Why aren't all Cosmic Ray energies 'effective'?

[Michel_vdg]

On the Ultra-high-energy cosmic rays Wikipedia page there is an explanation that the 'effective' energy of cosmic rays differs from the actual(?) energy; and that 'only a small fraction is available for interaction'. How can that be, why isn't all energy 'effective'?

"The energy of this particle is some 40 million times that of the highest energy protons that have been produced in any terrestrial particle accelerator. However, only a small fraction of this energy would be available for an interaction with a proton or neutron on Earth, with most of the energy remaining in the form of kinetic energy of the products of the interaction.

The effective energy available for such a collision is the square root of double the product of the particle's energy and the mass energy of the proton, which for this particle gives 7.5×10¹⁴ eV, roughly 50 times the collision energy of the Large Hadron Collider."

Source: http://en.wikipedia.org/wiki/Ultra-high-energy_cosmic_ray

 

[mfb]

The center-of-mass energy is lower, as the center of mass is not the frame of Earth (=where we measure the energy).

It is a bit like hitting a table tennis ball with a truck. Sure the truck has a lot of energy, but the collision is not more violent than hitting a truck with a table tennis ball (with a tiny energy).

 

[Michel_vdg]

So is 'effective energy available' the same as the limited amount of energy that can be ... diffused / released / lost / converted / chopped off ... from that energetic proton during a collision because the other proton is so weak; while at the lhc during a full frontal collision all energy is converted into new matter?

 

[mfb]

Collisions in the LHC at 7 TeV/proton look the same as cosmic rays with ~50 PeV (50000 TeV) hitting a proton at rest: the same particles are produced with the same probability and so on, because in the center of mass frame for the cosmic ray collision, both the cosmic ray proton and the atmospheric proton have an energy of 7 TeV.

 

[Michel_vdg]

~50 PeV = 5×10¹⁶ that's a 1000 smaller than UHECR's who have an energy of 5×10¹⁹ eV

The Wikipedia article mentions 'effective energy available' of 7.5×10¹⁴ eV and roughly 50 times the collision energy of the LHC.

How come the difference ... is there more subtraction when both are more equal?

 

[mfb]

See the article, the center of mass energy grows with the square root of the cosmic ray energy. To increase the collision energy by a factor of 50 you have to take an incoming particle with 2500 times the 50 PeV, or 1.25*10²⁰ eV. Which is not so far away from the energy of the oh-my-god-particle.

 

[Michel_vdg]

Alright, thanks!

 

[ChrisVer]

Isn't it the same reason you want two colliding beams instead of an 1 beam hitting a target in colliders like LHC?

The target setup:
If your proton has some momentum $p^\mu_1 = \begin{pmatrix} E \\ p \end{pmatrix}$ and hits a stationary target proton $p_2 = \begin{pmatrix} m_p \\ 0 \end{pmatrix}$ then :
$(p_1 + p_2)^2 = (E+m_p)^2 -p^2= E^2 + m_p^2 + 2 E m_p -p^2 = 2 m_p^2 + 2 E m_p$

Whereas at the CM frame (2 beams colliding):
$(p_1' + p_2')^2 = (E_1 + E_2)^2 = (2 E_{cm} )^2 = 4 E_{cm}^2$

Since $E_1 = E_2 = E_{cm} = \sqrt{p_{cm}^2 +m_p^2 }$ (they have the same momentum magnitude).

Setting the above equal (to see when the CM collision of two beams can give the same results as a stationary target collision), you obtain: $E= \frac{2 E_{cm}^2 - m_p^2}{m_p}$
So the lab-available energy is less than the CM energy. So for the same results in a lab-frame you need larger energies.

If the CM energy is let's say approximately 7TeV (like in LHC), if you wanted to achieve the same energy with a fixed target collider, you'd need:

$E \sim \frac{ 49 TeV^2 - 1 GeV^2}{GeV} \sim 49 \times 10^6 ~GeV \sim 50~PeV$

See that the relation is quadratic (the mass division is a fixed parameter to keep the dimensionality correct), so at some point $E$ starts raising way faster than $E_{cm}$.

So a particle from cosmic rays with let's say 1PeV energy, when it collides with a stationary target [that is on earth], their collision won't take all the 1PeV, but it will take only a smaller fraction of it. Their CM energy is less (it goes as the square root of E).

 

[mfb]

Isn't it the same reason you want two colliding beams instead of an 1 beam hitting a target in colliders like LHC?

