Calc Excess Rad on Earth in Curved 3D Space | Feynman Lectures 6-2

[Shirish]

I'm reading the Feynman lectures chapter on "Curved Space", section 6-2. Say we're trying to figure out a way to measure average curvature on Earth. We know that 3d space is curved if Euclidean geometry rules don't work - e.g. the ratio of circumference and radius of a circle isn't $2\pi$, or angles in a triangle don't sum up to $\pi$, etc. The described method is:

We can specify a sphere by taking all the points that are the same distance from a given point in space. Then we can measure the surface area by laying out a fine-scale rectangular grid on the surface of the sphere and adding up all the bits of area. According to Euclid the total area $A$ is supposed to be $4\pi$ times the square of the radius; so we can define a "predicted radius" as $\sqrt{A/4\pi}$. But we can also measure the radius directly by digging a hole to the center and measuring the distance. Again, we can take the measured radius minus the predicted radius and call the difference the radius excess. $$r_{excess}=r_{meas}-\bigg(\frac{\text{measured area}}{4\pi}\bigg)^{1/2}$$

Then later on to measure curvature near Earth,

The rule that Einstein gave for the curvature is the following: If there is a region of space with matter in it and we take a sphere small enough that the density $\rho$ of matter inside it is effectively constant, then the radius excess for the sphere is proportional to the mass inside the sphere. $$\text{Radius excess}=r_{meas}-\sqrt{\frac{A}{4\pi}}=\frac{G}{3c^2}.M$$ where $M=4\pi\rho r^3/3$ is the mass of the matter inside the sphere.

...

(Earth assumed to have uniform density) Suppose we were to measure the surface of the earth very carefully, and then dig a hole to the center and measure the radius. From the surface area we could calculate the predicted radius we would get from setting the area equal to $4\pi r^2$. When we compared the predicted radius with the actual radius, we would find that the actual radius exceeded the predicted radius by the amount given in the above equation.

My confusion is: to calculate $M$ we're using $4\pi\rho r^3/3$. But didn't we just argue that standard geometry rules may not work in curved space? For example the surface area formula sure doesn't seem to work. So why should the sphere volume rule work? Just as we measured the surface are by laying out a fine-scale rectangular grid on the surface, wouldn't we be forced to measure volume by laying out fine-scale cubical grid inside the Earth? Is there any assumption or mathematical result that I'm missing?

 

[Andrew Mason]

My confusion is: to calculate $M$ we're using $4\pi\rho r^3/3$. But didn't we just argue that standard geometry rules may not work in curved space? For example the surface area formula sure doesn't seem to work. So why should the sphere volume rule work? Just as we measured the surface are by laying out a fine-scale rectangular grid on the surface, wouldn't we be forced to measure volume by laying out fine-scale cubical grid inside the Earth? Is there any assumption or mathematical result that I'm missing?

I don't think the statement:

Radius excess $= r_{meas} - \sqrt{\frac{A}{4\pi}} = \frac{G}{3c^2}M$

can be correct because it suggests that the curvature of space is independent of proximate mass.
Can you provide a link to the lecture to which you are referring?

 

[Ibix]

The metric for the interior of a constant density sphere is given at Wikipedia. You can read off directly that the measured radius is$$\begin{eqnarray*}
r_{\mathrm{meas}}&=&\int\sqrt{g_{rr}}dr\\
&=&\int_0^{r_g}\frac{r_g^{3/2}}{\sqrt{r_g^3-r_Sr^2}}dr
\end{eqnarray*}$$which isn't a difficult integral if you want to confirm the result.

Of interest is that the gravitational mass $M$ inside a sphere of uniform density $\rho$ and radius $r_g$ is exactly $4\pi\rho r_g^3/3$. You can confirm that by setting $r=r_g$ in the metric linked above and matching it to an exterior Schwarzschild metric at that point and reading off $M$ from that. Although the volume inside the sphere is slightly larger than $4\pi r_g^3/3$ (which you can attribute to the curvature of your spatial slices), the "extra" rest energy of the matter is offset by a negative contribution to $M$ from the internal pressure.

