27
$\begingroup$

Photons do not have (rest) mass (that's why they can move at the speed of "light").

So my question is: how can the gravity of a classical$^1$ black hole stop light from escaping?

--

$^1$ We ignore quantum mechanical effects, such as, Hawking radiation.

$\endgroup$

6 Answers 6

22
$\begingroup$

Black holes affect the causal structure of spacetime in such a manner that all future light cones within a black hole lie within the event horizon of it.

Although photons are massless they have energy and have to obey the geometry of a curved spacetime. Since all future lies within the event horizon, photons are trapped inside the black hole. enter image description here

$\endgroup$
3
  • 3
    $\begingroup$ Right! Asking why a photon can't escape from a black hole is exactly like asking why a photon can't travel from here to last Thursday: both are travel into the past. $\endgroup$
    – Ted Bunn
    Apr 12, 2011 at 14:17
  • 1
    $\begingroup$ I've seen the shifted light cones in books for decades and they still don't answer the question. If I am in Earth's gravitational field, I am accelerating, not moving toward the center of mass. Consider an accelerating spaceship in flat space for simplicity - someone with a special relativity background should expect light in the + and - direction both move at c, although there is an asymptotic event horizon due to long term behavior. Note, this is inconsistent with a linearly 'tilted' light cone. $\endgroup$ Jun 4, 2011 at 5:33
  • $\begingroup$ If gravity prevents light from escaping a black hole while it is still "moving at c" by changing spacetime, it it slowing time to zero? $\endgroup$ Mar 2, 2023 at 18:45
6
$\begingroup$

Even though photons have no mass, they are still affected by gravity. That's how we can see black holes - by the way they distort the light going near them.

The reason nothing can escape a black hole is because within the event horizon, space is curved to the point where all directions are actually pointing inside.

The escape velocity from within a black hole's event horizon is faster than the speed of light, hence light cannot go at that speed and thus cannot escape.

$\endgroup$
4
  • 2
    $\begingroup$ ""Even though photons have no mass, they are still affected by gravity. "" Silly statement. By what means would "gravity" affect a photon, if not by mass? Photon have a mass, what they haven't is rest mass! $\endgroup$
    – Georg
    Apr 12, 2011 at 8:56
  • 10
    $\begingroup$ For roughly the last fifty years, the vast majority of physicists have used the word "mass" to mean "rest mass." $\endgroup$
    – Ted Bunn
    Apr 12, 2011 at 14:16
  • 2
    $\begingroup$ @Georg, gravity affects photons because gravity actually curves space. It's the same reason a feather and bowling ball would fall at the same speed (neglecting air resistance). Photons don't have mass, they have energy. That's a different point and indicates that the photon actually has its own gravitational field. $\endgroup$ Dec 11, 2015 at 12:25
  • 2
    $\begingroup$ Escape velocity is a terrible concept to use to explain this. There is no such thing as escape velocity for light. Escape energy would make more sense. $\endgroup$
    – B T
    Mar 21, 2017 at 1:16
3
$\begingroup$

An account "from scratch".

Intervals and light cones

An "event" in relativity is defined as something with a set of coordinates in 4-dimensional spacetime. Events are separated by an interval, which in a flat spacetime, could be written as $$ ds^2 = -c^2 dt^2 + dx^2 + dy^2 + dz^2\ , $$ where the implicit notation that $ds^2 \equiv (ds)^2$ is used. This interval is fundamental in relativity because it is an invariant that all observers agree on.

The interval for a light beam is always zero. That means, for example, if we send a light beam along the x-axis, where $dy=dz=0$, then $cdt/dx = \pm 1$. Thus if we drew a graph of $x$ vs $ct$ (on the y-axis) then light would travel along paths with a gradient of $\pm 45$ degrees from their starting point. Since light is the fastest signal you can send (much faster than a carrier pigeon), then these lines will bound all possible events in the future that you could influence. They also contain the coordinates of all possible events that you could be at in the future, no matter how fast you travelled. This is known as the future light cone.

A similar cone can be extended backwards in time to bound the set of coordinates from which a light beam (or anything slower) could have reached you; and this is known as the past light cone.

The picture below illustrates these ideas. There is an event at A, with events B-G plotted in the same diagram. The future and past light cones are labelled. Events E, F, and G are in the future light cone of A. Events B, C, and D are in the past light cone of A.

