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Probability of interaction between two particles tends to wane with increasing energy. Technically, the cross section of most interactions falls off with increasing velocity.

$$\sigma(v) \propto \frac{1}{v}$$

This begs a fun question. Since interaction probability diminishes with increasing relative velocity, if you impart enough energy to a particle, might it just mostly pass through solid matter? Might solid matter pass through solid matter (with some "radiation damage" of course)? The above relation, of course, is lacking a great deal. We are really interested in the mean path length, as well as the linear rate of energy deposition. Let's consider the problem in the context of two chunks of solid matter moving at each other really fast (like space jousting). We have several requirements for a survivable experiment.

  • Average path length for any given nucleus & electron from spaceship A moving through spaceship B must be much greater than spaceship B's length
  • The energy deposition as a result of the passing must be small enough such that they don't explode like a nuclear bomb right after passing

The question also becomes highly relativistic, and I want to hear commentary from people who have knowledge of interactions in high energy accelerators.

Let's say you're on the Star Trek Enterprise, and the captain proposes an alternative to navigating the densely packed matter in the approaching galaxy by increasing speed to just under the speed of light, and not worrying about obstacles (because you'll pass through them). What arguments would you use to convince him this might not be the best idea?

EDIT: this video seems to make the claim that super high energy protons may pass through the entire Earth. Do the answers here contradict the claim?

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This is nonsense. Strong interactions grow with energy, as does gravity. –  Ron Maimon Oct 26 '11 at 16:14
To answer the Star Trek version: "Captain, we can't go through solid matter at Warp, what makes you think we can while moving slower?" –  Izkata Oct 26 '11 at 18:05
The probability of collision will never be negligible as vividly shown by the fact that cosmic rays interact quite non negligibly with the atmosphere even at the highest energies. –  whistles Oct 26 '11 at 23:04
Related:… –  Incnis Mrsi Nov 24 '14 at 16:36
The claim in the video is wrong, as clarified later: COMMENT FROM TONY: "My comments at 1.30 were a bit hasty. Cosmic rays do not pass through the earth. Most collide in the upper atmosphere. I was actually thinking of the LHC safety debate. If they exist, mini blackholes produced by cosmic ray collisions on earth would slip straight through because the earth isn't dense enough. But not the protons themselves." –  mmc Apr 30 at 1:19

3 Answers 3

up vote 3 down vote accepted

It doesn't look possible, because the stopping power starts growing for very high energies even for non-strongly interacting particles (see Fig. 27.1 in Passage of particles through matter).

To do a best case estimate, let's consider two cubic spaceships colliding, each of them with $1\,{\rm mm^3}$ of volume, a density of $1\,{\rm g\cdot cm^{-3}}$ and a number density of $10^{21}\,{\rm cm^{-3}}$. In general the atoms of the colliding spaceship will appear as a shower of nuclei and electrons but, for our purposes of lower bounding the deposited energy, let's consider the colliding spaceship as a shower of muons, one for each atom.

Then we will have the equivalent of $10^{18}$ muons ($10^{21}\ {\rm cm^{-3}}\cdot 10^{-3}\ {\rm cm^3}$) colliding with a target with an areal density of $0.1\ {\rm g\cdot cm^{-2}}$. Using the minimum ionization energy we get an stopping power greater than $1\ {\rm MeV\cdot cm^2\cdot g^{-1}}$, making each muon deposit approximately $100\ {\rm keV}$ of energy.

As the number of particles in both spaceships is equal, the expected temperature of both spaceships after the collision will be around the energy deposited by each particle, $100\ {\rm keV} \approx 10^9\ K$.

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The minimum ionization energy concept has great utility for answering this question. That puts a very definable limit on the minimum interaction. This should convince the captain if he has any sense! –  Alan Rominger Oct 26 '11 at 17:59

The short answer is no. Ron is right.

Have a look at plot 41 and on of total cross sections for particles, in this particle data link.

Your 1/v assumption might hold water until about 5GeV center of mass system energy, though it is affected at the particle level by resonances etc. From where the arrow starts, particle cross sections rise, they do not fall. This means that once relativistic velocities are reached and keep increasing, your two blobs of matter will be seeing each other grow bigger and bigger, that is what cross section means. If they are on a collision course they will burst into a plethora of particles and disappear in a particle radiation bomb.

They certainly cannot go through each other.

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This law you are giving isn't valid. Nuclear cross sections grow as a very small power, probably asymptoting to logarithmic growth. At very very high energies, gravity takes over, and the collision produces a decaying black hole. This cross section grows as the mass squared, since it is determined by the ratio of the impact parameter to the Schwarzscild radius.

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logarithmic cross section with respect to what variable? How does that scale compared to the other factors? To bring up the inevitable point, the size of the spaceship will play a role, as I imagine that large molecules, at least, could manage this in some sense. –  Alan Rominger Oct 26 '11 at 15:26
In s, but logaritmic is logaritmic with any polynomial of energy –  Ron Maimon Oct 26 '11 at 16:09
There is no way to do what you are asking. The question is nonsense. –  Ron Maimon Oct 26 '11 at 16:10
What is "In s"? The question is sufficiently complete from the title. Statements like "cross sections grow as a very small power" are incomplete. You're welcome to ask for more clear elaboration on any point I've made. I try to be as articulate as possible. Maybe you're the type who believes "bad" questions exist from the very core of the question? –  Alan Rominger Oct 26 '11 at 16:21
@Zassounotsukushi He is speaking of $s = (p_1 + p_2)^2$, the center of mass energy and one of the Mandelstam variables. –  mmc Oct 26 '11 at 16:33

protected by Qmechanic May 15 at 8:35


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