5 ms·

Photons cooled down to room temperature, made into a blob of matter

91 points by juiceandjuice 16y ago

djcapelis 16y ago
Should it be called matter if it's mass-less? Or is the article wrong and did this have a mass?
- gowiththeflow 16y ago
  Not a physics major, but my guess is E=mc^2 still applies.
  - harshpotatoes 16y ago
    Yes and no. Yes, E=mc^2 does still apply, but this refers only to a particles rest energy. So in the rest frame of a photon, the photon has no energy, and thus has no mass (or at least that's the theory, in principle more difficult to test and for all intents and purposes it is very nearly zero and most probably is exactly zero). In our rest frame, the photon moves at the speed of light, so... E = mc^2 + kintetic energy + potential energy -> which basically reduces to E = hf (plank's constant * frequency of the light).
  - jules 16y ago
    A physics major, this is correct, in the sense that if the photons in the "matter of photons" have total energy E, then it acts like an object of mass m (under the condition that the total momentum is zero, which I presume is approximately true because the photons are moving in random directions?). However, photons do not have rest mass. The formula for photons is E = pc, where p is the momentum of the photon. The full form is E^2 = (mc^2)^2 + (pc)^2, where m is the rest mass and p is the momentum. Taking p = 0 for an object at rest gives E = mc^2, taking m = 0 for a photon gives E = pc.
- juiceandjuice 16y ago
  Given that the definition of "matter" is slightly ambiguous, it's not really wrong, but it's not really correct either. Traditionally, matter usually has to have mass, but if you just think of matter as a volume of particles, then a collection of photons could be defined loosely as "matter".
- lmkg 16y ago
  The photons that make up the condensate have no rest mass, which is why they travel at the speed of light whenever they have a non-zero amount of kinetic energy. Anything that has any form of energy has mass, since those are just two different words for the same property. Because of this, the photons can be called "massless" (i.e. no rest mass) but still have mass, and the condensate could in theory have mass without any of its constituents having mass. The photons have mass from their individual kinetic energies, and the condensate could have mass beyond that from any potential energy required to get it into that state. Although, if the condensate is stable in any sense, it probably has less mass than the sum of the energies of its photons, with the difference being the loss of potential energy it gains from being stable. As a comparison, atomic nuclei are composed of particles with non-zero rest mass, and they tend to have less mass than the sum of their components because of their stability. The difference between the mass of the nuclei and its components is how much energy you have to dump into the nuclei to split it (plus enough kinetic energy so that they fly away before they reform). As to the question of what constitutes "matter"... I'm not sure that's actually defined in a way that's 100% free of contention. My personal opinion would by any configuration of particles that's stable. This new stuff could qualify.
  - alecst 16y ago
    Photons have momentum. But is it not misleading to say that they have mass? You also say that E and m are just different words for the same property. This, I believe, is not fair. E (J) differs from m (kg) by a conversion factor of c (m/s) ^2. To give an analogy, I think it would be misleading to say that an electric field E(r) is a different word for some charge Q, because without one you cannot have the other, and they only differ by some conversion factor. Could you explain a little bit more thoroughly what you mean?
- Estragon 16y ago
  I don't really know what it means, but my understanding is that photons actually have a mass. E=mc^2 => m=E/c^2, and photons have energy...
  - rorrr 16y ago
    Photons don't have a mass. If they did, they wouldn't be able to travel at the speed of light (it takes infinite energy to make any mass travel at the the speed of light). http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_p... E = mc^2 applies to matter, not photons.
iwr 16y ago
What does "temperature" mean with regard to photons? A photon has an energy level given by its wavelength or frequency. But temperature is a macroscopic phenomenon (equivalent to the kinetic energy of a group of particles).
- iwr 16y ago
  So reading more about this: http://en.wikipedia.org/wiki/Photon_gas http://en.wikipedia.org/wiki/Photon_gas , it looks like a scenario where a blackbody container was created, with a Boltzmann "temperature" peak equal to the energy level of the incident photons. Thus, all absorbed photons would be re-emitted at with the same distribution as the blackbody walls. And apparently, this phenomenon was tested in an ultra-low temperature scenario. Still, I wonder. Given the low temperatures, the Boltzmann distribution suggests photons in the microwave or lower frequencies.
  - Qz 16y ago
    Actually the article says the photons were in a room-temperature scenario.
- harshpotatoes 16y ago
  It's hard for me to say without reading the actual research article, but my guess is that temperature is referring to energy of each of the photons relative to one another. Meaning, if you were shining a laser beam down the z axis, typically the photons would also have some small velocity in the x and y directions (this would mean the light beam is diverging). So looking in a frame of reference moving along the z axis at the speed of the photons in the z direction, you would notice that most of the photons would be stationary in the z direction but are slowly moving in the x and y directions. The temperature is probably referring to the average kinetic energy of the photons in the x and y directions. So in order to get the photons all into a single state, you would want their x and y velocities to all be zero. That's my guess, I know sometimes temperature can refer to similar scenarios with a beam of atoms. Like with the LHC, they often refer to the atoms as having a very low temperature. However, the atoms are moving around in circles at very nearly the speed of light. With such a high kinetic energy, you would expect the atoms to have a high temperature. But really, the atoms are moving very slowly relative to one another, and thus temperature is referring to this kinetic energy.
1053r 16y ago
A slightly more informative article, which references the paper published in nature (unfortunately, I couldn't find a copy of that that wasn't behind a paywall). http://www.nature.com/news/2010/101124/full/news.2010.630.html http://www.nature.com/news/2010/101124/full/news.2010.630.ht... Note that the photons weren't made into matter. They were made into a Bose-Einstein Condensate, which is where all of the component particles share quantum properties and interact with the rest of the universe in "lockstep". http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensate http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensat...
- juiceandjuice 16y ago
  To be a little more clear though (from the Nature article): "Bose-Einstein condensates (BECs) are a bizarre quantum phase of matter. " I can email people the copy if they want, I won't post it though.
  - Shamiq 16y ago
    Is it readable by those of us who fall under the "layman" category?
- mchouza 16y ago
  I believe that the preprint is at: http://arxiv.org/abs/1007.4088 http://arxiv.org/abs/1007.4088
epynonymous 16y ago
tele-transportation is just years away!
powrtoch 16y ago
"The physics behind the Bose-Einstein condensation is the transition from a particle-like behavior at high temperatures to a wave-like behavior at cold temperatures" I'm no physicist, but isn't this backwards?
danparsonson 16y ago
Noobie question: how do they know that this actually happened? I would have thought that having some sort of detector inside the tiny mirror box would disturb the experiment? How do you measure something like this?
- danparsonson 16y ago
  Should've read the other article linked on here (http://www.nature.com/news/2010/101124/full/news.2010.630.html http://www.nature.com/news/2010/101124/full/news.2010.630.ht...): The team could tell when the transition had occurred because the small number of photons in the BEC formed an intense beam of yellow light — like a laser — in the centre of the cavity, surrounded by the 'gas' of remaining normal photons. To double-check that they were seeing a BEC of light, the researchers repeated the experiment with different numbers of photons. In each case, once the transition had taken place, they measured the spectrum of light leaking from the cavity and found that it matched theoretical predictions for the corresponding BEC. (BEC == Bose-Einstein Condensate)
aerique 16y ago
I'm probably a very shallow individual for asking but are there pictures of this?

