# Real Earth Diameter

1. Dec 4, 2014

### keinstein

Hello,

Gravitational lenght contraction inside Earth means, the measuring rod is also shortened and in consequence inside is more space than according to the mathematical circumference of an empty geoid. I´d like to know something about this discrepance: mathematical diameter - physical diameter (Earth)

Which lenght has the real inside Earth diameter according to the SRT ?

2. Dec 4, 2014

### A.T.

You have to integrate the interior Schwarzschild metric to get an approximation based on a uniform density non-rotating sphere. But the actual Earth is of course neither of those things, so there are only numerical approximations possible.

3. Dec 4, 2014

### Jonathan Scott

For the weak field approximation, which certainly applies to anything in the solar system, the fraction by which time is dilated and rulers are effectively shortened relative to infinity is given by the Newtonian gravitational potential in dimensionless units, which is $-Gm/rc^2$ at distance $r$ from the middle of mass $m$. By entering "(G * mass of earth) / (radius of earth * c^2)" into Google, I get that the fractional change in clocks and rulers is about $6.9530712 \times 10^{-10}$. Multiplying that by the radius of the earth, that means that the sort of difference that makes is equivalent to about 4.4mm in the radius.

You can't say exactly what the difference is without choosing your coordinate system, which is effectively your choice of how you want to map curved local space to a flat coordinate system, as this is not determined by the physics but rather by choice of convention. However, the above calculation indicates the order of magnitude of the effect.

4. Dec 4, 2014

### A.T.

That is the potential outside of the Earth.

If I understand the question correctly, it is about the proper-circumference of the Earth / 2PI vs. the proper-radius of the Earth. This should be independent of coordinates.

5. Dec 4, 2014

### Staff: Mentor

I think that the issue here is not so much choice of coordinate system as it is how carefully the problem is phrased. OP is asking for the ratio between the proper distances along two different paths between the same two points; that ratio is clearly not coordinate-dependent.

The coordinate-dependent tricky part is that we have to specify the two points at the same time, and that requires a simultaneity convention (or equivalently, a choice of time coordinate). I expect that OP was assuming that the selection of the two events was so obvious as not to need mentioning.

6. Dec 4, 2014

### A.T.

You mean because the Earth rotates? Can't we just layout ideal rulers along the circumference and diameter, to get a coordinate independent answer?

7. Dec 4, 2014

### Jonathan Scott

I didn't find the question that clear, but the theoretical difference between the proper radius and the radius is still of the order of magnitude of 4.4mm. If we assume uniform density for the earth, use Schwarzschild coordinates and use the weak field approximation to assume that coordinate $dr$ is approximately $(1+Gm/rc^2)$ times local $dr$, where $m$ is the mass inside radius $r$, assumed to be proportional to $r^3$, then I think the factor is 1/3, giving about 1.48mm as the answer.

Edit: This is wrong. I've failed to take into account the effect on the metric of the mass outside the given radius, which means the factor is at least 1.

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

### Staff: Mentor

That works, and is doubtless what OP would say if pressed to completely specify what he was asking for.

I was referring to a more general problem: given two worldlines representing the endpoints of something, what is the length of that something? You have to choose a point on each worldline "at the same time". In this case of course there's a simultaneity convention so natural that it would be perverse to use any other.

9. Dec 4, 2014

### Jonathan Scott

OK, trying again.

For a hypothetical uniform sphere (which the earth isn't really) of mass $m$ and radius $r_e$ I make it that the potential at radius $r$ is as follows: $$- \frac{G m}{r_e} \left ( \frac{3}{2} - \frac{1}{2} \frac{r^2}{{r_e}^2} \right )$$
That means that the difference between proper and coordinate radius is approximately given by $$\int_0^{r_e} \frac{Gm}{r_e c^2} \left ( \frac{3}{2} - \frac{1}{2} \frac{r^2}{{r_e}^2} \right ) dr = \frac{Gm}{r_e c^2} \left [ \frac{3}{2} r - \frac{1}{6}\frac{r^3}{{r_e}^2} \right ]_{r=0}^{r=r_e}$$ $$= \frac{Gm}{c^2} \left ( \frac{3}{2} - \frac{1}{6} \right ) = \frac{4}{3} \frac{Gm}{c^2}$$

Feel free to check my maths.

10. Dec 4, 2014

### pervect

Staff Emeritus
To get the interior Schwarzschild geometry, you'll need to consider m(r). If you look up the interior Schwarzschild metric development, you'll see an equation of the sort $m(r) = 4 \pi r^2 \rho$, which we don't see in JS's calculations. In general $\rho$ could be a function of depth, too, the usual textbook assumption is the unrealistic one that the density is independent of depth (equivalently, that the equation of state that gives density vs pressure is independent of pressure).