 

[Shirish]

I don't think the statement:

Radius excess $= r_{meas} - \sqrt{\frac{A}{4\pi}} = \frac{G}{3c^2}M$

can be correct because it suggests that the curvature of space is independent of proximate mass.
Can you provide a link to the lecture to which you are referring?

Here's the link: https://www.feynmanlectures.caltech.edu/II_42.html

Ref. sections 42-2 and 42-3

 

[Shirish]

The metric for the interior of a constant density sphere is given at Wikipedia. You can read off directly that the measured radius is$$\begin{eqnarray*}
r_{\mathrm{meas}}&=&\int\sqrt{g_{rr}}dr\\
&=&\int_0^{r_g}\frac{r_g^{3/2}}{\sqrt{r_g^3-r_Sr^2}}dr
\end{eqnarray*}$$which isn't a difficult integral if you want to confirm the result.

Of interest is that the gravitational mass $M$ inside a sphere of uniform density $\rho$ and radius $r_g$ is exactly $4\pi\rho r_g^3/3$. You can confirm that by setting $r=r_g$ in the metric linked above and matching it to an exterior Schwarzschild metric at that point and reading off $M$ from that. Although the volume inside the sphere is slightly larger than $4\pi r_g^3/3$ (which you can attribute to the curvature of your spatial slices), the "extra" rest energy of the matter is offset by a negative contribution to $M$ from the internal pressure.

Thanks for the answer! I'm really sorry but I'm entirely new to GR (hence reading about it from Feynman lectures to start things off), so I didn't understand the majority of what you said.

What I got from your explanation is that: " it just so happens, once you read GR in detail you'll find that the earth mass should be $4\pi\rho r_g^3/3$, and this $r_g$ is different from both $r_{meas}$ (measured directly by digging to earth center) and $r_{pred}=\sqrt{A/4\pi}$ ". Is that about right? Maybe there is no easier way of explaining this and I'll need to understand GR much better first.

 

[Ibix]

Almost. $r_g$ is what Feynman's calling $r_{pred}$. It's $\sqrt{A/4\pi}$, which would be unambiguously the radius of the sphere in a Euclidean space. But space inside the mass is not Euclidean, it's slightly curved, so the distance to the center of the sphere of a given area is ever so slightly larger (in this case, it can be smaller in other cases) than you would expect from blindly applting Euclid.

The quantity $\sqrt{A/4\pi}$ is often called the "areal radius" since it's derived from the area of the sphere. It's a useful quantity because it uniquely identifies concentric spherical surfaces, but it is not in general the distance to the center.

 

[Shirish]

Almost. $r_g$ is what Feynman's calling $r_{pred}$. It's $\sqrt{A/4\pi}$, which would be unambiguously the radius of the sphere in a Euclidean space. But space inside the mass is not Euclidean, it's slightly curved, so the distance to the center of the sphere of a given area is ever so slightly larger (in this case, it can be smaller in other cases) than you would expect from blindly applting Euclid.

The quantity $\sqrt{A/4\pi}$ is often called the "areal radius" since it's derived from the area of the sphere. It's a useful quantity because it uniquely identifies concentric spherical surfaces, but it is not in general the distance to the center.

That's very interesting! So my takeaway from the Feynman had phrased it was: $4\pi r^2$ (Euclidean surface area of a sphere) isn't the correct formula for the surface area, so $4\pi r^3/3$ (Euclidean vol. of a sphere) should also be useless for calculating the volume.

But as it turns out, it just so happens it's not the Euclidean formula that's useless - it's the parameter that's causing problems. i.e. $4\pi r_{actual}^2$ is wrong, $4\pi r_{pred}^2$ is correct. Similarly, $4\pi r_{actual}^3/3$ is wrong, but $4\pi r_{pred}^3/3$ is correct.

