Back off, man. I’ve been cyber-flirting with Karen (about once a year) since last century! I bet you don’t even know that she’s a penguinist. Only her biggest fans remember that kind of stuff.
Gosh. I sound a tad stalker-ish, don’t I?
<Movie-Trailer-Voiceover-Voice>
In space, it’s about 2.7K outside.
Space… is about to heat up…
</Movie-Trailer-Voiceover-Voice>
In space, no one can hear you pant.
Just to pile on with the pointless nitpickery, Merriam-Webster shows “kelvin” as the unit of temperature for the Kelvin scale and “Kelvin” as an adjective referring to the Kelvin scale.
Well, hubba, hubba! I got her attention…!
So, Karen…just what do you like to talk about, besides how cold it is in space?? Maybe we could warm it up a bit…
(…geez, intellectually, you are SO out of my league…)
…and I suppose I should stop now, so as not to hijack your nice, clean comments thread any further…
Gosh, you mean you actually take a company’s rules–soley to protect its trademark–as a serious issue?
Good grief. I can guarantee you that, within a few years, both “photoshopping” and “googling” will show up in the latest Oxford’s. And Adobe’s rules be hanged.
It’s called a registered trademark, and even The Great One believes. As for showing up in the Oxford English Dictionary (I don’t think it’s trademarked, but “Oxford’s” is a rather cheeky way to refer to such an austere reference), This is how “Band-Aid” shows up:
Note that
(a) it is clearly marked out as proprietary,
(b) the listed alternate meanings are not directly related to the use of literal Band-Aid brand adhesive strips,
(c) the OED tries walking a fine line between descriptivism and prescriptivism anyhow, chronicling usage but also indicating nonstandard or improper usage as it does. That is: the fact that “to photoshop” might be listed in the OED doesn’t mean that even the editors of the OED think it’s a correct usage.
And as an addendum: I didn’t say I follow their rules. I was using it as a comparative to the “proper” use of “Kelvin”, as prompted by Hail Ants. I.e.: Informally I say an image was photoshopped because it’s the quickest and easiest way to communicate what I mean. Formally I know that it’s technically incorrect.
Now, if a staffer were to write an article saying she “googled” something and a doper were to call her on the technicality (especially calling it an “unbelievable rookie mistake”), we’d all agree that the doper in question was just being an insufferable prick. All I was doing was drawing the analogy to attempt to shame **Ants into having respect for his betters :D.
Please forgive my cheekiness regarding Oxfo… I mean, The Oxford English Dictionary. (No sarcasm.) I had no idea that anyone might object to that usage. And doubtless I could have found better ways to say what I said.
However, the gist of my point is that, to the dismay of many corporations, language is going to evolve regardless of anyone’s rules. People are still going to apply a “band-aid” manufactured by, say, Curad. People are still going to “google” for information. They’ll still ask for a “kleenex” when they sneeze. They’ll still ask for a “coke” when an establishment offers only Pepsi products, and drink what they’re given. (Okay, maybe not that last one. 
)
I fully understand why companies try to protect their trademarks, and would do so myself if I were in their position. But you can’t change human behavior. To wit:
This is precisely how such a word becomes inextricably embedded in our language. Well, okay–one way. But’s a big way, ain’t it? (Irony alert!)
And I probably shouldn’t have invoked the OED in the first place–they have their own criteria for deciding what words are valid, and the population tends to have its own.
Oh–and permit me a brief addendum, too (dang it, I wish we could edit posts!)
Referring to the Great One’s column you mentioned, he saved me quite a bit of googling (oops!) by mentioning no less than nineteen trademarked names that have fallen victim to common use:
What more can I say?
As it stands, however, “Photoshop” is still a legal trademark of Adobe. There are technically right and technically wrong usages, but to insist on technical accuracy of such minutiae in a (relatively) informal setting intended to illuminate rather than expound would be, as I said, rather prickish. In fact, if someone responded to my hypothetical staffer’s article that used “photoshop” as a verb by pointing out the error, the most likely conclusion is that the responder is showboating – trying to impress the forum with his knowledge of such trivia.
I do hope my subtext is clear enough that I needn’t connect these dots for anyone reading.
