<blockquote>
  <p>If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon?</p>
</blockquote>

<p>I think so. I also think it will be destroyed before tidal forces have any effect.  </p>

<blockquote>
  <p>If so, why?</p>
</blockquote>

<p>Because it would otherwise end up falling faster than the speed of light.</p>

<p>Sounds strange I know, but take a look at the <a href="https://en.wikipedia.org/wiki/Shapiro_delay" rel="nofollow noreferrer">Shapiro delay</a>: <em>"the speed of a light wave depends on the strength of the gravitational potential along its path"</em>. Or see Professor Ned Wright's <a href="http://www.astro.ucla.edu/~wright/deflection-delay.html" rel="nofollow noreferrer">Deflection and Delay of Light</a>:  <em>"in a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light"</em>:</p>

<p><img src="https://i.sstatic.net/Jiogu.gif" alt="enter image description here"></p>

<p>Alternatively see <a href="http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html" rel="nofollow noreferrer">this PhysicsFAQ article</a> by Don Koks: <em>"this difference in speeds is precisely that referred to above by ceiling and floor observers".</em> He's referring to the way Einstein said light curves because the speed of light varies with position. See the <a href="http://einsteinpapers.press.princeton.edu/" rel="nofollow noreferrer">Einstein digital papers</a> for examples of that. <a href="http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22" rel="nofollow noreferrer">Here's</a> one from 1920, see the second paragraph: </p>

<p><img src="https://i.sstatic.net/8KXbI.jpg" alt="enter image description here"></p>

<p>Light doesn't curve because spacetime is curved. Einstein never actually said that. It curves because the speed of light is lower at a lower altitude, rather like the way <a href="http://fas.org/man/dod-101/navy/docs/es310/SNR_PROP/snr_prop.htm" rel="nofollow noreferrer">sonar</a> waves curve: </p>

<p><img src="https://i.sstatic.net/sNuHt.gif" alt="enter image description here"><br>
As to why this isn't common knowledge, I don't know. There's this myth that Einstein gave up on a varying speed of light in 1911, but he didn't, see <a href="https://en.wikipedia.org/wiki/Variable_speed_of_light#Einstein.27s_VSL_attempt_in_1911" rel="nofollow noreferrer">this Wikipedia article</a> and <a href="http://einsteinpapers.press.princeton.edu/vol4-trans/271?highlightText=%22writer%20of%20these%20lines%22" rel="nofollow noreferrer">this example from 1914</a>. I don't know why the reason matter falls down isn't common knowledge either. You know about pair production and electron diffraction and the wave nature of matter, just think of an electron as a wave going round a closed path, then simplify it to a square path, like this:</p>

<p><img src="https://i.sstatic.net/ATcJA.jpg" alt="enter image description here"></p>

<p>The horizontals bend downwards, so the electron falls down. Could it <em>be</em> any simpler? Anyway, your TV camera falls down because the speed of light is reducing with altitude. We tend to call this the "coordinate" speed of light nowadays, even though Einstein just called it the speed of light. But regardless of what we call it, you don't have to be the Brain of Britain, or the Brain of France, to work out that half down, there's some kind of problem. At this point the camera will be falling as fast as the "coordinate" speed of light at that location. And it isn't going to slow down. Things always fall faster, not slower. But matter can't go faster than light, because of the wave nature of matter. When matter is made of waves, there is no way that matter can move faster than the speed of those waves. </p>

<p>So what's going to happen? I can see no other option: that wave has to <em>break</em>. Again it sounds strange, but take a look at an old version of the firewall article on <a href="https://en.wikipedia.org/w/index.php?title=Firewall_(physics)&amp;oldid=587309921" rel="nofollow noreferrer">Wikipedia</a>. Follow the reference 7 link to Friedwardt Winterberg's <a href="http://www.znaturforsch.com/aa/v56a/56a0889.pdf" rel="nofollow noreferrer">Gamma-Ray Bursters and Lorentzian Relativity</a>: <em>"If the balance of forces holding together elementary particles is destroyed near the event horizon, all matter would be converted into zero rest mass particles which could explain the large energy release of gamma ray bursters".</em> I think this is what happens to your TV camera. <em>Flash!</em> It turns into a gamma-ray burst. It would be like an atom bomb, but a lot more efficient. So make sure you drop it from a safe distance. </p>

CC BY-SA 3.0
2016-01-02T12:39:09.833Z
Would the downvoters care to comment?
John Duffield
CC BY-SA 3.0
2016-01-01T13:49:20.357Z
2016-01-01T13:49:20.357Z
John Duffield
<p>If the black hole has no surrounding matter, so that there is no violent radiation generated by accretion or the like, then it still depends on the mass of the black hole. If it is in the thousands of solar masses or more, it is possible for a realistic camera to survive free-fall to the horizon. This is essentially a non-issue for supermassive black holes. On the other hand, smaller black holes are quite a bit more punishing.</p>