I wonder if the above assertion is true in general: Euclidean surface area and volume formulas still carry over to the curved space scenario - just with modified parameters. e.g. cylinder volume would be $\pi (r+\delta r)^2 (h+\delta h)$, where $r,h$ are the actual measured radius and height of cylinder respectively.

 

[Ibix]

I think you are reading more into this than there is.

The Euclidean formulas don't relate distance from the center to the area or the volume of a sphere except in the Euclidean case (which is not correct here). However, you can always measure the area of a sphere and calculate $\sqrt{A/4\pi}$, whether it is physically meaningful or not - even in cases like black holes that don't even have centers to measure a distance from.

Nevertheless, one sphere inside another has a smaller area, and it's occasionally mathematically useful to have a parameter that is smaller for the smaller sphere and can be defined even in strange geometries where there's no center. That's what the areal radius does for us.

Separate from this,there's a coincidence with the gravity of a sphere of uniform density (I think it only applies to the iniform density case). The volume of the sphere is ever so slightly larger than a Euclidean calculation would suggest, so naively you would expect the mass of the Earth to be slightly larger than $4\pi\rho r_g^3/3$. But in GR there are more sources of gravity than the Newtonian "how much stuff there is there" idea. The only relevant one here is the internal pressure of the sphere supporting itself against its own gravity - and that happens to be exactly the same size but opposite sign as the "extra" mass. So it's not that one is correct to use naive Euclidean calculations. Rather, it's that the two mistakes you make bu doing so happen to cancel out. As I say, I suspect that cancellation is a special feature of a uniform density sphere, but I'm not completely sure.

 

[PeterDonis]

The only relevant one here is the internal pressure of the sphere supporting itself against its own gravity - and that happens to be exactly the same size but opposite sign as the "extra" mass.

No, that's not quite right. Pressure as an additional source of gravity actually increases the mass, but that is exactly canceled by the decrease in mass due to negative gravitational binding energy, which shows up as a factor less than $1$ due to the $g_{tt}$ metric coefficient.

See this Insights article for more details:

https://www.physicsforums.com/insights/is-pressure-a-source-of-gravity/

 

[Shirish]

I think you are reading more into this than there is.

...

So you mean the last paragraph in my last post is overreaching - it's not correct of me to generalize like that. It just so happens to be true (i.e. $4\pi r_{pred}^3/3$ being the correct volume of the sphere despite the geometry not being Euclidean) in the case of a uniform density sphere, due to whatever reasons you or Peter mentioned?

 

[PeterDonis]

It just so happens to be true

It's true for the special case of a spherically symmetric object. But exact spherical symmetry is very rare. To be exactly spherically symmetric, the object must not be rotating at all, and its structure must be determined entirely by hydrostatic equilibrium, i.e., it must be made of a perfect fluid or something that behaves equivalently to a perfect fluid under the appropriate conditions.

 

[PeterDonis]

$4\pi r_{pred}^3/3$ being the correct volume of the sphere

No, it isn't. The correct mass of the sphere for the case of uniform density is $4 \pi r_{pred}^3 \rho / 3$, but that does not mean the correct volume of the sphere is $4 \pi r_{pred}^3 / 3$. It's not. That's what the geometry of space not being Euclidean means.

 

[Vanadium 50]

I may be coming at this from the wrong direction, but it seems to me that we have two scales in the problem: m )the Swarzchild radius) and R (the Earth radius). I am using units where m has dimensions of length.

So your answer is (within terms of order 1), m, m²/R, mⁿ⁺¹/Rⁿ or something like that on dimensional arguments alone. So the questions are "what is n?", and "what are the terms of order 1"?

The suggestion is that they are 0 and 1/3 respectively. Could be. The thing I don't quite like about this is the limit of being arbitrarily far away the spatial curvature is still detectable. That does not match my intuition.

 

[Ibix]

No, that's not quite right.

Thanks - that does make more sense.

