<p><strong>A) Don't take Newton's law too seriously in the relativistic regime.</strong></p>

<blockquote>
  <p>According to Newton's Law, when it reaches the event horizon it must be traveling at $c$, because $c$ is the escape velocity at the event horizon.</p>
</blockquote>

<p>This is true, assuming the particle freefalls from rest at infinity, but the details are quite different from Newtonian theory. For orbits in Schwarzschild spacetime, the <a href="http://en.wikipedia.org/wiki/Effective_potential" rel="nofollow">effective potential</a> is
$$V = -\frac{GM}{r} + \frac{l^2}{2r^2} - \frac{GMl^2}{c^2r^3}\text{,}$$
where $l\equiv L/m$ is the specific angular momentum of the orbit. The last term offers relativistic corrections that have no analogue in Newtonian gravity, so in general Newton's law cannot be counted on when near the horizon of the black hole.</p>

<p>But for radial freefall, $l = 0$, and we a total orbital specific energy:
$$\mathcal{E} = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 - \frac{GM}{r}\text{,}$$
which has exactly Newtonian form. In particular, a particle freefalling from rest at infinity will have
$$\left|\frac{\mathrm{d}r}{\mathrm{d}\tau}\right| = \sqrt{\frac{2GM}{r}}\text{,}$$
so at as it nears the horizon ($r\to 2GM/c^2$), we have $|\mathrm{d}r/\mathrm{d}\tau|\to c$, just as Newtonian theory would predict.</p>

<p>However, the motion is only roughly analogous to what's prescribed by Newton's law, because there are very important differences in the interpretation of those quantities. Specifically, (1) the $\tau$ refers to the proper time of the orbiting particle, not coordinate time $t$, and certainly not any absolute time as Newtonian theory proper, and (2) the $r$ refers to the Schwarzschild radial coordinate, which is defined such that a sphere at this coordinate have surface area $4\pi r^2$, and this is <strong>not</strong> the radial distance to the center.</p>

<p><strong>B) Coordinates have no intrinsic meaning.</strong></p>

<blockquote>
  <p>Therefore, after it passes the event horizon, it must continue to accelerate past $c$, yet theory of relativity states that it's never gonna happen. </p>
</blockquote>

<p>Suppose I fling a particle along a straight ruler, and that ruler is marked in feet. So if, for example, my particle goes five marked units along the ruler in a nanosecond, that's clearly a problem, because the particle must be superluminal ($c\approx 1\,\mathrm{ft}/\mathrm{ns}$).</p>

<p>Now suppose the ruler is not marked in feet, and I don't tell you how it's marked. Is it a problem that the particle goes five units in a nanosecond? Without knowing what the marks are, the statement of this sort of "coordinate speed" is not even meaningful. Finally, suppose I tell you how the ruler is marked, but its marks do not actually measure the distance along the ruler. Then is it a problem that it goes through five of those marks in a nanosecond?</p>

<p>The moral is simply this: by themselves, coordinates are simply labels to identify events. One can pick coordinates such that the coordinate speed is arbitrarily large or small, and it doesn't matter, because coordinates are not physical things. The universe does not come with its own coordinates; they're just labels we put on things.</p>

<p>However, what can give coordinates meaning is the <em>metric tensor</em>, which relates coordinates to actual lengths or durations, so that, for example, the Schwarzschild radial coordinate has a set meaning in terms of surface areas. Finally, the Schwarzschild coordinates are pathological near the horizon. They are not defined across the horizon (technically, the Schwarzschild coordinates are actually two entirely different coordinate charts disconnected by the horizon). You can see this in the Schwarzschild metric (units of $G = c = 1$):
$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2M}{r}\right)^{-1}\,\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{,}$$
where the metric coefficients in the Schwarzschild chart are undefined at the horizon $r = 2M$.</p>

