Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

I`ve just stumbled about a sentence which says that high curvature of spacetime implies that any matter present is at high temperature.

This somehow confuses me, so my probably dumb question(s) are:

1) How is this general (?) relationship between the temperature of matter and curvature of spacetime derived or explained? Just looking at Einsteins field equations, I dont see why there cant be some cold mass just "sitting there" leading to large curvature ...

2) Is this relationship generally valid or is its domain of applicability restricted somehow? For example has the curvature of spactime to be so large that quantum effects kick in and it has something to do with the uncertainty principle?

share|improve this question
1  
Please dont shoot me if this question and my considerations are too crazy, I`m sure my confusion is curable ... :-) –  Dilaton Mar 25 '12 at 17:13
    
I'm not a GR expert, but this seems to answer your first question.. Temperature is defined as $\frac{\partial U}{\partial S}$. Since internal energy increases with curvature, temperature will also increase provided that entropy does not increase too much. –  Manishearth Mar 25 '12 at 17:47
1  
Please give the reference. The temperature of an event horizon certainly increases as the curvature increases (i.e. as the black hole size decreases) but this seems only a partial answer to your question. –  John Rennie Mar 25 '12 at 17:55
    
@John Rennie The big bang is mentioned as an example of a high curvature situation. But I`ve never heard that the curvature of spacetime should be the reason for the high temperature in this case. What really confuses me is that the sentence says that any matter (independent from its density?) inside the region considered and not (at) the boundary should be at high temperature. –  Dilaton Mar 25 '12 at 18:16
1  
In fact, an idealized neutron star sort of assumes that the neutron star/white dwarf is at a low temperature state, so that the neutrons/electrons are repelling each other due to Fermi exclusion –  Jerry Schirmer Mar 26 '12 at 4:53
add comment

3 Answers

up vote 10 down vote accepted

There's no way to generally identify (some function of) curvature with temperature. Ron Maimon (below) is secretly eager to give examples illustrating this pont.

There is a nice "dictionary" relating black holes and thermodynamics, where the surface area of the event horizon is entropy and black hole "surface gravity" is temperature. In its classic form it applies only to black holes, but it's incredibly interesting and important, so it's worth studying:

http://en.wikipedia.org/wiki/Black_hole_thermodynamics

http://en.wikipedia.org/wiki/Surface_gravity#Surface_gravity_of_a_black_hole

http://www.physics.umd.edu/grt/taj/776b/lectures.pdf

Naturally people have tried to generalize these ideas from black holes to other solutions of Einstein's equations. Ted Jacobson has an interesting argument that derives Einstein's equations from the equation dQ=TdS connecting heat, entropy, and temperature:

http://arxiv.org/abs/gr-qc/9504004

As he notes, "The key idea is to demand that this relation hold for all the local Rindler causal horizons through each spacetime point, with dQ and T interpreted as the energy flux and Unruh temperature seen by an accelerated observer just inside the horizon."

A Rindler horizon is different from an event horizon; roughly speaking, it's the boundary separating the part of spacetime an accelerating observer can see from the part he will never see. Simplest example: if you're in a rocket in Minkowski spacetime and you fire your thrusters so that you feel a constant g force in the same direction, your velocity will approach the speed of light but never get there, yet some photons coming from behind you will never catch up with you, so there will be some objects behind you that you'll never get to see, even if you look over your shoulder. The imaginary surface separating the spacetime points you'll see from those you won't is called the Rindler horizon. But Jacobson is considering a subtler 'local' version of the Rindler horizon, and considering it in curved spacetime.

Since Erik Verlinde's work on 'entropic gravity' there have been further attempts to relate general relativity and thermodynamics. Some of these are reviewed in this paper by T. Padmanabhan:

http://arxiv.org/abs/0911.5004

I haven't really read this paper, though.