 

[Shirish]

No, it isn't. The correct mass of the sphere for the case of uniform density is $4 \pi r_{pred}^3 \rho / 3$, but that does not mean the correct volume of the sphere is $4 \pi r_{pred}^3 / 3$. It's not. That's what the geometry of space not being Euclidean means.

But I thought that since the mass density is uniform, then the volume would be mass/density. So the volume would be $4 \pi r_{pred}^3 / 3$?

 

[Ibix]

I may be coming at this from the wrong direction, but it seems to me that we have two scales in the problem: m )the Swarzchild radius) and R (the Earth radius). I am using units where m has dimensions of length.

So your answer is (within terms of order 1), m, m²/R, mⁿ⁺¹/Rⁿ or something like that on dimensional arguments alone. So the questions are "what is n?", and "what are the terms of order 1"?

The suggestion is that they are 0 and 1/3 respectively. Could be. The thing I don't quite like about this is the limit of being arbitrarily far away the spatial curvature is still detectable. That does not match my intuition.

The explicit calculation in my #3 comes out to a measured radius (in the sense of a tunnel to the center) of $$\sqrt{\frac{r_g^{3}}{r_S}}\sin^{-1}\left(\sqrt{\frac{r_S}{r_g}}\right)= r_g+\frac 16r_S+O\left(\frac{r_S^2}{r_g}\right)$$if that helps. That matches Feynman.

 

[PeterDonis]

I thought that since the mass density is uniform, then the volume would be mass/density

No. It's not. Welcome to GR.

 

[Ibix]

But I thought that since the mass density is uniform, then the volume would be mass/density. So the volume would be $4 \pi r_{pred}^3 / 3$?

The fun (for slightly masochistic values of fun) of GR is that all sorts of things you thought were nice simple additive quantities, aren't.

 

[Shirish]

No. It's not. Welcome to GR.

Oof

The fun (for slightly masochistic values of fun) of GR is that all sorts of things you thought were nice simple additive quantities, aren't.

Okay so underneath the "simple" explanation in Feynman lectures, there seem to be a whole host of details and pitfalls. Just the simple fact that we take the mass of earth as $4\pi\rho r_{meas}^3/3$ is a coincidence due to spherical symmetry and uniform density - and even then the volume isn't obtainable in a simple way.

 

[PeterDonis]

Just the simple fact that we take the mass of earth as $4\pi\rho r_{meas}^3/3$ is a coincidence due to spherical symmetry and uniform density

Not a coincidence: it's what the applicable laws of physics say about the case of spherical symmetry and uniform density. (Note that the actual Earth is neither spherically symmetrical nor uniform in density.) A result that comes from the laws of physics is not a coincidence.

 

[PeterDonis]

underneath the "simple" explanation in Feynman lectures, there seem to be a whole host of details and pitfalls

Only if you go beyond what the lectures actually say. Feynman, as far as I can tell, never makes the claim about volume and density that you make. The only thing he says is a consequence of spatial curvature is the "radius excess"--the increase in the actual, physical radius over the "areal radius" $\sqrt{A / 4 \pi}$ that we would calculate assuming Euclidean geometry.

 

[Shirish]

Not a coincidence: it's what the applicable laws of physics say about the case of spherical symmetry and uniform density. (Note that the actual Earth is neither spherically symmetrical nor uniform in density.) A result that comes from the laws of physics is not a coincidence.

In that case "coincidence" is the wrong word to use - I meant to say that the conditions in the example are just right for the mass formula to be like $4\pi\rho r_{meas}^3/3$

Only if you go beyond what the lectures actually say. Feynman, as far as I can tell, never makes the claim about volume and density that you make. The only thing he says is a consequence of spatial curvature is the "radius excess"--the increase in the actual, physical radius over the "areal radius" $\sqrt{A / 4 \pi}$ that we would calculate assuming Euclidean geometry.