<p><strong>C) "So, in this condition, what will the object do?"</strong></p>

<p>It will cross the event horizon and meet its end at the singularity.</p>

<p>Here's the same Schwarzschild geometry in <a href="http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates" rel="nofollow">Gullstrand-Painlevé coordinates</a>:
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \left(\mathrm{d}r + \sqrt{\frac{2M}{r}}\,\mathrm{d}t\right)^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{.}$$
In this coordinate chart, space is moving towards the singularity at the estate velocity $\sqrt{2GM/r}$, and the particles are carried along with it. In a related coordinate chart, <a href="http://en.wikipedia.org/wiki/Lema%C3%AEtre_coordinates" rel="nofollow">Lemaître coordinates</a>, particles freefalling from rest at infinity have zero coordinate speed, with $\mathrm{d}r/\mathrm{d}t = 0$, where here $r$ is the Lemaître radial coordinate. (Edit: corrected example mistakenly attributing this property to the GP chart.)</p>

<p>You can pick any coordinate chart that's well-behaved at the horizon, and any such chart will show it moving across the horizon with no trouble. But the Schwarzschild coordinates do not give such a chart, because they're undefined at the horizon. </p>

CC BY-SA 3.0
2014-08-16T02:37:46.227Z
2014-08-16T03:48:45.680Z
2014-08-16T03:48:45.680Z
Stan Liou
<p>Newton's Law of Universal Gravity states that an object will accelerate constantly as it falls. Let's assume an object falling into a black hole. According to Newton's Law, when it reaches the event horizon it must be traveling at <code>c</code>, because c is the escape velocity at the event horizon. Therefore, after it passes the event horizon, it must continue to accelerate past <code>c</code>, yet theory of relativity states that it's never gonna happen. <strong>So, in this condition, what will the object do?</strong></p>

CC BY-SA 3.0
2014-08-15T09:29:09.190Z
I have no idea where I heard this, and this is going to be over-simplified and missing a lot of pieces, but when I asked a similar question, the response I was given led me to believe that as you are affected by the space-time distortion of the gravitational force of the black hole, you effectively "shrink". That is, it will take you an infinite amount of time to reach the center of the black hole because as you are traveling toward it, the space between you gets *seems to* become greater. Also, have a look at https://suite.io/isaac-m-mcphee/k2f2y0
MegaMark
<black-hole>
Contradiction between Theory of Relativity and Newton's Law of Universal Gravitation?
ashu
# Contradiction between Theory of Relativity and Newton's Law of Universal Gravitation?

(Asked by: username_0 on 2014-08-15)

Newton's Law of Universal Gravity states that an object will accelerate constantly as it falls. Let's assume an object falling into a black hole. According to Newton's Law, when it reaches the event horizon it must be traveling at `c`, because c is the escape velocity at the event horizon. Therefore, after it passes the event horizon, it must continue to accelerate past `c`, yet theory of relativity states that it's never gonna happen. **So, in this condition, what will the object do?**

**Tags:** black-hole

> **Comments:**
> **username_1**: I have no idea where I heard this, and this is going to be over-simplified and missing a lot of pieces, but when I asked a similar question, the response I was given led me to believe that as you are affected by the space-time distortion of the gravitational force of the black hole, you effectively "shrink". That is, it will take you an infinite amount of time to reach the center of the black hole because as you are traveling toward it, the space between you gets *seems to* become greater. Also, have a look at https://suite.io/isaac-m-mcphee/k2f2y0  

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## 1 answers

### Answer by username_2 on 2014-08-16. Score: 4 | Accepted answer

**A) Don't take Newton's law too seriously in the relativistic regime.**

> According to Newton's Law, when it reaches the event horizon it must be traveling at $c$, because $c$ is the escape velocity at the event horizon.

This is true, assuming the particle freefalls from rest at infinity, but the details are quite different from Newtonian theory. For orbits in Schwarzschild spacetime, the [effective potential](http://en.wikipedia.org/wiki/Effective_potential) is
$$V = -\frac{GM}{r} + \frac{l^2}{2r^2} - \frac{GMl^2}{c^2r^3}\text{,}$$
where $l\equiv L/m$ is the specific angular momentum of the orbit. The last term offers relativistic corrections that have no analogue in Newtonian gravity, so in general Newton's law cannot be counted on when near the horizon of the black hole.

