Can the Oh-My-God particle kill an astronaut?


by petergreat
Tags: astronaut, kill, ohmygod, particle
petergreat
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#1
Aug13-10, 07:28 PM
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The Oh-My-God cosmic ray particle is a single ultra-relativistic proton (?) carrying 50 joules of energy. Is it sufficiently energetic to kill an astronaut in the unlikely event of hitting the person? More precisely, I'm asking:

i) Does most of the energy get absorbed by the human body, or penetrate the body without doing harm? (Since protons are highly ionizing, I guess a significant amount of energy would be dumped inside the body with a shower of secondary particles, but I have no idea exactly how much it will be.)

ii) Is the release of energy from this particle localized to a small area, or spread out in the human body? (Even if the radiation dose is large, the person can survive if it's localized. For example, the radiation could destroy one of his fingers leaving other part of the body intact.
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Norman
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#2
Aug13-10, 08:23 PM
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At those energies (about 10^20 eV or 100 EeV) I don't think you can call it a proton.

Anyways, at those high energies the LET (linear energy transfer-a measure of the energy deposited along a track by a particle) from a proton-like particle is pretty small. Is it going to ionize some cells? Likely. Will that damage result in acute exposure problems. Absolutely not. Will there be some increased risk of cancer due to possible damage to these cells and its daughters? Yes, but it is going to be very small and very.

At these ultra high energies, we actually don't have a clue what the LET looks like in tissue. I am making an educated guess based on what the LET curve looks like at high energies. The stopping power (which I am actually going to use as a surrogate at these high energies) is likely fairly flat above 10 GeV.

You should have a look around the web on ionizing dose, linear energy transfer, stopping power and biological dose to get some background on this. All your questions are all basically answered with the fact that a single dose event from a single particle (no matter the energy or the particle type) will cause immediate death. Not enough energy transferred to the person.
bcrowell
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#3
Aug13-10, 09:01 PM
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If *all* the energy was deposited in the body, then it wouldn't really be an issue of radiation dose. 50 joules is simply enough heat energy to do serious harm. It really comes down to a question of how much of its energy it loses while passing through the body.

I believe dE/dx has a minimum at some energy and then comes back up, so I think this is not actually a minimum-ionizing particle. At very high energies, I think the dominant process is Bremsstrahlung. WP has a formula for stopping of electrons by Bremsstrahlung: http://en.wikipedia.org/wiki/Radiation_length I don't know if this particular formula can be accurately extrapolated to these insane energies, but the answer comes out to be on the order of a meter, which suggests the OMG particle could deposit quite a bit of its energy. (I don't think it matters if the OMG particle is a proton, not an electron, does it? It's got one unit of charge, just like an electron, and in both cases the inertia is dominated by the KE, not the rest mass.)

[EDIT] Hmm...the WP article on Bremsstrahlung says the radiated power goes like [itex]m^{-6}[/itex], so apparently my reasoning in parens above was wrong. This makes me think that a 50-joule electron could kill you, but a 50-joule proton couldn't. I also don't understand how the energy-independence of the formula at http://en.wikipedia.org/wiki/Radiation_length is consistent with the apparent energy-dependence of the equations at http://en.wikipedia.org/wiki/Bremstrahlung -- unless the smallness of [itex](d\beta/dt)^2[/itex] makes up for the largeness of [itex]\gamma^4[/itex].

Norman
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#4
Aug16-10, 09:22 AM
P: 922

Can the Oh-My-God particle kill an astronaut?


A quick and dirty, back of the envelope calculation follows:

Take worst case scenario stopping power of:
[tex] \frac{dE}{dx} = 10^3 \frac{\mbox{MeV cm^2}}{\mbox{g}}[/tex]

Take an average ray length in the body of
[tex]x = 30 \frac{\mbox{g}}{\mbox{cm^2}} [/tex] (this is approx. one foot in tissue).

