<p>Probably no. It all depends on definition of photon location as photons have both wave and particle properties. In terms of macrocosmos location of photons are well (enough) defines, but as we speak lengths closer to its wavelength (Planck length being in the very extreme of this), it is simply difficult to pinpoint whether the photon is on this or that side. </p>

<p>I would assume that it needs to be at least one wavelength away from the EH in order to be able to escape, but then again - this is highly theoretical as we also need something that’ll emit that photon. I would be very interested in hearing some real professional opinions on this one!</p>

CC BY-SA 4.0
2019-04-23T19:43:19.840Z
2019-04-23T19:43:19.840Z
tuomas
<p>Yes it should be possible.</p>

<p>Every body has a Schwarzschild radius which defines the radius at which light would not be able to escape:</p>

<p><span class="math-container">$$r_s= \frac{2GM}{c^2}$$</span></p>

<p>By this formula, if a black hole had the mass of the Earth its radius would be about 9mm, but as the Earth has a bigger radius than this we do not see light being captured.</p>

<p>A black hole is an object smaller than its Schwarzschild radius.</p>

<p>Also I found an article on Wikipedia which says that photons will orbit the black hole at <span class="math-container">$\frac{3}{2}r_s$</span>. Within that <a href="https://en.m.wikipedia.org/wiki/Photon_sphere" rel="nofollow noreferrer">photon sphere</a> no stable orbit is possible:</p>

<blockquote>
  <p>"Any free fall orbit that crosses it from the outside spirals into the black hole. Any orbit that crosses it from the inside escapes to infinity."</p>
</blockquote>

CC BY-SA 4.0
2019-04-23T16:53:25.160Z
2019-04-24T23:16:28.543Z
2019-04-24T23:16:28.543Z
Danish 
<p>It depends on the direction the photon travels. </p>

<p>If the photon is directed straight away from the black hole, it will escape (but will be <em>very</em> redshifted). </p>

<p>Obviously all directions towards the event horizon will lead to capture. But since the spacetime is so curved, this also includes many directions that point away from the hole: the photon trajectory will curve around and hit the event horizon anyway. Closer to the event horizon the cone of directions where photons can escape gets smaller and smaller, until it is essentially just the direction straight away. (<a href="https://www.zarm.uni-bremen.de/fileadmin/user_upload/space_science/gravitational_theory/tarusa.pdf" rel="nofollow noreferrer">See slide 6 in this presentation</a>)</p>

<p>See also  <a href="http://spiro.fisica.unipd.it/~antonell/schwarzschild/live/" rel="nofollow noreferrer">this interactive web page</a>; try it for a fixed position looking along the "orbit". The higher up in the picture the closer you get to the horizon. I have found that one can get past the <span class="math-container">$1.5R_s$</span> limit (the photon sphere, where exactly half of the sky is directions leading into the hole) by moving to a different browser tab, scrolling, and returning; the effect between 1 and 1.5 <span class="math-container">$R_s$</span> is to make the sky look like an ever smaller ball - photons emitted in any other direction will be captured.</p>

CC BY-SA 4.0
2019-04-23T22:47:30.937Z
2019-04-23T22:47:30.937Z
Anders Sandberg
<p>I understand that one could theoretically replace the sun with a blackhole without affecting the orbits of the planets. This leads me to believe that a blackhole isn't a "cosmic vacuum cleaner" so much as it is a concentration of mass that has the strongest gravitational field known.</p>

<p>Intuitively, this must mean that there is a threshold beyond which escaping the blackhole's gravity becomes impossible even at light speed, which is the event horizon. Since a Planck length is the smallest unit known at which quantum gravitational effects become apparent, would it be reasonable to assume that a photon that is one Planck length away from the event horizon of a blackhole can escape its gravitational pull?</p>

CC BY-SA 4.0
2019-04-24T23:14:06.107Z
FWIW, there is no deep physical significance to the Planck length (or other Planck units), although we do expect quantum gravity effects to become relevant *somewhere* around that scale. (OTOH, the Hawking-Bekenstein entropy formula does become very simple when expressed in Planck units).
PM 2Ring
CC BY-SA 4.0
2019-04-25T03:52:02.860Z
It gets very tricky when you mix general relativity (black holes) and quantum mechanics (Planck length). In truth, we don't know how to do this completely. Regardless, if the photon is outside the event horizon, no matter how close, then it will escape. You are correct that black holes are not cosmic vacuum cleaners.
Florin Andrei
<black-hole><gravity><general-relativity><event-horizon>
If a photon were 1 Planck length away from the event horizon of a blackhole, would it be able to escape the blackhole's gravitational pull?
Dharmesh Tarapore
# If a photon were 1 Planck length away from the event horizon of a blackhole, would it be able to escape the blackhole's gravitational pull?