Exactly. Fixed-target experiments are impractical in the multi-GeV range.

 

[Michel_vdg]

Regarding a Fixed-taret, is it negligible for the energy of such protons and the whole collision if they are bonded within an atom (Carbon), within a crystal (Graphite), within a loose dust particle ... or in a rock on earth? Does that effect the calculation very much?

btw perhaps a stupid question, but can one say that the effective energy available for the 'stationary' proton is ~100%

 

[ChrisVer]

btw perhaps a stupid question, but can one say that the effective energy available for the 'stationary' proton is ~100%

I don't understand what you mean by "effective energy available [...] is ~100%" ...

Regarding a Fixed-taret, is it negligible for the energy of such protons and the whole collision if they are bonded within an atom (Carbon), within a crystal (Graphite), within a loose dust particle ... or in a rock on earth? Does that effect the calculation very much?

I don't think it matters much... Different matterial will behave differently in fixed-target collisions because they have different numbers of protons,neutrons or electrons.

 

[Michel_vdg]

I don't understand what you mean by "effective energy available [...] is ~100%" ...

The stationary proton has almost no kinetic energy vs. the UHECR, so all it's energy is effectively converted during the collision (100%) while for the UHECR it is only a fraction. The same for the protons at the LHC where the energy goes completely into the collision.

I don't think it matters much... Different material will behave differently in fixed-target collisions because they have different numbers of protons,neutrons or electrons.

ok. I just thought that chemical bonding would also be a sort of mass that goes into the equation, but I guess it only collides with light helium atoms with not so much bonding ... what if it is with a more dense atom such as the heavy lead ions at the LHC wouldn't that increase the effective energy as the bonding is much stronger?

 

[mfb]

The stationary proton has almost no kinetic energy vs. the UHECR, so all it's energy is effectively converted during the collision (100%) while for the UHECR it is only a fraction.

I don't think it makes sense to talk about conversion of energy of individual protons in asymmetric collisions.

Chemical bonds are completely negligible. Proton/nucleus collision (for larger nuclei) look a bit different from proton-proton collisions. The binding energy of the nuclei is still negligible, however.

 

[Michel_vdg]

Proton/nucleus collision (for larger nuclei) look a bit different from proton-proton collisions.

Does that mean that some UHECR's could also be iron atoms traveling roughly at the same velocity as a proton at the LHC, colliding with a static iron atom (dust) particle in the atmosphere, considering that their atomic number is 26 giving a total of ~50

 

[ChrisVer]

Does that mean that some UHECR's could also be iron atoms traveling roughly at the same velocity as a proton at the LHC, colliding with a static iron atom (dust) particle in the atmosphere, considering that their atomic number is 26 giving a total of ~50

Cosmic rays iron component has energy around ~1-10GeV per nucleon. So no.
But again that's not what mfb meant. I think we are confusing this too much.
What mfb said is that you get a bit different results from proton-proton collisions than you get from proton-nucleus (a fixed target of the X matterial of large nuclei for example) collision, and that's not because of the chemical bonds.

 

[Michel_vdg]

Cosmic rays iron component has energy around ~1-10GeV per nucleon.

Mh, but if it would be traveling at the same velocity as the particles of the LHC it would have 23 times the Kinetic energy, and add to this the larger energy of the fixed iron nucleus which has the same nucleus ... increasing the effective energy when those two collide.

 

[mfb]

If you get an iron nucleus at the same speed, hitting an iron nucleus in the atmosphere (not very realistic, we don't have an iron atmosphere) the total center of mass energy is ~56 times the energy a proton-proton collision would have. Sure, but where is the point? The energy would also be more spread out as the nuclei are larger.

 

[Michel_vdg]

If you get an iron nucleus at the same speed, hitting an iron nucleus in the atmosphere (not very realistic, we don't have an iron atmosphere) the total center of mass energy is ~56 times the energy a proton-proton collision would have.

No we don't have an iron atmosphere, but there are meteorites in the upper atmosphere that usually contain nickel and irons, so it's not impossible that such collisions could occur.

Sure, but where is the point? The energy would also be more spread out as the nuclei are larger.

As a final question I guess, it comes down to understanding how the Pierre Auger Observatory can measure the difference between a cosmic ray shower emerging out an UHECR collision consisting of a proton on proton vs. one from iron on iron.

 

[mfb]

No we don't have an iron atmosphere, but there are meteorites in the upper atmosphere that usually contain nickel and irons, so it's not impossible that such collisions could occur.

