5 ms·

CERN’s Large Hadron Collider achieves the hottest manmade temperatures ever

70 points by ananyob 14y ago

pif 14y ago
> to momentarily create a quark gluon plasma, a sub-atomic soup It's even a sub-nuclear soup. It's not obvious until you think to it, but an atom is just enormous when you're talking at nuclear scale: a typical nucleus in its atom is like a tennis ball in a stadium! http://public.web.cern.ch/public/en/research/MinChall-en.html http://public.web.cern.ch/public/en/research/MinChall-en.htm...
- Retric 14y ago
  What I find mind blowing is there at energy levels where making these mater mater vs mater energy collisions is practicably meaningless. At 7Tev a collision between 2 protons has enough energy to create over 3,500 proton anti-proton pairs. And honestly anything that makes antimatter collisions seem like small potatoes is just insane.
ctdonath 14y ago
What constitutes "hot" at such a small scale? Years ago I addressed "hot" in terms of molecular speed, applying relativistic limits, calculating an upper limit on temperature and finding a large, but not inconceivable, maximum. Alas I've lost the paper (which was probably wrong anyway, but it was a fun exercise). The question has long bugged me: is there a real upper limit on temperature?
- theorique 14y ago
  There's certainly a relativistic limit - i.e. if the mean velocity of randomly oscillating particles in the gas / liquid approaches c. (I leave aside the question of what is used to confine such highly energetic matter)
  - windsurfer 14y ago
    I believe the limit is energy density. As the mass of the particle gets high enough (e=mc^2), its gravity gets stronger until it becomes a black hole and then explodes in a puff of Hawking radiation.
    - jessriedel 14y ago
      FYI, the temps accessible at the LHC are no where near high enough that this is a consideration.
    - ars 14y ago
      It's more complicated than that. The force of gravity is counteracted by the pressure caused by the temperature, and it would not become a black hole.
  - Jabbles 14y ago
    That in itself wouldn't limit the temperature. The average speed of the particles could approach c and their average energy would continue increasing. Molecules will break apart above a few thousand degrees. If you take a gas (say Helium) above a certain temperature (probably ~1e4 K) the collisions between the atoms will knock the electrons off and you'll be left with a plasma. Confining a plasma could either be done with magnets (e.g. a fusion reactor), or with more plasma! (in the case of stars, where the gravity keeps itself from exploding). Above 1e7 K you start to get fusion of Hydrogen, and I think above 1e11 K pretty much any nucleus will fuse/break down. So it's difficult to say there's a maximum temperature, as it's quite hard to define. What's your maximum temperature? About 41'C... above that and you'll die...
    - polysaturate 14y ago
      That last line worries me...I live in phoenix and most summer days are above 110' F...41'C is 105.8' F :/ Hopefully there is some clarification to explain how I won't die...
      - 14y ago
- cygx 14y ago
  is there a real upper limit on temperature? Technical limitations aside, not really. However, there's the Planck temperature of 1.42×10^32 K where our physical models break down.
  - glhaynes 14y ago
    Is there any simple way of explaining why our model would break down at that point? The first (utterly naive) guess that springs to my mind would be that at that point the vibration would be sufficiently fast that when sampled at a Planck time interval, there'd be no motion at all. Like overflow of an unsigned int. This is almost surely wrong. :)
    - twiceaday 14y ago
      Interesting read: http://www.pbs.org/wgbh/nova/physics/absolute-hot.html http://www.pbs.org/wgbh/nova/physics/absolute-hot.html
- deleted 14y ago
  [deleted]
- jessriedel 14y ago
  The abstract definition of temperature is that you have some set of modes (mode = possibly-occupied quantum states with fixed energy) and that there is enough interaction between these modes such that their probability of being occupied has a characteristic dependence on their respective energy. This characteristic dependence is a thermal distribution (which is either Bose-Einstein or Fermi-Dirac statistics). According to wikipedia (heh), quark-gluon plasma is not really a plasma so much as a liquid. Presumably, that means the modes are still roughly momentum modes, and you can literally think of a hotter QGP as faster quarks and gluons sliding around and slamming into one another. It's a Fermi liquid, though, which means the exclusion principle has significant effect. (I'm a physics grad student, but not a high-energy theorist, so take that for what it's worth.)
- ars 14y ago