Space is not a perfect vacuum. The number of particles per unit volume can be calculated. You might have done this back in physics classes. The particle population density varies with proximity to stars, planets, comet trails, astronaut feces, etc. One way to define temperature is a measure of the kinetic energy of the particles that make up whatever it is you are measuring the temperature of. The temperature of space can also be calculated. The extremely low particle population density dictates an extremely low temperature. That’s all there is to it.
Unless of course you use some other definition of temperature such as the crappy one in my dictionary which says it is “the degree of hotness or coldness of something as shown by a thermometer.” This definition would make Heisenberg gag since the thermometer has mass and can influence the temperature of the substance being measured. The lower the mass of the substance the more the influence. So a thermometer would not be an appropriate device for measuring the temperature of space since the mass of a unit volume of space is so low.
The radiation in space does influence the kinetic energy of the particles in space but the overiding factor is the low number of particles. The radiation could however influence the temperature of a thermometer making it even less useful for this application.
OK, I’m done.
Yeah, but what’s with that website? 
Nothing but a picture of a penguin and a link back here (well, the Straight Dope HomePage).
Granted, the penguin is nekkid, but there’s way better penguin porn to be found on the Internet.
This isn’t a problem, since the volume of space is so large. The thermometer will change its temperature until it is in equilibrium, and the temperature of space will not be detectably different.
This could be a problem, but more because if the “temperature” of the particles and the “temperature” of the radiation are different, then the “temperature” of space isn’t really well defined in the first place. Different thermometers could give different readings. If you have radiation and particles in thermal equilibrium, then all thermometers will give the same reading.
This statement is simply wrong. The number of particales has nothing to do with what their temperature is.
Did you ever consider that perhaps she is a penguin? I hope you didn’t offend her.
OK, ya got this little thermometer - let’s say about 3 grams worth. Then ya got this volume of space - let’s say 1 cubic kilometer. The space has about 1 particle per cubic centimeter. Most of the particles are hydrogen atoms. So the cubic kilometer of space has about 1 x 10^15 particles or about 1.66 x 10^-8 grams. To match 3 grams you would need 181 million cubic kilometers. Remember this is for 3 grams. You would need many times this much to reduce the effect of the thermometer to undetectable levels depending on your desired accuracy. Now the kenetic energy is transferred during collisions, particle - thermometer collisions and particle - particle collisions. To reach equilibrium you would need lots and lots of collisions. The effect of the thermometer would very slowly spread in a spherical pattern. The closer you get to equilibrium the slower the process gets. We could calculate the time but I bet your grandchildren will lose their teeth first. Don’t know about you but I call this a problem.
No, the problem is more the effect of the radiation on the thermometer like I said. I think we do agree the radiation is a messy factor though. The thermometer is just a bigger target than the nuclei of a few particles and absorbs more energy. This is very much like trying to measure the temperature of our air while the thermometer is in direct sunlight. Hint - it always reads high.
You are talking about the temperature of a particle, I am talking about the temperature of a system. This depends on definitions and gets kind of semantical. Lets say temperature is the measure of the hotness or coldness of something. Actually there is no such thing as coldness, it is really just the lack of kenetic energy or absence of heat. The particles in space have their own individual levels of kenetic energy and you could say this is the temperature of the particle. I tend to agree but the particle is not the something we are looking at here, space is the something we are looking at. Lets say you have a container filled with something warm. The energy of the particles in the something is transferred to the wall of the container by collisions with the wall. If the something is melted butter you will have many collisions of butter particles[sic] with the wall and the temperature will be a warm temperature because much energy is transferred to the wall. Now lets say the something in the container is space. Each of the particles in the space has the same kenetic energy as one particle of butter. There will be very few collisions with the wall because there are very few particles. There will be a definite absence of energy and heat. Regardless of the ‘temperature’ of the individual particles, there just are not enough of them. When the particle population is so extremely low the temperature of the system has everything to do with the number of particles.
I made no comment about her penguin.
No, you don’t need any collisions. Even with no particles present, the thermometer will emit blackbody radiation until it is equilibrium. If it were out in space between the galaxies, it would drop down to 2.7 K, assuming it’s hotter than that to begin with. If it were colder, it would absorb energy until it rose to 2.7 K.
The problem with radiation is that if the radiation and the particles aren’t in thermal equilibrium, temperature isn’t well defined.
You seem to be confusing energy and heat with temperature.
I neither said nor implied you did.