It is also not impossible that shards of a cup of tea spontaneously reassemble to an intact cup, but the chance is completely negligible.

As a final question I guess, it comes down to understanding how the Pierre Auger Observatory can measure the difference between a cosmic ray shower emerging out an UHECR collision consisting of a proton on proton vs. one from iron on iron.

They don't, as no such collisions occur. Iron on nitrogen happens. Those collisions produce more particles and a different angular spread compared to protons on nitrogen. In total, the particle shower looks a bit different.

 

[Michel_vdg]

It is also not impossible that shards of a cup of tea spontaneously reassemble to an intact cup, but the chance is completely negligible.

Every day about 100 tons of meteoroids -- fragments of dust and gravel and sometimes even big rocks – enter the Earth's atmosphere.
http://science.nasa.gov/science-news/science-at-nasa/2011/01mar_meteornetwork/

They don't, as no such collisions occur.

Mh, I found in this thesis 'The Propagation of Ultra High Energy Cosmic Rays' (pdf - 983 Kb):

"The question of the composition of ultra high energy cosmic rays remains unresolved, with the range of possibilities leading to quite different results in both the secondary fluxes of particles produced through cosmic ray energy loss interactions en route, and the arriving cosmic ray spectra at Earth. A large range of nuclear species are considered in this work, spanning the range of physically motivated nuclear types ejected from the cosmic ray source."

and

"... the variation in the models of the CIB considered lead to a few % difference in the CR spectrum observed at Earth for the case of CR Iron nuclei."

Iron on nitrogen happens. Those collisions produce more particles and a different angular spread compared to protons on nitrogen. In total, the particle shower looks a bit different.

ok. Thanks.

 

[mfb]

Every day about 100 tons of meteoroids -- fragments of dust and gravel and sometimes even big rocks – enter the Earth's atmosphere.
http://science.nasa.gov/science-news/science-at-nasa/2011/01mar_meteornetwork/

The mass of the atmosphere is 5*10¹⁵ tons.

Mh, I found in this thesis 'The Propagation of Ultra High Energy Cosmic Rays' (pdf - 983 Kb):

I don't see any mentioning of iron-iron collisions there.

 

[ChrisVer]

I have started losing the point... What is your current question/discussion topic?

 

[Michel_vdg]

I have started losing the point... What is your current question/discussion topic?

My original question has been answered. A side question now was if an UHECR (event) could also be an iron nucleus with the same velocity as the protons at the LHC colliding with a static iron nucleus (meteoric dust particle) in the upper atmosphere.

 

[ChrisVer]

iron nucleus with the same velocity of the protons at the LHC colliding with a static iron nucleus (meteoric dust particle) in the upper atmosphere.

It could happen but it's ultra-higly unlikely.
How could that help you?
I mean a few of the cosmic rays could also strike some meteorites outside the Earth's atmosphere. So?

 

[Michel_vdg]

How could that help you?

It helps to understand that a graph like the one below isn't only about the energy of individual protons of the Cosmic Rays but also what kind of particles it are:

"The actual particles that being observed as cosmic rays are thought to be protons and atomic nuclei (with ratios of each chemical species roughly reflecting the Galaxies own chemical abundances- ie. protons heavily dominant, followed by Helium nuclei, Carbon, Nitrogen, Oxygen and a fair amount of Iron nuclei). " link

 

[ChrisVer]

Again this diagram cannot be seen as a whole, since at different energy scales you have different sources and so can have different components.
As for where it comes from, it comes from nuclei in general... that means protons, alpha particle, etc etc...however the proton is the over-dominant component .

 

[Michel_vdg]

Again this diagram cannot be seen as a whole, since at different energy scales you have different sources and so can have different components.

Exaclty. Thanks.

 

[ChrisVer]

Still I think I should change the "components" to "abundances" ...
Also for the detection, you can have a look at this:
http://icecube.wisc.edu/tev/proceedings/Wakely/029_TEVPA2.pdf

 

[mfb]

It helps to understand that a graph like the one below isn't only about the energy of individual protons of the Cosmic Rays but also what kind of particles it are:

Discussing meteorites in that context certainly does not help, because it leads to confusion about incoming and target particles.

A small fraction of the particles in that diagram are heavier nuclei.

 

[Michel_vdg]

Discussing meteorites in that context certainly does not help, because it leads to confusion about incoming and target particles.