  > applying relativistic limits, calculating an upper limit on temperature and finding a large, but not inconceivable, maximum. There is no relativistic limit to temperature. You probably calculated temperature by using the speed of the particle (and assumed it to max out a c), but at relativistic speeds this is wrong, you should use the energy of the particle instead. The question is: Does particle A impart energy to particle B, or does it take away energy? Whichever is faster imparts energy to the slower one. It makes no difference that the speed is limited - one particle can always have more energy than the other, and can therefor impart energy to it. It's not possible to actually reach that speed, so no matter how fast a particle is moving, you can always increase its speed. Time dilation should cause some very interesting artifacts though, I have no idea how to go about calculating what would happen. Length dilation is even more interesting - you may find your particles are so small that they pass each other and don't collide. However it's not actually the particle that collides, it's the electric fields that collide. But electric fields travel at the speed of light - so the fields won't really have time to notice each other. And finally at these energies a collision is unlikely to bounce, instead it would create some new particles, using up the energy and cooling rapidly.
rplnt 14y ago
Did they just totally ignore units in the whole article? (Yes, they did). Anyway, in the comments author said it's in degrees Celsius, if anyone was wondering (as at that scale the difference is quite big).
- aroberge 14y ago
  What did you expect: Farenheit?... Any scientific paper will refer to temperature either in Kelvin or Celsius ... and the difference between the two at that scale is quite small.
  - ElectroPrime 14y ago
    Kelvin and Celsius use the same scale, except the Kelvin scale's zero point is absolute zero.
  - rplnt 14y ago
    Yeah, K/°C difference is small, but I didn't expect Kelvin as it isn't a "degree". And I don't know that site/portal/.. If it were US newspaper, I would expect it to be in degrees F without mentioning it. It happens. Also, any scientific paper should use units. But to be fair, even though the difference between Farenheit and Celsius is huge at this scale, I can't really imagine it or compare it.
netvarun 14y ago
Out of curiosity. How do they handle such heat? I mean won't it melt anything around it? Also, how did they measure it? [I am assuming it was measured using a sophisticated thermal imaging camera, though I would like to think that some brave soul stuck in a huge badass thermometer inside the cauldron =)]
- Jabbles 14y ago
  They handle the heat by having an inconceivably small amount of matter at that "temperature". They measure it by looking at the resulting particles and working out how much energy the collision had, and converting between units of energy and temperature. There are some problems with describing the energy of a particle accelerator as a "temperature", but it makes the article more accessible and understandable, which is surely a good thing.
  - forensic 14y ago
    Should we be making shitty news articles more accessible or understandable, or should we be making the public more literate and intelligent? At a certain point, dumbing stuff down strips it of all meaning. Our species is going to have to start taking collective responsibility for its collective intelligence if we want a high tech future.
  - jessriedel 14y ago
    I don't think there's much of a problem with calling it a temperature. They don't use that term when describing merely one particle scattering off another, they only use it when they are colliding whole nuclei and there are enough particles to observe a thermal distribution. The key part about having a temperature is to have a distribution of modes (e.g. free particle momentum states) and that there is enough mixing around between these modes to generate a thermal distribution.
- jessriedel 14y ago
  The temperature is achieved by slamming two nuclei together. There is enough time for the various particles inside to scatter off each other many times and establish a thermal distribution of energies between themselves, but not enough time for any of them to interact with anything outside the two nuclei. After the collision takes place, theses particles will stream out of the interaction region just like in normal proton-on-proton collisions, going through decays and physically traveling through various detector elements. Presumably, the temperature is inferred by measuring the various energies of the outcoming particles (just like in a normal collision) and noting that they are distributed thermally (i.e. according to Bose-Einstein or Fermi-Dirac statistics). This distribution is tell-tale evidence that the particles exchanged energy through scattering many times. I guess you could call this thermal imagining, although "thermal imaging camera" denotes something that looks at infrared radiation and measures its distribution (in order to infer the temperature of the matter that radiated it). In this case, the particles themselves (and their decay products) are exploding outwards and having their energies measured.
web007 14y ago