This gives the an energy deposited of
[tex]E=1000 \mbox{MeV}.[/tex]

From track structure, we can approximate the volume of tissue in which the particle deposits its energy. Take a cylinder of radius 25 cm (remember back of the envelope- not exact but ballpark and is likely larger than 25 cm, so this will over estimate dose), of length one foot (about 30 cm).
Then the volume over which the energy is deposited is:
[tex]V \approx 6 \times 10^{-4} \mbox{cm^3}[/tex]
The density of water (substituted for tissue) is:
[tex]\rho = 10^{-3} \frac{\mbox{kg}}{\mbox{cm^3}}[/tex]
which gives a mass of tissue in which the energy is deposited as:
[tex]M=10^{-7} \mbox{kg}[/tex]
Then the Dose (energy per unit mass) is:
[tex] D = \frac{E}{M} \approx 10^{10} \frac{\mbox{MeV}}{\mbox{kg}} \approx 10^{-3} \mbox{Gy} [/tex]

Assume a worst case weighting factor for this particle of 5 and this dose is then a dose equivalent of 5 mSv. This is less than a CT scan, but more than a PET scan.
Bob S
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#5
Aug16-10, 05:21 PM
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Quote Quote by Norman View Post
A quick and dirty, back of the envelope calculation follows:

Take worst case scenario stopping power of:
[tex] \frac{dE}{dx} = 10^3 \frac{\mbox{MeV cm^2}}{\mbox{g}}[/tex]

Take an average ray length in the body of
[tex]x = 30 \frac{\mbox{g}}{\mbox{cm^2}} [/tex] (this is approx. one foot in tissue).

This gives the an energy deposited of
[tex]E=1000 \mbox{MeV}.[/tex]

From track structure, we can approximate the volume of tissue in which the particle deposits its energy. Take a cylinder of radius 25 cm (remember back of the envelope- not exact but ballpark and is likely larger than 25 cm, so this will over estimate dose), of length one foot (about 30 cm).
Then the volume over which the energy is deposited is:
[tex]V \approx 6 \times 10^{-4} \mbox{cm^3}[/tex]
The density of water (substituted for tissue) is:
[tex]\rho = 10^{-3} \frac{\mbox{kg}}{\mbox{cm^3}}[/tex]
which gives a mass of tissue in which the energy is deposited as:
[tex]M=10^{-7} \mbox{kg}[/tex]
Then the Dose (energy per unit mass) is:
[tex] D = \frac{E}{M} \approx 10^{10} \frac{\mbox{MeV}}{\mbox{kg}} \approx 10^{-3} \mbox{Gy} [/tex]

Assume a worst case weighting factor for this particle of 5 and this dose is then a dose equivalent of 5 mSv. This is less than a CT scan, but more than a PET scan.
A single session of prostate cancer radiation treatment is about 160 rads at the prostate (the radiation source rotates 360 degrees around the patient so the dose to the surrounding tissue is less). A complete prostate cancer radiation treatment program is about 42 sessions, or 6000 to 7000 rads over ~9 weeks.

1 rad = 100 ergs per gram, or 10-5 joules per gram, so 160 rads per session (lasts less than 1 minute) represents ~1.6 x 10-4 joules per gram (=0.16 joules/Kg = 0.16 Gy per session). This is insufficient to produce a noticeable rise in temperature (you won't feel it).

Bob S
bcrowell
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#6
Aug16-10, 07:16 PM
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Quote Quote by Norman View Post
Take worst case scenario stopping power of:
[tex] \frac{dE}{dx} = 10^3 \frac{\mbox{MeV cm^2}}{\mbox{g}}[/tex]
But where does this figure come from? How do you know it's not 10^13 instead of 10^3? If my understanding is correct (and it may very well not be), then this is not a minimum-ionizing particle, and the stopping power, which is dominated by Bresmsstrahlung, may be many orders of magnitude higher than you'd expect based on more ordinary energies.
Norman
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#7
Aug16-10, 08:42 PM
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Quote Quote by bcrowell View Post
But where does this figure come from? How do you know it's not 10^13 instead of 10^3? If my understanding is correct (and it may very well not be), then this is not a minimum-ionizing particle, and the stopping power, which is dominated by Bresmsstrahlung, may be many orders of magnitude higher than you'd expect based on more ordinary energies.
I got the 10^3 term from the PSTAR database from NIST (protons in liquid water): http://physics.nist.gov/PhysRefData/...ext/PSTAR.html. PSTAR calculates the sum of the nuclear and electronic stopping powers, but not the radiative stopping (Bremsstrahlung) power. I took the peak value as a worst case scenario.

