Consider a photon source emitting photons near the surface of a Schwarzschild black hole. What angle, as a function of the source's radius from the event horizon, must the photons be emitted at such that they can escape to an observer at infinity?

  • 1
    $\begingroup$ when you say the source's radius, are you asking at what angle a photon can be emitted and still escape as a function of distance from the event horizon? $\endgroup$
    – Jim
    Commented May 27, 2013 at 13:39
  • $\begingroup$ Yes, with the angle being measured from outward radial direction. $\endgroup$
    – Andyb
    Commented May 27, 2013 at 13:44
  • $\begingroup$ Comment to the question (v4): The tag escape-velocity is usually only for massive particles. $\endgroup$
    – Qmechanic
    Commented May 27, 2013 at 20:07
  • $\begingroup$ This is one of the more interesting and challenging homework like question, so I think it should not get closed even though there are some closevotes. $\endgroup$
    – Dilaton
    Commented May 31, 2013 at 9:12
  • $\begingroup$ @Dilaton Agreed, though the OP should show what they've tried and where they are getting stuck. $\endgroup$
    – Michael
    Commented May 31, 2013 at 9:21

4 Answers 4


At the Schwarzschild radius, a photon must be emitted exactly normal to the surface in order to escape. As you travel outwards, the angle of emission decreases such that just above 1.5 times the Schwarzschild radius (i.e. the Photon Sphere) the photon can be emitted parallel to the tangent of the horizon and still escape.

According to this source (which is also a good source for all fun things regarding Schwarzschild black holes), there is a critical emission angle for photons from a stationary source some radius, $R$, from the black hole. Note that this equation technically should work for radii less than the Schwarxschild radius (the event horizon radius, $r_s=\frac{2GM}{c^2}$), but it'll give you negative angles because photons can't escape. Also note that all angles are given relative to the radial direction. $\theta=0$ means directed radially outwards and $\theta=\pi$ is radially inwards.

Inside the photon sphere, $R\le{3\over2}r_s$, the angles at which photons can escape are given by: $$\theta\le\arcsin\left[\frac{\sqrt{27}r_s}{2R}\sqrt{1-\frac{r_s}{R}}\right]$$

Outside the photon sphere, $R\ge{3\over2}r_s$, the escape angles are: $$\theta\le\pi-\arcsin\left[\frac{\sqrt{27}r_s}{2R}\sqrt{1-\frac{r_s}{R}}\right]$$

To get correct angles, just assume that arcsin always results in values between $-\pi/2$ and $\pi/2$. You'll note that for $R=r_s$, you find that $\theta=0$, which means only photons directed radially outwards escape. For $R=\frac{3}{2}r_s$, $\theta=\frac{\pi}{2}$ as my first paragraph stated. And that a radially inward photon ($\theta=\pi$) is always absorbed (this angle is asymptotically approached as $R\to\infty$).

  • $\begingroup$ The link is broken (which is why you shouldn't reference a link like that, though I've probably done it myself). $\endgroup$
    – ProfRob
    Commented Dec 4, 2021 at 17:26
  • $\begingroup$ @ProfRob thanks for letting me know. This is why I always summarize the stuff in a link. Never know when it dies $\endgroup$
    – Jim
    Commented Dec 7, 2021 at 15:27

As always in GR when you ask questions like this you have to specify who is doing the measuring, or equivalently, just which frame of reference you are using.

For example, if you plot the escape trajectories of light in a Schwarzschild coordinate $r,\phi$ diagram, then it turns out that the angle from the radial direction of the emitted photon (emitted just above $r=r_s$) is $\simeq \tan^{-1} (b/r_s)$, where $b$ is the impact parameter of the light (proof shown in Appendix 1, which also gives a general formula for $\theta(b, r)$, as requested in your question.)

For the light to escape, $b < 3\sqrt{3}r_s/2$, since only if $b$ is less than this value would the light avoid a turning point inside the photon sphere at $r=1.5r_s$. Equivalently you can say this is the maximum value of $b$ which enables light to exceed the effective potential maximum at $r=1.5r_s$. Putting $b < 3\sqrt{3}r_s/2$ leads to the constraint for escape as $$\theta(r) < \tan^{-1}\left[\frac{r_s}{r}\left( \frac{4}{27} - \left(1-\frac{r_s}{r}\right)\frac{r_s^2}{r^2}\right)^{-1/2}\right]\ , $$ which approaches $\theta < \tan^{-1}(3\sqrt{3}/2) = 68.9^\circ$ as $r\rightarrow r_s$.