Anyone know how this compares to the NIF in terms of temperature? They talk about terawatts and megajoules versus degrees, and as a non-physics geek I can't equate the two. NIF press release, 522.6TW and 1.855MJ - https://lasers.llnl.gov/newsroom/project_status/index.php https://lasers.llnl.gov/newsroom/project_status/index.php
- aroberge 14y ago
  Nor should you (be able to equate the two). Very roughly speaking, temperature is akin to a measure of the average energy density. Joule is a measure of energy and watt is a measure of energy per unit time. So, they give you a measure of how fast they delivered the energy (in terawatts), how much energy was delivered (in megajoules) but this does not tell us anything about the temperature at which the sample was heated since we don't know the size of the target.
- jessriedel 14y ago
  The temperature at the LHC and RHIC are many orders of magnitude larger than at NIF. As aroberge mentions, temperature is roughly a measure of energy density. (Here, you should probably think of it as energy per particle rather than per spatial volume.) NIF is applying a large amount of energy to a small but macroscopic amount of matter. The LHC and RHIC are colliding mere nuclei, so the amount of matter is roughly 22 orders of magnitude less. The LHC and RHIC collisions are vastly higher temperature (energy per particle) than NIF, but the total energy is vastly less.
unreal37 14y ago
Before reading the article, I wondered how hot that would be, like 5000 degrees C or something. For reference, the surface of the Sun is approx 10000 F/5500 C. They're claiming 5.5 trillion degrees Celsius. My mind is unable to comprehend that. A billion times hotter than the surface of the Sun...
- aroberge 14y ago
  Don't confuse the temperature at the surface of the Sun (5000 C or so, which is quite cool) with the temperature in the interior of the Sun (15 000 000 C or so ... quite a bit hotter). Still, it pales in comparison with the temperatures achieved in that experiment.
Create 14y ago
just for the record and to please the prospective downvoting mob, here are my experimental observations consistent with the cern experimental domain in order to warn any non-westerners: "The cost [...] has been evaluated, taking into account realistic labor prices in different countries. The total cost is X (with a western equivalent value of Y) [where Y>X] source: LHCb calorimeters : Technical Design Report ISBN: 9290831693 http://cdsweb.cern.ch/record/494264 http://cdsweb.cern.ch/record/494264 about integrity: http://cdsweb.cern.ch/record/1127343?ln=en http://cdsweb.cern.ch/record/1127343?ln=en
- cygx 14y ago
  From looking at the abstracts and your quote, I don't see anything particularly unreasonable. What exactly do you want to tell us? Capitalism is evil and public relations morally wrong?
  - scott_s 14y ago
    Check his history. He posts this on every article that mentions CERN.
  - Create 14y ago
    Propaganda meets reality: none of the statements below are actually true: "Message from the Director-General CERN is an Equal Opportunities employer. Our Equal Opportunities policy is founded on four main pillars: in recruitment and career development, the Organization will not discriminate against any applicant or employee on the grounds of sex, ethnic origin, disability, sexual orientation, religion, nationality or age; respect and dignity for all in the workplace; support for working parents; work-life balance." -- http://equal-opportunities.web.cern.ch/equal-opportunities/Welcome.php http://equal-opportunities.web.cern.ch/equal-opportunities/W...
cshimmin 14y ago
Late to the party, but I thought I'd clarify on what temperature really means. Many people are familiar with temperature as characterized by "average thermal velocity" of, say, air molecules. However, in statistical mechanics (the physics of many particle systems), temperature is defined in terms entropy and energy. Specifically, temperature is the rate-of-change of energy with respect to entropy. (In math: T = dU/dS.) This arises as a natural definition if you think of temperature as "the thing that is equal between two systems in thermal equilibrium". Think about it this way: A cup of water and an ocean have VASTLY different energies. Similarly, the ocean has an enormous entropy compared to the cup. However, if they're at the same temperature, the unit change of entropy per energy added is the same. Coming back to the article, the temperature being reported by ALICE is derived from the entropy and energy of the collisions. Specifically, they're looking at quark-gluon plasma from the heavy-ion collision at LHC. Under normal conditions, quarks and gluons are confined to hadrons like protons and neutrons. Quark-gluon plasma occurs only at extreme energies, and in these collisions it is a essentially a chaos of quarks and gluons. The chaos (entropy) is not a thermal property but rather largely comes from the nature of the strong force. So, changing the energy of a quark-gluon plasma doesn't have a huge effect on its entropy. Conversely, you could say it takes a lot of energy to change the entropy, and so by the definition of temperature, you get a very large value!