(Asked by: username_0 on 2019-04-23)

I understand that one could theoretically replace the sun with a blackhole without affecting the orbits of the planets. This leads me to believe that a blackhole isn't a "cosmic vacuum cleaner" so much as it is a concentration of mass that has the strongest gravitational field known.

Intuitively, this must mean that there is a threshold beyond which escaping the blackhole's gravity becomes impossible even at light speed, which is the event horizon. Since a Planck length is the smallest unit known at which quantum gravitational effects become apparent, would it be reasonable to assume that a photon that is one Planck length away from the event horizon of a blackhole can escape its gravitational pull?

**Tags:** black-hole, gravity, general-relativity, event-horizon
> **Comments:**
> **username_1**: FWIW, there is no deep physical significance to the Planck length (or other Planck units), although we do expect quantum gravity effects to become relevant *somewhere* around that scale. (OTOH, the Hawking-Bekenstein entropy formula does become very simple when expressed in Planck units).  
> **username_2**: It gets very tricky when you mix general relativity (black holes) and quantum mechanics (Planck length). In truth, we don't know how to do this completely. Regardless, if the photon is outside the event horizon, no matter how close, then it will escape. You are correct that black holes are not cosmic vacuum cleaners.  

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## 3 answers

### Answer by username_5 on 2019-04-23. Score: 3 | Accepted answer

It depends on the direction the photon travels.

If the photon is directed straight away from the black hole, it will escape (but will be *very* redshifted).

Obviously all directions towards the event horizon will lead to capture. But since the spacetime is so curved, this also includes many directions that point away from the hole: the photon trajectory will curve around and hit the event horizon anyway. Closer to the event horizon the cone of directions where photons can escape gets smaller and smaller, until it is essentially just the direction straight away. ([See slide 6 in this presentation](https://www.zarm.uni-bremen.de/fileadmin/user_upload/space_science/gravitational_theory/tarusa.pdf))

See also [this interactive web page](http://spiro.fisica.unipd.it/~antonell/schwarzschild/live/); try it for a fixed position looking along the "orbit". The higher up in the picture the closer you get to the horizon. I have found that one can get past the $1.5R\_s$ limit (the photon sphere, where exactly half of the sky is directions leading into the hole) by moving to a different browser tab, scrolling, and returning; the effect between 1 and 1.5 $R\_s$ is to make the sky look like an ever smaller ball - photons emitted in any other direction will be captured.


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### Answer by username_3 on 2019-04-23. Score: 0

Probably no. It all depends on definition of photon location as photons have both wave and particle properties. In terms of macrocosmos location of photons are well (enough) defines, but as we speak lengths closer to its wavelength (Planck length being in the very extreme of this), it is simply difficult to pinpoint whether the photon is on this or that side.

I would assume that it needs to be at least one wavelength away from the EH in order to be able to escape, but then again - this is highly theoretical as we also need something that’ll emit that photon. I would be very interested in hearing some real professional opinions on this one!


---

### Answer by username_4 on 2019-04-23. Score: 0

Yes it should be possible.

Every body has a Schwarzschild radius which defines the radius at which light would not be able to escape:

$$r\_s= \frac{2GM}{c^2}$$

By this formula, if a black hole had the mass of the Earth its radius would be about 9mm, but as the Earth has a bigger radius than this we do not see light being captured.

A black hole is an object smaller than its Schwarzschild radius.

Also I found an article on Wikipedia which says that photons will orbit the black hole at $\frac{3}{2}r\_s$. Within that [photon sphere](https://en.m.wikipedia.org/wiki/Photon_sphere) no stable orbit is possible:

> "Any free fall orbit that crosses it from the outside spirals into the black hole. Any orbit that crosses it from the inside escapes to infinity."


---

