<p>It's a consequence of using a measurement system where time and space have different units. In relativity, spacetime has a unified geometry. If you use different units for time and space in your formulae, <span class="math-container">$c$</span> automatically shows up.</p>
<p>If you use a &quot;natural&quot; relativistic unit system with <span class="math-container">$c=1$</span>, the <span class="math-container">$c$</span> factors disappear.</p>

CC BY-SA 4.0
2024-06-25T14:21:11.637Z
Bingo. It's all one manifold, like a volume. You can rotate vectors in it. That's why they call it "spacetime." On a null surface, the vector is just pointing orthogonal to the time direction, that's all. And as usual, I ask myself why that didn't occur to me and I feel silly.
Miss Understands
CC BY-SA 4.0
2024-06-25T18:21:08.823Z
But it does have a single unit: the Planck Interval. It's the unit vector of spacetime and it defines a teeeeny little hypersphere. Photons and electrons and other strings are blobs of that size. Human busybodies break it into vectors with different units. I don't understand those humans.
Miss Understands
CC BY-SA 4.0
2024-06-25T19:13:37.773Z
@Miss_Understands That's a hypothesis. How would you test it experimentally?
John Doty
CC BY-SA 4.0
2024-06-25T19:48:34.143Z
@Miss_U Ok, we *expect* that a quantum gravity theory will say that spacetime quantum effects (of some kind) will become significant in the vicinity of the Planck length, but please see https://physics.stackexchange.com/a/185943/123208
PM 2Ring
CC BY-SA 4.0
2024-06-26T05:59:40.650Z
@JohnDoty I know I don't keep up with science as much as I should, but I believe that this newfangled mapping between quantum field theory and string theory does that.
Miss Understands
CC BY-SA 4.0
2024-06-26T11:58:06.220Z
@Miss_Understands How would you experimentally test that mapping?
John Doty
CC BY-SA 4.0
2024-06-26T13:52:36.470Z
@Miss_Understands, `this newfangled mapping between quantum field theory and string theory` which mapping? Nothing about string theory has ever been accessible to experiment or falsification. You cannot say `Photons and electrons and other strings are`, there is no indication that string theory is anything more than an aesthetically pleasing mathematical building.
AnoE
CC BY-SA 4.0
2024-06-26T14:48:16.147Z
@AnoE What mapping? ==[ I only know what I watched on youtube vids because I'm too stupid to do the math. But apparently if you're willing to believe that space is negatively curved, then specific strings map directly to specific quantum objects, like photons and electrons. https://laces.web.cern.ch/laces09/notes/dbranes/lezionilosanna.pdf  I am the only person I know who thinks spacetime is negative  curved (because of the (3,1) pseudometric signature).
Miss Understands
CC BY-SA 4.0
2024-06-26T15:59:23.137Z
Ever play around with the idea that c might take a different (constant) value?
Joshua
CC BY-SA 4.0
2024-06-26T16:30:09.087Z
@Joshua The trouble with that notion is that to measure $c$ you need a clock and a ruler. Right now, our best rulers are calibrated using clocks and a fixed, arbitrary value for $c$. So, for $c$ to vary you need to come up with a different clock/ruler pair and make the case that this is a better foundation for measurement than our current scheme. That's very difficult.
John Doty
CC BY-SA 4.0
2024-06-26T16:43:07.803Z
@JohnDoty: If you change c too far, gold isn't yellow anymore. I don't need a very good clock/ruler pair to consider what happens when c is multiplied by ten. The fixed point is the limit as c goes to infinity reduces the relativistic kinematic equations to the classical variants.
Joshua
CC BY-SA 4.0
2024-06-26T17:03:14.437Z
@Joshua What standard ruler and standard clock are you using to define your value of $c$?
John Doty
CC BY-SA 4.0
2024-06-26T17:34:04.363Z
@JohnDoty: If you hold the artifact kilogram constant (which was the definition up into the star of the 21th century); the effect of c on h becomes obvious.
Joshua
CC BY-SA 4.0
2024-06-26T18:03:09.997Z
@Joshua Except $c$ cannot change in the SI *by definition*, so it cannot have any effect on $h$. If you throw away the definition, you no longer have any means to measure speed, so speaking of a changing $c$ is impossible.
John Doty
CC BY-SA 4.0
2024-06-26T18:11:25.563Z
@JohnDoty: This is a thought exercise. What does physics look like if the relationship between c and 1AU were very different from the current value; but the sun's light and heat distribution on earth didn't change.
Joshua
CC BY-SA 4.0
2024-06-26T18:19:00.363Z
@Joshua You cannot sensibly reason from a counterfactual.
John Doty
CC BY-SA 4.0
2024-06-26T19:39:21.173Z
@JohnDoty: On the contrary; the history of physics contains the line of reasoning within. We started out with terrestrial and then astronomical measurements before we could even measure c as non-infinite. Natural units occasionally hides a coupling constant.
Joshua
CC BY-SA 4.0
2024-06-25T13:44:58.837Z
2024-06-25T13:44:58.837Z
John Doty
<p>The Schwarzschild's radius describes the radius of no return for objects, specifically the fastest objects we know about - photons, which travel at a certain speed - to escape a non-rotating, electromagnetically neutral black hole. Fundamentally, you'd expect that if these objects (photons) went faster, they could escape the black hole from closer, and if they went slower, they'd have to start further away to escape.</p>
<blockquote>
<p>... what in the world does the speed of light have to do with the
amount of mass it takes to bend space 90 degrees(in an imaginary angle)?</p>
</blockquote>
<p>A better question would be 'what does the Schwarzschild's radius have to do with the amount of mass ...' and the short answer is there's a linear relationship that's only meaningful for black holes. The longer answer is that <span class="math-container">$R_s$</span> is a convenient grouping of constants that appears in a lot of places.</p>
<p>The idea that the Schwarzschild's radius depends on units is ludicrous. The number describing the distance depends on units, but the radius is a very specific length. Choosing <span class="math-container">$c=1$</span> does not change the length, nor the importance of the speed of those fast objects that define the Schwarzschild's radius.</p>

