Does a vacuum have a temperature?

[harp AP 2010]

Does a vacuum have a temperature?

 

[OmCheeto]

Does a vacuum have a temperature?

no.

a vacuum is an absence of matter.
temperature is vibration of matter.

 

[K^2]

Depends on how you want to look at it. I would say that electromagnetic field, being a component of electromagnetic vacuum, should be considered part of vacuum. And electromagnetic field, when excited, can have a temperature. Of course, that would mean you have photons in your vacuum, and it really depends on whether you define space with some photons in it as still being vacuum.

 

[OmCheeto]

Depends on how you want to look at it. I would say that electromagnetic field, being a component of electromagnetic vacuum, should be considered part of vacuum. And electromagnetic field, when excited, can have a temperature. Of course, that would mean you have photons in your vacuum, and it really depends on whether you define space with some photons in it as still being vacuum.

I don't believe photons, nor electromagnetic fields, have what would be defined as a "temperature" component.

They might indicate that the originators of the photon or e-m field had a certain temperature, but they themselves don't.

 

[K^2]

It's a stat-mech thing. A single photon doesn't have a temperature, but there is no such thing as a single physical photon. A mixed state, however, does have a temperature, because it has an energy distribution. So any physical EM field will have a temperature.

(To clarify, here "single photon" means a photon of one specific frequency. You can have a superposition of these excited so that you will always measure exactly one photon, but there is still going to be an energy distribution in the state itself, and that still means a finite temperature.)

 

[OmCheeto]

It's a stat-mech thing. A single photon doesn't have a temperature, but there is no such thing as a single physical photon. A mixed state, however, does have a temperature, because it has an energy distribution. So any physical EM field will have a temperature.

(To clarify, here "single photon" means a photon of one specific frequency. You can have a superposition of these excited so that you will always measure exactly one photon, but there is still going to be an energy distribution in the state itself, and that still means a finite temperature.)

Well, you are obviously more educated than I am, as I would not be able to us some of the terms/phrases you've used(stat-mech, superposition, energy distribution in the state) in a sentence.

But perhaps you could clarify; "there is no such thing as a single physical photon."

I was under the impression that it was a single photon that was responsible for the photo-electric effect. Also, when an electron jumps from a higher orbit to a lower orbit, a single photon is emitted.

I'm just a layman in this area, so please break it to me gently if I've had a gross misconception of reality for the last 30 years.

 

[Danger]

As a high-school dropout, I agree with you Om. How can a single photon not exist in the face of electron orbital shifts and the photoelectric effect? We might both be missing something, but you make good sense to me.

 

[OmCheeto]

As a high-school dropout, I agree with you Om. How can a single photon not exist in the face of electron orbital shifts and the photoelectric effect? We might both be missing something, but you make good sense to me.

Well, as K^2 said, it's a "stat-mech" thing, and I have no knowledge of that.

Someone edumacate me!

 

[K^2]

Ah, the confusion here is that "photon" can mean one of two things. It can be a normal mode in the EM field, or it can be a single quantum of energy.

This whole thing is only going to make sense in context of quantum mechanics. I don't think there is a point in trying to explain everything in high detail. So let's try dropping all of the complications.

Basically, forget what I said above. The main point is that there is no way that EM field is going to have just one specific energy. If you try to measure energy, you'll end up with some sort of a probability to measure different values. That probability distribution is going to give you the temperature.

EM field is a boson field, so there are some additional QM complications, but in general, if you forget all the QM stuff, the probability of finding a particle with energy E at temperature T is given by exp(-E/kT)/Z, where Z is such that the total probability is 1.

The real trick here is that even if you take a single photon, in the photoelectric experiment sense, the energy of that photon is still undetermined, and will have a distribution to it. It's not going to be the above distribution, but it's still going to be a distribution where we can talk about the temperature of that photon, and it will be related to the temperature of the particle that has emitted it.

 

[OmCheeto]

Ah, the confusion here is that "photon" can mean one of two things. It can be a normal mode in the EM field, or it can be a single quantum of energy.

This whole thing is only going to make sense in context of quantum mechanics. I don't think there is a point in trying to explain everything in high detail. So let's try dropping all of the complications.

Basically, forget what I said above. The main point is that there is no way that EM field is going to have just one specific energy. If you try to measure energy, you'll end up with some sort of a probability to measure different values. That probability distribution is going to give you the temperature.

EM field is a boson field, so there are some additional QM complications, but in general, if you forget all the QM stuff, the probability of finding a particle with energy E at temperature T is given by exp(-E/kT)/Z, where Z is such that the total probability is 1.

The real trick here is that even if you take a single photon, in the photoelectric experiment sense, the energy of that photon is still undetermined, and will have a distribution to it. It's not going to be the above distribution, but it's still going to be a distribution where we can talk about the temperature of that photon, and it will be related to the temperature of the particle that has emitted it.

bolding mine...

I do believe, that that is what I originally said.

I am old, and only by a twist of fate, am I not your grandmother...

remember that the next time you try and correct your grandmother, just for the fun of it.
Cuz' remember, she wiped your mom's butt...

My apologies if I come off as somewhat rude. I'm venting a national rage at the moment.

I should crawl off to chat or something...

 

[K^2]

Oh, I can explain it. I'd have to start way, way, way back, and it'd essentially be a series of lectures. I hope I don't have to explain why I'd be disinclined to do so. I can point you to a few textbooks, of course.

 

[Danger]

What confuses me most is the contention of scientists that sound is not transmitted in a vacuum. I accidentally vacuumed up the ex-wife's parakeet a couple of years back, and I could hear the damned thing squawking for 3 days.