70 points by ananyob 14y ago

pif 14y ago

> to momentarily create a quark gluon plasma, a sub-atomic soup It's even a sub-nuclear soup. It's not obvious until you think to it, but an atom is just enormous when you're talking at nuclear scale: a typical nucleus in its atom is like a tennis ball in a stadium! http://public.web.cern.ch/public/en/research/MinChall-en.html http://public.web.cern.ch/public/en/research/MinChall-en.htm...

Retric 14y ago
What I find mind blowing is there at energy levels where making these mater mater vs mater energy collisions is practicably meaningless. At 7Tev a collision between 2 protons has enough energy to create over 3,500 proton anti-proton pairs. And honestly anything that makes antimatter collisions seem like small potatoes is just insane.

ctdonath 14y ago

What constitutes "hot" at such a small scale? Years ago I addressed "hot" in terms of molecular speed, applying relativistic limits, calculating an upper limit on temperature and finding a large, but not inconceivable, maximum. Alas I've lost the paper (which was probably wrong anyway, but it was a fun exercise). The question has long bugged me: is there a real upper limit on temperature?

theorique 14y ago
There's certainly a relativistic limit - i.e. if the mean velocity of randomly oscillating particles in the gas / liquid approaches c. (I leave aside the question of what is used to confine such highly energetic matter)
- windsurfer 14y ago
  I believe the limit is energy density. As the mass of the particle gets high enough (e=mc^2), its gravity gets stronger until it becomes a black hole and then explodes in a puff of Hawking radiation.
  - jessriedel 14y ago
    FYI, the temps accessible at the LHC are no where near high enough that this is a consideration.
  - ars 14y ago
    It's more complicated than that. The force of gravity is counteracted by the pressure caused by the temperature, and it would not become a black hole.
- Jabbles 14y ago
  That in itself wouldn't limit the temperature. The average speed of the particles could approach c and their average energy would continue increasing. Molecules will break apart above a few thousand degrees. If you take a gas (say Helium) above a certain temperature (probably ~1e4 K) the collisions between the atoms will knock the electrons off and you'll be left with a plasma. Confining a plasma could either be done with magnets (e.g. a fusion reactor), or with more plasma! (in the case of stars, where the gravity keeps itself from exploding). Above 1e7 K you start to get fusion of Hydrogen, and I think above 1e11 K pretty much any nucleus will fuse/break down. So it's difficult to say there's a maximum temperature, as it's quite hard to define. What's your maximum temperature? About 41'C... above that and you'll die...
  - polysaturate 14y ago
    That last line worries me...I live in phoenix and most summer days are above 110' F...41'C is 105.8' F :/ Hopefully there is some clarification to explain how I won't die...
    - 14y ago
cygx 14y ago
is there a real upper limit on temperature? Technical limitations aside, not really. However, there's the Planck temperature of 1.42×10^32 K where our physical models break down.
- glhaynes 14y ago
  Is there any simple way of explaining why our model would break down at that point? The first (utterly naive) guess that springs to my mind would be that at that point the vibration would be sufficiently fast that when sampled at a Planck time interval, there'd be no motion at all. Like overflow of an unsigned int. This is almost surely wrong. :)
  - twiceaday 14y ago
    Interesting read: http://www.pbs.org/wgbh/nova/physics/absolute-hot.html http://www.pbs.org/wgbh/nova/physics/absolute-hot.html
deleted 14y ago
[deleted]
jessriedel 14y ago
The abstract definition of temperature is that you have some set of modes (mode = possibly-occupied quantum states with fixed energy) and that there is enough interaction between these modes such that their probability of being occupied has a characteristic dependence on their respective energy. This characteristic dependence is a thermal distribution (which is either Bose-Einstein or Fermi-Dirac statistics). According to wikipedia (heh), quark-gluon plasma is not really a plasma so much as a liquid. Presumably, that means the modes are still roughly momentum modes, and you can literally think of a hotter QGP as faster quarks and gluons sliding around and slamming into one another. It's a Fermi liquid, though, which means the exclusion principle has significant effect. (I'm a physics grad student, but not a high-energy theorist, so take that for what it's worth.)
ars 14y ago
> applying relativistic limits, calculating an upper limit on temperature and finding a large, but not inconceivable, maximum. There is no relativistic limit to temperature. You probably calculated temperature by using the speed of the particle (and assumed it to max out a c), but at relativistic speeds this is wrong, you should use the energy of the particle instead. The question is: Does particle A impart energy to particle B, or does it take away energy? Whichever is faster imparts energy to the slower one. It makes no difference that the speed is limited - one particle can always have more energy than the other, and can therefor impart energy to it. It's not possible to actually reach that speed, so no matter how fast a particle is moving, you can always increase its speed. Time dilation should cause some very interesting artifacts though, I have no idea how to go about calculating what would happen. Length dilation is even more interesting - you may find your particles are so small that they pass each other and don't collide. However it's not actually the particle that collides, it's the electric fields that collide. But electric fields travel at the speed of light - so the fields won't really have time to notice each other. And finally at these energies a collision is unlikely to bounce, instead it would create some new particles, using up the energy and cooling rapidly.

