How many tons of lead is needed to curve space 1 nanometer?

Question

I'm not sure, I think I mean the diameter of a circle around the mass which is really the length of the curve of the well in 2D. How should one word the question? — Jitter, Sep 09 '15 at 01:20
The problem is that measuring space curvature by Nano-meters is very unspecific. Angle of curvature for a light ray passing just outside is more meaningful or, perhaps more accurate, the sum of 3 angles in a triangle. If you want some much more detailed answers, look here: http://physics.stackexchange.com/questions/109731/how-to-measure-the-curvature-of-the-space-time There are ways to measure distance sometimes but you need specifics, for example, Mercury appears to "jump" to the observer and a distance of that "Jump" can be measured, but that's a pretty specific scenario. — userLTK, Sep 09 '15 at 01:48
Oh, you mean, the circumference to radius ratio and enough gravity to add 1 nano-meter to the radius or remove 1 nano-meter from the circumference (2 different answers by the way), but due to gravity the radius to circumference ratio is no longer Pi. The Math gets a little complicated for me. — userLTK, Sep 09 '15 at 01:53
The measurement I am after is the one that would run through a hole through the center of a lead ball. Then I wanted to know Would the curvature make the diameter out by 1 nanometer if you measured it with a rod? — Jitter, Sep 09 '15 at 02:17
@UserLTK. Yes that is what I was after. Then I wanted to know if anyone has ever tried to test it. — Jitter, Sep 09 '15 at 02:22
Related question by OP: http://physics.stackexchange.com/q/205950/2451 — Qmechanic, Sep 09 '15 at 06:29

score 4 · Accepted Answer · edited Apr 13 '17 at 12:39

If you use this answer and its references, you should be able to estimate this easily.

In the notation of that answer, if you measure out a length $\mathrm{d} R$ radially with a measuring rod of length $\mathrm{d} R$, the increase in radial co-ordinate $r$ is given by:

$$\mathrm{d} R = \frac{\mathrm{d} r}{\sqrt{1-\frac{2\,G\,\mathcal{M}(r)}{c^2\,r}}}$$

where $\mathcal{M}(r)$ is the mass contained inside a sphere defined by the radial co-ordinate $r$. In the case of a constant density $\rho$ sphere, the discrepancy between $r$ and a total $R$ measured by you by laying out measuring rods end to end is given by (I'm using the binomial theorem to approximate):

$$R\approx \int_0^r (1+\frac{4}{3\,c^2}\, G\, \pi\, \rho\, {r^\prime}^2)\,\mathrm{d}\,r^\prime$$

so that the discrepancy is estimated by:

$$R-r \approx \frac{4}{9\,c^2}\, G\, \pi\, \rho\, r^3 = \frac{1}{3}\,\frac{G\,M}{c^2}$$

or one sixth the Schwarzschild radius for a mass $M$, meaning you will need quite a few elephants, or tonnes of lead, to see the discrepancy of the magnitude you seek.

You can also glean the same formula from Feynman's elementary introduction to GR in chapter 42 of the second volume of the lectures (equation 42.3). Incidentally, the discrepancy works out to about 1.5mm for the Earth at its surface.

How many tons of lead is needed to curve space 1 nanometer?

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