Material strain from spacetime curvature

Let's say that you moved an object made of rigid materials into a place with extreme tidal forces. Materials have a modulus of elasticity and a yield strength. Does the corresponding 3D geometric defect from curvature of spacetime directly cause material strain? How could you calculate that?

An obvious first step for practical geometries would be to take the Gaussian curvature of Flamm's paraboloid (thus avoiding much more GR specific knowledge), luckily, a book already did that for us. Here is the curvature:

$$K = -\frac{r_s}{2 r^3 }$$

The negative indicates hyperbolic geometry.

The problem in concrete terms:

Imagine a location where tidal forces are equal to roughly $(10 \mathrm{m}/\mathrm{s}^2)/(2 \mathrm{m})=5 \mathrm{s}^{-2}$. I pick this value for relationship to human experience. An astronaut could wedge themselves between a ceiling and a floor $2 \mathrm{m}$ apart and experience about $\frac{1}{2}$ Earth gravity pulling their feet and hands in opposite directions. Furthermore, imagine this location as the Innermost Bound Circular Orbit of a black hole. These two constraints determine exactly the necessary mass of the black hole, which is 25,000 solar masses. Not uncommon. The distance of the knife-edge orbit from the singularity would be about the radius of the sun. This would also be perfectly accessible due to the orbital physics of a black hole. With conventional rockets you could throw the ISS into that orbit and retrieve it later, given that your initial trajectory had sufficient accuracy. What I want to know is: would the astronauts' bones break due to tidal forces?

There is another important implication of this question - whether spaghettification is a relevant phenomenon for a rigid object falling into a black hole. Obviously, an unraveled string (oriented in a line intersecting the singularity) falling into a black hole would be spaghettified. But for any object that is roughly spherical (meaning length isn't much greater than width), would the hyperbolicity of space itself exceed the material limits first? Could we construct a criteria that would determine if something is spaghettified or strained to death? We already have the geometric curvature. The object properties of length, width (assuming a coke can shape), modulus of elasticity, yield strain or stress, and black hole mass would seem to be sufficient. Then a radius of spaghettification breaking would follow, along with a radius for the strain breaking. You could then obviously say which one is larger.

Perhaps this approach is fundamentally mistaken in separating geometric stress from gravitational forces. But I don't see how. Consider:

• For tidal force breaking (spaghettification), I only need material yield stress
• For geometric defect strain I only need material yield strain

By dimensional analysis, these two things must be distinct, but maybe there is some other deep reality of general relativity that isn't obvious to me.

• oh no, there was an edit that was lost. I want to include the changes that were suggested, but since the edit was (somehow) rejected I don't know how to get it back. – Alan Rominger Feb 24 '13 at 21:03
• try hitting the back button in your browser – raindrop Feb 24 '13 at 21:04
• your edit seems to have loaded, edited 5 minutes ago by you – raindrop Feb 24 '13 at 21:05
• @Raindrop I know, but I don't want my edit, I want your edit, and it disappeared. It had some math formatting. – Alan Rominger Feb 24 '13 at 21:06
• Related – PyRulez Mar 5 at 9:26

Starting with the basic concept of Gaussian curvature, we look to the metric given in the question, and try to identify some concept of strain from that. This is what hyperbolic curvature looks like for 2D in 3D, and I believe it should extend to the present problem of 3D to 4D without any material change.

From this picture we could imagine calculating the circumference of a circle given the radius from the center point. The basic idea of hyperbolic geometry is that we should find $C'>2 \pi t$, where $t$ is the radius of a circle in the plane. I will assume that $t$ is constant, so that $C'$ is the circumference after moving into the curved region of space. From basic visual inspection, I will suppose that the elevation of the surface follows a sin form as a function of the angle. With this, I can use the formula for arc length of a function to get $C'$. I also substitute $x=t \theta$ to put the integral in terms of the angle.

$$C' = \int_{a}^{b} \sqrt { 1 + [f'(x)]^2 }\, dx = t \int_{0}^{2 \pi} \sqrt { 1 + [f'(\theta)/t]^2 }\, d\theta$$

By integrating this arc length multiplier we can put the circumference in terms of the maximum elevation (I'll call $h$) of the surface at the radius $r$. The integral can't be evaluated easily so I introduce the assumption $t \gg h$ and use a Taylor series of the expression.