[add]That does the non-rotating case, I don't know how you'd do a rotating case. You could try some form of linearized gravity maybe.

This calculation is a bit too messy for me to want to do, though I can say that if you embed the geometry in the equatorial plane, you get a spherical geometry inside the Earth, and Flamm's paraboloid outside the Earth.

I think a lot of the assumptions made by the original poster are highly suspect, in particular the existence of any truly meaningful form of "gravitational length contraction". We see this term a lot in posts, by people who assume it exists for some reason, but I've never seen it in the literature. Possibly I just haven't read the correct literature, it's not impossible I could have missed something. Howver, I continue to view the term with much suspicion until I see a literature reference though.

I believe that AT's and JS's reformulation of the OP's question in terms of proper circumference and proper radius makes the question answerable. It remains to be seen if that reformulated question was actually what the OP wanted to ask.

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

### Jonathan Scott

I assumed the Newtonian potential as within a uniform sphere, with constant $\rho$. One way of calculating it is to note that the field at a given radius can be calculated trivially by ignoring everything outside that radius (it's simply proportional to the radius) so one can integrate that to get the potential relative to the surface. The more long-winded way is to integrate the potential due to stuff outside radius r (which is equal to the potential at the origin if there was a hole of radius r, so it can be integrated over the spherical shells) and add it to the potential due to the stuff inside radius r (which is equal to the potential if all of the mass inside radius r were at the center). As I'm a bit error-prone I calculated it both ways and checked that they gave the same result and that they passed sanity checks at the surface and center, but that doesn't necessarily guarantee that I got it right.

12. Dec 8, 2014

### keinstein

I thought it was more than an ants length. As not an nomenclature expert, Sorry of this wrong formed term of a non-existing "gravitational lenght contraction" I meant "length" and maybe distortion of spacetime metrics through the influence of gravity or whatever else - it can´t even affect GPS (standard accuracy 15m ;)

Thanks a lot!

13. Dec 9, 2014

### Staff: Mentor

Your result for the potential looks the same as the result given in MTW for a uniform density sphere.

14. Dec 16, 2014

### Staff: Mentor

In the course of another thread, I revisited this computation, and realized that I should have pointed out two items:

(1) The value of $G m / c^2$ for the Earth is about 9 mm, so 4/3 of that is about 12 mm, or 1.2 cm.

(2) More importantly, in order to calculate the proper radius of the Earth, the key metric coefficient is not $g_{tt}$ , it's $g_{rr}$ . Numerically, this doesn't affect the result much, but it's the principle that counts. ;) I posted a brief estimate using $g_{rr}$ in the other thread.

15. Dec 16, 2014

### Jonathan Scott

I think you've taken the Schwarzschild radius which is $2 G m/c^2$, a factor of 2 high.

16. Dec 16, 2014

### Staff: Mentor

Oops, you're right. In my other post I was just estimating the order of magnitude, so there was probably more than a factor of 2 slop in the numbers anyway. But your value of about 6 mm in this thread is correct for what you were calculating (based on the potential difference).

17. Dec 16, 2014

### Jonathan Scott

Of course the change in diameter was twice that anyway; that was obviously what you were calculating instead!

18. Dec 18, 2014

### Jonathan Scott

My first calculation used the integral of the potential due to the mass inside the given radius, which as it happens is the same as the factor from $g_{rr}$, so I think my previous two mistakes cancel and my first answer of 1.48mm was correct. Unless, of course, I've just made a third mistake.

19. Dec 18, 2014

### Staff: Mentor

You mean 4.4mm, correct? That's what you gave in post #3. This is basically the same rough estimate that I gave in the other thread (except that I rounded the fractional change of about $7 x 10^{-10}$ up to $10^{-9}$ to get an even rougher estimate than yours).

20. Dec 18, 2014

### Jonathan Scott

No, what I meant is with the factor of 1/3, as follows (but still assuming I haven't made another mistake):

The exact radial factor is $1/\sqrt{1-2Gm(r)/rc^2}$ where $m(r)$ is the mass inside radius $r$. In this case, this can be approximated as $1+Gm(r)/rc^2$. Assuming (somewhat inaccurately) uniform density, the mass inside radius $r$ can be represented as $M (r^3/R^3)$ where $M$ is the overall mass of the earth and $R$ is its radius. This means that the proper radius is as follows: $$\int_0^R \left ( 1 + \frac{G (M r^3/R^3)} { rc^2 } \right ) dr$$ $$= \int_0^R \left ( 1 + \frac{GM}{R^3 c^2} r^2 \right ) dr$$ $$= \left [ r + \frac{GM}{R^3 c^2} \frac{r^3}{3} \right ]_0^R$$ $$= R + \frac{GM}{3 c^2}$$

(I don't know how to line up the "=" signs for continued equations here).