Is there such a thing as the highest temperature ?

[Starman]

No, no, just no. Alright?

1. You can't think of temperature as the kinetic energy of bound systems like atoms.

I'm not talking about the internal KE of the bound system, but the KE of the entire system with respect to the external coordinate system. Not the same thing.

2. Atoms do have a certain amount of energy even when they're in their ground state (zero-point energy), but what you implied about not being able to reach absolute zero is nonsense. Atoms have a discrete amount of energy, higher than 0, however you can get arbitrarily close to absolute zero. The kinetic energy of the atoms is regarded as the kinetic energy of the entire atom (nucleus and electrons combined, thought of as a single object). That can be arbitrarily close to 0.

This is classical physics. In quantum physics the "wobble" of the nucleus appears as the KE of the entire atom. Granted that this is a very small amount of energy, but, since the electron cannot have less energy than it does in the ground state, the energy of the "wobble" motion cannot be removed. Since it is a KE of the entire atom, not just a part of it, a collection of atoms (or molecules, the same principle applies) will have some vibrational energy which cannot be removed. Since a vibration contains KE, and temperature is the KE of the system of atoms, there will be a temperature below which the collection of particles gannot go.

3. You can't think of an atom like a solar system, with the electron whizzing around and causing the nucleus to wobble like a star.

Yes you can. Energy/momentum is always conserved. For any energy term in a classical expression of a system, there will be a corresponding term in the QM wave function, so analogies between classical and quantum energies will always exist. You may notice that I used the term "analogous".

There is a tiny correction to the energies of all of the electron levels in the hydrogen atom due to the classical wobble. This shows up as differences in the observed wavelengths of the hydrogen emission lines from the predicted wavelengths where the "wobble" motion energy is not taken into account. The hydrogen atom is the simplest case, of course, but this effect appears in all atomic spectra.

4. Photons don't just turn into particles spontaneously. Never! Ever!

Yes, they do. This has been observed many times.

If that were the case, just about any photon flying from the Sun to Earth could turn into a particle-antiparticle pair.

Nonsense. In order to produce a particle/antiparticle pair, the photon must have sufficient energy to account for the mass-equivalence of the two particles. Since essentially none of the photons coming from the Sun have anywhere near the required energy, none of them can pair-produce.

The only way for a photon to turn into matter is to interact with something - hit a particle, for example. Even so, the amount of energy the photon carries is irrelevant once it reaches high enough energy for certain heavy particles to form. A 511 keV photon has enough energy to turn into an electron and that's only X-ray, not some bizarrely insane energy density.

Sorry, this is incorrect. A .511 MeV photon has only half the energy required, since the energy of both the electron and the positron (the same energy) must be present. One gamma rays exceeding this energy (1.022 MeV) appear, positron/electron pairs also appear. The same goes for gamma-rays of photon energies greater than 1.862 GeV, twice the mass-energy of protons (931 MeV); proton/antiproton pairs appear. This has been observed many times.

5. Even if the state you described existed, it wouldn't limit the upwards temperature.

I'm not clear about what you mean by "upwards temperature". Please define this.

6. "Higgs field conditions"? Can you translate that into English for me, please?

The Higgs field is a condition where energy densities are so high that no distinction can be made between electromagnetic fields and "color" fields between quarks. In this state, distances and times can no longer be defined, and no particles exist. This is thought to be the state of the universe immediately after the Big Bang.

Finally, a freindly admonishment: before you use vehement language in a scientific discussion, you might find it helpful to know with whom you are speaking. I've been teaching physics at the university level for decades.

 

[Fizyk]

Nonsense. In order to produce a particle/antiparticle pair, the photon must have sufficient energy to account for the mass-equivalence of the two particles.

A single photon in vacuum can't produce a pair, even if it has sufficient energy. One of the created particles would have to be a tachyon.

Two photons can create a particle/antiparticle pair (of course if they have enough total energy), but I don't know what is the cross-section for that reaction, probably not very big, so it also won't happen very often.

 

[Starman]

A single photon in vacuum can't produce a pair, even if it has sufficient energy. One of the created particles would have to be a tachyon.

Two photons can create a particle/antiparticle pair (of course if they have enough total energy), but I don't know what is the cross-section for that reaction, probably not very big, so it also won't happen very often.

One photon, not two. Photons do not interact with one another.

That depends on what you mean by a "vacuum".