Completely fair, but I think it's still good to go beyond - in this case, the mass was stated as $4\pi\rho r_{meas}^3/3$, which led me to incorrectly assume that that the volume is this divided by $\rho$, which in turn made me think that the "volume being $4\pi r_{meas}^3/3$ contradicts the fact that Euclidean formulas generally don't work in curved space". Because of that you guys were able to clarify the wrong assumption of vol = mass/density, and the nice form of the mass is due to specific conditions. And I was able to learn a few new things

 

[PeterDonis]

the mass was stated as $4\pi\rho r_{meas}^3/3$

No; Feynman doesn't use your notation of $r_{meas}$ vs. $r_{pred}$. He just uses $r$. If you work through the actual math, it turns out that $r$ in that formula has to be $r_{pred}$, not $r_{meas}$. But that formula only works for the mass (for the case of constant density). It doesn't work for the volume--see below.

the "volume being $4\pi r_{meas}^3/3$

That is not the volume. The volume doesn't obey Euclidean formulas no matter which r you use. You have to obtain it by doing an integral.

 

[Shirish]

No; Feynman doesn't use your notation of $r_{meas}$ vs. $r_{pred}$. He just uses $r$. If you work through the actual math, it turns out that $r$ in that formula has to be $r_{pred}$, not $r_{meas}$. But that formula only works for the mass (for the case of constant density). It doesn't work for the volume--see below.

Ah! My bad - typo on my end. Yes it's $r_{pred}$

That is not the volume. The volume doesn't obey Euclidean formulas no matter which r you use. You have to obtain it by doing an integral.

I know now that that's not the volume - I'm saying that's the wrong impression I had before posting this thread and then you guys clarified it. I mentioned in my last post that the statement in quotes was the wrong impression I had due to not knowing GR specifics (since I just started on the subject).

 

[Vanadium 50]

if that helps

It does. In my notation, the answer is m/6 or about a millimeter for the earth.

I am surprised that the spatial difference is independent of distance: if you were to go far from Earth and measure the difference in circumference between two nearby circular paths around the Earth, you'd measure the mass. And the circles wouldn;t have to be complete. That strikes me as unexpected.

 

[DrGreg]

if you were to go far from Earth and measure the difference in circumference between two nearby circular paths around the Earth, you'd measure the mass.

But

... a measured radius ... of $$\sqrt{\frac{r_g^{3}}{r_S}}\sin^{-1}\left(\sqrt{\frac{r_S}{r_g}}\right)= r_g+\frac 16r_S+O\left(\frac{r_S^2}{r_g}\right)$$

implies that $\Delta r_\text{meas}= \Delta r_g \left( 1 + O\left(r_S^2 / r_g^2 \right) \right)$

 

[Vanadium 50]

That's exactly right. The effect on space (not spacetime) is a constany very far away from the source. It never gets arbitrarily flat.

This bugs me, but it's not unprecedented. Like time dilation, it says that the space distortion goes as the potential and not any of its derivatives.

 

[Shirish]

The explicit calculation in my #3 comes out to a measured radius (in the sense of a tunnel to the center) of $$\sqrt{\frac{r_g^{3}}{r_S}}\sin^{-1}\left(\sqrt{\frac{r_S}{r_g}}\right)= r_g+\frac 16r_S+O\left(\frac{r_S^2}{r_g}\right)$$if that helps. That matches Feynman.

I asked the same question on stackexchange here, and one answer does agree with what you're saying. The top rated answer takes a different approach though that involves a small perturbation to the radius. Just out of curiosity, is that approach also correct?

 

[A.T.]

The effect on space (not spacetime) is a constany very far away from the source. It never gets arbitrarily flat.

The local space very far away from the source gets arbitrarily flat. But as long you consider an area of space that still includes the source the effect persists.

This is similar to the curvature of a cone, which is zero unless you include the apex in the considered area.

 

[Ibix]

I asked the same question on stackexchange here, and one answer does agree with what you're saying. The top rated answer takes a different approach though that involves a small perturbation to the radius. Just out of curiosity, is that approach also correct?

Yes. It's just doing the approximationin a different way.