Interstellar Medium Shielding

It might surprise you that you need to shield your ship from the interstellar medium, especially as velocities approach c. This is a result of interstellar space being filled with a diffuse medium of mostly hydrogen, which when relative to a ship at high enough velocities, comes to increasingly resemble ionizing radiation. To boot, the medium also bears a not insignificant component of dust grains, making up 1% of the total mass of the medium on average.

The main dangers are particle-induced heating and erosion from dust grains. Erosion from particles like typical hydrogen atoms is utterly insignificant -- enough that a 1 cm thick carbon shield can go 25,000 light-years (at a speed regime of 30% of c). However, heating proves to be a significant concern, and erosion from dust grains even more so!

Interstellar Medium Density

To begin with, the interstellar medium density varies greatly, ranging from 10^-4 particles per cubic centimeter in the coronal gas component of the galactic halo of the Milky Way, to 10⁶ particles per cubic centimeter in molecular clouds.

This is important in calculating the flux that the forward portion of the ship will receive at a particular velocity.

Particle Density Table

(In units of particles per cubic centimeter)

Component	Particle Density
Molecular clouds	102-106
H II regions	102-104
Cold neutral medium	20-50
Warm neutral medium	0.2-0.5
Warm ionized medium	0.2-0.5
Coronal gas (Hot ionized medium)	10-4-10-2

The local neighborhood around the sun is assumed to have a particle density of 1 particle per cubic centimeter on average.

Interstellar Medium Composition

By mass, the interstellar medium is 70% hydrogen, 28% helium and 2% heavier elements.
By number of atoms, the interstellar medium is 91% hydrogen, 8.9% helium and 0.1% heavier elements.

There is also a dust component to the interstellar medium; the dust is considerably more dangerous than the diffuse gases as the particles are much larger. In the interstellar medium immediately around the Solar System, the mass of dust is ~0.5% of the mass of the gas, with the bulk of the particles ranging from 1E-18 to 1E-14 kg; however, the population of less-numerous but larger particles which pose the greatest hazard is not yet well known.

Erosion from particles

Particle-induced erosion is not taken to be a significant component of the danger in interstellar shielding. For example, at 30% of c a ship's forward shield will encounter 1E+18 ISM particles per square centimeter per light-year traveled (ignoring differences in ISM density through the journey).
A light year contains 946.1 quadrillion centimeters. In that length, there are thus 946.1 quadrillion cubic centimeters, and assuming a particle density of one per cubic centimeter, there are 9.467E+17 particles in that volume, rounding up to 1E+18.

If each impact displaces 2 atoms from the shield, every light year traveled will cause the loss of 2E+18 atoms per square centimeter. For carbon shields, this is a loss of 40 micrograms per light year per square centimeter, ignoring that not all particles displaced will be lost to space, instead landing back on the shield.

To get the mass loss rate, 2E+18 times the atomic mass of carbon gives 40 micrograms.

This means that a 1 cm thick shield, can survive a trip of 56,250 light-years before being worn through. At high relativistic velocities however, space-time contraction is significant enough that the effective ISM density increases.

The density of carbon, times the length, divided by mass loss rate gives the max trip length due to erosion.

Note: In the Daedalus report, a number of other mass loss factors and average of a variety of material choices gave a mass loss rate of 80 milligrams per cubic centimeter per light year at a speed of 25% of c.

According to , at a certain velocity regime (5% to 20% of the speed of light), impacting particles may have additional erosive effects by leaving "ion tracks". These tracks are essentially trails of damaged material left in the wake of the ion, which has penetrated deeply into the material. However, there are numerous issues with the paper, according to M. Karlusic's comment , and the ion track effect may not even apply if the shield is made out of conductive metals.

Another paper, , argues that hydrogen and helium atoms at relativistic velocity implant themselves in the material, becoming slowly diffusing gas atoms. These then cause damage through bubble formation, blistering and exfoliation.

However, even in both cases, the erosion is limited to on the order of a millimeter depth every 4 light years travelled (as these papers generally cover the case of a Breakthrough Starshot spacecraft journeying to Proxima Centauri, the closest star). For relatively large starships, these concerns may not matter much. Extrapolating from the rate, for a meter of material to be eroded the starship would need to travel 4,000 light years.

Dust Collisions

Interstellar dust grain density ranges from a few hundred to a few thousand grains per cubic kilometer. For the rest of this section, we'll assume a density of 500 grains/km³. This translates to $5\cdot 10^{-7}$ grains per cubic meter. Note that the interstellar dust cloud which Earth is moving through, has an order of magnitude higher density -- $10^{-6}$ dust grains per cubic meter.

A light year contains 9.454 quadrillion cubic meters, so for our ship there would be ~4,727,127,478 dust grains per square meter per light year.