91 points by juiceandjuice 16y ago

djcapelis 16y ago

Should it be called matter if it's mass-less? Or is the article wrong and did this have a mass?

gowiththeflow 16y ago
Not a physics major, but my guess is E=mc^2 still applies.
- harshpotatoes 16y ago
  Yes and no. Yes, E=mc^2 does still apply, but this refers only to a particles rest energy. So in the rest frame of a photon, the photon has no energy, and thus has no mass (or at least that's the theory, in principle more difficult to test and for all intents and purposes it is very nearly zero and most probably is exactly zero). In our rest frame, the photon moves at the speed of light, so... E = mc^2 + kintetic energy + potential energy -> which basically reduces to E = hf (plank's constant * frequency of the light).
- jules 16y ago
  A physics major, this is correct, in the sense that if the photons in the "matter of photons" have total energy E, then it acts like an object of mass m (under the condition that the total momentum is zero, which I presume is approximately true because the photons are moving in random directions?). However, photons do not have rest mass. The formula for photons is E = pc, where p is the momentum of the photon. The full form is E^2 = (mc^2)^2 + (pc)^2, where m is the rest mass and p is the momentum. Taking p = 0 for an object at rest gives E = mc^2, taking m = 0 for a photon gives E = pc.
juiceandjuice 16y ago
Given that the definition of "matter" is slightly ambiguous, it's not really wrong, but it's not really correct either. Traditionally, matter usually has to have mass, but if you just think of matter as a volume of particles, then a collection of photons could be defined loosely as "matter".
lmkg 16y ago
The photons that make up the condensate have no rest mass, which is why they travel at the speed of light whenever they have a non-zero amount of kinetic energy. Anything that has any form of energy has mass, since those are just two different words for the same property. Because of this, the photons can be called "massless" (i.e. no rest mass) but still have mass, and the condensate could in theory have mass without any of its constituents having mass. The photons have mass from their individual kinetic energies, and the condensate could have mass beyond that from any potential energy required to get it into that state. Although, if the condensate is stable in any sense, it probably has less mass than the sum of the energies of its photons, with the difference being the loss of potential energy it gains from being stable. As a comparison, atomic nuclei are composed of particles with non-zero rest mass, and they tend to have less mass than the sum of their components because of their stability. The difference between the mass of the nuclei and its components is how much energy you have to dump into the nuclei to split it (plus enough kinetic energy so that they fly away before they reform). As to the question of what constitutes "matter"... I'm not sure that's actually defined in a way that's 100% free of contention. My personal opinion would by any configuration of particles that's stable. This new stuff could qualify.
- alecst 16y ago
  Photons have momentum. But is it not misleading to say that they have mass? You also say that E and m are just different words for the same property. This, I believe, is not fair. E (J) differs from m (kg) by a conversion factor of c (m/s) ^2. To give an analogy, I think it would be misleading to say that an electric field E(r) is a different word for some charge Q, because without one you cannot have the other, and they only differ by some conversion factor. Could you explain a little bit more thoroughly what you mean?
Estragon 16y ago
I don't really know what it means, but my understanding is that photons actually have a mass. E=mc^2 => m=E/c^2, and photons have energy...
- rorrr 16y ago
  Photons don't have a mass. If they did, they wouldn't be able to travel at the speed of light (it takes infinite energy to make any mass travel at the the speed of light). http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_photon_mass http://en.wikipedia.org/wiki/Photon#Experimental_checks_on_p... E = mc^2 applies to matter, not photons.

