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This is a follow-up question after this one:

Yes, the curvature is enough to bend lasers. In fact, there is an interesting feature in Schwarzschild spacetime that might answer your question well. but in principle you can arrange light to orbit the black hole indefinitely. I should add that this does not necessarily means the curvature is large. For very big black holes, the curvature at the horizon can be as small as the curvature in your living room (or smaller).

Is the curvature so extreme at the event horizon, that you could see curved laser beams?

One thing I do not comprehend is where Níckolas Alves's comment says the curvature is small (for large black holes, curvature is like in the living room), but light still cannot escape (it will orbit). So the fact that light cannot escape is not connected to the extreme curvature itself. Then what is it connected to and how can you have small curvature but still orbiting light? Naively said, my living room has no bent light rays (and none orbiting Earth), so what is it exactly that is different in my living room and the event horizon that makes light orbit (I assume now it is not the extreme curvature)?

Or, as another solution, light is always bent in any gravitational field, it is just that here on Earth in the living room light is only bent slightly (so little that our apparatus cannot detect it), but around the black hole light's path is bent in an obvious (visible with "naked-eye") way.

So curvature is a complex thing, a tensor, has many components, and some of it are actually smaller at the event horizon (of a large black hole) than my living room, and some components are the other way around (acceleration). But which one is responsible for the extent to which light rays are bent? If the curvature is the same, then why are light rays bent around a large black hole, but not in my living room?

So black holes bend light rays' path, we know that. But what about other gravitational fields? Does Earth's gravitational field bend light rays like that too, just not to that extent? In other words, is this a qualitative or quantitative difference, is bending light rays a boolean (yes, no), or not?

Question:

  1. Are light rays always bent in any gravitational field, or just in a strong one like a black hole?
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    $\begingroup$ Spacetime curvature or gravity is not required to observe the bending of light. For example, the Rindler spacetime is flat, but it has the horizon that locally is the same as Schwarzschild, so a Rindler observer would see laser beams curved exactly like near a black hole. A Rindler observer is anyone who is in an accelerating spaceship in a completely flat ordinary (Minkowski) spacetime with no spacetime curvature whatsoever. And yet he can observe laser beams curved. $\endgroup$
    – safesphere
    Commented Mar 30 at 4:24
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    $\begingroup$ This is exactly how Einsteins theory of relativity was tested, using the gravitational deflection of light passing near the sun. $\endgroup$
    – Triatticus
    Commented Mar 30 at 5:07

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Light is bent in all gravitational fields. Assuming that your living room is small enough to be considered “local”, then the primary determinant of the amount of deflection is the proper acceleration of the living room. Close to the horizon of a supermassive black hole there may not be a lot of curvature, but the proper acceleration of the living room becomes arbitrarily large. On the other hand, here on earth $9.8 \mathrm{\ m/s^2}$ doesn’t produce much deflection in the living room.

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Light follows null geodesics. Those geodesics are curved near any concentration of mass / energy, but spacetime curvature is usually quite small, except in extreme circumstances.

There's a difference between gravitational field strength and gravitational potential. The field strength is basically the gravitational acceleration, so it determines how heavy you feel when you're standing on a planet, or hovering above it. But the potential controls things like orbital motion and escape velocity.

In Newtonian gravity, the acceleration due to a point source or uniform sphere is given by $$g = GM/r^2$$

The potential is given by $$\Phi = -GM/r$$

Note that the field strength is the derivative of the potential, $$g = \frac{d\Phi}{dr}$$

The circular orbit velocity is given by $$v^2 = GM/r$$ and the escape velocity is given by $$v_e^2 = 2GM/r$$

Combining the last equation with the first acceleration equation we get $$v_e^2 = 2gr$$

So two planets with identical surface gravity $g$ but different density can have very different escape velocities. If you were on a planet with Earth gravity but four times the radius (so much lower density), the escape velocity would be twice as high, and rockets would need four times the kinetic energy to escape. And of course escape velocity also affects what gases a planet can retain in its atmosphere.

I should also mention that the tidal stress is determined by the derivative of the field strength, so it's proportional to $1/r^3$.


It may seem counter-intuitive that for two planets with identical surface gravity that the one with lower density has the higher escape velocity. Density is mass per unit volume, so a sphere of radius $r$ and mean density $\rho$ has mass $$M = \left(\frac{4\pi}{3}\right) r^3 \rho$$

Plugging that into our previous equations, we get $$g = \left(\frac{4\pi G}{3}\right) r \rho$$

and $$v_e^2 = \left(\frac{8\pi G}{3}\right) r^2 \rho$$

So for a fixed $g$, $r$ and $\rho$ are inversely proportional. If we multiply $r$ by some constant $k$ and divide $\rho$ by $k$ then $g$ remains the same, but $v_e^2$ is multiplied by $k$.

Rearranging the previous equation for $g$, we get $$r = \left(\frac3{4\pi G}\right)g/\rho$$

Plugging that into $v_e^2 = 2gr$ we get $$v_e^2 = \left(\frac3{2\pi G}\right)g^2/\rho$$ or $$v_e = \sqrt{\frac3{2\pi G}}\left(\frac{g}{\sqrt{\rho}}\right)$$

So escape velocity is proportional to $g$, but inversely proportional to the square root of the mean density.


