Temperature.

Discussion in 'Physics & Math' started by Daecon, Aug 16, 2015.

  1. Daecon Kiwi fruit Valued Senior Member

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    I've just come across the concepts of "absolute hot" such as Planck and Hagedorn temperatures, and it got me thinking, is it mathematically possible to determine the relationship between a given temperature and the speed at which that atom is vibrating?

    Or is the nature of "speed" in this instance not applicable? For example, would it be nonsense to construct a thought experiment where an atom is vibrating at relativistic speeds?

    From what little I understand, the Planck threshold is where the kinetic energy of the atom would produce a gravitational effect comparible to the other three fundamental forces, and the Hagedorn threshold is where the atoms shake themselves apart on the elementary particle level?
     
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  3. James R Just this guy, you know? Staff Member

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    Take an ideal gas of atoms at temperature \(T\). The temperature is a measure of the average kinetic energy of the atoms. Specifically:

    \(\frac{3}{2}k_B T = \frac{1}{2}mv^2\)
    where \(k_B\) is a constant, \(m\) is the atomic mass, and \(v\) is the average speed of each atom in the gas.

    If you want to go relativistic with this, replace the Newtonian kinetic energy by the relativistic one.

    I have basically described a monatomic gas here, with no internal degrees of freedom such as vibration. When such additional degrees of freedom are added, the relationship between temperature and the various speeds (e.g. frequency of vibration) changes a little, but there's still a relationship.
     
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  5. danshawen Valued Senior Member

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    Check the thread on the fastest neutrino ever detected / OMG particle. 3,000 TeV for the neutrino, and 3 x 10*8 TeV iron nucleus OMG particle, respectively. The energies of both of these particles make the LHC upgrade energy look trifling.
     
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  7. Daecon Kiwi fruit Valued Senior Member

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    Is that related to speed rather than temperature?

    Is there any correlation between the two? Kinetic energy from temperature vs. kinetic energy from motion through space. Would a relativistic particle cool down or heat up?

    I'm a little reluctant to ask such a question, I don't remember exactly but I get the feeling something along these lines was asked before, and I think the thread descended into a crackpot fest. I tried to search for that thread but the forum came back with 200 results.

    [edit] Anyway, the reason for asking is that I was wondering if the more energy a particle has for accelerating through space, the less it will have available for vibration. Or whether there was a phenomenon similar to time dilation at work - from the perspective of an observer, as time appears to slow down for the particle, its vibrational speed might also appear to slow down, suggesting a cooling. But I guess that's irrelevant because from the "perspective" of the particle itself, nothing has changed. Unfounded assumptions on my part, I admit.
     
    Last edited: Aug 18, 2015
  8. James R Just this guy, you know? Staff Member

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    Daecon:

    Temperature is a measure of the collective average motion of a set of particles in system with reference to the centre of mass (or something like that).

    If you put a brick in the fridge and cool it to -270 degrees Celcius, then you'll have a cold brick. If you then throw that brick out into space at 99.99% of the speed of light, you'll then have a fast cold brick.

    The force to accelerate the particle (and hence the kinetic energy) comes from outside the particle. The vibrational energy is internal.

    Your line of reasoning suggests that the temperature of an object is best measured in its rest frame.
     
    Last edited: Aug 19, 2015
  9. exchemist Valued Senior Member

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    The force to accelerate the particle (and hence the kinetic energy) comes from outside the particle. The vibrational energy is internal.


    Your line of reasoning suggests that the temperature of an object is best measured in its rest frame.[/QUOTE]

    I suppose one obvious example is red or blue shift of stars, which would tend to make their temperature seem lower or higher than they actually are, wouldn't it?

    Or, thinking about it……perhaps the shape of the black body envelope would tell you what the true temperature must be, actual wavelength readings notwithstanding…..I'm not an astronomer.
     
  10. James R Just this guy, you know? Staff Member

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    With a moving star, we look at the spectrum and identify certain spectral lines. This is how the red shift is measured - we know where certain sets of lines would be if the star was at rest. Once the red shift is known it can be removed to recover the "true" blackbody spectrum of the star and hence determine its temperature.
     
  11. QuarkHead Remedial Math Student Valued Senior Member

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    As a matter of idle curiosity, how do "we" know this (assuming we cannot be at rest relative to this star)?
     
  12. Maxila Registered Senior Member

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    James, Just to clarify (I thought Daecon brought up an interesting question), if the brick (or star) were traveling at a relativistic velocity to an observer they would appear to be lower in temperature than the temperature measured from their rest frame, correct?
     
