# Diameter of the earth with GRT

1. Dec 16, 2014

### exmarine

I think I read somewhere that Feynman once said that if we could measure the earth’s diameter, it would be maybe an inch short of what we would expect from the circumference and Euclid. Is that correct? (Did he actually say that?) It seems to me that it would be a tad longer?? I see a clock lower than me running slower than my clock. So in 10 seconds, it measures or indicates only 9 seconds say. Since the metric matrix coefficient of the radial coordinate is the reciprocal of the time coefficient, then wouldn’t a (radially oriented) 10 inch ruler below me measure or indicate say 11 inches? I can watch clocks measure time from a distance, but how do you compare distances from a distance?

2. Dec 16, 2014

### A.T.

3. Dec 16, 2014

### A.T.

I'm not sure how a 10 inch ruler can indicate 11 inches

You don't. You compare the circumference and diameter, both measured with local rulers laid along them.

Last edited: Dec 16, 2014
4. Dec 16, 2014

### Staff: Mentor

No. The proper diameter (the diameter we would measure with rulers) is longer than the circumference divided by $\pi$. A brief calculation follows.

Looking back at the thread A. T. linked to, I realized that, while Jonathan Scott's formula for the potential (of a uniform density sphere) is correct, his assumption that the potential is what you need to calculate the proper diameter is wrong. What you need is the metric coefficient $g_{rr}$; the potential is $g_{tt}$ .

Fortunately, the metric coefficient $g_{rr}$ is actually simpler to work with, particularly if we are willing to make a rough estimate. This metric coefficient is given by

$$g_{rr} = \frac{1}{1 - \frac{2 G m(r)}{c^2 r}}$$

where $m(r)$ is the "mass inside radius $r$ ". The square root of this value gives the ratio of proper distance to coordinate distance at radius $r$. Since we always have $g_{rr} \ge 1$ for $r \ge 0$ , we can already see that proper distance will always be at least as large as coordinate distance, and will be strictly larger for any $r$ greater than zero and less than infinity. But the order of magnitude of my estimate should be correct.

Since $m(r)$ is zero at $r = 0$ , and is just $M$ , the total mass of the Earth, at $r = r_e$, we can do a quick rough estimate by just averaging the two values of $\sqrt{g_{rr}}$. One value is just 1; the other value, plugging in numbers, comes out to about $1 + 2 \times 10^{-9}$. So averaging tells us that the proper diameter of the Earth is larger than the coordinate diameter (i.e., the circumference divided by $\pi$ ) by about $10^{-9}$ times the coordinate diameter, or about 1 centimeter. Note that this is a rough estimate; a more accurate calculation would have to first obtain the function $m(r)$ by integrating the density from $r = 0$ outward, and then evaluate the integral $\int_0^{r_e} \sqrt{g_{rr}} dr$ using the function $m(r)$ that was obtained.

5. Dec 16, 2014

### A.T.

This is $g_{rr}$ of the exterior metric. But we want to know the diameter measured inside of the mass. Shouldn't we use the interior metric for this? What is wrong with the approach I proposed in the other thread (intergrate the interior $g_{rr}$ from 0 to R)?

Here the interior metric given in an old thread:

6. Dec 16, 2014

### Staff: Mentor

Not just the exterior. Note that I put in $m(r)$, the "mass inside radius $r$ " function. With that included, the $g_{rr}$ I wrote down is valid everywhere in a spherically symmetric spacetime. MTW shows this in one of their sections on the Schwarzschild metric; that's the derivation I'm most familiar with.

This gives the same $g_{rr}$ I wrote down, but with the appropriate form of $m(r)$ for a sphere of uniform density: $m(r) = M r^3 / R^3$, where $M$ is the total mass and $R$ is the surface radius. You could certainly integrate this from $r = 0$ to $r = r_e$ to get a more accurate result; I was just too lazy to write down the complicated expression for the integral. ;)

7. Dec 17, 2014

### Jonathan Scott

Yes, I agree. Thanks for pointing that out.

8. Dec 17, 2014

### exmarine

That's what I thought. My question was actually simpler than that other thread: Is the diameter longer or shorter than we would expect with Euclid?

Thanks!

9. Dec 17, 2014

### A.T.

With Euclid (no curvature) we would expect diameter = circumference/pi. The spatial curvature inside a massive sphere is positive, so diameter > circumference/pi

Last edited: Dec 17, 2014