<p>For Newtonian gravity with potential $\Phi$, in a free-falling frame, a particle at $x^k$ near the origin of the frame will be accelerated at
$$\frac{\mathrm{d}^2x^j}{\mathrm{d}t^2} = -\frac{\partial\Phi}{\partial x^j} = -\frac{\partial^2\Phi}{\partial x^j\partial x^k}x^k\text{,}$$
where the the second derivatives of the potential, $\Phi_{,jk}$, form the so-called <em>tidal gravitational field</em>. Since $\Phi = -GM/r$ for a point-source, you should expect the tidal forces on a free-falling object to be proportional to $GM/r^3$ times the size of the object. Thus, at the Schwarzschild radius, this is on the order of $c^6/(GM)^2$.</p>

<p>Of course, black holes are not Newtonian. However, it turns out that for the non-rotating, uncharged (Schwarzschild) black hole, radial free-fall of a test particle has the same form as in Newtonian theory, except in Schwarzschild radial coordinate (not radial distance) and proper time of the particle (not universal time), so the above is essentially correct even for Schwarzschild black holes.</p>

<p>To be relativistically correct, the tidal forces on an free-falling object are described by the equation of <a href="https://en.wikipedia.org/wiki/Geodesic_deviation" rel="noreferrer">geodesic deviation</a>, in which the gravitoelectric part of the Riemann curvature provides forms the tidal tensor:
$$\frac{\mathrm{D}^2x^\alpha}{\mathrm{d}\tau^2} = -R^\alpha{}_{\mu\beta\nu} u^\mu u^\nu x^\beta\text{.}$$
In Schwarzschild spacetime, this turns out to be $+2GM/r^3$ in the radial direction, stretching the free-falling object, and $-GM/r^3$ in the orthogonal directions, squeezing it. This stretching-and-squeezing due to gravitational tidal forces is sometimes called <a href="https://en.wikipedia.org/wiki/Spaghettification" rel="noreferrer">spaghettification</a>.</p>

<p>Some example numbers: say the camera size is on the order of $0.1\,\mathrm{m}$. The the following are the approximate tidal accelerations near the horizon for black holes of different multiples of solar masses:</p>

<ul>
<li>$M\sim 10\,\mathrm{M_\odot}$: $\sim 10^6$ Earth gravities;</li>
<li>$M\sim 10^4\,\mathrm{M_\odot}$: $\sim 1$ Earth gravity;</li>
<li>$M\sim 10^6\,\mathrm{M_\odot}$: $\sim 10^{-4}$ Earth gravities.</li>
</ul>

<p>The curvature around a rotating black hole is more complicated, but the moral of the story is basically the same.</p>

CC BY-SA 3.0
2015-12-31T23:08:04.637Z
@Ricky feel free to contribute an answer that uses no equations.
ProfRob
CC BY-SA 3.0
2015-12-31T23:33:28.983Z
@RobJeffries: Oh, and each time I hear the word "spacetime," I think "phlogiston."
Ricky
CC BY-SA 3.0
2016-01-01T01:14:03.287Z
@Ricky I'm not going to discuss the validity of evidence for the existence of black holes in these comments (it's irrelevant here), but I will note that your attitude is completely backwards. One must first understand at least the bare basics of how black holes *would* behave (and a question like this counts) in order to be in any position to even start *evaluating* evidence of their existence, regardless of whether one accepts the evidence or not.
Stan Liou
CC BY-SA 3.0
2016-01-01T01:39:47.703Z
@RobJeffries: I'd love to understand the basics of black hole behavior, but I'm still working on how phlogiston behaves. As for accepting the evidence: I find it hard to accept anything that doesn't exist. There are fields of human activity where blind faith is called for; science isn't one of them. I'm sorry if this upsets you.
Ricky
CC BY-SA 3.0
2016-01-01T09:45:59.250Z
@Ricky, we've already established that you don't understand what the scientific method is, and here you are demonstrating your medieval take on things again. Black holes are a natural consequence of theories of General Relativity and Stellar evolution; both of which have been extensively tested in thousands of "experiments". These theories make predictions about phenomena (such as those posed by this question) and then others search for evidence that may falsify the theory. *This* is science.
ProfRob
CC BY-SA 3.0
2016-01-01T14:06:22.777Z
@Ricky : we have good evidence that black holes exist, for example we know from the [motion of stars near the centre of the galaxy](http://www.skyandtelescope.com/wp-content/uploads/smaller-sgr-a.gif) that there's something very _very_ massive there. But you're right about spacetime. The Earth is surrounded by space, not spacetime. There's no motion through spacetime. It's an abstract "mathematical space", and the map is not the territory.
John Duffield
CC BY-SA 3.0
2015-12-31T15:28:15.313Z
2015-12-31T19:04:09.913Z
2015-12-31T19:04:09.913Z
Stan Liou
<p>This question came to my mind after reading <a href="https://astronomy.stackexchange.com/questions/13017/a-camera-and-time-dilation/13020?noredirect=1#comment18789_13020">this.</a></p>

