Radius Excess of Curvature in Relativity: Explained

[italicus]

TL;DR Summary

A formula by R. Feynman about the radius excess of curvature

I think the best place to put this post is the section on special and general relativity. Reading Feynman’s lecture n.42 , volume II here linked :
https://www.feynmanlectures.caltech.edu/II_42.html
I’ve met the following formula 42.3 for the radius excess of curvature, that Feynman attributes to Einstein :

$$\text {Radius excess} = R{measured} – \sqrt {\frac {A} {4\pi} } = \frac {GM} {3c^2} $$

Feynman says the space is curved, according to Einstein. The first observation here is that spacetime is curved by matter, not only space. Anyway, let us go on, some expert will clarify this. I have understood that, if we take a sphere of matter small enough that density can be considered constant inside, and A is the surface area, then the measured radius is grater than $\sqrt {\frac {A} {4\pi} }$ , and that this difference is given by $\frac {GM} {3c^2}$ .

My question is : where does the last equality come from ? I have studied relativity on several books , Schutz, Rindler, D’Inverno...but haven’t found that formula. I know that in a curved geometry “r” is a coordinate and not a distance, but I’m perplexed. Perhaps there is something i haven’t well understood

Thank you for the attention.

 

[PeterDonis]

Feynman says the space is curved, according to Einstein.

As you note, it should be spacetime, not just space, in general. However, Feynman is discussing one particular case, the spacetime geometry of a spherically symmetric massive body surrounded by vacuum, in which there is a natural way to split up spacetime into "space" and "time", and he is talking about the curvature of space (which is the same at every moment of time) for this case.

where does the last equality come from ?

It comes from the known exact solution for a spherically symmetric massive object of constant density surrounded by vacuum. Feynman doesn't give any details of that solution, probably because doing so would require a more in-depth discussion of GR than he wants to give. But the solution is discussed in various GR textbooks (MTW, for example, discusses it in Box 23.2). Feynman is actually implicitly using an approximation in which the correction to Euclidean spatial geometry is very small.

 

[italicus]

Thank you Peter.

However, Feynman is discussing one particular case, the spacetime geometry of a spherically symmetric massive body surrounded by vacuum, in which there is a natural way to split up spacetime into "space" and "time"

So, first of all, one has to study a lot of GR , until the spacetime geometry of Schwarzschild at least ( I guess this is the solution you referred to , right? )
Then you split spacetime into space and time: I am curious about this procedure, and to see details of how one arrives at the said “Radius excess”. Probably, one has to put “dt = 0” in the Sch. solution, but not only.
Could you explain me (briefly !) what is the way to proceed?
I have found this in your archive :
https://www.physicsforums.com/threads/what-is-feynmans-excess-radius-for-a-black-hole.322567/

But this applies to a BH only.

We must only accept the Feynman issue, the way it is.

 

[PeterDonis]

first of all, one has to study a lot of GR , until the spacetime geometry of Schwarzschild at least ( I guess this is the solution you referred to , right? )

It's the Schwarzschild interior geometry (i.e., inside a massive spherically symmetric object with uniform density), which is not the one usually called the "Schwarzschild geometry" (which describes the geometry of the vacuum region outside such an object).

Then you split spacetime into space and time

Yes, and this is done by choosing coordinates such that the geometry of a spacelike slice of constant coordinate time is the same for all coordinate times. This can be done for any spacetime that is stationary. Most GR textbooks discuss this.

Probably, one has to put “dt = 0” in the Sch. solution

Once you have chosen appropriate coordinates, yes, the geometry of "space" is what you get when you set $dt = 0$ in the line element.

I have found this in your archive :
https://www.physicsforums.com/threads/what-is-feynmans-excess-radius-for-a-black-hole.322567/

But this applies to a BH only.

No, actually, the solution given in that thread by @George Jones is the solution for the interior of a constant density sphere, not a black hole. The OP of the thread asked about black holes, but unfortunately never got an answer to whether the method used was applicable to black holes. It turns out that it isn't, because a black hole is not an ordinary object with an "interior" like a star or planet, and there is no meaningful notion of "distance to the center of the hole", the way there is a meaningful notion of distance to the center of an ordinary object like a star or planet.

 

[italicus]

It's the Schwarzschild interior geometry (i.e., inside a massive spherically symmetric object with uniform density), which is not the one usually called the "Schwarzschild geometry" (which describes the geometry of the vacuum region outside such an object).
...
Most GR textbooks discuss this.

Once you have chosen appropriate coordinates, yes, the geometry of "space" is what you get when you set $dt = 0$ in the line element.
No, actually, the solution given in that thread by @George Jones is the solution for the interior of a constant density sphere, not a black hole. The OP of the thread asked about black holes, but unfortunately never got an answer to whether the method used was applicable to black holes. It turns out that it isn't, because a black hole is not an ordinary object with an "interior" like a star or planet, and there is no meaningful notion of "distance to the center of the hole", the way there is a meaningful notion of distance to the center of an ordinary object like a star or planet.

Thank you again Peter.
Yes, i have studied mathematically gr until the Sch. BH , and know that the “interior “ is something very different from what we think is the interior of a planet or a star: space becomes time and viceversa, when one trespass the event horizon, which is not a physical singularity but depends on the choice of coordinates. Really, the true singularity is at r=0 ; and there are solutions which extend coordinates beyond the horizon, without problems (exception made for mathematics!).
But what happens to matter falling into a BH ? Let’s consider a Sch. BH for simplicity; I think matter should go to the center, reaching the singularity at r=0 , so the inside of the BH is essentially empty.
So, I can take the solution by Jones as the answer to my question ( sorry, I have never understood the difference between answer and reply, ahahah), for a sphere of constant density, not a BH.
I am curious about the box by MTW...but perhaps the solution can be found in other books, f.i. Cheng, Sthephani, Hartle, Mc Callum , or others , isn’t it?
I’am grateful for your clarifications . See soon.

 

[PeterDonis]

space becomes time and viceversa

No, that's not what happens inside a black hole's horizon. One particular choice of coordinates leads to the coordinate called $t$ being spacelike and the coordinate called $r$ being timelike inside the horizon, but (a) that's only true in those particular coordinates, and (b) that's not the same as "space becomes time and vice versa".

what happens to matter falling into a BH ?

It ends up hitting the singularity and being destroyed.

I suggest reading a series of Insights articles I wrote on the Schwarzschild geometry, plus an article I wrote on black holes specifically:

https://www.physicsforums.com/insights/schwarzschild-geometry-part-1/

https://www.physicsforums.com/insights/black-holes-really-exist/

These articles clear up a number of common misconceptions.

 

[italicus]

For the first “No “ , I didn’t realize that it depends on the choice of particular coordinates, now it’s clear.
I will read carefully your articles, thank you once more.