rplnt 14y ago

Did they just totally ignore units in the whole article? (Yes, they did). Anyway, in the comments author said it's in degrees Celsius, if anyone was wondering (as at that scale the difference is quite big).

aroberge 14y ago
What did you expect: Farenheit?... Any scientific paper will refer to temperature either in Kelvin or Celsius ... and the difference between the two at that scale is quite small.
- ElectroPrime 14y ago
  Kelvin and Celsius use the same scale, except the Kelvin scale's zero point is absolute zero.
- rplnt 14y ago
  Yeah, K/°C difference is small, but I didn't expect Kelvin as it isn't a "degree". And I don't know that site/portal/.. If it were US newspaper, I would expect it to be in degrees F without mentioning it. It happens. Also, any scientific paper should use units. But to be fair, even though the difference between Farenheit and Celsius is huge at this scale, I can't really imagine it or compare it.

netvarun 14y ago

Out of curiosity. How do they handle such heat? I mean won't it melt anything around it? Also, how did they measure it? [I am assuming it was measured using a sophisticated thermal imaging camera, though I would like to think that some brave soul stuck in a huge badass thermometer inside the cauldron =)]

Jabbles 14y ago
They handle the heat by having an inconceivably small amount of matter at that "temperature". They measure it by looking at the resulting particles and working out how much energy the collision had, and converting between units of energy and temperature. There are some problems with describing the energy of a particle accelerator as a "temperature", but it makes the article more accessible and understandable, which is surely a good thing.
- forensic 14y ago
  Should we be making shitty news articles more accessible or understandable, or should we be making the public more literate and intelligent? At a certain point, dumbing stuff down strips it of all meaning. Our species is going to have to start taking collective responsibility for its collective intelligence if we want a high tech future.
- jessriedel 14y ago
  I don't think there's much of a problem with calling it a temperature. They don't use that term when describing merely one particle scattering off another, they only use it when they are colliding whole nuclei and there are enough particles to observe a thermal distribution. The key part about having a temperature is to have a distribution of modes (e.g. free particle momentum states) and that there is enough mixing around between these modes to generate a thermal distribution.
jessriedel 14y ago
The temperature is achieved by slamming two nuclei together. There is enough time for the various particles inside to scatter off each other many times and establish a thermal distribution of energies between themselves, but not enough time for any of them to interact with anything outside the two nuclei. After the collision takes place, theses particles will stream out of the interaction region just like in normal proton-on-proton collisions, going through decays and physically traveling through various detector elements. Presumably, the temperature is inferred by measuring the various energies of the outcoming particles (just like in a normal collision) and noting that they are distributed thermally (i.e. according to Bose-Einstein or Fermi-Dirac statistics). This distribution is tell-tale evidence that the particles exchanged energy through scattering many times. I guess you could call this thermal imagining, although "thermal imaging camera" denotes something that looks at infrared radiation and measures its distribution (in order to infer the temperature of the matter that radiated it). In this case, the particles themselves (and their decay products) are exploding outwards and having their energies measured.