The theory is that a photon requires interaction with a Coulomb field in order to pair-produce: a nucleus has a Coulomb field, of course, but such fields can (and do) exist independently of the presence of matter. It is possible that fields other than the Coulomb field can trigger it as well. For example: the hypothesized of pair-production at the event-horizon of a Black Hole. in that case, the gravitational field suffices.

It ought to be noted that the theory of particle-production are require ad hoc connections between low and high energy state wavefunctions. Such "bridges" can be useful in describing particular situations, but are suspect when applied to situation different than the ones under consideration. For a complete description of pair-production, we will likely have to wait for a viable unified-field theory.

It's not clear that tachyons exist.

 

[Fizyk]

One photon, not two. Photons do not interact with one another.

That depends on what you mean by a "vacuum".

The theory is that a photon requires interaction with a Coulomb field in order to pair-produce: a nucleus has a Coulomb field, of course, but such fields can (and do) exist independently of the presence of matter. It is possible that fields other than the Coulomb field can trigger it as well. For example: the hypothesized of pair-production at the event-horizon of a Black Hole. in that case, the gravitational field suffices.

Yes, one photon can create pairs in the presence of an external field, but that is exactly what I call "not a vacuum"

Photons don't interact with one another directly, but they can interact indirectly:

If time flows to the right, we have two photons, which produce an electron and a positron.

It's not clear that tachyons exist.

My point exactly.

 

[Linguofreak]

One photon, not two. Photons do not interact with one another.

That depends on what you mean by a "vacuum".

The theory is that a photon requires interaction with a Coulomb field in order to pair-produce: a nucleus has a Coulomb field, of course, but such fields can (and do) exist independently of the presence of matter.

For instance, in electromagnetic waves. And a photon is just a quantized electromagnetic wave, and thus has associated electric (Coulomb) and magnetic fields. So the photons interact with each other's electric fields to produce the pair.

Also, as far as I understand it, the structure of quantum electrodynamics implies that if the reaction "positron + electron -> 2*gamma" exists (an electron/positron pair annihilating in a vacuum, which needs the two photons to conserve momentum), then the opposite reaction also has to exist.

 

[RisingFury]

I'm not talking about the internal KE of the bound system, but the KE of the entire system with respect to the external coordinate system. Not the same thing.

Exactly. And that can be arbitrarily low.

This is classical physics. In quantum physics the "wobble" of the nucleus appears as the KE of the entire atom. Granted that this is a very small amount of energy, but, since the electron cannot have less energy than it does in the ground state, the energy of the "wobble" motion cannot be removed. Since it is a KE of the entire atom, not just a part of it, a collection of atoms (or molecules, the same principle applies) will have some vibrational energy which cannot be removed. Since a vibration contains KE, and temperature is the KE of the system of atoms, there will be a temperature below which the collection of particles gannot go.

No no no.

For one, any energy the electrons and protons have cannot be considered the temperature of the system. Only the movement of the atom relative to the external frame of reference can be.

Second, you can't think of an atom as an electron orbiting the nucleus. It's impossible. Any accelerated charge gives off electromagnetic waves and loses energy. In such a system, the electron would spiral into the nucleus very quickly.

An additional consideration is the uncertainty principle. You can't know where a particle is, you can just assign a probability for any moment in time. You can't know its velocity, the best you can know is the movement of expected position and uncertainty.

Yes, they do. This has been observed many times.

Yea, and in what universe would that be?

Nonsense. In order to produce a particle/antiparticle pair, the photon must have sufficient energy to account for the mass-equivalence of the two particles. Since essentially none of the photons coming from the Sun have anywhere near the required energy, none of them can pair-produce.

Absolutely not. A photon alone won't ever produce pairs. It needs something to interact with.

Besides, photon energy can be very low and pair generation can still exist. For example, if a high energy particle hits a low energy photon, then in the frame of reference of the particle, the photon has high energy.

Sorry, this is incorrect. A .511 MeV photon has only half the energy required, since the energy of both the electron and the positron (the same energy) must be present. One gamma rays exceeding this energy (1.022 MeV) appear, positron/electron pairs also appear. The same goes for gamma-rays of photon energies greater than 1.862 GeV, twice the mass-energy of protons (931 MeV); proton/antiproton pairs appear. This has been observed many times.

Yea, you're right, it needs double the energy and in most cases some more because conservation of momentum will dictate that usually particles created after the fact will have to move...

I'm not clear about what you mean by "upwards temperature". Please define this.

The system you described wouldn't have a maximum highest temperature. Even if you get to the point where particles don't exist or exist for very short periods of time, temperature can still be defined. You can still assign temperature to a photonic gas...

