Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

B Temperature of an accelerating system

  1. Aug 18, 2016 #1
    if i take a box and fill with solid metal that has a temperature T .
    what if the box is accelerated to the speed of light . would T -> 0 ?
     
  2. jcsd
  3. Aug 18, 2016 #2

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    You can't accelerate a box to the speed of light. Might you be thinking of the Unruh effect? https://en.wikipedia.org/wiki/Unruh_effect

    Note that as far as the Unruh effect goes, it's the acceleration that matters. Speed is irrelevant to the effect.

    If you are not thinking of the Unruh effect, you might want to give us a source for what made you ask the question.
     
  4. Aug 18, 2016 #3

    Nugatory

    User Avatar

    Staff: Mentor

    As asked, those question is meaningless, because there can be no such thing as a box of solid metal that has been accelerated to the speed of light. Do you mean a box that has been accelerated to a speed near that of light?
     
  5. Aug 18, 2016 #4
    yes , let it be 80% the speed of light or any high speed , i just said "the speed of light" as an extreme , whould T decrease ?
     
  6. Aug 18, 2016 #5

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    By definition, and I cannot stress this important insight enough, nowadays temperature is defined as a scalar quantity, i.e., it is independent of the motion of the body and it is measured with a thermometer co-moving with the matter. It is so important to keep this in mind, because in the early days of relativity there was a big confusion in the community of physicists, and sometimes temperature was defined as a quantity which changes with the frame of reference, but this is not a very clever idea, and for some decades now one defines temperature as a scalar (field).

    For an irrotational ideal gas in local thermal equilibrium the distribution function is the Boltzmann-Jüttner distribution function,
    $$f(t,\vec{x},\vec{p})=\exp \left (-\frac{p \cdot u-\mu}{T} \right),$$
    where ##T=T(t,\vec{x})## is the local temperature, ##\mu=\mu(t,\vec{x})## the chemical potential (related to any conserved charge; if there is no conserved charge for the particles of the gas in question, ##\mu=0##), and ##u=\gamma(1,\vec{v}/c)=u(t,\vec{x})## is the fluid's four-velocity field. Since the phase-space distribution is (also by definition) a scalar, this means that ##T## and ##\mu## are scalars too since it's written in manifestly covariant form.

    For details on these issues of thermodynamic quantities in relativity, see my lecture notes on the relativistic Boltzmann Equation:

    http://th.physik.uni-frankfurt.de/~hees/publ/kolkata.pdf
     
  7. Aug 18, 2016 #6
    If the box is in space and collides with the interstellar medium then it heats up from the collisions.
     
  8. Aug 18, 2016 #7

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Sure, but the temperature of the box is always measured in its rest frame by definition. Thus it's a scalar quantity. Of course this quantity can change when somehow heat is transferred to the body.
     
  9. Aug 18, 2016 #8
    If I place a metal box in a vacuum chamber with a thermometer attached to it, am I reading the true temperature of the box or the temperature of the thermometer plus the box once they are in equilibrium?
     
  10. Aug 18, 2016 #9

    Nugatory

    User Avatar

    Staff: Mentor

    You are reading the true temperature of the box after the thermometer and the box have reached equilibrium. This may be slightly different than the true temperature of the box before you attached the thermometer to it.
     
  11. Aug 18, 2016 #10
    thank you
     
  12. Aug 18, 2016 #11

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    @vanhees71: Hendrik, to be able to write notes on this topic, you've done a lot of reading, I am sure. Tell me, is Tolman's famous 1934 book https://www.amazon.com/Relativity-T...e=UTF8&qid=1471556363&sr=1-48&keywords=Tolman stil valid, I mean the physics in it is it still correct, or you can value it today only from a historical viewpoint, as the subsequent research in the field superseded the knowledge level of 1934? [In a similar vein, Pauli's famous GR notes/book from 1921 I believe to still be accurate to a large extent, the same for Eddington's book]
     
    Last edited by a moderator: May 8, 2017
  13. Aug 19, 2016 #12

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    I've once had a look on this book, and I found it totally outdated, particularly about the very point of thermodynamic quantities. The same holds for Pauli's otherwise great review.
     
  14. Aug 20, 2016 #13

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    As I stressed before, nowadays temperature is defined as a scalar. The Planck distribution for an arbitrarily moving observer with written in covariant form as the Bose distribution of a gas of massless particles by
    $$f_{\text{B}}(\vec{p})=\frac{g_{\gamma}}{\exp(\beta u \cdot p)-1}.$$
    Here ##\beta=1/(k_{\text{B}} T)##, ##g_{\gamma}=2## is the helicity degeneration factor for photons, ##u=\gamma(1,\vec{\beta})## the four-velocity of the observer relative to the restframe of the cavity containing the black-body radiation, and ##p=(|\vec{p}|,\vec{p})## the four-momentum vector of the photon. Since the phase-space distribution function is a scalar, it is immediately clear that ##T##, the temperature of the black-body radiation, is a scalar.
     
  15. Aug 20, 2016 #14
    So essentially the defined temperature is similar to proper time, right (something every reference frame agrees on that is measured in its rest frame)? If so, is there something similar to coordinate time for temperature? Call it "coordinate temperature," and if so what is it?
     
