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I'm wondering what it would be like to touch or interact with very dense degenerate matter like this found at white dwarfs. I understand that white dwarfs are initially very hot, but for the sake of this question, let's assume that we can get close enough to a white dwarf to "test" its matter. I also understand that at these temperatures and densities, the matter is in plasma state, so my question here is basically about how "hard" a very dense plasma is.

My intuition tells me that more density should mean more hardness (act more like a solid). But that doesn't work once I think about trying to stick something like a wooden rod once into Mercury (Hg) and once into Iron. Because although Mercury has higher density than Iron, it doesn't "feel" as hard as Iron when I try to penetrate them with the rod, and so I concluded that density might be irrelevant when determining hardness.

Now, let's imagine we have a cube of degenerate matter extracted from a WD that keeps all its properties of density, degeneracy, pressure and temperature although it is outside the WD, and let's say it is as dense as the average density of a WD, which is $10^6-10^7$ $g/cm^3$.

So my quesitons are:

1- How would matter in this cube react when I try to penetrate it with the rod ? Would it be easy or difficult to penetrate it ?

2- How much softer or harder would it feel compared to something like Steel ?

ProfRob
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Abanob Ebrahim
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1 Answers1

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Whilst the question is I think largely a matter of conjecture, as I explain below, I think the answer is much harder than the hardest steel.

Matter exists in basically two phases inside a white dwarf: The outer part of a white dwarf is a gas, albeit a very dense one, consisting of a degenerate gas of electrons and a non-degenerate gas of completely ionised ions. The inner part can (if it has cooled sufficiently to below a few million K) be a degenerate electron gas with a crystalline form of an ionic solid.

In either case, the pressure exerted by the electron gas is vast compared with our every day experience of gases. For example, at densities of $10^{10}$ kg/m$^3$, the electron gas in a white dwarf exerts a pressure of $\sim 10^{23}$ Pa. The kinetic energies of particles in the gas are so high, that your question is almost like asking what does it feel like to touch an explosion. The material would have to be contained, if not by a white dwarfs gravity, then by some sort of super-container to stop it just exploding. Any "surface" to the material would just instantly explode, so you wouldn't be able to touch white dwarf material, even if it were cold.

However in terms of pushing a rod through it - how hard can you push? At those pressures, if the rod had a cross-section of 1 square cm, then the gas is pushing back with a force of $10^{19}$ N.

Another way of thinking about it - the "compressibility" of a substance is defined as the inverse of the bulk modulus $K$; where $$K^{-1} = \frac{1}{\rho} \frac{d\rho}{dP}$$

In the lower density regions of a white dwarf (assume carbon), the equation of state (for non-relativistically degenerate electrons) is $$ P_{deg} = \frac{h^2}{20m_e}\left(\frac{3}{\pi}\right)^{2/3} \left(\frac{\rho}{2 m_u}\right)^{5/3} \simeq 3.2\times 10^{6} \rho^{5/3},$$ in SI units, and $$ K = \rho \frac{dP}{d\rho} \simeq 5.4 \times 10^{6} \rho^{5/3}\ \ Pa$$

So at white dwarf densities of $\sim 10^{10}$ kg/m$^3$, the compressibility is $\sim 4\times 10^{-24}$ Pa$^{-1}$.

By comparison, the compressibility of steel is about $6 \times 10^{-12}$ Pa$^{-1}$.

White dwarf material is about 12 orders of magnitude less compressible than steel.

In response to Chris White's comment: It might be thought that the shear modulus would be a more appropriate thing to consider. The shear modulus of a perfect fluid is zero. However, if you push a fluid aside, it has to go somewhere else. Imagine a container of mercury, with shear modulus approximately zero. If you push a rod into it, this is not effortless because you have to displace the mercury in the Earth's gravitational field. The opposing force is $\rho g$ times the volume of rod inserted, which is the depth into the mercury times the cross-sectional area of the rod. i.e. It is $h \rho g$ (the pressure in the mercury) times the cross-sectional area of the rod.

OK, but you might argue let's do this somewhere with $g=0$. But you can't get away with that because the white dwarf material has to be contained and under pressure, even at it's "surface", otherwise it would explode. So you therefore have to have it in a fixed volume or constrained by an immense gravitational field. Either way, even though the shear modulus might be zero, the incompressibility ensures that you need a massive force to insert a rod into it.

ProfRob
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