But for radial freefall, $l = 0$, and we a total orbital specific energy:
$$\mathcal{E} = \frac{1}{2}\left(\frac{\mathrm{d}r}{\mathrm{d}\tau}\right)^2 - \frac{GM}{r}\text{,}$$
which has exactly Newtonian form. In particular, a particle freefalling from rest at infinity will have
$$\left|\frac{\mathrm{d}r}{\mathrm{d}\tau}\right| = \sqrt{\frac{2GM}{r}}\text{,}$$
so at as it nears the horizon ($r\to 2GM/c^2$), we have $|\mathrm{d}r/\mathrm{d}\tau|\to c$, just as Newtonian theory would predict.

However, the motion is only roughly analogous to what's prescribed by Newton's law, because there are very important differences in the interpretation of those quantities. Specifically, (1) the $\tau$ refers to the proper time of the orbiting particle, not coordinate time $t$, and certainly not any absolute time as Newtonian theory proper, and (2) the $r$ refers to the Schwarzschild radial coordinate, which is defined such that a sphere at this coordinate have surface area $4\pi r^2$, and this is **not** the radial distance to the center.

**B) Coordinates have no intrinsic meaning.**

> Therefore, after it passes the event horizon, it must continue to accelerate past $c$, yet theory of relativity states that it's never gonna happen.

Suppose I fling a particle along a straight ruler, and that ruler is marked in feet. So if, for example, my particle goes five marked units along the ruler in a nanosecond, that's clearly a problem, because the particle must be superluminal ($c\approx 1\,\mathrm{ft}/\mathrm{ns}$).

Now suppose the ruler is not marked in feet, and I don't tell you how it's marked. Is it a problem that the particle goes five units in a nanosecond? Without knowing what the marks are, the statement of this sort of "coordinate speed" is not even meaningful. Finally, suppose I tell you how the ruler is marked, but its marks do not actually measure the distance along the ruler. Then is it a problem that it goes through five of those marks in a nanosecond?

The moral is simply this: by themselves, coordinates are simply labels to identify events. One can pick coordinates such that the coordinate speed is arbitrarily large or small, and it doesn't matter, because coordinates are not physical things. The universe does not come with its own coordinates; they're just labels we put on things.

However, what can give coordinates meaning is the *metric tensor*, which relates coordinates to actual lengths or durations, so that, for example, the Schwarzschild radial coordinate has a set meaning in terms of surface areas. Finally, the Schwarzschild coordinates are pathological near the horizon. They are not defined across the horizon (technically, the Schwarzschild coordinates are actually two entirely different coordinate charts disconnected by the horizon). You can see this in the Schwarzschild metric (units of $G = c = 1$):
$$\mathrm{d}s^2 = -\left(1-\frac{2M}{r}\right)\mathrm{d}t^2 + \left(1-\frac{2M}{r}\right)^{-1}\,\mathrm{d}r^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{,}$$
where the metric coefficients in the Schwarzschild chart are undefined at the horizon $r = 2M$.

**C) "So, in this condition, what will the object do?"**

It will cross the event horizon and meet its end at the singularity.

Here's the same Schwarzschild geometry in [Gullstrand-Painlevé coordinates](http://en.wikipedia.org/wiki/Gullstrand%E2%80%93Painlev%C3%A9_coordinates):
$$\mathrm{d}s^2 = -\mathrm{d}t^2 + \left(\mathrm{d}r + \sqrt{\frac{2M}{r}}\,\mathrm{d}t\right)^2 + r^2(\mathrm{d}\theta^2+\sin^2\theta\,\mathrm{d}\phi^2)\text{.}$$
In this coordinate chart, space is moving towards the singularity at the estate velocity $\sqrt{2GM/r}$, and the particles are carried along with it. In a related coordinate chart, [Lemaître coordinates](http://en.wikipedia.org/wiki/Lema%C3%AEtre_coordinates), particles freefalling from rest at infinity have zero coordinate speed, with $\mathrm{d}r/\mathrm{d}t = 0$, where here $r$ is the Lemaître radial coordinate. (Edit: corrected example mistakenly attributing this property to the GP chart.)

You can pick any coordinate chart that's well-behaved at the horizon, and any such chart will show it moving across the horizon with no trouble. But the Schwarzschild coordinates do not give such a chart, because they're undefined at the horizon.


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