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.

 

[mfb]

@Michel_vdg: Every cosmic ray hits one nucleus of the atmosphere at random (the collision products then hit other nuclei and so on, but the initial collision is a single nucleus). The chance to hit a nucleus of a specific type is (approximately) proportional to the amount of nuclei in the atmosphere. Out of 50000000000000 atoms in the atmosphere, about 39000000000000 are nitrogen, 10000000000000 are oxygen, 500000000000 are Argon, ..., and 1 atom is from a meteorite. The chance that a specific high-energetic particle from space hits an atom from a meteorite instead of one of the 50000000000000 atoms from the remaining atmosphere is tiny.
Sure, given the large number of cosmic rays, it happens sometimes, but the rate is completely negligible.

 

[Michel_vdg]

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/

According to the graph on the linked-to page it seems to be the same; protons, alpha particles (helium nuclei), electrons, Carbon nuclei and Iron nuclei (Fe) on the inside, they are only at each stage exponentionally less dense:

... 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.

Yes that's very rare but so are UHECR's in general:

"These particles are extremely rare; between 2004 and 2007, the initial runs of the Pierre Auger Observatory detected 27 events with estimated arrival energies above 5.7×10¹⁹ eV, i.e., about one such event every four weeks in the 3000 km² area surveyed by the observatory." - Wiki

The question of course is how much collisions does the observatory process, billions ... I guess they only look for those above a certain threshold generating enough Cherenkov radiation ...

Anyway I think they would specifically say if it there are iron on iron collisions, and like 'mfb' already pointed out it's not really mentioned. I guess it is a difficult field of research with a limited amount of precision, measuring collisions up to 50 km in the sky that are initially happening at the nano meter scale.

 

[ChrisVer]

Yes that's very rare but so are UHECR's in general:

Then even worse, since the flux is itself low... but still when you are looking for UHECR your flux is fixed (either they hit the A nucleus in the atmosphere or the iron, the flux is still the same). What that means is that the rate of events is low by itself, and asking for a particular struke target which is very "rare", will make the events you want to look into even less...

My "billion" number was just an example.

There is no need to mention the iron target [I don't think they have to mention the target at all]... it's totally negligible and it's not a difficult field of research... I'd call it a waste of money research (even if you had the extreme sci-fi technology to distinguish the event)...because there's nothing to study out of it that can help you understand the nature, composition and different characteristics of the UHECR.

 

[Michel_vdg]

There is no need to mention the iron target [I don't think they have to mention the target at all]... it's totally negligible and it's not a difficult field of research... I'd call it a waste of money research (even if you had the extreme sci-fi technology to distinguish the event)...because there's nothing to study out of it that can help you understand the nature, composition and different characteristics of the UHECR.

That's like saying that the LHC is useless because we don't learn anything about the characteristics of protons. Of course there's little use for that, the importance here is the interaction between the different sorts of Cosmic rays and the different targets. As an example Parity Violation was discovered by shooting Beta rays at a cobalt target; or even more basic the nucleus was discovered by shooting Alpha rays at a metal target. Ray and target go hand in hand.

 

[ChrisVer]

There is some difference between in-laboratory experiments and cosmic rays ones...Otherwise we wouldn't need accelerators since we have the cosmic ray interactions.
One big difference is that in the lab we have a pretty good picture of what is happening...

 

[mfb]

There is absolutely no way to find and identify an iron/iron collisions in 10^13 iron/nitrogen collisions. Also, you would have to cover a significant fraction of the surface of Earth with detectors to get 10^13 recorded events in the first place.
Separating the small rate of iron/nitrogen collisions from proton/nitrogen collisions is hard enough and those are way easier to distinguish.

 

[Michel_vdg]

There is absolutely no way to find and identify an iron/iron collisions in 10^13 iron/nitrogen collisions. Also, you would have to cover a significant fraction of the surface of Earth with detectors to get 10^13 recorded events in the first place.

The problem is indeed collecting enough data, but one has to be careful with saying 'absolutely no way', we can now discover exoplanets which was also unthinkable years ago.

Separating the small rate of iron/nitrogen collisions from proton/nitrogen collisions is hard enough and those are way easier to distinguish.

Mh, according to this article in Nature it shouldn't be too hard:

Cosmic-ray theory unravels

"... they are seeing small air showers that are indicative of iron nuclei, rather than the larger showers that point to protons."

and

"The group revealed new data that weaken the link between the high-energy particles and the AGN (active galactic nuclei) ... the team has found evidence that these highest-energy cosmic rays might be iron nuclei, rather than the protons that make up most cosmic rays."

 

[mfb]

The problem is indeed collecting enough data, but one has to be careful with saying 'absolutely no way', we can now discover exoplanets which was also unthinkable years ago.

It was not unthinkable years ago. The idea is as old as the insight that our sun is a star like others, and the two most successful methods of today were proposed decades ago.
This is orders of magnitude easier than finding one event in 10¹³. We are doing this for rare Higgs decays at the LHC, for example, I know how hard it is. And we have the detector all around the interaction point, and a Higgs decay looks completely different from most of the other collisions.

Mh, according to this article in Nature it shouldn't be too hard:

Cosmic-ray theory unravels

If by "not too hard", you mean 10 or more work years: yes sure, it is not too hard.
And the result is still "just" a statistical evidence. Given all events, they can tell that some come from heavy nuclei, but they cannot say "this event was an iron nucleus for sure".

 

[Michel_vdg]

That's all true.

An interesting article I found was this one:

How Astrophysicists Are Turning The Entire Moon Into A Cosmic Ray Detector
The $1.5 billion plan breaks ground in 2018 and should be complete by 2025
https://medium.com/the-physics-arxi...-moon-into-a-cosmic-ray-detector-6dd20a6acc62

"These cascades also generate another signal. The rapid acceleration and deceleration of charged particles produces radio waves. So another signature of the impact of an ultra-high energy cosmic ray is a brief burst of radio waves, known as the Askaryan effect after the Soviet-American physicist who proposed it in the early 1960s.

It is this signal that astronomers hope to pick up from the Moon. The idea is that ultrahigh energy cosmic rays should smash into the lunar surface generating a cascade of other particles and a short burst of radio waves less than a nanosecond long."

--

After that they can fly to the moon and dig up the collision sites.