  16. Aug 20, 2016 #15

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    In a sense yes. Take the above given Planck distribution. Let's rewrite this as a distribution function for the energy of photons and the angle ##\vartheta## relative to the observer's velocity against the restframe of the heat-bath. For the following I use natural units of many-body HEP, i.e., ##\hbar=c=k_{\text{B}}=1##, which simplifies the formulae somewhat. First of all we have for photons ##E=|\vec{p}|##. So we have for the photon-number distribution
    $$\mathrm{d} N = \frac{\mathrm{d}^3 \vec{p}}{(2 \pi)^3} \frac{1}{\exp(\gamma E(1-\beta \cos \vartheta)/T}.$$
    Now you have
    $$\mathrm{d}^3 \vec{p} = P^2 \mathrm{d} P \mathrm{d} [\cos \vartheta] \mathrm{d} \varphi=E^2 \mathrm{d} E \mathrm{d} [\cos \vartheta] \mathrm{d} \varphi.$$
    So in any direction the moving observer sees a Planck spectrum as if he were at rest with an "effective Temperature"
    $$T_{\text{eff}}=\frac{T \sqrt{1-\beta^2}}{1-\beta \cos \vartheta}.$$
    This is nothing else than the Doppler effect for thermal radiation. This can be used, e.g., to measure our velocity against the cosmic microwave background radiation (i.e., the usually used "fundamental frame" of cosmology). Indeed the maps of the Planck or WMAP satellites' measurement of the background-radiation temperature are only so uniform after taking out this effect of our motion relative to the fundamental coordinate system, which is defined as a reference frame at rest relative to the CMBR. In the angular power spectrum of the temperature variation this leads to a large dipole contribution. See, e.g.,

    http://www.astro.ucla.edu/~wright/CMB-DT.html

    Still, it's clear that we define only ##T## (a scalar!) as the temperature characterizing the radiation.

    A similar effect has to be taken into account in the physics of heavy-ion collisions. Also there photons are a very important probe to figure out the properties of the hot and dense medium (partially a Quark Gluon Plasma, partially a hot and dense gas of hadrons), which consists of a very hot and dense nearly thermalized fireball undergoing rapid expansion (collective flow) and cooling down. This fireball emits photons, which are distributed similarly as Planck radiation, although it is not Planck radiation, because the photons immediately decouple from the medium. Since, however this medium is close to thermal equilibrium, the radiation shows a exponential decay with the photons' energy as Planck radiation. The corresponding expoinential slope of the momentum (or transverse-momentum) spectra is not directly the temperature but it's Doppler blue shifted since the photons are emitted from the radially expanding medium. So when estimating the temperature of the fireball (or more precisely a space-time weighted average of the temperature over the entire fireball evolution) one has to take into account this Doppler shift by modelling the expansion of the medium accurately (e.g., using realtivistic hydrodynamics or transport theory).

    Another measure are invariant-mass spectra of dileptons (i.e., electron-positron or muon-antimuon pairs). There's a mass window between 1 and 2.5 GeV invariant mass, where the source is pretty unstructured (i.e., free of resonances), and thus the slope admits a direct estimate of the temperature of the medium, since as a function of invariant mass, there's no blue shift of the spectra due to the motion of the source, exactly because invariant mass is a scalar quantity. For more details on this, see the slides of my lectures at various graduate-school lecture weeks:

    http://th.physik.uni-frankfurt.de/~hees/hqm-lectweek14/index.html
    http://th.physik.uni-frankfurt.de/~hees/publ/qm14-lect.pdf

    http://th.physik.uni-frankfurt.de/~hees/publ/graz15-1.pdf
    http://th.physik.uni-frankfurt.de/~hees/publ/graz15-2.pdf

    http://th.physik.uni-frankfurt.de/~hees/publ/cbm16-lect.pdf
     
    Last edited: Aug 20, 2016
  17. Aug 20, 2016 #16
    Thanks a bunch for this. I love how the theory is so symmetrical in so many ways.
     
  18. Aug 20, 2016 #17

    pervect

    User Avatar
    Staff Emeritus
    Science Advisor

    Approaches also exist in which the inverse temperature, ##\beta##, is just one part of a 4-vector. See for instance Nakumura, "Covariant thermodynamics of an object with finite volume". In these approaches, one writes ##\Delta S = \beta \cdot \Delta Q## rather than ##\Delta S = \Delta Q/T##. ##\beta## here is a 4-vector, not just a simple scalar, and rather than write the non-covariant change in energy as a scalar, one gives the change in the energy-momentum 4-vector.

    So a bit of care is suggested in assuming what an author is defining tempreature as. I think it's also good practice (if a bit tedious) to define the frame in which the temperature is being measured at least briefely.
     
    Last edited: Aug 20, 2016
  19. Aug 21, 2016 #18

    vanhees71

    User Avatar
    Science Advisor
    2016 Award

    Yes, in my notation it's ##\beta u##, where ##u## is the four-velocity of the fluid cell with ##u \cdot u=1## (in west-coast convention). It's also clear from the preprint cited in #22 that this is simply another (unusual) convention of the author. Everything can (and in my opinion should) be defined in as simple a way as possible, and that's for sure by defining intrinsic quantities of matter as scalars by using the fact that in this case a physically preferred reference frame exists, namely the rest frame of the matter. This can also be a set of local rest frames as for a fluid, where one defines the local rest fluid rest frames in terms of the four-velocity and corresponding projections in its direction and Minkowski-transverse to it.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted