brightly are different things. A small patch of the surface tilted away from the observer looks fainter for two reasons: it presents a smaller projected area, the cosine factor that the four cosines at the edge of a photograph found darkening the corners of every lens’s image; and its light is spread over a larger or smaller solid angle by the bending, depending on how neighbouring rays diverge on the way out.

Both are captured by a single factor. For a small spot radiating evenly in every outward direction — a surface that looks equally bright from any angle, in the sense of Lambert — the light reaching a distant observer from a spot at angle $\psi$ is proportional to

$$\cos\alpha \, \frac{d(\cos\alpha)}{d(\cos\psi)},$$

the cosine of the angle at which its ray actually leaves, times the rate at which leaving angles map onto viewing angles. With straight rays $\alpha = \psi$ and the factor is $\cos\psi$, which falls to zero at the geometric limb. With bending the factor stays positive past 90° and only reaches zero at $\psi_{\max}$.

The curves are close to straight lines in $\cos\psi$ rather than cosines, and the reason is worth having. If the exact relation between $\alpha$ and $\psi$ is replaced by the approximation Andrei Beloborodov published in 2002,

$$1 - \cos\alpha = \left(1 - \frac{r_s}{R}\right)(1 - \cos\psi),$$

then $\cos\alpha$ is a linear function of $\cos\psi$, its derivative is the constant $1 - r_s/R$, and the brightness factor becomes, relative to the face-on value,

$$\frac{r_s}{R} + \left(1 - \frac{r_s}{R}\right)\cos\psi.$$

A spot at the geometric limb, $\cos\psi = 0$, keeps a fraction $r_s/R$ of its face-on brightness: 34 per cent for a $2.9\,r_s$ neutron star, where straight rays would have given none. The exact factor at the limb is 34 per cent at $3\,r_s$ and 51 at 2, against the rule’s 33 and 50, and the shape of the whole curve is close to the rule’s line. The approximation is not exact anywhere except in the weak-field limit, and the earlier figure shows where it drifts: within a few degrees down to three Schwarzschild radii, and overshooting the tangential ray by 27° at two. Its virtue is that it turns an integral into a line.

One more factor multiplies everything without changing the shape. Light climbing out of the field loses energy and arrives spread over a longer time, and both are fixed by the surface redshift, the same at every point of a spherical star. For a $2.9\,r_s$ star the received power from every spot is reduced by $(1 - r_s/R)^2$, a factor of 0.43. It dims the whole picture uniformly and leaves the share of the surface in view and the relative brightness of its parts exactly as drawn.

Two spots that stop flickering #

Neutron stars rarely glow evenly. A star with a strong magnetic field funnels the gas falling onto it, or the particles accelerated above it, onto its magnetic poles, and the poles become hot spots a few hundred metres across, typically two of them, roughly opposite each other. As the star spins, each spot swings round, alternately towards the observer and away. With straight rays the result is a strong pulsation: when one spot faces the observer the other is hidden, and in between both can be near the limb and dim.

With bending, less of each spot’s swing is hidden, and the linear brightness factor has a consequence that looks like a conspiracy. Put two identical spots exactly opposite each other, at angles $\psi$ and $180^\circ - \psi$. In Beloborodov’s form their brightnesses are $r_s/R + (1 - r_s/R)\cos\psi$ and $r_s/R - (1 - r_s/R)\cos\psi$, and the sum is $2r_s/R$ — independent of $\psi$, and so independent of the rotation, as long as both spots are in view.

The exact paths do not quite keep the conspiracy, because the exact factor is not quite linear, but they come close. At five Schwarzschild radii the two spots’ combined light varies by 60 per cent over a rotation; at three, by 34; at two, by only 7. At $1.8\,r_s$, where the whole surface is visible and so both spots are always in view, the sum has a flat top and residual dips where each spot passes through the region seen twice, and the variation is 12 per cent. In no case does the star stop varying entirely, but the trend is the point: the more compact a star, the less its light pulses, because its gravity shows the observer more of what a straight-ray view would hide.

That trend is a measurement waiting to be made. The amplitude and shape of a spinning neutron star’s pulses depend on the star’s compactness, $r_s/R$, through exactly this geometry, and the mass enters only through $r_s$. Fitting the pulse shapes of X-ray hot spots with the paths drawn here, corrected for rotation, is how the radius of a neutron star of known mass has been estimated in the past decade, and it is one of only two routes to that number that do not depend on a model of the star’s interior.

Where the geometry stops being enough #

Every figure here assumes a star that is spherical and does not rotate, so that the geometry outside it is Schwarzschild’s and every ray stays in one plane. That is exact for the static star and wrong for every real neutron star in the ways that matter most for its light.

Rotation enters four times. A pulsar spinning 600 times a second has a surface moving at a tenth of the speed of light. Its light is Doppler-shifted and beamed forward on the approaching side and dimmed on the receding side, which makes the two halves of a pulse unequal. Its surface is flattened at the poles by a few per cent. Its rotation drags the surrounding geometry round, which twists the rays slightly out of their planes. And different parts of the surface are at different distances from the observer, so light from them arrives at different times. Each correction is a few per cent to tens of per cent at high spin, and none changes the conclusion that more than half the star is visible.

The surface is not a Lambert radiator. A neutron star’s atmosphere, a few centimetres of hot hydrogen or helium, emits more strongly in some directions than others; light leaving near the vertical comes from deeper, hotter layers. That reshapes the brightness factor without moving the edge of the visible region.

The spots have size. A spot a kilometre across on a star of twelve straddles a range of $\psi$, and near $\psi_{\max}$ part of it can be visible while part is not. The point-spot curves describe the limit.

The figures say nothing about what is inside. They depend on the star only through its mass and its radius. That is the strength of the method — it measures $R/M$ without knowing the equation of state — and its limit: a star of a given radius and mass shows the same fraction of itself whether its centre is neutrons, hyperons or free quarks.

Still open: how large a neutron star is #

The radius of a neutron star of given mass is set by the stiffness of matter at densities several times that of an atomic nucleus, and no experiment on Earth reaches those densities. The theories disagree by a few kilometres, and a few kilometres is the difference between a star of 2.4 and of 3.4 Schwarzschild radii — between showing 83 per cent of its surface and showing 71.

Three kinds of measurement now bear on it. The pulse profiles of hot spots, read through the geometry drawn here, have put the radii of two neutron stars near 12 to 13 kilometres with uncertainties of about a kilometre. The tidal deformation of neutron stars spiralling into each other, read from the shape of the gravitational waves they emit just before merging, gives an independent constraint, consistent with the first. And the heaviest neutron stars known, at about two solar masses, rule out every equation of state too soft to hold them up — the same competition between pressure and gravity that set the mass no cold matter can hold up for white dwarfs.

Whether those constraints converge on a single relation between mass and radius, and whether that relation shows the abrupt softening that a transition to quark matter in the core would produce, has not been settled. The light-bending part of the problem is not the uncertain part. The geometry outside a star is known exactly, and the angle through which a ray is carried on its way out — the same angle that, for light passing far from the Sun, comes to twice Newton’s estimate — is computed here to a fraction of a degree. What is uncertain is everything the light passes through and everything the star does while it is emitting. The far side of a neutron star can be seen; seeing it clearly enough to read off the star’s size is the work still being done.