You do have to keep in mind that meteors are not just small static dots in the atmosphere for in-flying High-Energy Cosmic rays to hit with a very very unlikely chance; the fact is that they do cover a lot of ground and evaporate / spreading out and leaving a trail which increases the chance of interaction significantly.

A small fraction of the particles in that diagram are heavier nuclei.

I think we all know what 'protons heavily dominant' means.

Anyway the paper linked to by 'ChrisVer' showed a beautiful example / spike of a 100 TeV cosmic ray iron nucleus, a bunch of particles of which the individual protons have an energy of ~4 Tev that's in the same region as the LHC.

 

[ChrisVer]

Pouring a drop of some liquid into the ocean, you don't expect the interactions of other particles with that ocean to change... Just because you get a tail that spreads out doesn't change the composition of the atmospheric gases much... Again I find it a weird topic to discuss, since we are not that much interested into the target (what the cosmic rays strike), since most of those highly energetic strikes will lead to a hadronization (pions, Kaons etc) and so an electromagnetic shower (muons, electrons, gamma rays).
What we care about is identifying those showers and reconstructing the initial UHECR particle's characteristics. For example in the Cherenkov radiation method, they say that the Cherenkov radiation yield signal depends on the initial particle's atomic number squared $Z^2$.

 

[mfb]

You do have to keep in mind that meteors are not just small static dots in the atmosphere for in-flying High-Energy Cosmic rays to hit with a very very unlikely chance; the fact is that they do cover a lot of ground and evaporate / spreading out and leaving a trail which increases the chance of interaction significantly.

It does not matter. Even if with perfectly uniform distribution, and even if all their components stay in the atmosphere for one day for whatever reason, the mass of the atmosphere is 50000000000000 times larger. The chance to hit one atom from a meteorite instead of one out of 50000000000000 in the atmosphere is about 1/50000000000000.

 

[Michel_vdg]

It does not matter. Even if with perfectly uniform distribution, and even if all their components stay in the atmosphere for one day for whatever reason, the mass of the atmosphere is 50000000000000 times larger. The chance to hit one atom from a meteorite instead of one out of 50000000000000 in the atmosphere is about 1/50000000000000.

That's a comparison that doesn't make a lot of sense. Every cosmic ray has a chance of 1/1 to collide with a particle that makes up the atmosphere. If you now have 1 or 1 billion of cosmic rays, the chance stays the same 1/1. Now let's say you have 1 million of iron nuclei vs. 1, than the chance already becomes 1/50.000.000 that one Iron nucleus hits an other iron nucleus coming from a meteoric dust particle.

Next you not only need to look at mass but more at surface as an atmosphere is layer upon layer of particles, and Cosmic Rays shoot right through top to bottom (for those that are being observed), so you should mainly look at the upper layer surface where the particles are the lightest (low mass) and the meteors are the most spread out.

 

[Michel_vdg]

What we care about is identifying those showers and reconstructing the initial UHECR particle's characteristics. For example in the Cherenkov radiation method, they say that the Cherenkov radiation yield signal depends on the initial particle's atomic number squared $Z^2$.

True. What surprises me a little is that as a 'showroom model' they use the spike of a 100 TeV cosmic ray iron nucleus, while one might think that much faster cosmic rays would have been obsereved, because with it's ~4 Tev / proton it isn't something very exceptional.

 

[ChrisVer]

Again I am not sure, but I think that the iron nuclei don't appear at very large energies...
https://alteaspace.wordpress.com/2011/11/27/galactic-cosmic-rays-gcr/
I am talking about a figure as the second above. I am not sure if it's exceptional, but its flux is extremely small. Again I don't remember pretty well, but iron nuclei in cosmic rays, originate/are accelerated by supernovae, and so they have a threshold of energies they can reach. For the even higher energies we are not sure about the source mechanisms and there is a large literature dealing with different candidates.

Also I think you have a misconception when it comes to particle interactions.
The # of interaction events depends on the cross section (probability of a given interaction) times a factor that we call integrated luminosity. The integrated luminosity in this case depends on the incoming flux (which is fixed for every such kind of interaction=flux of incoming cosmic rays) and the density of the target...
The # of events for hitting an iron will get suppressed by the factor mfb wrote (1/500more zeroes) because the particular density of the iron in the whole atmosphere is very low. Nobody says that this can't happen, but it's very unlikely to happen by the factor mfb wrote down = you get very few events... if for example you have 1billion events in total in the atmosphere, you will have less than 1 of them coming from hitting an iron nucleus.