<blockquote>
  <p>Can the gravity well be infinitely deep?</p>
</blockquote>

<p>Only if the spin parameter of a rotating black hole is maximal so that the rotation speed of the horizon would be the speed of light $\rm (a=M)$ the integral of the $g_{\rm r r}$ component blows up, and the isometric embedding of the radial plane goes infinite at the horizon (see <a href="http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&amp;bibcode=1972ApJ...178..347B&amp;letter=.&amp;classic=YES&amp;defaultprint=YES&amp;whole_paper=YES&amp;page=347&amp;epage=347&amp;send=Send%20PDF&amp;filetype=.pdf#page=7" rel="nofollow noreferrer">Bardeen 1972, Fig. 2</a>). Here is Flamm's paraboloid for all spin parameters, as long as the spin is subextremal the distances are finite (when $\rm a=0$ the exterior Schwarzschild solution is recovered):</p>

<p><a href="http://org.yukterez.net/flamms.paraboloid.for.the.kerr.metric.html" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iSRUK.gif" alt="enter image description here"></a></p>

CC BY-SA 3.0
2018-03-05T07:20:20.363Z
2018-03-05T07:26:09.067Z
2018-03-05T07:26:09.067Z
Yukterez
<p>Newtonian gravity of a point-source can described by a potential $\Phi = -\mu/r$. If we suppress one spatial dimension and use it to graph the value of this potential instead, we get something that looks very close to this illustration, and is indeed infinitely deep at the center--at least, in the idealization of a point-mass. And farther away from the center, it goes flat, just as many illustrations like this have it.</p>

<p>Illustrations like this are fairly common, and I'm guessing that they're ultimately inspired by the Newtonian potential, because they have almost nothing to do with with the spacetime curvature.</p>

<p>Here's an isometric embedding of the Schwarzschild geometry at an instant of Schwarzschild time, again with one dimension supressed:</p>

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

<p>Above the horizon (red circle), the surface is a piece of a paraboloid (the <em>Flamm paraboloid</em>). Unlike the potential, it does not go flat at large distances.</p>

<p>Being isometric means that it correctly represents the spatial distances above the horizon at an instant of Schwarzschild time. Below the horizon, the embedding isn't technically accurate because the Schwarzschild radial coordinate does not represent space there, but time. Although if we pretend it's spacelike below the horizon, that would be the correct embedding. Picture the below-horizon part as having one-directional flow into the singularity.</p>

<p>Since we've only represented space and not time, the embedding is not enough to reconstruct the trajectories of particles in this spacetime. Still, it is a more accurate representation of a <em>part</em> of the spacetime curvature of point-source--specifically the spatial part.</p>

<hr>

<blockquote>
  <p>The velocity of the object from this perspective, would seem to increase, until a point - where the velocity in x,y coordinates starts to decrease due to most of the motion happening "down" the time dimension. Is this also correct? Would a photon seem to slow down when moving down the well, if seen from above?</p>
</blockquote>

<p>The above is an embedding of a slice of spatial geometry, and is not a gravity well. The mathematical form of the paraboloid above the horizon is best described in <a href="http://en.wikipedia.org/wiki/Cylindrical_coordinates" rel="noreferrer">cylindrical coordinates</a> as
$$r = 2M + \frac{z^2}{8M}\text{.}$$
Here the vertical $z$ coordinate doesn't mean anything physically. It's purely an artifact of creating a surface of the same intrinsic curvature in Euclidean $3$-space as the $2$-dimensional spatial slice of Schwarzschild geometry.</p>

<p>For the Schwarzschild spacetime, radial freefall is actually exactly Newtonian in the Schwarzschild radial coordinate and proper time, i.e. time experienced by the freefalling object, rather than Schwarzschild time. So the Newtonian gravity well isn't actually a bad picture for the physics--it's just not the geometry and so is not a good representation of how any part of spacetime is curved. For non-radial orbits, the effective potential is somewhat different that than the Newtonian one, but ignoring the effects of angular momentum gets us the Newtonian form.</p>

