This question is not a duplicate of Experimental measurement of the radial excess or How to derive exterior and/or interior Schwarzschild solution using Feynman excess radius equation?. The linked questions ask something different even though they refer to the same section in Feynman lectures.

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 radio 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?


2 Answers 2


The method Feynman is employing here is a standard way to do something called perturbation theory. This is a very general idea, with application across physics, not just GR. The concept is that you have a situation where the equations are complicated, but they are close to some simpler equation which you can solve. The idea is that the answer to the problem under investigation will be given by that simpler solution plus a small correction. So how to you find the small correction? Let's use the GR problem as an example.

The simpler model here is the one given by Euclidean geometry. The result is that the surface area is $4 \pi r^2$ so the radius excess is zero.

Next we need the correction. It is given by $G M / 3 c^2$ and $M$ is given by $$ M = \frac{4}{3} \pi(r + \delta r)^3 \rho $$ where the $\delta r$ is related to the curved space correction which we are trying to calculate. This $\delta r$ is not itself the radius excess, but it is a small quantity whose size is of similar order of magnitude as the radius excess. So we put this into the formula for radius excess: \begin{eqnarray} \text{radius excess} &=& \frac{G}{3c^2} \frac{4}{3} \pi(r + \delta r)^3 \rho \\ &=& \frac{G}{3c^2}\frac{4}{3} \pi \left(r^3 + 3 r^2 \delta r + 3 r \delta r^2 + \delta r^3 \right) \rho \\ &\simeq & \frac{G}{3c^2}\frac{4}{3} \pi r^3 \rho . \end{eqnarray} It is the last step here which is what the question is all about. The point is that as long as $\delta r \ll r$ then the approximation here is very good, and in the case of planet Earth it is good at the level of parts per billion.

  • $\begingroup$ So just to clarify, we know that the Euclidean formula for volume is wrong - i.e. $4\pi r_{meas}^3/3$ is NOT the correct volume for earth. BUT since we think that the deviation of the actual volume formula from the Euclidean volume is small, so the actual volume formula is approximated by plugging in $r_{meas}$ plus correction in the Euclidean formula. Is that correct? $\endgroup$
    – Shirish
    Jan 19, 2023 at 16:02
  • $\begingroup$ Yes that's correct. $\endgroup$ Jan 19, 2023 at 16:35
  • $\begingroup$ @Ghoster oh thanks, I see I had several mistakes (wrote it a bit hastily). Thanks for pointing it out. Should be correct now. $\endgroup$ Jan 20, 2023 at 23:26

If we assume uniform density for an appriximation we can use the inner Schwarzschild metric with

$\rm g_{rr}=1/(1-r_s \ r^2/r_g^3)$

then the physical radius $\rm R$ measured with stationary rulers is

$\rm R=\int_0^{r_g} \surd{g_{rr}} \ dr = r_g^{3/2} \ arcsin \sqrt{r_s/r_g} \div \surd{r_s}$

so for the earth with $\rm r_s=0.00887 \ m , \ r_g=6371000 \ m$

we get $+2.32 \cdot 10^{-8} \%$:

$\rm R-r_g \approx 0.001478 \ m$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.