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Temperature of space

  1. Apr 14, 2005 #1

    Pengwuino

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    Ok i heard that the temperature of space was like 2.6K or something. I thought space was a vacuum and a big ol void of nothing (besides the planets). Can someone throw me the rest of the information so it makes sense :)
     
  2. jcsd
  3. Apr 14, 2005 #2
    The temperature of the CMB, the putative afterglow of the big-bang which presumingly fills all of space, is 2.725 K.
     
  4. Apr 14, 2005 #3
    remember light does not need anything to propogate threw and this cbr is just very low energy light which we see as 2.725K this is how space can have a tempature even if it is a vacume but space is not a perfect vacume either even if it is much better then anything we can create today it still had a very low particle density
    fish
     
  5. Apr 14, 2005 #4

    SpaceTiger

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    The kinetic temperature of the gas in space (what we usually mean by "temperature" on earth) can vary quite a lot, anywhere from ten to ten million kelvin. Meanwhile, the temperature of one of the ambient radiation fields (the CMB) is exactly what was stated earlier in the thread. There is yet another radiation field that's given by the combined effect of all the non-dark objects in the universe, but this can't be said to be at a specific temperature. This all just goes to show that the interstellar and intergalactic media are not in thermodynamic equilibrium.
     
  6. Apr 15, 2005 #5

    Chronos

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    Space has no 'temperature', only matter. What you are saying essentially boils down to a big picture that includes an 'aether'. I don't buy that.
     
  7. Apr 15, 2005 #6

    hellfire

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    One can assign a temperature to every kind of event horizon. If we are located inside one (cosmological) event horizon, then space would be in a thermal bath of photons. The corresponding temperature is very, very small.
     
  8. Apr 15, 2005 #7

    Chronos

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    And I have no disagreement with that. I'm only stuck upon the notion we need to equivocate temperature in terms of the average kinetic energy of mass per unit volume of free space.
     
  9. Apr 15, 2005 #8

    Nereid

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    SpaceTiger mentioned something that doesn't get much airtime in the popular media - the 'temperature' of some blob of matter implies an equilibrium, in turn interactions ... how long does/would it take for common blobs to reach equilibrium (assuming a steady state); for example, the IGM, the ISM, the dense core of a giant molecular cloud, (stars in) a globular cluster, (stars in) elliptical galaxies, (galaxies in) a rich galaxy cluster, the outer atmosphere of a red giant, the wind from a Wolf-Rayet, a pulsar wind, ...? Given that many of these are plasmas, how different would the (steady-state) electron temperature be from that of the 'neutrals' (unionised gas atoms or molecules)?
     
  10. Apr 16, 2005 #9
    Nereid, you mentioned something called a "Wolf-Rayet" and the wind it comes from. If you don't mind the curiousity of a student, what is a Wolf-Rayet and how does it have wind?
     
  11. Apr 16, 2005 #10
    Spacetiger mentioned something earlier about the universe not being in thermal equilibrium, which is true. The reason space has such a low 'temperature' is because the 'temperature' measured is just an average kinetic energy measure of all the matter in space. If the universe was in thermal equilibrium, it is referred to as 'heat death' in which the universe reaches thermal equilibrium and would mean that everthing would cease to exsist. There was a thread I started about it a while back titled "Heat Death". It might explain a little bit if you went through and read a few (or all if you like) of the posts.
     
  12. Apr 16, 2005 #11

    Janus

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    Wolf-Rayet's are types of stars. They are very hot stars, with temps in the order of 100,000°.
     
  13. Apr 16, 2005 #12
    That is quite the star. How do they produce the 'wind' Nereid mentioned? Would said wind be produced when the star explodes into a super nova (or is it just called nova? I'm pretty new to astronomy so I really don't know. :blushing: I hope that's ok.) Would a Wolf-Rayet be more prone to a type 1 or type 2 super nova? How old are these stars roughly? How do they get to be so hot!?

    Sorry for smacking you with all those questions all at once. :redface: I hope that's ok.
     
  14. Apr 17, 2005 #13

    Garth

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    misskitty Every star produces a wind - our Sun is no exception. This is because when you solve the five equations that describe the gravitational 'bottle', which contains the high temperature and pressure plasma that makes up the Sun, you find it is leaky, the outer pressure cannot be zero - it has to have a small value, so any surface around the Sun always leaks, no matter how strong the gravity is.