<p>In Schwarzschild time, yes, a photon (or anything else) does slow down as gets near the horizon. In fact, in Schwarzschild time it never reaches the horizon, which is one indication that the Schwarzschild coordinates are badly behaved at the horizon. The coordinate acceleration actually becomes repulsive close to the horizon--and for a fast enough infalling object, is <em>always</em> repulsive. This can be understood as the particle moving to places with more and more gravitational time dilation. In proper time of any infalling observer, however, close to the horizon the acceleration is always attractive.</p>

CC BY-SA 3.0
2014-01-15T14:49:27.397Z
Thank you; this is just a followup regarding part two of the question; "motion" down the well. When viewing it from above you would only be able to observe two physical dimensions. The velocity of the object from this perspective, would seem to increase, until a point - where the velocity in x,y coordinates starts to decrease due to most of the motion happening "down" the time dimension. Is this also correct? Would a photon seem to slow down when moving down the well, if seen from above?
frodeborli
CC BY-SA 3.0
2014-01-15T12:43:07.740Z
2014-01-15T22:22:16.047Z
2014-01-15T22:22:16.047Z
Stan Liou
<p>When a particle travels through spacetime, this is sometimes illustrated moving in flat grid with various depths. How accurate is this way of imagining gravity?</p>

<p>Can the gravity well be infinitely deep?</p>

<p><a href="http://blakemaybank.com/wp-content/uploads/2012/12/Gravity_well.gif" rel="noreferrer">Gravity well http://blakemaybank.com/wp-content/uploads/2012/12/Gravity_well.gif</a></p>

<p>Looking at this illustration, it suggests that there is an amount of distance going down the well. Is movement <em>down the well</em> limited by the speed of light? Remember, I'm aware that it is an illustration, and the question is about the accuracy of the illustration.</p>

<p>Is all gravitational wells infinitely deep at the exact center? So that if a particle could be made small enough to fit within the center, it would become a black hole.</p>

CC BY-SA 4.0
2021-05-28T13:13:54.587Z
The link to your illustration is dead. Could you please provide an updated link?
Donald.McLean
<gravity><general-relativity>
Spacetime curvature illustration accuracy
frodeborli
# Spacetime curvature illustration accuracy

(Asked by: username_0 on 2014-01-15)

When a particle travels through spacetime, this is sometimes illustrated moving in flat grid with various depths. How accurate is this way of imagining gravity?

Can the gravity well be infinitely deep?

[Gravity well http://blakemaybank.com/wp-content/uploads/2012/12/Gravity\_well.gif](http://blakemaybank.com/wp-content/uploads/2012/12/Gravity_well.gif)

Looking at this illustration, it suggests that there is an amount of distance going down the well. Is movement *down the well* limited by the speed of light? Remember, I'm aware that it is an illustration, and the question is about the accuracy of the illustration.

Is all gravitational wells infinitely deep at the exact center? So that if a particle could be made small enough to fit within the center, it would become a black hole.

**Tags:** gravity, general-relativity

> **Comments:**
> **username_1**: The link to your illustration is dead. Could you please provide an updated link?  

---

## 2 answers

### Answer by username_3 on 2014-01-15. Score: 8 | Accepted answer

Newtonian gravity of a point-source can described by a potential $\Phi = -\mu/r$. If we suppress one spatial dimension and use it to graph the value of this potential instead, we get something that looks very close to this illustration, and is indeed infinitely deep at the center--at least, in the idealization of a point-mass. And farther away from the center, it goes flat, just as many illustrations like this have it.

Illustrations like this are fairly common, and I'm guessing that they're ultimately inspired by the Newtonian potential, because they have almost nothing to do with with the spacetime curvature.