According to , the mean mass of a dust grain is $3\cdot 10^{-16}$ kilograms.

Calculating Collision Effects

Note: For collisions, there are various regimes which govern the response of the material after being hit by an impactor. At low velocities (below many kilometers per second), the regime is hydrodynamic. For "hypervelocity" impacts, these matters are entirely governed by the crater regime - in which the impactor leaves a crater in the material. On the extreme end -- the ultra-relativistic regime, impactors are so penetrating that they end up being more like big, mega-bunches of particles leaving cones of primarily radiation and thermal damage.

In the purview of this article, velocities tend to fall between 1% to a hair under c. However, we have entirely no idea of what the intermediate case is like -- when you're already many thousands of km/s, but still below the relativistic regime? We can only guess -- so bear in mind the following is pure and utter conjecture.

Now, dust grain impacts might end up as a hybrid between the hypervelocity crater and the ultra-relativistic "cone". With increasing velocity, the dust grains will penetrate deeper and deeper, along with secondary showers and exotic effects at such high kinetic energies. The original, roughly hemispherical/parabolic crater shape might change to resemble more that of a cone with increasing velocity.

Let's assume that the hypervelocity crater regime still holds for the most part, bearing in mind that up to some high fraction of c, or more generously, when gamma is a large multiple of 1, it will cease to be even slightly accurate.

An approximation for calculating the excavated volume from a dust grain impact is:

$V_{exc}\approx {\frac {E_{k}}{3\cdot \sigma Y}}$

• where $V_{exc}$ is the volume excavated by the impact
• where $E_{k}$ is the kinetic energy of the dust grain
• where $\sigma Y$ is the yield strength of the material being impacted.

If we approximate the crater as a hemisphere, we can calculate the depth:

$D_{c}\approx {\sqrt[{3}]{\frac {0.159\cdot E_{k}}{\sigma Y}}}$

• where $D_{c}$ is the depth of the impact crater.

The yield strength is the amount of tensile stress you can put on a material until it permanently deforms.
Note: Some materials do not have a yield strength that is below their ultimate tensile strength -- in which case, you may substitute the earlier for the latter.

Prospects for Shield Materials

Contrary to popular discussions of dust impacts -- fanciful imaginations conjuring up multi-tonne TNT explosions on the shield, kinetic energies for these dust grains are much lower than thought. Even at 90% of the speed of light, grain impacts are barely comparable to a bullet. In other words, it's as if someone was firing a bullet at you every tenth of an AU (from dust density). Although the yield strength of ice isn't particularly well measured, it should be roughly 0.1 MPa. Given this, and a velocity of 0.1 c (the velocity for which ice is still a solid, see latter sections):

~32.1 meters / light year travelled $\approx {\sqrt[{3}]{\frac {0.159\cdot 0.136J}{0.1MPa}}}\cdot 4727127478$ grains/m³/ly $/190385$ craters/m²

In other words, the ice shield erodes by 32.1 meters amount of thickness for every light year travelled. What does that $190385$ craters/m² figure mean? Well, if this weren't there - the shield erosion would be astronomically thick - about 6.1 million meters. Fortunately, that's if every impact was being concentrated on the exact same point - and here, assuming an uniform distribution, it would be spread over the whole square meter. That 190,385 number is just measuring the amount of craters that can fit in one square meter.

We can get this from crater depth with the formula of $\pi \cdot D_{c}\,^{2}$, the classic formula for circle area. Then we just divide 1 square meter by that, and there you go. Craters per square meter.

For our very best modern carbon fiber, the results are much much better:

0.046 meters / light year travelled $\approx {\sqrt[{3}]{\frac {0.159\cdot 0.136J}{7GPa}}}\cdot 4727127478$ grains/m³/ly $/15009474$ craters/m²

It takes a trip of 24 light years just to get a 1 meter to be eroded. Carbon fiber really is an amazing material.

Now of course, keep in mind we are making oversimplifying assumptions like the crater ejecta not landing back on the shield, a point raised in the section about particle erosion. Also, we are not accounting for Lorentz length contraction, which will increase the grain density as velocity approaches the speed of the light. Likewise, erosion will only get worse as velocity increases.

Calculating the Heat Flux

Before we can begin calculating the flux, the mass density of the interstellar medium first be known.
The mass density is given by:

$\rho =mp$

• where $\rho$ is the mass density of the interstellar medium
• where $m$ is the mass of the particle
• where $p$ is the particle density of the interstellar medium

Since the interstellar medium is not homogeneous, a weighted average must be done per the composition of the interstellar medium. We can assume that all of the heavier elements are iron atoms as an approximation.