iwr 16y ago

What does "temperature" mean with regard to photons? A photon has an energy level given by its wavelength or frequency. But temperature is a macroscopic phenomenon (equivalent to the kinetic energy of a group of particles).

iwr 16y ago
So reading more about this: http://en.wikipedia.org/wiki/Photon_gas http://en.wikipedia.org/wiki/Photon_gas , it looks like a scenario where a blackbody container was created, with a Boltzmann "temperature" peak equal to the energy level of the incident photons. Thus, all absorbed photons would be re-emitted at with the same distribution as the blackbody walls. And apparently, this phenomenon was tested in an ultra-low temperature scenario. Still, I wonder. Given the low temperatures, the Boltzmann distribution suggests photons in the microwave or lower frequencies.
- Qz 16y ago
  Actually the article says the photons were in a room-temperature scenario.
harshpotatoes 16y ago
It's hard for me to say without reading the actual research article, but my guess is that temperature is referring to energy of each of the photons relative to one another. Meaning, if you were shining a laser beam down the z axis, typically the photons would also have some small velocity in the x and y directions (this would mean the light beam is diverging). So looking in a frame of reference moving along the z axis at the speed of the photons in the z direction, you would notice that most of the photons would be stationary in the z direction but are slowly moving in the x and y directions. The temperature is probably referring to the average kinetic energy of the photons in the x and y directions. So in order to get the photons all into a single state, you would want their x and y velocities to all be zero. That's my guess, I know sometimes temperature can refer to similar scenarios with a beam of atoms. Like with the LHC, they often refer to the atoms as having a very low temperature. However, the atoms are moving around in circles at very nearly the speed of light. With such a high kinetic energy, you would expect the atoms to have a high temperature. But really, the atoms are moving very slowly relative to one another, and thus temperature is referring to this kinetic energy.

1053r 16y ago

A slightly more informative article, which references the paper published in nature (unfortunately, I couldn't find a copy of that that wasn't behind a paywall). http://www.nature.com/news/2010/101124/full/news.2010.630.html http://www.nature.com/news/2010/101124/full/news.2010.630.ht... Note that the photons weren't made into matter. They were made into a Bose-Einstein Condensate, which is where all of the component particles share quantum properties and interact with the rest of the universe in "lockstep". http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensate http://en.wikipedia.org/wiki/Bose%E2%80%93Einstein_condensat...

juiceandjuice 16y ago
To be a little more clear though (from the Nature article): "Bose-Einstein condensates (BECs) are a bizarre quantum phase of matter. " I can email people the copy if they want, I won't post it though.
- Shamiq 16y ago
  Is it readable by those of us who fall under the "layman" category?
mchouza 16y ago
I believe that the preprint is at: http://arxiv.org/abs/1007.4088 http://arxiv.org/abs/1007.4088

epynonymous 16y ago

tele-transportation is just years away!

powrtoch 16y ago

"The physics behind the Bose-Einstein condensation is the transition from a particle-like behavior at high temperatures to a wave-like behavior at cold temperatures" I'm no physicist, but isn't this backwards?

danparsonson 16y ago

Noobie question: how do they know that this actually happened? I would have thought that having some sort of detector inside the tiny mirror box would disturb the experiment? How do you measure something like this?

danparsonson 16y ago
Should've read the other article linked on here (http://www.nature.com/news/2010/101124/full/news.2010.630.html http://www.nature.com/news/2010/101124/full/news.2010.630.ht...): The team could tell when the transition had occurred because the small number of photons in the BEC formed an intense beam of yellow light — like a laser — in the centre of the cavity, surrounded by the 'gas' of remaining normal photons. To double-check that they were seeing a BEC of light, the researchers repeated the experiment with different numbers of photons. In each case, once the transition had taken place, they measured the spectrum of light leaking from the cavity and found that it matched theoretical predictions for the corresponding BEC. (BEC == Bose-Einstein Condensate)

aerique 16y ago

I'm probably a very shallow individual for asking but are there pictures of this?