Finally, a freindly admonishment: before you use vehement language in a scientific discussion, you might find it helpful to know with whom you are speaking. I've been teaching physics at the university level for decades.

Great, you warn me of my language and use an appeal to authority in the same sentence. Teaching Physics at university level doesn't make you immune to incorrect information...

 

[RisingFury]

Also, as far as I understand it, the structure of quantum electrodynamics implies that if the reaction "positron + electron -> 2*gamma" exists (an electron/positron pair annihilating in a vacuum, which needs the two photons to conserve momentum), then the opposite reaction also has to exist.

Two or three. Besides conservation of energy and momentum, you have to conserve angular momentum. A photon has a spin of 1 and electron and positron have spin of 1/2, so you can have two cases:
1/2 + 1/2 = 1 + 1 - 1 for three photons or
1/2 - 1/2 = 1 - 1 for two photons.

 

[Linguofreak]

Two or three. Besides conservation of energy and momentum, you have to conserve angular momentum. A photon has a spin of 1 and electron and positron have spin of 1/2, so you can have two cases:
1/2 + 1/2 = 1 + 1 - 1 for three photons or
1/2 - 1/2 = 1 - 1 for two photons.

I wasn't excluding the possibility of more when I said two were needed.

And couldn't you also have:

4 photons: 1/2 - 1/2 = 1 + 1 - 1 - 1
5 photons: 1/2 + 1/2 = 1 + 1 + 1 - 1 - 1

and so forth, or is there something I don't know about that prevents that?

 

[RisingFury]

I wasn't excluding the possibility of more when I said two were needed.

And couldn't you also have:

4 photons: 1/2 - 1/2 = 1 + 1 - 1 - 1
5 photons: 1/2 + 1/2 = 1 + 1 + 1 - 1 - 1

and so forth, or is there something I don't know about that prevents that?

Yes, you can have more photons like that, just the probability of it decreases quite quickly.

As long as charge, energy, linear and angular momentum are conserved...

 

[Starman]

Exactly. And that can be arbitrarily low.

No, it cannot. The zero-point energy has a minimum value, inherent in the electron being bound to the nucleus. It is not arbitrarily low, it is proportional to the energy of the ground-state.

No no no.

For one, any energy the electrons and protons have cannot be considered the temperature of the system. Only the movement of the atom relative to the external frame of reference can be.

The oscillations of the system ARE relative to the external frame of reference.

Second, you can't think of an atom as an electron orbiting the nucleus. It's impossible. Any accelerated charge gives off electromagnetic waves and loses energy. In such a system, the electron would spiral into the nucleus very quickly.

Again, this is classical physics, falsified by quantum physics. An electron in a bound state will not emit EM waves unless the energy emitted corresponds to a change in energy state. Only free particles emit when accelerated (e.g. Compton scattering).

An additional consideration is the uncertainty principle. You can't know where a particle is, you can just assign a probability for any moment in time. You can't know its velocity, the best you can know is the movement of expected position and uncertainty.

This is not a correct statement of the Uncertainty Principle. You CAN know where a particle is, and you CAN know its momentum, energy, etc., as long as the product of the uncertainties are greater than Planck's Constant. In a bound system like the atom, the volume of space enclosed by the uncertainty is the same size as the atom itself.

But, now that you have invoked Uncertainty, it ought to be noted that the Uncertainty Principle forbids energy in any bound system from being "arbitrarily low". The product of the energy uncertainty and the time uncertainty always must be greater than Planck's Constant; the result of this is that an arbitrarily small energy, with a correspondingly small uncertainty in the energy, wold take an unlimited amount of time to measure. Measuring an energy exactly zero would be impossible, but the Uncertainty Principle dictates some kinetic energy must always be present.

Besides, photon energy can be very low and pair generation can still exist. For example, if a high energy particle hits a low energy photon, then in the frame of reference of the particle, the photon has high energy.

Sorry, but photons don;t hit other photons. If you have an example of such a [phenomenon, you ought to share it.

Yea, you're right, it needs double the energy and in most cases some more because conservation of momentum will dictate that usually particles created after the fact will have to move...

To clarify: photons carry both energy AND momentum. A positron-electron pair emerging from pair-production can have zero momentum (if they're moving away from each other, the net momentum is zero) but the photon;s momentum must be conserved, so the two particles will be moving with VERY relativistic speeds, unless there is another particle involved. A proton would absorb the momentum of the positron-electron photon without reaching relativistic speeds, also providing a perfectly good Coulomb field. There are indications that this is what provides the energy in active-galaxies.

