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

1. May 12, 2015

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

2. May 12, 2015

Staff: Mentor

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).

3. May 12, 2015

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?

4. May 12, 2015

Staff: Mentor

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.

5. May 12, 2015

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?

6. May 12, 2015

Staff: Mentor

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.

7. May 12, 2015

Michel_vdg

Alright, thanks!

8. May 12, 2015

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).

Last edited: May 12, 2015
9. May 12, 2015

Staff: Mentor

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

10. May 13, 2015

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%

Last edited: May 13, 2015
11. May 13, 2015

ChrisVer

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

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.

12. May 13, 2015

Michel_vdg

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.

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?

13. May 13, 2015

Staff: Mentor

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.

14. May 13, 2015

Michel_vdg

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

15. May 13, 2015

ChrisVer

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.

16. May 13, 2015

Michel_vdg

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.

17. May 13, 2015

Staff: Mentor

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.

18. May 13, 2015

Michel_vdg

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.

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.

19. May 13, 2015

Staff: Mentor

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

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.

20. May 14, 2015

Michel_vdg

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/

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."

ok. Thanks.