CC BY-SA 4.0
2024-06-26T16:30:45.013Z
2024-06-27T16:16:01.707Z
2024-06-27T16:16:01.707Z
user121330
<p>The equation of the Schwarzschild radius is really simple, just twice the mass over <span class="math-container">$c^2$</span> (times a unit conversion constant that goes away in natural units).
<span class="math-container">$$ R = \frac{2GM}{c^2}$$</span></p>
<p>The question occurred to me, what in the world does the speed of light have to do with the amount of mass it takes to bend space 90 degrees (in an imaginary angle)?</p>
<p>Yes, the escape velocity at the horizon is c, but isn't there something special about that space that is independent of time? Like, <span class="math-container">$z$</span> (radial) distance between events shrinks to zero at the horizon. That fact doesn't depend on time or the phase of anything.</p>
<p><a href="https://jila.colorado.edu/%7Eajsh/bh/schwp.html" rel="nofollow noreferrer">All the definitions of distance in &quot;Schwarzschild geometry space&quot;</a> involve movement. But distance between objects at a particular snapshot of time can be expressed in purely spatial terms. Specifically, the <span class="math-container">$ct^2$</span> term vanishes, leaving only real (spacelike) distance.</p>
<p>Yet there is a real circumference of the BH who's value depends on how fast something moves (photons).</p>
<p>I know the horizon is a <a href="https://en.wikipedia.org/wiki/Null_hypersurface" rel="nofollow noreferrer">null surface</a> in 4 dimensions, like the photon. But that just means it doesn't experience time, just space, so it says even less about why time should be involved.</p>
<p>I'm hoping someone will understand and explain in the comments why the distance between 2 pixels on a photo of a BH sitting unmoving in space depends on something's speed.</p>