You are simply choosing to ignore what space is. Space is a low concentration of baryons in a volume. This space also happens to have microwave radiation passing through it. The radiation is passing through space, it is not space.
Actually the thermometer will emit thermal radiation not blackbody radiation.
No, you do not need equilibrium to have a “well defined” temperature. That would imply that all temperatures are constant and dT/dt = 0. This just isn’t true. The entropy in the universe is always increasing and we really won’t have thermal equilibrium until it stops but that hardly prevents us from having well defined temperatures.
Just what are you using for your definition of “temperature”? I tend to like the theoretical definitions best.
“Temperature is a measure of the tendency of an object to spontaneously give up energy to its surroundings. When two objects are in thermal contact, the one that tends to spontaneously lose energy is at the higher temperature.”
Schroeder, Daniel V., An Introduction to Thermal Physics, Addison Wesley, 2000.
Or my favorite:
“Temperature is that property of matter which is equal in two systems at thermal equilibrium with each other.”
Long live penguins.
Well this is your problem. Space is not the baryons in a volume, space is the volume that the baryons are in. Space contains baryons, just as space contains the radiation passing through it. As Karen Lingel said way back in August,
When discussing the temperature of space, the only sensible meaning is the temperature of a volume of space, including whatever particles and radiation are in it. If you want to talk about the temperature of just the baryons, fine, but don’t claim that that is the temperature of space.
If I have a jar, and in that jar is a glass of water at 40 C and an otherwise identical glass of water at 0 C, what is the temperature of that jar? This is entirely analogous to asking for the temperature of space, with the baryons at one temperature and the radiation at another.
Let’s try to use your definitions of temperature for the jar with the two glasses of water. If I put an object at, say, 20 C next to the side of the jar with 40 C glass of water, heat will flow into the object from the jar, so by the first definition you quoted, the jar is at a higher temperature than the object. On the other hand, if the object is placed on the side of the jar with the 0C glass of water, heat will flow into the jar, so, again by your first definition, the jar is colder than the object. Thirdly, I can put the object next to the jar at the right location and have no heat flow, so the jar and object are at the same temperature by your second definition. Clearly the object can’t be hotter, colder, and at the same temperature as the jar all at once. The resolution is that your definitions of temperature implicitly assume that an object with a temperature is at thermal equilibrium.
Try not to be so itsy bitsy - I did not mean space is baryons. Saying space is baryons in a volume means the same as space is a volume with baryons in it. Just like saying sound is compression waves in a medium means the same as sound is a medium with compression waves in it or like saying 3 is 2 plus 1 means the same as 3 is 1 plus 2. I was merely giving a casual description of space to include the paticles and I should have included the photons too. This was not meant in any way to be a formal definition.
“Temperature is a measure of the tendency of an object to spontaneously give up energy to its surroundings. When two objects are in thermal contact, the one that tends to spontaneously lose energy is at the higher temperature.”
Schroeder, Daniel V., An Introduction to Thermal Physics, Addison Wesley, 2000.
Ohforgodssake, even you can’t believe what you are saying. Your jar as a system is not at thermal equilibrium. Big deal. It still has temperature. It just has different temperatures in different regions. The glass with the higher temperature is a region in the jar which is giving up energy to it’s surroundings. Some of it’s surrounding are inside the jar and some are outside. Haven’t you ever heard of a temperature gradient? Take a metal rod and heat one end. You will have different temperatures all along the rod but you will still have temperatures. The definition does not imply or assume equilibrium at all, try reading it. There are so many real things that are not in thermal equilibrium and still have temperature. How is it possible you could have missed them? Your own body is not at thermal equilibrium, your feet are at a lower temperature than your kidneys. But if you shove a thermometer up your ass you will find it still has temperature.
I think you just want to argue and I am tired of it. Good luck in your next marriage. Bye.
You have temperatures (plural) at different points along the rod. The question you can’t answer is what is the temperature (singular) of the rod.
As I said before, this is entirely analogous to asking what the temperature (singular) of space is. You wrote “I was merely giving a casual description of space to include the paticles and I should have included the photons too.” (bolding mine), so you agree that you should have included the radiation when talking about the temperature of space. If the particles in space are at one temperature, and the radiation (photons) are at another, then what is the temperature of space?
No, I stand by this statement.