<p><strong>1. Assumptions</strong></p>
<ul>
<li>1.a: gravity propagates at <em>c</em> (maximum)</li>
<li>1.b: gravitational fields cause time dilation</li>
<li>1.c: escape velocity at the event horizon is <em>c</em></li>
<li>1.d: gravity &quot;warps&quot; space-time, but in the absence of it, space-time &quot;snaps back&quot; to flat</li>
</ul>
<p><strong>2. The paradox</strong></p>
<p>I think you might already see where this is going.<br />
Consider a passing planet. It has a gravitational effect but once it has passed, the effect weakens. Space-time around you is &quot;snapping back&quot; towards a flatter geometry.</p>
<p>Now consider matter falling into a black hole:<br />
<em>Once it is beyond the event horizon, how can its gravitational effect continue to be felt outside?</em><br />
After all, it can only travel at <em>c</em> and therefore will be unable to exit the black hole, similar to light.</p>
<p>Therefore, black holes should appear massless.</p>
<p><strong>3. Possible resolution</strong></p>
<p>As gravity wells cause time dilation, time &quot;stops&quot; at the event horizon. For an outside observer, matter never really crosses the horizon, just asymptotically approaches it. So its gravity can still be felt outside, similar to its light.</p>
<p><strong>4. Outstanding questions</strong></p>
<ul>
<li>4.a: is there a Doppler-shift effect on gravity, similar to light?</li>
<li>4.b: what happens to time beyond the event horizon? Does it reverse? If so, what happens when the escape velocity exceeds <em>2c</em>?</li>
</ul>
<hr />
<p>Please let me know if any of this makes sense, which assumptions and conclusions are wrong, etc.</p>

CC BY-SA 4.0
2023-09-11T20:45:41.630Z
Your assumption 1.d is wrong.  Gravitational waves do propagate at light speed (recently measured with optical counterpart to GW detection).  But for a black hole to cause a gravitational acceleration, *nothing* has to propagate.  For simplicity, assume a spherically symmetric star collapsing to a blackhole with no supernova or anything.  Distant observers will notice the light going out, but there will be no change to the gravitational acceleration from the mass that was the star; same mass, same distance.  Nothing has to "escape" the black-hole to maintain the same curvature of spacetime.
antlersoft
CC BY-SA 4.0
2023-09-11T21:21:39.633Z
It's hard to answer this succinctly.  The misconception seems to be that gravity is like a "tractor beam" that mass emits, which shines on other mass and pulls the two together.  But gravity really isn't like that.  It is described by the equations of general relativity. And when you use general relativity you find that there is no problem here to be solved. The answer to your questions 4a.b is "no. gravity isn't a tractor beam, and you need to use GR to understand it".  But do please look at the link that PM2ring gives. I think this will help you.
James K
CC BY-SA 4.0
2023-09-11T21:57:50.380Z
@antlersoft This is why I gave the example of a passing planet. What maintains the curvature of space-time around it is its gravitational field which propagates at the speed of light. Once it's moved away, the space it once occupied is not forever warped. In the same way, once a star's matter falls into the black hole, it disappears from our universe, so it cannot exert a force through the event horizon. But I think the solution is that - from our perspective - it never actually crosses the EH.
P Varga
CC BY-SA 4.0
2023-09-12T01:50:42.200Z
From https://physics.stackexchange.com/questions/937/how-does-gravity-escape-a-black-hole#comment567153_943 The information that escapes about the mass never actually reaches the horizon. It's encoded in the curvature of space around the black hole (this includes the part very close to the horizon, but a lot of the information isn't close to the horizon). – Peter Shor
PM 2Ring
<black-hole><gravity><general-relativity><space-time>
Question about how it's possible for black holes to have gravity and a possible resolution
P Varga
# Question about how it's possible for black holes to have gravity and a possible resolution

(Asked by: username_0 on 2023-09-11)

**1. Assumptions**

* 1.a: gravity propagates at *c* (maximum)
* 1.b: gravitational fields cause time dilation
* 1.c: escape velocity at the event horizon is *c*
* 1.d: gravity "warps" space-time, but in the absence of it, space-time "snaps back" to flat

**2. The paradox**

I think you might already see where this is going.  
Consider a passing planet. It has a gravitational effect but once it has passed, the effect weakens. Space-time around you is "snapping back" towards a flatter geometry.

Now consider matter falling into a black hole:  
*Once it is beyond the event horizon, how can its gravitational effect continue to be felt outside?*  
After all, it can only travel at *c* and therefore will be unable to exit the black hole, similar to light.

Therefore, black holes should appear massless.

**3. Possible resolution**

As gravity wells cause time dilation, time "stops" at the event horizon. For an outside observer, matter never really crosses the horizon, just asymptotically approaches it. So its gravity can still be felt outside, similar to its light.

**4. Outstanding questions**

* 4.a: is there a Doppler-shift effect on gravity, similar to light?
* 4.b: what happens to time beyond the event horizon? Does it reverse? If so, what happens when the escape velocity exceeds *2c*?

---

Please let me know if any of this makes sense, which assumptions and conclusions are wrong, etc.

**Tags:** black-hole, gravity, general-relativity, space-time

> **Comments:**
> **username_1**: Your assumption 1.d is wrong.  Gravitational waves do propagate at light speed (recently measured with optical counterpart to GW detection).  But for a black hole to cause a gravitational acceleration, *nothing* has to propagate.  For simplicity, assume a spherically symmetric star collapsing to a blackhole with no supernova or anything.  Distant observers will notice the light going out, but there will be no change to the gravitational acceleration from the mass that was the star; same mass, same distance.  Nothing has to "escape" the black-hole to maintain the same curvature of spacetime.  
> **username_2**: It's hard to answer this succinctly.  The misconception seems to be that gravity is like a "tractor beam" that mass emits, which shines on other mass and pulls the two together.  But gravity really isn't like that.  It is described by the equations of general relativity. And when you use general relativity you find that there is no problem here to be solved. The answer to your questions 4a.b is "no. gravity isn't a tractor beam, and you need to use GR to understand it".  But do please look at the link that PM2ring gives. I think this will help you.  
> **username_0**: @username_1 This is why I gave the example of a passing planet. What maintains the curvature of space-time around it is its gravitational field which propagates at the speed of light. Once it's moved away, the space it once occupied is not forever warped. In the same way, once a star's matter falls into the black hole, it disappears from our universe, so it cannot exert a force through the event horizon. But I think the solution is that - from our perspective - it never actually crosses the EH.  
> **username_3**: From https://physics.stackexchange.com/questions/937/how-does-gravity-escape-a-black-hole#comment567153_943 The information that escapes about the mass never actually reaches the horizon. It's encoded in the curvature of space around the black hole (this includes the part very close to the horizon, but a lot of the information isn't close to the horizon). – Peter Shor  

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## 0 answers