    A Wolf-Rayet star is so powerful that its gravitational field is even worse at keeping the hot plasma in, and a very strong wind is observed so W-R stars are surrounded by a hot and often beautiful nebula.

    Garth
     
    Last edited: Apr 17, 2005
  15. Apr 17, 2005 #14

    turbo

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    Here is a paper on just that topic. The author's treatment of gravitation and thermodynamics leads him to the conclusion that we may be able to model gravitation as a quantum-dynamical effect.

    http://citebase.eprints.org/cgi-bin/citations?id=oai:arXiv.org:gr-qc/0311036
     
  16. Apr 17, 2005 #15

    SpaceTiger

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    Unfortunately, I don't think there's a simple answer to any of these questions. They all depend on the conditions you assume. If you're asking how long it takes for them to reach equilibrium with each other, then the problem is far too complicated for any kind of quick calculation. In fact, I would venture to guess that we couldn't do such a calculation with what we currently know...


    The ISM and IGM aren't generally two-temperature plasmas, but there are special cases, particularly in the presence of magnetic fields (e.g. ambipolar diffusion), where this can change. Again, these are a bit too complicated for the current discussion.

    My posts at the beginning of this thread address some of these issues in more detail.
     
  17. Apr 17, 2005 #16

    SpaceTiger

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    Type II is the core collapse of a massive star and that's what a Wolf-Rayet is.


    Millions of years, usually. They go supernova very quickly compared to most other stars.


    See my discussion of gas pressure in this thread. It's the first few paragraphs.
     
  18. Apr 17, 2005 #17

    Nereid

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    Thanks Space Tiger; I like this answer a lot :approve:

    Would it be OK everyone if we do some exploring, and assume some conditions? I'm interested to get some OOM answers.

    Can we start with the dense core of a giant molecular cloud perhaps? What sorts of things do we have make assumptions about? The ionisation state of the constituent gas? The density? The (gas) composition (which elements? what molecules? what relative abundances?), and dust composition? the size (e.g. so that it receives no significant - unreprocessed - input from the CMBR)?
    Indeed; I would welcome a continuation of that discussion here in this thread, without being sidetracked by crackpot ideas.
     
  19. Apr 18, 2005 #18

    SpaceTiger

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    I'll do my best. Before we get into specifics, it's worth mentioning a few standard numbers. A professor in this department is writing a book on the subject, so I'll take these from his class notes. There are several "phases" of the ISM:

    Coronal Gas: [tex]T\gtrsim 3\times 10^5 K[/tex], [tex]n \simeq 0.003~cm^{-3}[/tex], [tex]f \simeq 0.4[/tex]

    HII: [tex]T\simeq 10^4~K[/tex], [tex]n \simeq 0.3 - 10^4~cm^{-3}[/tex], [tex]f \simeq 0.1[/tex]

    HI Warm: [tex]T\simeq 6000~K[/tex], [tex]n_H \simeq 0.3~cm^{-3}[/tex], [tex]f \simeq 0.5[/tex]

    HI Cool: [tex]T\simeq 100~K[/tex], [tex]n_H \simeq 20~cm^{-3}[/tex], [tex]f \simeq 0.02[/tex]

    Diffuse H2: [tex]T\simeq 60~K[/tex], [tex]n_H \simeq 20 - 100~cm^{-3}[/tex], [tex]f \simeq 0.01[/tex]

    Dense H2: [tex]T\simeq 10 - 100~K[/tex], [tex]n_H \simeq 100 - 10^6~cm^{-3}[/tex], [tex]f \simeq 0.0005[/tex]

    where f is the fraction of the volume of the galaxy composed of that phase. We'll need these numbers to calculate some of the relevant timescales. Also, it's worth just reviewing the big picture for a moment.

    Except for the dense H2, most of the ISM is in approximate pressure balance (you can check for yourself by multiplying the temperatures and densities). Why is the H2 not in pressure balance? Because it's contained in self-gravitating clouds in which a pressure gradient is required for support.