In any case, the context here is temperature; at extremely high temperatures, where photon-energies of thermal radiation match mass-equivalence, there would be plenty of particles available to assist with pair-production.

The system you described wouldn't have a maximum highest temperature. Even if you get to the point where particles don't exist or exist for very short periods of time, temperature can still be defined. You can still assign temperature to a photonic gas...

Yes, but this is not a photonic gas, is it? By definition, we are discussing matter.

Statistical mechanics defines "temperatures" for many different types of systems. Sometimes statistical temperatures can even be negative.

The temperature at which pair-production is in equilibrium with the photon-field represents the maximum temperature; any further energy input simple means more matter is produced. An analogy: you can pump as much heat into ice as you like; its temperature will not increase until the ice is melted.

Great, you warn me of my language and use an appeal to authority in the same sentence. Teaching Physics at university level doesn't make you immune to incorrect information...

If you are going to pronounce someone's "information" to be "incorrect", you ought to have a thorough understanding of the subject yourself. Sorry, but you clearly do not have such understanding. Your description of the Uncertainty Principle exposes a very superficial, mass-media level of knowledge. You have a lot of physics-words in your vocabulary, this is clear, but you just don't connect the jargon and the concepts in a defendable way.

 

[Shadownailshot]

Soooo... back to the original question. Temperature is just an expression of the energy in a system. If we take one particle, it's maximum temperature would be the energy of that particle as close to c as possible, would it not?

 

[Linguofreak]

Soooo... back to the original question. Temperature is just an expression of the energy in a system. If we take one particle, it's maximum temperature would be the energy of that particle as close to c as possible, would it not?

Temperature is not defined for single particles

 

[Shadownailshot]

Temperature is not defined for single particles

But temperature is just energy in a system, so "max temperature" would be what ever the energy is when moving at/near c, right?

 

[Starman]

But temperature is just energy in a system, so "max temperature" would be what ever the energy is when moving at/near c, right?

Temperature is a measure of the average kinetic energy of particles in a system. The energy in the system depends on the number of particles in the system.

Also, as temperature increases to very high levels, radiation becomes a significant part of the system. Radiation pressure can actually be much greater than the pressure resulting from the motion of particles; this is the situation in the core of a star, for example.

As the speed of a particle approaches c, more and more energy must be supplied to increase the speed, since the mass of the particle increases as speed increases. So, at speeds close to c, you can add a lot of energy without increasing the speed very much. Most of the energy goes into the increase in mass.

 

[Shadownailshot]

That clears it up, thanks.

 

[orb]

ScienceNews: Hottest temperature ever measured is a negative one:

Coaxing a gas to a negative temperature on the kelvin scale has produced, paradoxically, the hottest temperature ever measured. The study, published in the Jan. 4 Science, will help physicists learn about quantum phenomena and perhaps even the strange form of energy that dominates the universe.

A negative kelvin temperature indicates that particles at high energies outnumber those at low energies.

{...}

Physicist Ulrich Schneider at the Ludwig Maximilians University of Munich set out to do something unusual: He wanted to cajole the particles within a substance to be confined to a very high amount of energy. In other words, instead of having the particles start at a minimum energy (corresponding to absolute zero) and spreading out toward higher energies, he wanted to start at a maximum energy and spread toward lower energies. By definition, such a substance would have a negative kelvin temperature.

His team achieved that with potassium atoms chilled to a few billionths kelvin above absolute zero. Through the use of lasers and magnets, the team managed to get the atoms to jump to a high-energy state. By creating a cluster of particles exclusively at high energies, Schneider and his colleagues had a gas at a few billionths negative kelvin.

This temperature is technically not below absolute zero, because negative on the kelvin scale (unlike that on the Fahrenheit or Celsius scale) is a construct that simply indicates something about the energy state of the particles involved. In fact, the new creation is extremely hot because of the high energies of the particles. Heat travels from hot to cold, Schneider says, and heat will always flow away from this gas. “It’s actually hotter than everything we know,” he says.

Despite the semantics involved, this experiment isn’t merely a fun physics trick. Scientists are fascinated by negative-temperature substances because they have other strange properties. The molecules in a typical gas spread out and exert a force on the walls of their container. But a negative-temperature gas also has negative pressure, meaning the particles tend to cave in rather than expand. “It wants to collapse into a single point,” Schneider says.

{...}