On the other hand, if we define the angle as the angle that would be measured by an observer hovering just above $r=r_s$, then the limiting angle changes because it must be measured in the local inertial frame of reference of the observer. In this case, the limiting angle $\theta_s$ measured in the local frame of the hovering observer approaches zero as $r \rightarrow r_s$ - as demonstrated in Appendix 2. i.e. The light must be emitted almost radially outwards for any value of $b$. More generally, for light to escape from a radius $r$ then $$\theta_s(r) < \sin^{-1} \left[ \frac{3\sqrt{3}r_s}{2r}\left(1 - \frac{r_s}{r}\right)^{1/2}\right]\ , $$ which becomes arbitrarily close to zero as $r \rightarrow r_s$.

Appendix 1

The equations of motion written in Schwarzschild coordinates are (with $c=1$) $$\frac{dr}{dt} = \pm \frac{b}{r_s}\left(1 - \frac{r_s}{r}\right)\left( \frac{r_s^2}{b^2} - \left(1-\frac{r_s}{r}\right)\frac{r_s^2}{r^2}\right)^{1/2}\ , $$ and $$ r\frac{d\phi}{dt} = \pm\frac{b}{r}\left(1 - \frac{r_s}{r}\right)\ . $$

If we construct a right angled triangle then the angle $\theta$ the trajectory makes with the radial direction is given by $$\theta(b,r) = \tan^{-1}\left[\frac{r d\phi/dt}{dr/dt}\right] = \tan^{-1}\left[\frac{r_s}{r}\left( \frac{r_s^2}{b^2} - \left(1-\frac{r_s}{r}\right)\frac{r_s^2}{r^2}\right)^{-1/2}\right]\ . $$

If $r=r_s$ then the angle is $$ \theta = \tan^{-1}\frac{b}{r_s}\ . $$

Appendix 2

The equations of motion are given in Appendix 1, but we need to shift into the local inertial frame of a "shell observer" - one who is at a fixed value of $r, \phi$ (which is only possible for $r > r_s$).

If we define a time interval on the shell observer's clock as $dt_s$, then from the Schwarzschild metric we have $$dt_s = \left(1 - \frac{r_s}{r}\right)^{1/2}\ dt\ . $$ The proper distance in the shell frame is also given by the Schwarzschild metric as $$dr_s = \left(1 - \frac{r_s}{r}\right)^{-1/2}\ dr\ . $$

Putting these together with the expression for $dr/dt$ from Appendix 1, we get $$\frac{dr_s}{dt_s} = \pm \frac{b}{r_s} \left( \frac{r_s^2}{b^2} - \left(1-\frac{r_s}{r}\right)\frac{r_s^2}{r^2}\right)^{1/2}\ = \pm \left(1 - \frac{b^2}{r^2}\left(1 - \frac{r_s}{r}\right)\right)^{1/2}\ . $$ Again, constructing a right-angled triangle in the local inertial frame of the shell observer, we can say that the hypotenuse of this triangle has a length $dt_s$ (recall $c=1$) and so the exression above is equal to $\cos \theta_s$, where $\theta_s$ is the angle from the radial direction measured in the shell observer's local inertial frame.

Using $\sin^2 \theta_s = 1 - \cos^2 \theta_s$, we finally arrive at the equation $$\theta_s(b,r) = \sin^{-1}\left[\frac{b}{r} \left(1 - \frac{r_s}{r}\right)^{1/2}\right]\ . $$ Thus to achieve any value of $b$ as $r \rightarrow r_s$ requires $\theta_s \rightarrow 0$. Alternatively, we can say that for values of $b < 3\sqrt{3}r_s/2$, then $$\theta_s < \sin^{-1} \left[ \frac{3\sqrt{3}r_s}{2r}\left(1 - \frac{r_s}{r}\right)^{1/2}\right]\ . $$


In the Schwarzschild metric, a photon emitted from the event horizon can escape if its impact parameter $ b $ is less than $ b_{crit}\left (=\frac{3\sqrt{3}}{2}R_s\right ) $ with $ R_s $ Schwarzschild radius $ \left (=\frac{2GM}{c^2}\right ) $.