<p>I think it is safe to assume that the supposedly thrown camera would not reach the event horizon in one piece. So my question is : </p>

<p>If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon, and if so, why?</p>

<gravity><black-hole>
Camera reaching the event horizon
Nico
# Camera reaching the event horizon

(Asked by: username_0 on 2015-12-31)

This question came to my mind after reading [this.](https://astronomy.stackexchange.com/questions/13017/a-camera-and-time-dilation/13020?noredirect=1#comment18789_13020)

I think it is safe to assume that the supposedly thrown camera would not reach the event horizon in one piece. So my question is :

If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon, and if so, why?

**Tags:** gravity, black-hole


---

## 2 answers

### Answer by username_2 on 2015-12-31. Score: 7 | Accepted answer

If the black hole has no surrounding matter, so that there is no violent radiation generated by accretion or the like, then it still depends on the mass of the black hole. If it is in the thousands of solar masses or more, it is possible for a realistic camera to survive free-fall to the horizon. This is essentially a non-issue for supermassive black holes. On the other hand, smaller black holes are quite a bit more punishing.

For Newtonian gravity with potential $\Phi$, in a free-falling frame, a particle at $x^k$ near the origin of the frame will be accelerated at
$$\frac{\mathrm{d}^2x^j}{\mathrm{d}t^2} = -\frac{\partial\Phi}{\partial x^j} = -\frac{\partial^2\Phi}{\partial x^j\partial x^k}x^k\text{,}$$
where the the second derivatives of the potential, $\Phi\_{,jk}$, form the so-called *tidal gravitational field*. Since $\Phi = -GM/r$ for a point-source, you should expect the tidal forces on a free-falling object to be proportional to $GM/r^3$ times the size of the object. Thus, at the Schwarzschild radius, this is on the order of $c^6/(GM)^2$.

Of course, black holes are not Newtonian. However, it turns out that for the non-rotating, uncharged (Schwarzschild) black hole, radial free-fall of a test particle has the same form as in Newtonian theory, except in Schwarzschild radial coordinate (not radial distance) and proper time of the particle (not universal time), so the above is essentially correct even for Schwarzschild black holes.

To be relativistically correct, the tidal forces on an free-falling object are described by the equation of [geodesic deviation](https://en.wikipedia.org/wiki/Geodesic_deviation), in which the gravitoelectric part of the Riemann curvature provides forms the tidal tensor:
$$\frac{\mathrm{D}^2x^\alpha}{\mathrm{d}\tau^2} = -R^\alpha{}\_{\mu\beta\nu} u^\mu u^\nu x^\beta\text{.}$$
In Schwarzschild spacetime, this turns out to be $+2GM/r^3$ in the radial direction, stretching the free-falling object, and $-GM/r^3$ in the orthogonal directions, squeezing it. This stretching-and-squeezing due to gravitational tidal forces is sometimes called [spaghettification](https://en.wikipedia.org/wiki/Spaghettification).

Some example numbers: say the camera size is on the order of $0.1\,\mathrm{m}$. The the following are the approximate tidal accelerations near the horizon for black holes of different multiples of solar masses:

* $M\sim 10\,\mathrm{M\_\odot}$: $\sim 10^6$ Earth gravities;
* $M\sim 10^4\,\mathrm{M\_\odot}$: $\sim 1$ Earth gravity;
* $M\sim 10^6\,\mathrm{M\_\odot}$: $\sim 10^{-4}$ Earth gravities.

The curvature around a rotating black hole is more complicated, but the moral of the story is basically the same.

