Interstellar Medium Shielding

It might surprise you that you need to shield your ship from the interstellar medium, specifically at velocities greater than 40% of 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.

The main danger is heating and not erosion – erosion is insignificant enough that a 1 cm thick carbon shield can go 25,000 light-years (at a speed regime of 0.3 c), ignoring that not all particles displaced will be lost to space, instead landing back on the shield.

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 10^6 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.

Erosion

Erosion is not taken to be a significant component of the danger in interstellar shielding. For example, at 0.3 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.

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:

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

• where $\rho$ is the mass density of the interstellar medium
• where v is the velocity of the ship
• where c is the speed of light
• where $\gamma$ is gamma, calculated with:

$\gamma =1/sqrt(1-v^{2})$

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/m^2.

Calculating the Temperature of the Forward Shield

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

$\phi _{e}=A_{d}\epsilon \sigma _{sb}T^{4}$

• where $\phi _{e}$ is the radiant power (interstellar medium flux)
• where $A_{d}$ is the radiating/absorbing surface area
• where $\epsilon$ is the emissivity of the radiating/absorbing material
• where $\sigma _{sb}$ is the stefan boltzmann constant
• where $T$ is the temperature of the material

Now we rearrange the equation to solve for temperature:

$T={\sqrt[{4}]{\phi _{e}}}/({\sqrt[{4}]{A_{d}}}{\sqrt[{4}]{\epsilon }}{\sqrt[{4}]{\sigma _{sb}}})$

Before we can solve the equation for temperature, we have to consider that the absorbing surface area may not be equal to the total surface area of the forward portion. Thus, we undergo an additional step in calculating the radiant power.

$\phi _{e}=IA$

• where $I$ is the interstellar medium flux
• where $A$ is the area exposed to the interstellar medium flux

Conclusions

A calculator for interstellar medium shielding is provided here:
[link?]

Below are two example tables:

Velocity and Flux Table

(assuming particle density of 1 particle per cubic centimeter)

Ship Velocity	Interstellar Medium Heat Flux
0.1c	20.122 w/m2
0.2c	167.283 w/m2
0.3c	603.483 w/m2
0.4c	1579.947 w/m2
0.5c	3549.648 w/m2
0.6c	7451.7 w/m2
0.7c	15,593.049 w/m2
0.8c	35,326.58 w/m2
0.9c	106,195.696 w/m2
0.99c	1,698,225.484 w/m2
0.999c	18,973,260.691 w/m2

Shield Temperature Table

Assuming a cylindrical shape, radius of 10 meters and thickness of 1 meter

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

https://en.wikipedia.org/wiki/Stefan%E2%80%93Boltzmann_law
Stefan Boltzmann Law

https://en.wikipedia.org/wiki/Interstellar_medium
Reference for ISM density and composition

https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29
Reference for ice sublimation

https://en.wikipedia.org/wiki/Carbon
Reference for graphite sublimation. I assume the point occurs at a lower temperature due to lower pressure.

Derivation of the Relativistic Flux Equation

1. The kinetic energy of an amount of mass is given by $(\gamma -1)mc^{2}$. To get the power, the kinetic energy is differentiated against time and thus assuming constant velocity, obtain that with ${\dot {m}}$ (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} .

1. 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.

1. 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 math alt=> A</math> in the earlier expression.