CC BY-SA 4.0
2024-06-25T14:05:56.890Z
FWIW, any mass has a Schwarzschild radius, not just black holes. Eg, for Earth, $r_s$ ~= 8.870056 mm.
PM 2Ring
CC BY-SA 4.0
2024-06-25T18:14:57.193Z
Yes, that's correct.
Miss Understands
CC BY-SA 4.0
2024-06-25T22:08:18.940Z
@PM2Ring the Schwarzschild radius is only actually interesting if it's larger than the object itself – there is no 8mm sphere inside earth where light wouldn't be able to escape, as the mass outside of the current height doesn't contribute to the attaction.
Paŭlo Ebermann
CC BY-SA 4.0
2024-06-25T23:42:24.853Z
@Paŭlo Sure, there's no event horizon inside the Earth, but that doesn't mean that its Schwarzschild radius ($r_s$) is *totally* useless. It's convenient to use $r_s$ when doing GR calculations, like gravitational time dilation, or deflection of starlight. Admittedly, both of those are pretty tiny for Earth. ;)
PM 2Ring
CC BY-SA 4.0
2024-06-27T03:06:45.437Z
Just an interesting observation...  Rearranging the equation would make it have a form like a = v^2/r, which in classical mechanics describes the acceleration of a mass orbiting in a circular path around some fixed point.  On the surface c^2=2GM/R looks like finding the radius where the gravitational acceleration would be sufficient to keep an object moving at the speed of light moving in a circular path rather than escaping.  Except for that there is a factor of two in there...
user4574
CC BY-SA 4.0
2024-06-27T11:46:59.540Z
@user4574 That factor of 2 also appears in the Newtonian equation for escape velocity (which is $\sqrt2$ times the circular orbit speed). It's a nice coincidence that the Newtonian equation gives the Schwarzschild radius, see https://physics.stackexchange.com/q/403461/123208 & links therein.
PM 2Ring
<black-hole><event-horizon>
Why does c show up in Schwarzschild's equation for the horizon radius?
Miss Understands
# Why does c show up in Schwarzschild's equation for the horizon radius?

(Asked by: username_0 on 2024-06-25)

The equation of the Schwarzschild radius is really simple, just twice the mass over $c^2$ (times a unit conversion constant that goes away in natural units).
$$ R = \frac{2GM}{c^2}$$

The question occurred to me, what in the world does the speed of light have to do with the amount of mass it takes to bend space 90 degrees (in an imaginary angle)?

Yes, the escape velocity at the horizon is c, but isn't there something special about that space that is independent of time? Like, $z$ (radial) distance between events shrinks to zero at the horizon. That fact doesn't depend on time or the phase of anything.

[All the definitions of distance in "Schwarzschild geometry space"](https://jila.colorado.edu/%7Eajsh/bh/schwp.html) involve movement. But distance between objects at a particular snapshot of time can be expressed in purely spatial terms. Specifically, the $ct^2$ term vanishes, leaving only real (spacelike) distance.

Yet there is a real circumference of the BH who's value depends on how fast something moves (photons).

I know the horizon is a [null surface](https://en.wikipedia.org/wiki/Null_hypersurface) in 4 dimensions, like the photon. But that just means it doesn't experience time, just space, so it says even less about why time should be involved.

I'm hoping someone will understand and explain in the comments why the distance between 2 pixels on a photo of a BH sitting unmoving in space depends on something's speed.

**Tags:** black-hole, event-horizon
> **Comments:**
> **username_1**: FWIW, any mass has a Schwarzschild radius, not just black holes. Eg, for Earth, $r_s$ ~= 8.870056 mm.  
> **username_0**: Yes, that's correct.  
> **username_2**: @PM2Ring the Schwarzschild radius is only actually interesting if it's larger than the object itself – there is no 8mm sphere inside earth where light wouldn't be able to escape, as the mass outside of the current height doesn't contribute to the attaction.  
> **username_1**: @Paŭlo Sure, there's no event horizon inside the Earth, but that doesn't mean that its Schwarzschild radius ($r_s$) is *totally* useless. It's convenient to use $r_s$ when doing GR calculations, like gravitational time dilation, or deflection of starlight. Admittedly, both of those are pretty tiny for Earth. ;)  
> **username_3**: Just an interesting observation...  Rearranging the equation would make it have a form like a = v^2/r, which in classical mechanics describes the acceleration of a mass orbiting in a circular path around some fixed point.  On the surface c^2=2GM/R looks like finding the radius where the gravitational acceleration would be sufficient to keep an object moving at the speed of light moving in a circular path rather than escaping.  Except for that there is a factor of two in there...  
> **username_1**: @user4574 That factor of 2 also appears in the Newtonian equation for escape velocity (which is $\sqrt2$ times the circular orbit speed). It's a nice coincidence that the Newtonian equation gives the Schwarzschild radius, see https://physics.stackexchange.com/q/403461/123208 & links therein.  

---

## 2 answers

### Answer by username_4 on 2024-06-25. Score: 21

It's a consequence of using a measurement system where time and space have different units. In relativity, spacetime has a unified geometry. If you use different units for time and space in your formulae, $c$ automatically shows up.

If you use a "natural" relativistic unit system with $c=1$, the $c$ factors disappear.