  13. Billy T Use Sugar Cane Alcohol car Fuel Valued Senior Member

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    No. Every wave length would be "blue shifted" (assuming emitter is traveling towards you). If it is a black body radiator at temperature T in its rest frame, the peak of the observed black body radiation would be at a shorter wave length (the observed T ' > T).

    If the brick is lead and T ' is greater than lead's melting point, the interesting question is what shape do you see the brick as having? My guess is it looks like a thin flat sheet about the same size as the flat front surface of the brick, certainly not a melted sphere. IE, if true then from POV of your frame, both temperature AND density have gone up.
     
    Last edited: Aug 20, 2015
  14. Maxila Registered Senior Member

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    Silly me, I didn't specify the objects were receding and observing them in infrared (I pictured James description of throwing a cold brick and based the question on that).
     
  15. James R Just this guy, you know? Staff Member

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    Because each element produces a set of spectral lines with known spacings. Therefore, we can identify an element not from the absolute value of its emitted wavelengths/frequencies, but from the relative spacings between them. And once we know, for example, "this line corresponds to such-and-such line in the Balmer series for hydrogen", then we know what the frequency of that line would be if the star were at rest. Hence we can calculate the red shift.
     
  16. James R Just this guy, you know? Staff Member

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    The peak of the observed blackbody curve would appear at a longer wavelength, so if you made a naive calculation based on that one piece of data then it would appear to be at a lower temperature, I guess. However, if we're a bit smarter we can determine the red shift and hence the actual temperature.
     
  17. Maxila Registered Senior Member

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    That's exactly what I wanted to know. I knew we could determine red shift I just never gave any thought to red or blue shift in regards to the observation of temperature before this thread, thank you.
     
  18. Fednis48 Registered Senior Member

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    Good points about the brick moving toward/away from the observer, but I think the OP raises another question that still has me confused. Say instead of a brick, we have a long, smooth metal rod zipping by the observer at a relativistic velocity. The observer is focusing on a very narrow slice of the rod, at its point of closest approach, such that the rod's velocity is perpendicular to its displacement from the observer at the point of observation. Because of the perpendicular velocity, the blackbody spectrum should not be directly red- or blue-shifted. But time dilation still occurs, so it seems like the metal atoms' thermal velocities would look slower to the observer. Does this mean that the blackbody radiation would still get redder, even without a Doppler redshift? Similarly, say the observer puts her hand on the (frictionless) rod, and lets it sweep across her fingers. Would it feel colder than in the rod's rest frame?
     
  19. brucep Valued Senior Member

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    Think in terms of local and remote measurements. The local measurements is invariant while the remote measurement is frame dependent. All observers agree with the local proper measurement after transforming there remote measurement to the local proper frame. This makes the measurement touching the rod a local proper frame measurement. The red and blue shift is a remote coordinate dependent measurement. Locally there is no red shift or blue shift.
     
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  20. Fednis48 Registered Senior Member

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    I'm not convinced. The measurement touching the rod may be local, but it's not proper because the rod is sliding past the observer. (If the observer actually grabbed the rod and held on to a fixed segment of it, I would agree with you.) Because the observer is still in motion with respect to the rod's local proper frame, I would expect her measurements of the metal atoms' velocities - and therefore their temperature - to disagree with the local proper measurement.
     
  21. brucep Valued Senior Member

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    The measurement is conducted in the local proper frame. The rod is in relative inertial motion and can be considered at rest with respect to a moving measurement device. Stuff doesn't lose temperature just because it can be modeled at rest or in relative motion. When the measurement is made both the rod and the measurement device are in the local proper frame. You can't change the temperature of the rod just by choosing different coordinates. That means the rod must change temperature due to some physical process other than being in relative motion.
     
    Last edited: Aug 23, 2015
  22. Q-reeus Banned Valued Senior Member

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    There is such a thing as transverse Doppler shift, and yes, overall there is a redshift. A few SR subtleties involved but nothing too complex or counter-intuitive once thought through:
    https://en.wikipedia.org/wiki/Relativistic_Doppler_effect#Transverse_Doppler_effect
    [An extreme variant of this is encountered in particle-accelerator ultra-relativistic synchrotron radiation: https://en.wikipedia.org/wiki/Synchrotron_radiation
    , where if one checks the formula, blueshift is confined to a very narrow forward lobe, with redshift elsewhere.]
     
    Last edited: Aug 23, 2015
  23. James R Just this guy, you know? Staff Member

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    Fednis48:

    You've discovered the tranverse Doppler effect. There's a Doppler shift even for an object moving transversely to the observer. It is due entirely to time dilation in that case.
     

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