    What about the filling factors? Well, the numbers above would indicate that the majority of the volume is made up of HI and coronal gas, while the majority of the mass is in a combination of HI and molecular gas. It would make sense that the molecular gas isn't a large fraction of the volume, being so dense.

    Anyway, one of the most important questions when we're talking about equilibrium is the time it takes to reach a Maxwellian velocity distribution. In absence of this, you can't say much of anything about the gas, so we hope that the timescales are short (enough). It turns out that the timescale for electrons to thermalize from scattering off protons is:

    [tex]t_s = \frac{3.8 \times 10^5 sec}{ln \Lambda}(\frac{T}{10^4 K})^{3/2}(\frac{cm^{-3}}{n_p})[/tex]

    The term with the natural log is called the "Coulomb logarithm" and is calculated based on the Debye length in the ISM. [tex]ln\Lambda[/tex] typically has values of order 20-30, meaning that most of the electrons in the ISM will thermalize in less than a year. Even the longest timescale (~100 years for the coronal gas) is much shorter than the timescales on which conditions in the ISM typically change.

    That's only for the electrons, however. If photoionization is the primary heating mechanism, then in order to thermalize the protons, you have to exchange energy from the electrons. This happens during scattering and the timescale is given roughly by:

    [tex]t_{s,p}=\frac{m_p}{m_e}t_{s,e}[/tex]

    This will make the timescale longer by about a factor of a thousand, maximizing at about 105 years for coronal gas. Even that is less than the typical timescale for dynamical change in the ISM, so most of the gas we see should be in an approximately Maxwellian distribution.

    I know I didn't answer all of your questions directly and there's a lot more, but let's just start there. Does this answer any of your questions? What else would you like to know?
     
    Last edited: Apr 18, 2005
  20. Apr 18, 2005 #19
    So what does all of that basically mean? My mathematical background is not advanced enough yet to understand what it all means.
     
  21. Apr 19, 2005 #20

    Nereid

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    1) most of the volume in 'the galaxy'* is occupied by two types of ISM ('interstellar medium'), "HI Warm" (to an OOM, atomic hydrogen), and "Coronal gas" (to an OOM, a hot, fully ionised plasma - mostly protons and electrons; it's called 'coronal' because it's similar to the plasma that comprises the corona of our Sun).

    2) a 'Maxwellian velocity distribution' is (crudely) 'what proportion of the constituent particles are going at what speeds? Plot them, and the curve looks like {insert nice graph here}'. If you've studied the 'ideal gas law', you'll have some idea where this comes from (here is a nice java applet that might help)

    3) the bit about 'Debye lengths' refers to plasmas; in a plasma there are electrons and there are positive ions (I'm talking about ISM plasmas; negative ions don't apply) - the former are much less massive than the latter (>1:1000), so 'equilibrium' is more complicated than for an ideal gas; in the absence of ('external') magnetic or electric fields, when a plasma 'settles down', the velocity distribution of its electrons and ions will have 'thermalized' ... you can describe that distribution with a single parameter, its 'temperature' (hence 'thermalize' :wink:)

    4) protons, being more massive, take longer to 'thermalize' than electrons. However, to get a handle on how long all this takes, you need to know what's keeping the plasma 'hot' (otherwise, if left 'all alone', surely it would end up being 2.7K, right?) - there are several mechanisms that heat ISM plasmas; Space Tiger chose to illustrate using just one ('hot photons' - i.e. UV, X-rays, ...).

    In answer to my question, SpaceTiger gave us the basic formulae for calculating the time it would take for the first two types of ISM to reach equilibrium (at the OOM, or half-OOM level) - i.e. the plasmas - and kindly crunched some numbers for us (you can plug them in and repeat the calculation for yourself). The bottom line, for these two types of ISM, is that equilibrium will happen much faster than any significant chunk of these ISM types will change (due to some 'external' change).

    From this we could examine what difference it would make, if (for example) a significant portion of the ISM were composed of He, or C/N/O ...

    We can also look at the other types of ISM; my favourites are the H2 clouds, and the extent to which dust and constituent gases other than molecular H make a difference :smile:

    *i.e. the Milky Way, but these numbers should be pretty OK (OOM-wise) for spirals and irregulars too.
     
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