$1.475\cdot 10^{-27}$ kg (average mass) = $(0.7\cdot H+0.28\cdot He+0.015\cdot Fe)/3$

• $H,\,\,He,\,\,Fe$ respectively refer to the atomic masses of hydrogen, helium and iron.

Now we can finally calculate the flux with the relativistic flux equation ^[7]:

$\phi =\gamma \rho vc(\gamma -1)c^{2}$

• where $\phi$ is the interstellar medium flux
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \rho} is the mass density of the interstellar medium
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle v} is the velocity of the ship
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle c} is the speed of light
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma} is gamma, calculated with:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma = 1/\sqrt{1-v^2}}

Assuming a particle density of 1 particle per cubic centimeter, at 41% of the speed of light, the flux is comparable to what Earth receives from the sun, already enough for ice to begin melting. At 80% of the speed of the light, the flux is 35,327 W/m².

Calculating the Temperature of the Forward Shield

The temperature of the forward portion is given by the Stefan Boltzmann Law :

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_e = A_d \epsilon \sigma_{sb} T^4 }

• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_e } is the radiant power
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A_d} is the radiating/absorbing surface area
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \epsilon} is the emissivity of the radiating/absorbing material
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \sigma_{sb}} is the stefan boltzmann constant
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T} is the temperature of the material

Now we rearrange the equation to solve for temperature:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle T= \sqrt[4]{\phi_e} / (\sqrt[4]{A_d} \sqrt[4]{\epsilon} \sqrt[4]{\sigma_{sb}})}

Before we can solve the equation for temperature, the radiant power must be obtained from the interstellar medium flux, given by:

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \phi_e = IA}

• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle I} is the interstellar medium flux
• where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} is the area exposed to the interstellar medium flux

Conclusions

A calculator for interstellar medium shielding (heat flux only) is provided here:
Interstellar shielding calculator

Below are two tables:

Example Required Shield Thickness Table

(Assuming mean mass of Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle 3 \cdot 10^{-16}} kilograms.)

Ship Velocity	Dust Grain Kinetic Energy (J)	Ice (m/ly t)	Carbon Fiber (m/ly t)
0.1c	0.136	32.1	0.046
0.2c	0.556	Beyond sublimation point, see shield temperature table	0.188
0.3c	1.302		0.439
0.4c	2.456		0.829
0.5c	4.171		1.408
0.6c	6.741		2.276
0.7c	10.793		3.644
0.8c	17.975		6.069
0.9c	34.894		11.782
0.99c	164.17		55.432
0.999c	576.09		194.518

Example Shield Temperature Table

Assuming 1 particle per cubic centimeter, a cylindrical shape, radius of 10 meters and thickness of 1 meter

Ship Velocity	Interstellar Medium Heat Flux (W/m2)	Ice Temperature (K)	Graphite Temperature (K)
0.1c	20.122	113.558	123.207
0.2c	167.283	192.824 Beyond sublimation point	209.208
0.3c	603.483	265.744	288.325
0.4c	1579.947	338.033 Too hot even at standard pressure	366.756
0.5c	3549.648		449.017
0.6c	7451.7		540.481
0.7c	15,593.049		650.054
0.8c	35,326.58		797.521
0.9c	106,195.696		1050.131
0.99c	1,698,225.484		2099.982
0.999c	18,973,260.691		3839.303 Beyond sublimation point

Notes: Ice has an emissivity of 0.97, while Graphite has an emissivity of 0.7

The parameters vary with changing exposed area, area and emissivity, flux. What is clear here is that the interstellar medium flux can present a significant danger at high enough velocities as to sublimate (in the vacuum of space) ice, and at ever increasing velocity, even graphite.

Therefore, care must be taken to shield your interstellar spacecraft from the flux if it is moving at a velocity high enough to heat the spacecraft with disastrous consequences.

Additional References

Derivation of the Relativistic Flux Equation

1. The kinetic energy of an amount of mass is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\gamma -1)mc^2} . To get the power, the kinetic energy is differentiated against time and thus assuming constant velocity, obtain that with Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \dot{m}} (the mass flow rate); the power is given by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle (\gamma -1)\dot{m} c^2} .
2. The mass flow is given by the mass per volume encountered every second, doing this in the reference frame of the ship, the ISM density is length contracted to to Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \gamma \rho} and multiply by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle Av} where Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} is area.
3. This yields Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle P = \gamma \rho Av(\gamma -1)c^2} , to obtain the flux per unit area divide by Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} and thereby cancel the Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle A} in the earlier expression.

Reference for the Derivation of the Relativistic Flux Equation

Credit

To Tshhmon for writing the article

• To lwcamp for helping with particle erosion calculation
• To Rocketman1999 for helping with dust erosion distribution
• To Kerr for the relativistic flux equation and derivation