I agree with you that one would expect bremsstrahlung to matter, so why no radiative stopping included? My first thought is that the stopping power goes something like 1/M^2 for bremsstrahlung. I think the reason PSTAR does not give the radiative portion for stopping power is it is simply not important at the energies they list.

After thinking about it for a while, I convinced myself that you are correct and that bremsstrahlung must matter at these high energies. So I poked around the internet for a bit and fell upon this which was linked at the PDG website: http://pdg.lbl.gov/2009/AtomicNuclea...ties/adndt.pdf

It is for muons, but I think it demonstrates quite well. So, at the extremely high energies considered for this particle, I would say it definitely matters. How much, I don't know, but considering the dose is directly proportional to the stopping power it might be interesting.

Is it enough to kill a person immediately (acute radiation sickness)? Probably not. Why? Typically to cause acute radiation sickness (ARS), as the thread author stated, the radiation must affect a large portion of the body and cause systemic failure. This is unlikely from a single particle. Is it possible? I suppose so if it made the proper shot through the chest cavity.
minerva
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#8
Aug25-10, 05:43 AM
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Even if those particles were delivering a substantial radiation dose, I doubt they would kill you, because they're passing through such a narrow, tiny localised region of the body, not the whole body.

For example: http://en.wikipedia.org/wiki/Anatoli_Bugorski

Oh-my-god particles hit *us*, on the surface of the Earth. So it's not just an astronaut specific phenomenon. Cosmic rays are hitting us right now. If they could kill you, they would be killing people now.
bcrowell
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#9
Aug25-10, 04:18 PM
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Quote Quote by minerva View Post
Even if those particles were delivering a substantial radiation dose, I doubt they would kill you, because they're passing through such a narrow, tiny localised region of the body, not the whole body.
I think you've got it backwards. The reason swords are sharp is to allow energy to be concentrated. The reason armor is effective against swords is that it spreads the energy out.

Quote Quote by minerva View Post
Oh-my-god particles hit *us*, on the surface of the Earth. So it's not just an astronaut specific phenomenon. Cosmic rays are hitting us right now. If they could kill you, they would be killing people now.
The flux of such particles is extremely small, and the density of human bodies on the surface of the earth is extremely low. I also don't think it's safe to assume that a particle with this energy reaches the surface of the earth, just because ordinary cosmic rays do. If it was minimum-ionizing, it would have a low dE/dx, so it would get through the atmosphere, it would also be harmless because it wouldn't lose any significant amount of energy in a human body. But it's not minimum-ionizing, because Bremsstrahlung increases with energy, so its dE/dx may be much higher. If dE/dx is high enough to dissipate a significant fraction of its energy in a human body, then it's also probably high enough that they wouldn't penetrate the atmosphere. Unless someone can come up with a dE/dx figure that we can feel confident is correct, I don't think we can conclude anything.
minerva
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#10
Aug29-10, 01:22 AM
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But we discovered, and detected, so-called oh-my-god particles, at approx. 10^20 eV, with detectors on the Earth, on the ground, and measured their energies at that level.
bcrowell
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Aug29-10, 10:13 AM
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Quote Quote by minerva View Post
But we discovered, and detected, so-called oh-my-god particles, at approx. 10^20 eV, with detectors on the Earth, on the ground, and measured their energies at that level.
But isn't it the shower of secondary particles in the sky that's actually measured from the ground?
Bob S
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#12
Aug29-10, 01:03 PM
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Quote Quote by bcrowell View Post
But isn't it the shower of secondary particles in the sky that's actually measured from the ground?
Absolutely right. The OMG particles above the atmosphere are most likely protons, which will create hadronic and electromagnetic showers within the first ~100 grams/cm2 of upper atmosphere. Ionizing cosmic rays at sea level are mostly muons, usually the result of pion decay in the upper atmosphere. Tables show the radiation length and nuclear interaction lengths to be ~ 37 and 90 grams/cm2 in air (roughly the same for H2O). The point at which such a shower starts is statistical in nature. It is quite possible for a single OMG proton to hit a tissue nucleus in the first 10 cm and create a shower with over 1000 pions, 1/3 of which will create an electromagnetic shower.