Here's an isometric embedding of the Schwarzschild geometry at an instant of Schwarzschild time, again with one dimension supressed:

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

Above the horizon (red circle), the surface is a piece of a paraboloid (the *Flamm paraboloid*). Unlike the potential, it does not go flat at large distances.

Being isometric means that it correctly represents the spatial distances above the horizon at an instant of Schwarzschild time. Below the horizon, the embedding isn't technically accurate because the Schwarzschild radial coordinate does not represent space there, but time. Although if we pretend it's spacelike below the horizon, that would be the correct embedding. Picture the below-horizon part as having one-directional flow into the singularity.

Since we've only represented space and not time, the embedding is not enough to reconstruct the trajectories of particles in this spacetime. Still, it is a more accurate representation of a *part* of the spacetime curvature of point-source--specifically the spatial part.

---

> The velocity of the object from this perspective, would seem to increase, until a point - where the velocity in x,y coordinates starts to decrease due to most of the motion happening "down" the time dimension. Is this also correct? Would a photon seem to slow down when moving down the well, if seen from above?

The above is an embedding of a slice of spatial geometry, and is not a gravity well. The mathematical form of the paraboloid above the horizon is best described in [cylindrical coordinates](http://en.wikipedia.org/wiki/Cylindrical_coordinates) as
$$r = 2M + \frac{z^2}{8M}\text{.}$$
Here the vertical $z$ coordinate doesn't mean anything physically. It's purely an artifact of creating a surface of the same intrinsic curvature in Euclidean $3$-space as the $2$-dimensional spatial slice of Schwarzschild geometry.

For the Schwarzschild spacetime, radial freefall is actually exactly Newtonian in the Schwarzschild radial coordinate and proper time, i.e. time experienced by the freefalling object, rather than Schwarzschild time. So the Newtonian gravity well isn't actually a bad picture for the physics--it's just not the geometry and so is not a good representation of how any part of spacetime is curved. For non-radial orbits, the effective potential is somewhat different that than the Newtonian one, but ignoring the effects of angular momentum gets us the Newtonian form.

In Schwarzschild time, yes, a photon (or anything else) does slow down as gets near the horizon. In fact, in Schwarzschild time it never reaches the horizon, which is one indication that the Schwarzschild coordinates are badly behaved at the horizon. The coordinate acceleration actually becomes repulsive close to the horizon--and for a fast enough infalling object, is *always* repulsive. This can be understood as the particle moving to places with more and more gravitational time dilation. In proper time of any infalling observer, however, close to the horizon the acceleration is always attractive.

> **Comments:**
> **username_0**: Thank you; this is just a followup regarding part two of the question; "motion" down the well. When viewing it from above you would only be able to observe two physical dimensions. The velocity of the object from this perspective, would seem to increase, until a point - where the velocity in x,y coordinates starts to decrease due to most of the motion happening "down" the time dimension. Is this also correct? Would a photon seem to slow down when moving down the well, if seen from above?  

---

### Answer by username_2 on 2018-03-05. Score: 2

> Can the gravity well be infinitely deep?

Only if the spin parameter of a rotating black hole is maximal so that the rotation speed of the horizon would be the speed of light $\rm (a=M)$ the integral of the $g\_{\rm r r}$ component blows up, and the isometric embedding of the radial plane goes infinite at the horizon (see [Bardeen 1972, Fig. 2](http://articles.adsabs.harvard.edu/cgi-bin/nph-iarticle_query?db_key=AST&bibcode=1972ApJ...178..347B&letter=.&classic=YES&defaultprint=YES&whole_paper=YES&page=347&epage=347&send=Send%20PDF&filetype=.pdf#page=7)). Here is Flamm's paraboloid for all spin parameters, as long as the spin is subextremal the distances are finite (when $\rm a=0$ the exterior Schwarzschild solution is recovered):

[![enter image description here](https://i.sstatic.net/iSRUK.gif)](http://org.yukterez.net/flamms.paraboloid.for.the.kerr.metric.html)


---