Light cones

If the interval $ds^2$ between two events is negative then these events could be causally connected, because one lies in the future light cone of the other. This is known as a timelike interval. If the interval is positive, it is called spacelike and the two events cannot be causally connected. That is, something at event A can influence, signal to or even travel to event E, but cannot influence or travel to event B, or any of the unlabelled events marked with squares in the "elsewhere" region, because they are not inside the future light cone of A.

Intervals and Light cones in General Relativity

These ideas translate to General Relativity where the spacetime can be curved. That means the status of the interval as an invariant is the same but the expression for the interval is more complicated - the coefficients multiplying $dt^2$ etc. may depend on $t, x, y, z$ or indeed other parameters like the mass or spin of a central object. $$ ds^2 = -f_t(t, x, y, z, ...)c^2dt^2 + f_x((t, x, y, z, ...)dx^2 + ... {\rm etc.}\ . $$ The ideas about future and past light cones are also valid in General Relativity. The light cones are also defined by the condition $ds^2 = 0$.

The complexity of the interval equation means however that the slope of the light cones could vary depending on the position of the originating event. On the face of it this seems to be saying that the speed of light may differ from $c$. Well, that is true in the sense that the speed of light expressed as a rate of change with respect to the spatial and temporal coordinates that are being used does change. But this speed is not the speed that would be measured by an observer who is local to the light beam. They would always say it travels at $c$.

Coordinates

In General Relativity, the coordinate system you express the interval equation can be become something of an issue and gets those new to GR into lots of trouble. In non-relativistic situations we are quite used to switching between say Cartesian and spherical polar coordinates to describe the location of an object. If we were measuring the distance between two points on the surface of a sphere, we could say that $dl^2 = dx^2 + dy^2 + dz^2$, but we could equally and perhaps more usefully use $dl^2 = r^2\sin^2\theta\ d\theta^2 + r^2d\phi^2$. Importantly, the choice of coodinates doesn't affect what anybody would actually measure. The same is true in General Relativity.

For non-spinning black holes, the interval equation can be written as $$ ds^2 = -\left(1 - \frac{2GM}{c^2r}\right)c^2dt^2 + \left(1 - \frac{2GM}{c^2r}\right)^{-1}dr^2 + r^2 \sin^2\ d\theta^2 + r^2 d\phi^2\ , $$ the last two terms being the same as for flat spacetime (because of spherical symmetry) and $M$ representing the mass of the black hole. The reason that the interval equation has this form is because this provides a valid (and in fact the unique) solution to the Einstein Field Equations of General Relativity for the spacetime outside a spherically symmetric, time-invariant mass.

The $t, r, \theta, \phi$ coordinates here are similar to spherical coordinates, but not quite. For example, because of the curvature of spacetime, $r$ is not the "radius" and $\Delta r$ is not the separation of two events at the same $t, \theta, \phi$. These coordinates are not suitable for dealing with motion across or inside the event horizon where $r = 2GM/c^2$. At this value of $r$, you can see that second term in the interval equation blows up to infinity. This does have a meaning. It means that somebody keeping track of events using this system of coordinates would never see an object reach (or cross) the event horizon in any finite $t$.

This does not mean that things cannot cross the event horizon. According to a falling observer, measuring time on their own clock (which is not equal to $t$), they do cross the event horizon and they do fall towards $r=0$ in a finite time. To handle this we can define a different coordinate system (and there are several to choose from), which are continuous at the event horizon. Perhaps the simplest is the Gullstrand-Painleve coordinate system, where the timelike coordinate $t$ is replaced by a new coordinate $T$ that is defined as $t - a(r)$, where $a(r)$ is defined so that an increment of $T$ is exactly the same as a tick of the clock carried by an observer that is freely falling into the black hole. This removes the coordinate problem at the event horizon and the interval equation is written $$ c^2 d\tau^2 = c^2\left( 1 - \frac{r_s}{r}\right)\,dT^2 -2c\left(\frac{r_s}{r}\right)^{1/2}\,dT\,dr - dr^2 - r^2\,d\theta^2 - r^2\sin^2\theta\,d\phi^2\, . $$

Light cones inside the black hole event horizon

By setting $ds^2$ to zero, and for simplicity considering radial paths where $d\theta=d\phi=0$, it is easy to see that the light cones are defined by the equation $$\left(\frac{dr}{dT}\right)^2 +2c\left(\frac{r_s}{r}\right)^{1/2}\,\frac{dr}{dT} - c^2\left( 1 - \frac{r_s}{r}\right) =0\ .$$ Solving a quadratic equation for $dr/dT$ and inverting the result, we have $$c\frac{dT}{dr} = -\frac{1}{\left(\frac{2GM}{c^2r}\right)^{1/2} \mp 1}\ , $$ where the $\mp$ sign applies to outgoing and ingoing light beams respectively.

