Quantcast
  • Register
PhysicsOverflow is a next-generation academic platform for physicists and astronomers, including a community peer review system and a postgraduate-level discussion forum analogous to MathOverflow.

Welcome to PhysicsOverflow! PhysicsOverflow is an open platform for community peer review and graduate-level Physics discussion.

Please help promote PhysicsOverflow ads elsewhere if you like it.

News

PO is now at the Physics Department of Bielefeld University!

New printer friendly PO pages!

Migration to Bielefeld University was successful!

Please vote for this year's PhysicsOverflow ads!

Please do help out in categorising submissions. Submit a paper to PhysicsOverflow!

... see more

Tools for paper authors

Submit paper
Claim Paper Authorship

Tools for SE users

Search User
Reclaim SE Account
Request Account Merger
Nativise imported posts
Claim post (deleted users)
Import SE post

Users whose questions have been imported from Physics Stack Exchange, Theoretical Physics Stack Exchange, or any other Stack Exchange site are kindly requested to reclaim their account and not to register as a new user.

Public \(\beta\) tools

Report a bug with a feature
Request a new functionality
404 page design
Send feedback

Attributions

(propose a free ad)

Site Statistics

205 submissions , 163 unreviewed
5,047 questions , 2,200 unanswered
5,345 answers , 22,709 comments
1,470 users with positive rep
816 active unimported users
More ...

  Why does large curvature of spacetime imply high temperature?

+ 9 like - 0 dislike
2475 views

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?

asked Mar 25, 2012 in Theoretical Physics by Dilaton (6,240 points) [ revision history ]
Please dont shoot me if this question and my considerations are too crazy, I`m sure my confusion is curable ... :-)

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Manishearth
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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user John Rennie
@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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
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

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Jerry Schirmer

3 Answers

+ 7 like - 0 dislike

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".

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user John Baez
answered Mar 26, 2012 by John Baez (405 points) [ no revision ]
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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Ron Maimon
Hi @JohnBeaz, thanks for this concise answer, I +1ed too :-)

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
@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...

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
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!

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Terry Bollinger
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!

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Manishearth
Dear @JohnBaez thanks a lot for these great additional explanations and very interesting links. Studying this will keep me busy and happy for a while :-)

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
+ 5 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Ron Maimon
answered Mar 26, 2012 by Ron Maimon (7,720 points) [ no revision ]
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)?

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
@RonMaimon, nice answer & interesting to read, thanks!

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Terry Bollinger
@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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Ron Maimon
+ 4 like - 0 dislike

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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Terry Bollinger
answered Mar 25, 2012 by Terry Bollinger (110 points) [ no revision ]
Most voted comments show all comments
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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Luboš Motl
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.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Luboš Motl
Thanks for these additional interesting comments @LubošMotl, as you suggest they point to an additional answer I´d appreciate too ...

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
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

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Terry Bollinger
Clarification on "tidal": Whenever an object is large enough in comparison to the geometry of the gravity field, it will encounter divergent acceleration vectors across all of its dimensions. I focused on the horizontal divergences, but the vertical divergences are the tidal forces. So, it's nice conceptually to realize that divergent acceleration vectors are at play in both cases. And this is all hugely simplified! I'm only trying to give some visualization for why curved gravity fields don't play well with big objects occupying space. See the John Baez and Ron Maimon answers for math leads.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Terry Bollinger
Most recent comments show all comments
Thanks @Terry Bollinger, this sounds quite reasonable to me (so +1). I think I`ve heard or read the term "tidal force" for the jostling due to a divergent gravitational field (?) and maybe this is exactly what I was missing. But we should wait and see if "real" relativity experts here have some objections or additional interesting comments.

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user Dilaton
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).

This post imported from StackExchange Physics at 2014-03-17 03:24 (UCT), posted by SE-user David Z

Your answer

Please use answers only to (at least partly) answer questions. To comment, discuss, or ask for clarification, leave a comment instead.
To mask links under text, please type your text, highlight it, and click the "link" button. You can then enter your link URL.
Please consult the FAQ for as to how to format your post.
This is the answer box; if you want to write a comment instead, please use the 'add comment' button.
Live preview (may slow down editor)   Preview
Your name to display (optional):
Privacy: Your email address will only be used for sending these notifications.
Anti-spam verification:
If you are a human please identify the position of the character covered by the symbol $\varnothing$ in the following word:
$\varnothing\hbar$ysicsOverflow
Then drag the red bullet below over the corresponding character of our banner. When you drop it there, the bullet changes to green (on slow internet connections after a few seconds).
Please complete the anti-spam verification




user contributions licensed under cc by-sa 3.0 with attribution required

Your rights
...