Would a relativistic ship really need much shielding?

Albertgauss
Gold Member
Messages
294
Reaction score
37
Hi all,

It doesn't seem that a relativistic ship traveling in deep space would need much shielding due to particles at rest in the space frame, appearing relativisitc now in the ship's frame. Here's the calculation I did:

Let's imagine a ship the size of a space shuttle traveling with γ of 7 (β = 0.99 ). I assume that when a society can build such a ship, it will be a decent size, comparable to our space shutttle, and use such values therein. I'll approximate the cockpit as a flat circle(normal vector parrallel to direction of travel) with a diameter of 8.7m, for an area of 59 m^2. Now, let's allow 1 proton per m^3 (over estimate) in deep space and sort them in an area of 59 m^2 so that, at a particular time, 30 protons or so will impact the cockpit of our ship. I know that 1 proton per m^3 is a volume density, so making such a number an area density is, again, an overestimate.

In the ship's frame, the ship will be at rest, and 30 protons will impact the cockpit. THe rest mass of a proton is 900 MeV, we have 30, and they each have γ=7, so the ship will encounter an energy of 1.89(10^11) eV worth of energy.

Compare this to the energy of a dental X-ray. E=hf, f = 10^17, so the energy of a photon here is 65eV. In a typical dental X-ray, there would be 10^26 particles or so, so we have about 10^27 eV in a dental X-ray, and ~10^11 eV from outerspace protons.

Thus, a ship traveling with a γ = 7 will encounter much less energy from protons in deep space than it would by bombarding it with a dental X-ray. It is true that the deep space protons themselves are far more energetic than a dental X-ray photon, but there are so few, would it even be worth much worry? I would guess there are more relativistic protons from our own atmosphere via solar wind that pass through a human body per second than a ship traveling at light speed would have to worry about in deep space.

Does this calculation make sense? Why is shielding from outer-space protons appearing relatvistic in the ship's frame so much of a problem for near light speed travel? Or maybe its not so big a deal for β=0.99 but is more of a concern for higher β's.
 
Physics news on Phys.org
Several major problems:

1) You don't compute dose over the whole journey. You compute a dose (even given all your figures) corresponding to the time it takes the ship to go one ship length. You need to mulitply this by, e.g. 100 light years / ship's length.

2) Your figures for the energy of an x-ray dose are way too high. See: http://en.wikipedia.org/wiki/X-ray

3) You ignore interstellar dust. You need to ask how much dust there is in ship's cross section X 100 light years. One speck could be devastating.

4) Your figures for overall density of interstellar medium are way off. See: http://en.wikipedia.org/wiki/Interstellar_medium. Note, you may encounter regions with 10^6 molecules per cubic centimeter.

5) You forget very rare events (that are not so rare over 100 light years). For example, a pinhead size piece of debris kicked into an interstellar region.
 
Thread 'Can this experiment break Lorentz symmetry?'
1. The Big Idea: According to Einstein’s relativity, all motion is relative. You can’t tell if you’re moving at a constant velocity without looking outside. But what if there is a universal “rest frame” (like the old idea of the “ether”)? This experiment tries to find out by looking for tiny, directional differences in how objects move inside a sealed box. 2. How It Works: The Two-Stage Process Imagine a perfectly isolated spacecraft (our lab) moving through space at some unknown speed V...
Does the speed of light change in a gravitational field depending on whether the direction of travel is parallel to the field, or perpendicular to the field? And is it the same in both directions at each orientation? This question could be answered experimentally to some degree of accuracy. Experiment design: Place two identical clocks A and B on the circumference of a wheel at opposite ends of the diameter of length L. The wheel is positioned upright, i.e., perpendicular to the ground...
In Philippe G. Ciarlet's book 'An introduction to differential geometry', He gives the integrability conditions of the differential equations like this: $$ \partial_{i} F_{lj}=L^p_{ij} F_{lp},\,\,\,F_{ij}(x_0)=F^0_{ij}. $$ The integrability conditions for the existence of a global solution ##F_{lj}## is: $$ R^i_{jkl}\equiv\partial_k L^i_{jl}-\partial_l L^i_{jk}+L^h_{jl} L^i_{hk}-L^h_{jk} L^i_{hl}=0 $$ Then from the equation: $$\nabla_b e_a= \Gamma^c_{ab} e_c$$ Using cartesian basis ## e_I...
Back
Top