web007 14y ago

Anyone know how this compares to the NIF in terms of temperature? They talk about terawatts and megajoules versus degrees, and as a non-physics geek I can't equate the two. NIF press release, 522.6TW and 1.855MJ - https://lasers.llnl.gov/newsroom/project_status/index.php https://lasers.llnl.gov/newsroom/project_status/index.php

aroberge 14y ago
Nor should you (be able to equate the two). Very roughly speaking, temperature is akin to a measure of the average energy density. Joule is a measure of energy and watt is a measure of energy per unit time. So, they give you a measure of how fast they delivered the energy (in terawatts), how much energy was delivered (in megajoules) but this does not tell us anything about the temperature at which the sample was heated since we don't know the size of the target.
jessriedel 14y ago
The temperature at the LHC and RHIC are many orders of magnitude larger than at NIF. As aroberge mentions, temperature is roughly a measure of energy density. (Here, you should probably think of it as energy per particle rather than per spatial volume.) NIF is applying a large amount of energy to a small but macroscopic amount of matter. The LHC and RHIC are colliding mere nuclei, so the amount of matter is roughly 22 orders of magnitude less. The LHC and RHIC collisions are vastly higher temperature (energy per particle) than NIF, but the total energy is vastly less.

unreal37 14y ago

Before reading the article, I wondered how hot that would be, like 5000 degrees C or something. For reference, the surface of the Sun is approx 10000 F/5500 C. They're claiming 5.5 trillion degrees Celsius. My mind is unable to comprehend that. A billion times hotter than the surface of the Sun...

aroberge 14y ago
Don't confuse the temperature at the surface of the Sun (5000 C or so, which is quite cool) with the temperature in the interior of the Sun (15 000 000 C or so ... quite a bit hotter). Still, it pales in comparison with the temperatures achieved in that experiment.

Create 14y ago

just for the record and to please the prospective downvoting mob, here are my experimental observations consistent with the cern experimental domain in order to warn any non-westerners: "The cost [...] has been evaluated, taking into account realistic labor prices in different countries. The total cost is X (with a western equivalent value of Y) [where Y>X] source: LHCb calorimeters : Technical Design Report ISBN: 9290831693 http://cdsweb.cern.ch/record/494264 http://cdsweb.cern.ch/record/494264 about integrity: http://cdsweb.cern.ch/record/1127343?ln=en http://cdsweb.cern.ch/record/1127343?ln=en

cygx 14y ago
From looking at the abstracts and your quote, I don't see anything particularly unreasonable. What exactly do you want to tell us? Capitalism is evil and public relations morally wrong?
- scott_s 14y ago
  Check his history. He posts this on every article that mentions CERN.
- Create 14y ago
  Propaganda meets reality: none of the statements below are actually true: "Message from the Director-General CERN is an Equal Opportunities employer. Our Equal Opportunities policy is founded on four main pillars: in recruitment and career development, the Organization will not discriminate against any applicant or employee on the grounds of sex, ethnic origin, disability, sexual orientation, religion, nationality or age; respect and dignity for all in the workplace; support for working parents; work-life balance." -- http://equal-opportunities.web.cern.ch/equal-opportunities/Welcome.php http://equal-opportunities.web.cern.ch/equal-opportunities/W...

cshimmin 14y ago

Late to the party, but I thought I'd clarify on what temperature really means. Many people are familiar with temperature as characterized by "average thermal velocity" of, say, air molecules. However, in statistical mechanics (the physics of many particle systems), temperature is defined in terms entropy and energy. Specifically, temperature is the rate-of-change of energy with respect to entropy. (In math: T = dU/dS.) This arises as a natural definition if you think of temperature as "the thing that is equal between two systems in thermal equilibrium". Think about it this way: A cup of water and an ocean have VASTLY different energies. Similarly, the ocean has an enormous entropy compared to the cup. However, if they're at the same temperature, the unit change of entropy per energy added is the same. Coming back to the article, the temperature being reported by ALICE is derived from the entropy and energy of the collisions. Specifically, they're looking at quark-gluon plasma from the heavy-ion collision at LHC. Under normal conditions, quarks and gluons are confined to hadrons like protons and neutrons. Quark-gluon plasma occurs only at extreme energies, and in these collisions it is a essentially a chaos of quarks and gluons. The chaos (entropy) is not a thermal property but rather largely comes from the nature of the strong force. So, changing the energy of a quark-gluon plasma doesn't have a huge effect on its entropy. Conversely, you could say it takes a lot of energy to change the entropy, and so by the definition of temperature, you get a very large value!