The equations are slightly more complicated in GR, but the same principle applies.

The Schwarzschild radius of a body is proportional to its mass $$r_s = 2GM/c^2$$

The photon sphere radius is just $1.5×r_s$.

Near a small black hole of a few solar masses, the field strength changes quickly with radial distance, so the tidal stress is extreme. For example, a $3 M_\odot$ (solar mass) BH, has $r_s \approx 8.860$ km. The tidal stress on a $1$ metre long A36 steel bar in vertical freefall is enough to rupture it at a distance of $112$ km. That makes it a bit tricky to observe curved light trajectories without a telescope. ;)

But around a SMBH the field strength changes slowly, so in freefall you can safely approach the vicinity of its photon sphere. You can even cross the event horizon without being spaghettified.

However, it's still not exactly easy to get close to a SMBH with safety. If you're falling towards the BH, or in an orbit, you have to be moving at relativistic speed. You can't simply hover: the field strength is extremely high, so your hovering rocket would need extreme acceleration. That requires a ridiculous amount of fuel, and the acceleration would crush you and your ship into an atom-thick paste.


I haven't actually said much about the gravitational deflection of light. I have some info about that in my recent answer to Are there actually any photons in orbit in the photon sphere of a black hole? which also links to a couple of other answers I've written on this topic.

On Astronomy.SE I wrote:

Here's a simple approximation for the deflection angle [of a light ray] which is valid when the deflection angle is small: $$\theta = \frac{2r_s}b$$ where $\theta$ is in radians, $r_s$ is the Schwarzschild radius of the lensing body, and $b$ is the impact parameter of the light ray, which is basically the minimum distance between the ray and the centre of the lensing body if you could turn gravity off so that the ray was just a straight line.

A more accurate equation for small deflections is $$\theta = \frac2{r/r_s - 71/73}$$ where $r$ is the minimum distance from the deflected ray to the centre of the lensing body. The exact equation relating $b$ and $r$ is $$b^2 = \frac{r^3}{r-r_s}$$

When $b$ is large, $r \approx b - r_s/2$.

When $b$ is small, we need to use an elliptic integral to compute the deflection. Here's a diagram of a ray deflected by 60°, with $b\approx 3.62r_s$, from the Physics.SE answer I linked above.

60° photon deflection

The diagram uses units where $r_s=1$, so it applies to any Schwarzschild black hole. The circle of radius $1$ is the event horizon, $b$ is is the perpendicular distance from the dashed black lines to the BH, and $r$ is the length of the dashed purple line where the trajectory makes its closest approach to the BH. Actually, the diagram applies to any spherically symmetrical body with negligible spin, but only black holes and neutron stars are smaller than their photon sphere.

For typical stars and planets we can calculate the deflection angle using the equation I gave earlier, $\theta = \frac2{r/r_s - 71/73}$. The relative error in using this equation to compute the deflection angle near the Sun is ~$8.21×10^{-11}$.

The Schwarzschild radius of the Sun is ~$2.95325$ km. The radius of its photosphere is ~$695700$ km. So the deflection of a ray of starlight grazing the Sun during a solar eclipse is ~$1.7511976$ arc-seconds or ~$1/117785$ radians. Thus the star's apparent position is shifted by ~$1.75$ arcsecs, and the deflected light ray drops by ~$1$ mm over a distance of $117.785$ metres.

The Schwarzschild radius of the Earth is only ~$8.870056$ mm. Its equatorial radius is ~$6378.137$ km. So the deflection of a light ray near the Earth's surface is only ~$0.000573704$ arcsecs or ~$1/359531949$ radians. That's very tiny, and totally undetectable in the atmosphere. If you have a light beam travelling through an evacuated horizontal pipe $359.5$ km long, the beam only drops $1$ mm.

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  • $\begingroup$ Good answer Sir +1 $\endgroup$
    – safesphere
    Commented Mar 30 at 4:26
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    $\begingroup$ This is fascinating! "If you were on a planet with Earth gravity but four times the radius (so much lower density), the escape velocity would be twice as high, and rockets would need four times the kinetic energy to escape.", this I would thought the opposite way, less dense, less escape velocity. Can you please elaborate on this? $\endgroup$ Commented Mar 30 at 7:41
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    $\begingroup$ @ÁrpádSzendrei To get a planet with the same surface gravity as on the Earth, but with surface being 4 times farther from the center, you need a 16 times higher mass of this planet. But, while surface gravity depends on a square of the radius, the mass, given a constant density, depends cubically on it. This means that this planet will have a density 4 times lower than that of the Earth. Remember that it's the total mass, not density, that defines external gravitational field of a spherically symmetric object. $\endgroup$
    – Ruslan
    Commented Mar 30 at 15:02
  • $\begingroup$ @Árpád I've expanded on the relation between radius, density, g and escape velocity. And I've added some info about deflection of light rays in the Schwarzschild metric. $\endgroup$
    – PM 2Ring
    Commented Mar 30 at 16:12

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