$$C' = t \int_{0}^{2 \pi} \sqrt { 1 + (\frac{h}{t} \cos{\theta})^2 }\, d\theta \approx t \int_{0}^{2 \pi} \left( 1+ \frac{1}{2} (h/t)^2 \cos^2{\theta} \right) d\theta$$

As the circle moves from flat spacetime region to curved region, the ratio of the new circumference to the old circumference is effectively the strain along that path.

$$\frac{C'}{C} = \frac{C'}{2 \pi r} = (h/t)^2/4 + 1$$

The definition of Gaussian curvature is the following, where $\kappa$ is a metric for curvature in a single direction, and the inverse of the radius of the circle that would fit the curvature there.

$$K = \kappa_1 \kappa_2 = \frac{1}{R_1} \frac{1}{R_2}$$

I still want to use the metric $K=-r_s/(2 r^3)$, but this doesn't give the breakdown of curvature in one axis versus the other. So I'll just assume they're the same, to give $K=\kappa^2$. Rearrange to find $R=1/\sqrt{K}$. Then, by the idea of a circle in the "extra" dimension, we find $h$ by multiplying the in-plane radius by the rise from moving that far along the edge of the big circle in the extra dimension.

$$h = R (1-\cos{(t/R)}) \approx R \left( \frac{t^2}{2R^2} \right)$$

$$h = \frac{ t^2}{2 R} = \frac{ t^2 \sqrt{ K} }{ 2 } = \frac{ t^2 \sqrt{r_s} }{ 2 \sqrt{2} r^{3/2} }$$

Now we can put the strain in terms of the known quantities. Here I will use the common definition of strain, $\epsilon$.

$$\frac{C'}{C} = \left( \frac{ t \sqrt{r_s} }{ 2 \sqrt{2} r^{3/2} } \right)^2 \frac{1}{4} + 1 = \epsilon + 1$$

$$\epsilon = \left( \frac{ t^2 r_s }{ 8 \times 4 r^{3} } \right) = \frac{1}{16} \frac{ t^2 r_s }{ 2 r^{3} }$$

So that's one part of the answer. As far as we care, all materials have a yield point and this point has a stress and strain associated with it. Practically, you can look up the yield strength and find the yield strain using Young's modulus. If you do this, you will find that many materials have a yield strain on the order of 1%, for things like bone. Ceramics are lower, but not by more than around 1 order of magnitude. Now we compare by doing a simple Newtonian calculation of specific strength (denoting $ss$) from tidal forces.

$$\text{field} = F = \frac{G M }{r^2} = \frac{r_s c^2 }{ 2 r^2}$$ $$\text{tidal} = T = \frac{ r_s c^2 }{ r^3 }$$

Now, let's imagine a constant tidal field, and a material at that location in space. We'll treat it as a linear bar for now, stretched out in the direction of growing tidal forces. If the center is at $t=0$ then the tidal field is $t T$. Integrate this field to find the gravitational potential difference, $1/2 t^2 T$, which is the limit that specific strength quantifies.

$$ss = 1/2 t^2 \frac{ r_s c^2 }{ r^3 } = \frac{ t^2 r_s }{ 2 r^3 } c^2$$

Specific strength has units of velocity squared as well. For bone I calculate about $130,000 m^2/s^2$. Now we can formulate the ultimate criteria for not breaking. As it turns out, the "winner" between strain and stress seems to be non-specific to the conditions. That is somewhat surprising since we had 3 degrees of freedom, the dimension of the object, the distance from the black hole, and the mass of the black hole. These are the criteria to prevent breaking.

$$\frac{ss}{c^2} > \frac{ t^2 r_s }{ 2 r^3 }$$

$$16 \epsilon > \frac{ t^2 r_s }{ 2 r^{3} }$$

$$\frac{ss}{c^2} = 1.45 \times 10^{-12}$$

$$\epsilon = 0.16$$

We see here that the tidal force requirement is much more restrictive. Actually, it's so much more restrictive that basically no thinkable value of Young's Modulus would change the balance.

• Oops! First big error, I think the equation I wrote to give $h$ a value should have $R$ as the front multiplier, not $t$. That's will change things. – Alan Rominger Feb 28 '13 at 20:12
• How the hell has this not got a load of upvotes? – Magpie Mar 4 '13 at 1:40