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. <ref>https://en.wikipedia.org/wiki/Interstellar_medium
 Reference for ISM density</ref>

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)
{| class="wikitable sortable"
|-
! 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. <ref>https://en.wikipedia.org/wiki/Interstellar_medium
 Reference for ISM composition</ref>

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. <ref>H. Kruger et. al., "Sixteen Years of Ulysses Interstellar Dust Measurements in the Solar System. I. Mass Distribution and Gas-to-Dust Mass Ratio", Astrophysical Journal, October 20, 2015. https://ui.adsabs.harvard.edu/link_gateway/2015ApJ...812..139K/PUB_PDF </ref>

=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. <ref>https://bis-space.com/shop/product/project-daedalus-demonstrating-the-engineering-feasibility-of-interstellar-travel/
 Reference for the Daedalus report figure.</ref>

=Dust Collisions=
{{WIPNotice}}
Interstellar dust grain density ranges from a few hundred to a few thousand grains per cubic kilometer<ref>https://openstax.org/books/astronomy/pages/20-1-the-interstellar-medium Reference for interstellar dust grain density.</ref>. For the rest of this section, we'll assume a density of 500 grains/km3. This translates to <math alt=>5 \cdot 10^{-7}</math> grains per cubic meter. Note that the interstellar dust cloud which Earth is moving through, has an order of magnitude higher density -- <math alt=>10^{-6}</math> dust grains per cubic meter<ref>https://ntrs.nasa.gov/citations/20050215611 Reference for the local interstellar dust density.</ref>.

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 <ref>https://link.springer.com/chapter/10.1007/978-1-4419-8694-8_5 Reference for the mean mass of interstellar dust grains.</ref>, the mean mass of a dust grain is <math alt=>3 \cdot 10^{-16}</math> kilograms.

==Calculating Collision Effects==
An approximation<ref>http://toughsf.blogspot.com/2016/03/electric-cannons-and-kinetic-impactors.htmlReference for the approximation equation.</ref> for calculating the excavated volume from a dust grain impact is:

<math alt=>V_{exc} \approx \frac{E_k}{3 \cdot \sigma Y}</math>
* where <math alt=>V_{exc}</math> is the volume excavated by the impact
* where <math alt=>E_k</math> is the kinetic energy of the dust grain
* where <math alt=>\sigma Y</math> is the yield strength of the material being impacted.

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

<math alt=>D_c \approx \sqrt[3]{\frac{0.159 \cdot E_k}{\sigma Y}}</math>
* where <math alt=>D_c</math> 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<ref>https://www.lpi.usra.edu/meetings/europa2004/pdf/7005.pdfReference for ice yield strength</ref>. 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 <math alt=> \approx \sqrt[3]{\frac{0.159 \cdot 0.136 J}{0.1 MPa}} \cdot 4727127478</math> grains/m3/ly <math alt=>/190385</math> craters/m2

In other words, the ice shield erodes by 32.1 meters amount of thickness for every light year travelled. What does that <math alt=>190385</math> 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 <math alt=>\pi \cdot D_{c}\,^2</math>, 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 <math alt=> \approx \sqrt[3]{\frac{0.159 \cdot 0.136 J}{7 GPa}} \cdot 4727127478</math> grains/m3/ly <math alt=>/15009474</math> craters/m2

It takes a trip of 24 light years just to get a 1 meter thick shield. 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:

<math alt=>\rho=mp</math>
* where <math alt=>\rho</math> is the mass density of the interstellar medium
* where <math alt=>m</math> is the mass of the particle
* where <math alt=>p</math> 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.

<math alt=>1.475 \cdot 10^{-27}</math> kg (average mass) = <math alt=>(0.7 \cdot H + 0.28 \cdot He + 0.015 \cdot Fe)/3</math>
* <math alt=>H,\,\,He,\,\,Fe</math> respectively refer to the atomic masses of hydrogen, helium and iron.

Now we can finally calculate the flux with the relativistic flux equation [[https://www.galacticlibrary.net/mediawiki-1.36.1/index.php?title=Interstellar_Medium_Shielding&action=submit#Derivation_of_the_Relativistic_Flux_Equation 7]]:

<math alt=>\phi=\gamma \rho vc(\gamma-1)c^2</math>
* where <math alt=>\phi</math> is the interstellar medium flux
* where <math alt=>\rho</math> is the mass density of the interstellar medium
* where <math alt=>v</math> is the velocity of the ship
* where <math alt=>c</math> is the speed of light
* where <math alt=>\gamma</math> is gamma, calculated with:
: <math alt=>\gamma = 1/\sqrt{1-v^2}</math>

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

<math alt=> \phi_e = A_d \epsilon \sigma_{sb} T^4 </math>
* where <math alt=> \phi_e </math> is the radiant power
* where <math alt=>A_d</math> is the radiating/absorbing surface area
* where <math alt=>\epsilon</math> is the emissivity of the radiating/absorbing material
* where <math alt=>\sigma_{sb}</math> is the stefan boltzmann constant
* where <math alt=>T</math> is the temperature of the material

Now we rearrange the equation to solve for temperature:

<math alt=> T= \sqrt[4]{\phi_e} / (\sqrt[4]{A_d} \sqrt[4]{\epsilon} \sqrt[4]{\sigma_{sb}})</math>

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

<math alt=> \phi_e = IA</math>
* where <math alt=>I</math> is the interstellar medium flux
* where <math alt=>A</math> is the area exposed to the interstellar medium flux

=Conclusions=

A calculator for interstellar medium shielding (heat flux only) is provided here:
 '''[https://www.desmos.com/calculator/1m8wnfd5dv Interstellar shielding calculator]'''

Below are two tables:

==Example Required Shield Thickness Table==
(Assuming mean mass of <math alt=>3 \cdot 10^{-16}</math> kilograms.)
{| class="wikitable sortable"
|-
! Ship Velocity !! Dust Grain Kinetic Energy (J)<ref>All calculations here were done with the relativistic kinetic energy equation.</ref> !! Ice (m) !! Carbon Fiber (m)
|-
| 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
{| class="wikitable sortable"
|-
! 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 <ref> https://en.wikipedia.org/wiki/Frost_line_%28astrophysics%29
 Reference for ice sublimation </ref> || 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/2 || || 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 <ref> https://en.wikipedia.org/wiki/Carbon
 Reference for graphite sublimation. I assume the point occurs at a lower temperature due to lower pressure. </ref>
|}
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 Reading=

=Additional References=
<references/>

==Derivation of the Relativistic Flux Equation==
<ref group="RFE">https://en.wikipedia.org/wiki/Kinetic_energy#Relativistic_kinetic_energy_of_rigid_bodies
 Reference for the relativistic kinetic energy equation.</ref>
# The kinetic energy of an amount of mass is given by <math alt=>(\gamma -1)mc^2</math>. To get the power, the kinetic energy is differentiated against time and thus assuming constant velocity, obtain that with <math alt=>\dot{m}</math> (the mass flow rate); the power is given by <math alt=>(\gamma -1)\dot{m} c^2</math>.
# 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 <math alt=>\gamma \rho</math> and multiply by <math alt=>Av</math> where <math alt=>A</math> is area.
# This yields <math alt=>P = \gamma \rho Av(\gamma -1)c^2</math>, to obtain the flux per unit area divide by <math alt=>A</math> and thereby cancel the <math alt=> A</math> in the earlier expression.

'''Reference for the Derivation of the Relativistic Flux Equation'''
<references group="RFE"/>

=Credit=
To Tshhmon for writing the article
* To lwcamp for helping with erosion calculation
* To Kerr for the relativistic flux equation and derivation

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