> **Comments:**
> **username_0**: Bingo. It's all one manifold, like a volume. You can rotate vectors in it. That's why they call it "spacetime." On a null surface, the vector is just pointing orthogonal to the time direction, that's all. And as usual, I ask myself why that didn't occur to me and I feel silly.  
> **username_0**: But it does have a single unit: the Planck Interval. It's the unit vector of spacetime and it defines a teeeeny little hypersphere. Photons and electrons and other strings are blobs of that size. Human busybodies break it into vectors with different units. I don't understand those humans.  
> **username_4**: @Miss_Understands That's a hypothesis. How would you test it experimentally?  
> **username_1**: @Miss_U Ok, we *expect* that a quantum gravity theory will say that spacetime quantum effects (of some kind) will become significant in the vicinity of the Planck length, but please see https://physics.stackexchange.com/a/185943/123208  
> **username_0**: @JohnDoty I know I don't keep up with science as much as I should, but I believe that this newfangled mapping between quantum field theory and string theory does that.  
> **username_4**: @Miss_Understands How would you experimentally test that mapping?  
> **username_5**: @Miss_Understands, `this newfangled mapping between quantum field theory and string theory` which mapping? Nothing about string theory has ever been accessible to experiment or falsification. You cannot say `Photons and electrons and other strings are`, there is no indication that string theory is anything more than an aesthetically pleasing mathematical building.  
> **username_0**: @AnoE What mapping? ==[ I only know what I watched on youtube vids because I'm too stupid to do the math. But apparently if you're willing to believe that space is negatively curved, then specific strings map directly to specific quantum objects, like photons and electrons. https://laces.web.cern.ch/laces09/notes/dbranes/lezionilosanna.pdf  I am the only person I know who thinks spacetime is negative  curved (because of the (3,1) pseudometric signature).  
> **username_6**: Ever play around with the idea that c might take a different (constant) value?  
> **username_4**: @Joshua The trouble with that notion is that to measure $c$ you need a clock and a ruler. Right now, our best rulers are calibrated using clocks and a fixed, arbitrary value for $c$. So, for $c$ to vary you need to come up with a different clock/ruler pair and make the case that this is a better foundation for measurement than our current scheme. That's very difficult.  
> **username_6**: @JohnDoty: If you change c too far, gold isn't yellow anymore. I don't need a very good clock/ruler pair to consider what happens when c is multiplied by ten. The fixed point is the limit as c goes to infinity reduces the relativistic kinematic equations to the classical variants.  
> **username_4**: @Joshua What standard ruler and standard clock are you using to define your value of $c$?  
> **username_6**: @JohnDoty: If you hold the artifact kilogram constant (which was the definition up into the star of the 21th century); the effect of c on h becomes obvious.  
> **username_4**: @Joshua Except $c$ cannot change in the SI *by definition*, so it cannot have any effect on $h$. If you throw away the definition, you no longer have any means to measure speed, so speaking of a changing $c$ is impossible.  
> **username_6**: @JohnDoty: This is a thought exercise. What does physics look like if the relationship between c and 1AU were very different from the current value; but the sun's light and heat distribution on earth didn't change.  
> **username_4**: @Joshua You cannot sensibly reason from a counterfactual.  
> **username_6**: @JohnDoty: On the contrary; the history of physics contains the line of reasoning within. We started out with terrestrial and then astronomical measurements before we could even measure c as non-infinite. Natural units occasionally hides a coupling constant.  

---

### Answer by username_7 on 2024-06-26. Score: 4

The Schwarzschild's radius describes the radius of no return for objects, specifically the fastest objects we know about - photons, which travel at a certain speed - to escape a non-rotating, electromagnetically neutral black hole. Fundamentally, you'd expect that if these objects (photons) went faster, they could escape the black hole from closer, and if they went slower, they'd have to start further away to escape.

> ... what in the world does the speed of light have to do with the
> amount of mass it takes to bend space 90 degrees(in an imaginary angle)?

A better question would be 'what does the Schwarzschild's radius have to do with the amount of mass ...' and the short answer is there's a linear relationship that's only meaningful for black holes. The longer answer is that $R\_s$ is a convenient grouping of constants that appears in a lot of places.

The idea that the Schwarzschild's radius depends on units is ludicrous. The number describing the distance depends on units, but the radius is a very specific length. Choosing $c=1$ does not change the length, nor the importance of the speed of those fast objects that define the Schwarzschild's radius.


---