The light cones corresponding to this eqution are plotted below (Figure adapted from Exploring Black Holes by Taylor, Wheeler & Bertschinger), where $r_s = 2GM/c^2$, the event horizon (where $r=r_s$ is marked with a red line and the future and past light cones are marked with an F and a P and are calculated according to the equation for $cdT/dr$ given above.

Light cones in G-P coordinates

The key point in this diagram, that answers the question posed here, is that at $r=r_s$, the light cones are defined by $$\frac{dT}{dr} = -0.5\ \ \ {\rm for\ ingoing\ light}\ , $$ $$\frac{dT}{dr} = \infty\ \ \ {\rm for\ outgoing\ light}\ .$$ And if $r<r_s$ then it is not possible for $dT/dr$ to be positive.

What this means is once at $r<r_s$ inside the event horizon, the light cones tip over towards decreasing $r$ and all future events inside the future light cones are at smaller values of $r$. Thus no signals emitted from inside the event horizon can ever be sent to larger $r$ and even a light beam that is directed outwardly will in fact move towards a smaller value of $r$. This is why it is called the event horizon.

$\endgroup$
1
$\begingroup$

Gravity is the force which bends the very fabric of Space time. During eclipse scientists have seen the light from distant stars which are near the Sun change their path. So it proves that light is affected by Gravity. Now that you know that light gets affected by gravity, you must also be knowing that the gravitational force of a Black-Hole is immense. As anything on earth needs to have a minimum velocity to overcome the gravitational pull of Earth (which is called escape velocity) is something that Man has been able to achieve, so our space ships and rockets reach Space. But the escape velocity required to overcome the gravitational pull of a black hole is greater than the speed of Light. And as we know that nothing travels faster than light, so Black Hole swallows anything and everything that comes near it including light.

$\endgroup$
2
  • 2
    $\begingroup$ The concept of escape velocity simply doesn't apply to photons. $\endgroup$
    – B T
    Mar 21, 2017 at 1:16
  • $\begingroup$ Okay. But light shows properties of both particle and waves. So shouldn't a particle have to overcome an escape velocity? $\endgroup$
    – MK Singh
    Aug 26, 2017 at 18:20
1
$\begingroup$

Here is a different explanation.

Due the equality principle standing on the surface of a planet and accelerating is equal.

Far from massive bodies, fixed proper acceleration leads to a hyperbolic trajectory in the space-time diagram. This hyperbola's asymptote is diagonal (approaching the speed of light.)

If you imagine this hyperbola you can see that if you shoot a beam of light towards the accelerating object beyond a certain distance, it will never reaches it. This is the Rindler-horizon, beyond it no light reaches you. If you accelerate with $a$ the Rindler horizon is at $c^2/a$ behind you.

The black hole's event horizon is analogous with this. If you hover over a black-hole you are in an accelerating reference frame, so Rindler-horizon exists at the event horizon (using the schwarzschild metric).

Rindler-horizon disappears if the observer stops acceleration. The observer near a black hole stops acceleration if it begins a free fall towards the black hole, it's movement becomes inertial so the event horizon should also disappear. But since gravity is not uniform, the event horizon won't disappear, but remains under the observer as it falls in.

$\endgroup$
-3
$\begingroup$

The above statement about a photon having no "rest" mass is actually quite contrary to reality. As an object approaches the speed C its mass becomes relatively smaller and smaller, until it reaches the asymptotic light speed and becomes virtually massless, timeless, and spaceless. The more energy the photon has, the smaller the wavelength, higher the frequency, and higher the rest mass. This equation can be given by mv=hf/c. Technically, you yourself (and this entire planet) resemble but a group of photons to some frame of reference (for example, a galaxy going in the opposite direction to ours at a very high speed). I think gravity effects photons because while they are massless (by definition, when it is moving at the speed of light it has no mass) is because they are, even massless, a photonic bundle of energy.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.