> **Comments:**
> **username_3**: @username_4 feel free to contribute an answer that uses no equations.  
> **username_4**: @RobJeffries: Oh, and each time I hear the word "spacetime," I think "phlogiston."  
> **username_2**: @username_4 I'm not going to discuss the validity of evidence for the existence of black holes in these comments (it's irrelevant here), but I will note that your attitude is completely backwards. One must first understand at least the bare basics of how black holes *would* behave (and a question like this counts) in order to be in any position to even start *evaluating* evidence of their existence, regardless of whether one accepts the evidence or not.  
> **username_4**: @RobJeffries: I'd love to understand the basics of black hole behavior, but I'm still working on how phlogiston behaves. As for accepting the evidence: I find it hard to accept anything that doesn't exist. There are fields of human activity where blind faith is called for; science isn't one of them. I'm sorry if this upsets you.  
> **username_3**: @username_4, we've already established that you don't understand what the scientific method is, and here you are demonstrating your medieval take on things again. Black holes are a natural consequence of theories of General Relativity and Stellar evolution; both of which have been extensively tested in thousands of "experiments". These theories make predictions about phenomena (such as those posed by this question) and then others search for evidence that may falsify the theory. *This* is science.  
> **username_1**: @username_4 : we have good evidence that black holes exist, for example we know from the [motion of stars near the centre of the galaxy](http://www.skyandtelescope.com/wp-content/uploads/smaller-sgr-a.gif) that there's something very _very_ massive there. But you're right about spacetime. The Earth is surrounded by space, not spacetime. There's no motion through spacetime. It's an abstract "mathematical space", and the map is not the territory.  

---

### Answer by username_1 on 2016-01-01. Score: -2

> If I dropped a TV camera into a black hole, will it be destroyed before it reaches the event horizon?

I think so. I also think it will be destroyed before tidal forces have any effect.

> If so, why?

Because it would otherwise end up falling faster than the speed of light.

Sounds strange I know, but take a look at the [Shapiro delay](https://en.wikipedia.org/wiki/Shapiro_delay): *"the speed of a light wave depends on the strength of the gravitational potential along its path"*. Or see Professor Ned Wright's [Deflection and Delay of Light](http://www.astro.ucla.edu/~wright/deflection-delay.html): *"in a very real sense, the delay experienced by light passing a massive object is responsible for the deflection of the light"*:

![enter image description here](https://i.sstatic.net/Jiogu.gif)

Alternatively see [this PhysicsFAQ article](http://math.ucr.edu/home/baez/physics/Relativity/SpeedOfLight/speed_of_light.html) by Don Koks: *"this difference in speeds is precisely that referred to above by ceiling and floor observers".* He's referring to the way Einstein said light curves because the speed of light varies with position. See the [Einstein digital papers](http://einsteinpapers.press.princeton.edu/) for examples of that. [Here's](http://einsteinpapers.press.princeton.edu/vol7-trans/156?highlightText=%22speed%20of%20light%22) one from 1920, see the second paragraph:

![enter image description here](https://i.sstatic.net/8KXbI.jpg)

Light doesn't curve because spacetime is curved. Einstein never actually said that. It curves because the speed of light is lower at a lower altitude, rather like the way [sonar](http://fas.org/man/dod-101/navy/docs/es310/SNR_PROP/snr_prop.htm) waves curve:

![enter image description here](https://i.sstatic.net/sNuHt.gif)  
As to why this isn't common knowledge, I don't know. There's this myth that Einstein gave up on a varying speed of light in 1911, but he didn't, see [this Wikipedia article](https://en.wikipedia.org/wiki/Variable_speed_of_light#Einstein.27s_VSL_attempt_in_1911) and [this example from 1914](http://einsteinpapers.press.princeton.edu/vol4-trans/271?highlightText=%22writer%20of%20these%20lines%22). I don't know why the reason matter falls down isn't common knowledge either. You know about pair production and electron diffraction and the wave nature of matter, just think of an electron as a wave going round a closed path, then simplify it to a square path, like this:

![enter image description here](https://i.sstatic.net/ATcJA.jpg)

The horizontals bend downwards, so the electron falls down. Could it *be* any simpler? Anyway, your TV camera falls down because the speed of light is reducing with altitude. We tend to call this the "coordinate" speed of light nowadays, even though Einstein just called it the speed of light. But regardless of what we call it, you don't have to be the Brain of Britain, or the Brain of France, to work out that half down, there's some kind of problem. At this point the camera will be falling as fast as the "coordinate" speed of light at that location. And it isn't going to slow down. Things always fall faster, not slower. But matter can't go faster than light, because of the wave nature of matter. When matter is made of waves, there is no way that matter can move faster than the speed of those waves.

So what's going to happen? I can see no other option: that wave has to *break*. Again it sounds strange, but take a look at an old version of the firewall article on [Wikipedia](https://en.wikipedia.org/w/index.php?title=Firewall_(physics)&oldid=587309921). Follow the reference 7 link to Friedwardt Winterberg's [Gamma-Ray Bursters and Lorentzian Relativity](http://www.znaturforsch.com/aa/v56a/56a0889.pdf): *"If the balance of forces holding together elementary particles is destroyed near the event horizon, all matter would be converted into zero rest mass particles which could explain the large energy release of gamma ray bursters".* I think this is what happens to your TV camera. *Flash!* It turns into a gamma-ray burst. It would be like an atom bomb, but a lot more efficient. So make sure you drop it from a safe distance.

> **Comments:**
> **username_1**: Would the downvoters care to comment?  

---