In short, there's a bunch of tantalizing relations between gravity and thermodynamics, but they don't proceed by saying "temperature is some function of curvature".

share|improve this answer
3  
Hi John Baez--- nice to see you here. +1 for a correct answer, but it would be nice to add a cold high-curvature example, like a tiny extremally rotating/charged BH. Also a high temeperature can only be uniform in a high curvature dS state, and otherwise will equilibrate away, but I am not sure if this is in the scope of the question. –  Ron Maimon Mar 26 '12 at 5:29
    
Hi @JohnBeaz, thanks for this concise answer, I +1ed too :-) –  Dilaton Mar 26 '12 at 9:15
    
@RonMaimon I`m interested in your counter example and your whole comment is in the scope of my interest. So I`d like it if you could expand it a bit... –  Dilaton Mar 26 '12 at 9:17
1  
Dr Baez, I am delighted to see you here! A brief answer from a deep subject matter expert is often the best answer of all, so I'm certainly giving a +1! –  Terry Bollinger Mar 26 '12 at 16:31
    
Note: I have approved your edit--you were not logged in it seems. Next time, when you want to edit your own post, ensure that you are logged in.. If you had not made the edit, please let me know (type @Manishearth to reply), and I'll reverse it. Thanks! –  Manishearth Mar 27 '12 at 11:17
show 1 more comment

This is an expansion to John Baez's answer, since he didn't do it.

You can have high curvature close to massive cold objects. The classical example is that extremal black hole, which has zero Hawking temperature, but whose curvature is arbitrarily large (up to the Planck limit) in the region outside the horizon. This is a black hole whose charge is equal to its mass in natural units (so that two of these neither repel or attract).

The metric for the charged Reissner Nordstrom black hole solution is the same as the polar Schwartschild form, with $f(r) = (1-M/r)^2$ replacing $(1-{2M\over r})$. Note the squaring. If you make M small, you have arbitrarily large curvature.

If you accelerate a detector in empty space, you see a thermal response which is arbitarily large. This is the same as going close to a non-extremal horizon, and is high-temperature blueshifted Hawking radiation. This is localized in a skin right next to the horizon, but the local temperature blows up without the curvature blowing up, without any curvature at all really.

If you have a high temperature that's the same at all points in space, you have a high energy density. A uniform energy density collapses the space. The way to have a uniform high temperature stable is to have a deSitter universe which is very small, then it has the deSitter temperature. This is the closest analog to a uniform thermal state in GR.

In a deSitter space, and only in this special case, the temperature and the curvature are proportional.

share|improve this answer
    
Thanks @RonMaimon, I`m not too familiar with extremal black holes so I have two potentially dumb queations about them: How can their Hawking temperature be zero (I thougt the Hawking temperature increases with decreasing mass of the Black hole ...)? And why would two opositely (electrically) charged ones not attract each other (I dont get this)? –  Dilaton Mar 26 '12 at 21:57
    
@RonMaimon, nice answer & interesting to read, thanks! –  Terry Bollinger Mar 27 '12 at 0:25
1  
@Nemo: Hawking temperature for neutral black hole increases with decreasing mass, for extremal black holes, it is just absolute 0. These black holes are not thermal, they have a definite quantum state, and this is why they are a good foundation for quantum gravity. The way to see this is that the surface gravity on the horizon diverges in the extremal limit. –  Ron Maimon Mar 27 '12 at 0:27
add comment

Here's an analogy, my own, that I checked out years ago with a very good GR expert. His comment was roughly "that may be one way to think of it," so please take it with some caution. I also immediately bow to anyone with deep tensor knowledge. Still, it is an analogy that is a lot easier to get hold of than most, so I think it's worth describing in answer to your question.

GR says that gravity has a deep equivalence to acceleration. Thus if you imagine a large, flat plane accelerating through space perpendicular to the surface, that flat area has an close mathematical similarity to a small, almost-flat section of a very large sphere with gravity acceleration comparable to that of the sheet.

Notice that for the flat sheet, the acceleration vectors for any point on the sheet are exactly parallel. That's important, because as with velocity vectors, acceleration vectors that are exactly parallel essentially hide the energy that they contain from each other. Thus if two cars are moving down the highway very close together at the same speeds and wit precisely parallel paths, someone can step from one to the other without any dangerous release of energy. The same is true if both cars are accelerating with identical vectors, although of course in that case you would need a ledge to walk on because of the gravity-like field.

Both cases are in sharp contrast to the opposite extreme of two cars with velocity or acceleration vectors that result in head-on collisions. In those cases, the "hidden" energy becomes very real indeed, and catastrophic for anyone inside the cars. In between those two extrema you have anywhere from a "tiny" bit of collision energy from slightly non-parallel car vectors to increasingly drastic energy releases as the parallelism fails more completely.

Now, let's go back to the accelerating sheet in space. There your acceleration vectors are perfectly parallel, so as with the example of the cars, the energy implied by those vectors is hidden from people on the sheet.

I also mentioned that the sheet is almost like a section of the surface of a world with gravity that gives the same acceleration, but of course there's an important difference: The world is a ball, so the gravity vectors on its similar sheet are not exactly parallel; they diverge by a tiny bit.

Now that has an odd implication, which is this: When mass-generated gravity fields curve around on themselves -- and of course all of them do -- they also create a bit of a mismatch in acceleration paths that evidences itself as a bit of energy added to any object that gets caught between its very slightly divergent acceleration paths. This "jostling" energy will be proportional to the area of the object that is orthogonal to the (average) acceleration path. It is of course astronomically tiny for gravity fields such as earth's, yet at the same time it is there and is inherent in the curved structure of the field.

Now finally, imagine that the radius of the object with the gravity field shrinks while the gravity field itself stays constant. In that case, the earlier approximation of the accelerating sheet gets mapped into an increasingly and in the end overtly curved subsection of the spherical surface of the object. At the same time, the acceleration vectors become so non-parallel that even for a small bit of matter, the acceleration vector at one edge of the matter becomes noticeably non-parallel to the other edge. The vectors now have a significant "jostling" factor that is inherent in space itself, and which will express itself on any matter within that space at a higher temperature.

As an addendum, I would add that tidal forces can be interpreted as the vertical version of the same effect. That's because the acceleration vectors will also change in magnitude (but not direction) from the top to the bottom of the object. The simplest heuristic for realizing that more analysis is needed is to look at the size of the object relative to the geometry of the gravity field. If field is strong and the object is large enough to "see" changes in the geometry or strength of the gravity field acceleration vectors, there's going to be trouble, and you need to do a precise mathematical analysis.

So, I think horizontal acceleration vector divergences give a reasonably accurate portrayal of why there is a temperature component to curved space. I'm also pretty sure that there is an equivalence to the concept of a temperature at the surface of a black hole, which as in the case I just described is also extremely dependent on how sharp the curvature is at the surface of the black hole.

And again, while I'd probably say that I'm better than average at four dimensional geometry problems, I am no GR expert and make no claim to be one. My only verification of my analogy was that quick look from an extremely good GR expert years ago who was kind enough to take a look at it.

share|improve this answer
1  
I wouldn't say the tidal force is the jostling - tidal force is more of an apparent pulling apart of the different pieces of object, but it occurs for the same reason (namely that closely separated acceleration vectors are not exactly identical). –  David Z Mar 25 '12 at 19:38
1  
The tidal force should be a bit different, I think. BTW, here's a slightly different way of thinking why the vector divergence is akin to jostling: Ask what kinds of real accelerators (think nano-rockets) would be required to keep an object at the same level in the gravity field if there is no solid surface. For large fields these nanoaccelerators all point in about the same direction, while for severely curved gravity fields the nanoaccelerators needed will try to blow even small objects apart. With confinement (the gravity sphere is small) the result is a pretty close analog to heat. –  Terry Bollinger Mar 25 '12 at 23:25
1  
Dr Bollinger's answer is on the right track but I believe that because he rejects the key concept of tidal forces, someone who is confident that the right buzzword is the "tidal forces" should post a full-fledged answer here (David?). The canonical example is the observer approaching a black hole singularity. He's torn apart by the tidal forces which are really the nonrelativistic equivalent of the spacetime curvature. Tidal forces rip him apart, big force, big temperature. –  Luboš Motl Mar 26 '12 at 5:13
2  
Even if someone sits on a surface of a neutron star, he may not be ripped apart and the situation may be stable. However, the phonons and other virtual particles inside him may undergo similar dynamics as the atmosphere and exhibit the adiabatic lapse rate, something that makes these quasiparticles inevitably warmer in the deeper levels of the gravitational field - the temperature gradient is again linked to the spacetime curvature. There are various effects and they may have different temperatures and there could possibly exist loopholes in which they're absent... Someone should answer fully. –  Luboš Motl Mar 26 '12 at 5:15
1  
Dr Baez was kind enough to send me a link an impressive Ted Jacobson tutorial on black hole thermodynamics. It's long, 40 pages, but impressive in its coverage, even at a first quick scan. Here's the site: physics.umd.edu/grt/taj/776b/lectures.pdf –  Terry Bollinger Mar 27 '12 at 0:21
show 3 more comments

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.