Bob S
Bob S
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#13
Aug30-10, 12:08 PM
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To really understand what happens when a OMG 50 Joule (3 x 1020 eV) proton hits a proton at rest, the lab energy has to be converted to the CM system (see Eq 38.3 in http://pdg.ihep.su/2009/reviews/rpp2...kinematics.pdf)

Ecm = (m1 +m2 + 2E1 lab m2)1/2 .

So a Elab = 3 x 1020 eV OMG proton hitting a proton at rest translates to ECM = ~5 x 1014 eV = 750 TeV. The LHC is now running at 7 TeV CM, and eventually at 14 TeV CM. How do multiplicities (average and extreme) scale up from 14 TeV to 750 TeV? If the extreme is >1 pion per GeV, this translates into a whole lot of pions sharing the 50 joules of the OMG particle. 2/3 of the pions will lose energy at ~2 MeV per gram/cm2 in tissue, and 1/3 (pi-zero's) in an electromagnetic shower.

Bob S
Dickfore
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#14
Aug30-10, 12:14 PM
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The kinetic energy of one of the weakest ammos for hand guns is 270 J, so I would say no.
Norman
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#15
Aug30-10, 02:25 PM
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Quote Quote by Dickfore View Post
The kinetic energy of one of the weakest ammos for hand guns is 270 J, so I would say no.
A cosmic ray is far from a ballistic projectile and making comparisons between bullets and cosmic rays is meaningless.
diggy
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#16
Aug31-10, 01:55 PM
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Somewhat redundant with other comments...

1) An astronaut couldn't stop any high energy particle if they tried. People have to use large (~meter) chunks of nice dense lead to stop anything around the 10^9eV range. 10^20 would be ridiculous to stop in a short distance.

2) A while back people were considering that thought was those very high energy particles were actually heavier elements like lead. Not sure if that's still the case.

3) DNA is double stranded to help protect you from these evil bullets of minimal destruction.

Its already been answered but i: Human body is basically transparent ii: It would be localized and the deposited energy would be much smaller than the full energy of what is hitting you (iii: You get hits with lots of just as dangerous stuff at all moments of your life.)
Daniel Batt
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#17
Jan11-12, 07:08 AM
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Okay, I know this is an old thread, but, if given a choice between and electron and a proton, it had to be a proton, because if it were an electron, we would have to add a few more .99's to the speed of light it was travelling.

While Utah were looking for cosmic rays, I don't see why it had to be in that class of particle to have caused the flash we saw. Astrophysical jets we know now accelerate plasma and perhaps other stuff from the poles of AGN at near light speed. Given that AGN cannibalise stars, their super dense cores and also neutron stars once in a while, I don't see why the odd bit of degenerate but charged matter doesn't get flung out at a significant proportion of c.

The GZK limit is for cosmic rays, and relies on the probability that energy will be decreased as distance increases. This 'limit' would also be a break on any heavy molecule or element or charged particle, I suppose. But when dealing with, say, what happens when a neutron star gets torn apart and ionized/charged as it approaches the event horizon, who can say?

Since we can't conceive what would accelerate a proton to 99.99999999999999999995% of c, it's worth thinking about what else might be hurtling our way, monopoles or other such mysteries, at much less a fraction of c but with just the right punch.


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