Since its emission direction $ \theta $ is given by $ \tan\left (\frac{\pi}{2}-\theta\right)=\frac{R_s}{b} $, the condition is $\tan\left (\frac{\pi}{2}-\theta\right)\gt\frac{R_s}{b_{crit}}=\frac{2}{3\sqrt{3}} $

which is: $$ \theta\le\frac{\pi}{2}-\arctan\left (\frac{2}{3\sqrt{3}}\right )\simeq 68.9^\circ $$ (it appears then that it does not depend from the mass of the black hole).

Hoping to have answered the question in a simple way,

Best regards.


$$\newcommand{\l}[0]{\left(1-\frac{2\mu}{r}\right)} \\ \newcommand{\lz}[0]{\left(1-\frac{2\mu}{r_0}\right)}$$

This was in the 2022 3rd year Cambridge exam, with no official solution so I came here.

In my opinion, the question is horribly vague because this angle is not well defined (the photon velocity is a null vector so you can't define the angle via a dot product). Nonetheless, if you switch your brain off and focus on passing the exam, you will hopefully do the following.

The Schwarzschild metric is $$\mathrm{d}s^2 = \l c^2 \mathrm{d}t^2 - \l^{-1}\mathrm{d}r^2-r^2\left(\mathrm{d}\theta^2 + \sin^2(\theta)\mathrm{d}\phi^2\right).$$ If we assume that all motion is in the plane $\theta = \frac{\pi}{2}$, then we may take the effective Lagrangian (see e.g. Hobson, Lasenby, Efstathiou) for a photon to be $$\dot{t}^2\l c^2 -\dot{r}^2\l^{-1} - r^2\dot{\phi}^2 = 0$$ where the derivatives are w.r.t. some affine parameter (say $\tau$). The Euler-Lagrange equations then lead us to $$\l \dot{t} = k \\ r^2 \dot{\phi} = h.$$

We can sub these into the Lagrangian to get $$\dot{r}^2 + V_{\text{eff}}(r) = c^2k^2 \\ V_{\text{eff}} = \frac{h^2}{r^2}\l.$$

If you sketch $V_{\text{eff}}$, you'll get this


The threshold for the photon to escape is that $\dot{r}=0$ at $r=3\mu$, which tells us $k^2c^2 = h^2/(3(3\mu)^2)$.

The Lagrangian is, up to a constant, equivalent to $\mathbf{p}^2$ (the photon momentum squared). The Lagrangian gives us a relation between $\dot{t}, \dot{r}$ and $\dot{\phi}$. The requirement that the photon only just escapes gave us a relation between $\dot{t}$ and $\dot{\phi}$ (via $k$ and $h$). So overall we have a relation between $\dot{r}$ and $\dot{\phi}$. Here's where we switch our brain off and do the 'obvious' thing. Let the components of the photon four-momentum be $$p^\nu = (p^0, p^1\cos\alpha, 0, p^1\sin\alpha).$$ This is actually pretty stupid because it doesn't correspond to the usual Euclidean projection. However, we can just treat it as a parameterization rather than ascribing a particular physical meaning.

Sub into the Lagrangian $(1-2\mu/r) \dot{t} = k$, taking $k^2c^2 = h^2/(3(3\mu)^2)$. Now let $\dot{r} = p^1\cos(\alpha)$ and $\dot{\phi} = p^1\sin(\alpha)$ and you'll find $$\cot^2 \alpha = \frac{{r_0}^4}{3(3\mu^2)}-{r_0}^2\lz.$$

This actually breaks down if $r_0 = 2\mu$ because we divided by $(1-2\mu/r)$. However, we can quickly see in this case $k=0$ implies $h=0$ so that $\alpha=0$.

The other extreme is at $r_0 = 3\mu$, where the equation tells us $\alpha=\pi/2$.

Notice that the angle is dimensionful. This is because of our brainless substitution and reflects the different dimensionality of the basis vectors.


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