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Difference between Thermal and Kinetic Energy

  1. Jul 1, 2009 #1
    Hi everyone,
    I'm new here, I have just discovered this forum and I think it will be very useful!
    First of all I would like to apologize for my english because I'm not a native english speaker (you are allowed to point out my errors :smile:).
    Anyway... this is the problem:
    when we speak about interstellar clouds or, in general, about thermodynamics in astrophysics, we meet whit 3 types of energy:
    - internal
    - thermal
    - kinetic

    I think that I have understood the difference between internal and thermal energy:
    the internal energy is the sum of "kinetics" contributions of the gas molecules (traslational, vibrational and rotational) + "zero point" energy. The thermal energy, that can be transmitted thru heat, is the sum of translational, vibrational and rotational energy.
    Hence thermal energy = internal energy - "zero point" energy

    Often, when we speak about gas clouds or about astrophysics phenomena, we clearly distinguish thermal and kinetic energy.
    Hence the question: the thermal energy is the average kinetic energy of the gas particles... and so what is kinetic energy?

    Let's take the simplest case: an hot intestellar cloud (T about 10^9 K) and so fully ionized gas (there is only translational contribution).
    I think that the thermal energy quantifies the level of kinetic "agitation" of the particles (in random and untidy motion), in fact we can associate thermal energy with temperature.
    Instead the kinetic energy quantifies an ordered flux of particles in a given direction.
    Hence, even if they represent the kinetic motion of the particles, we can say that: if kinetic energy prevails the cloud will modify its shape, elongating in a given direction.
    If thermal energy prevails there will not be the extension and my cloud will follow the usual thermodynamics laws.
    Hence we can also say that in a gas cloud the energy is mainly thermal or kinetic because if one of these prevails the other decreases.

    Is my interpretation correct?
    Where do I make mistake?
    Thank you!
    Bye
     
  2. jcsd
  3. Jul 1, 2009 #2
    I'm not in astro, but I believe you're mostly correct. Thermal energy is comprised of the translational, vibrational, and rotational energies of the particles. In this context, when you say kinetic energy, I believe this refers to NET motion of a massive collection of particles. A particle can be bouncing around inside a cloud. This would fall under thermal energy. However, when the entire mass of the cloud is moving in bulk, this is called kinetic energy. Someone please correct me if I'm wrong.
     
  4. Jul 3, 2009 #3
    Ok, but there is a problem with mean free path.
    In fact if mean free path decreases I should find more thermal energy than kinetic. This because particles can share their energy colliding and so increasing the motion agitation.
    The mean free path is proportional to 1/density and this is also intuitive.
    But I know that a stellar wind bubble interior, an hot gas with a very low density, has more thermal energy than kinetic and I don't understand the reason, because mean free path is very high and so I should find more kinetic energy than thermal...
     
  5. Jul 14, 2009 #4
    Hi FrankPlanck,

    Mean Free path, [itex]\lambda[/itex] for a plasma depends on Temperature ([itex]T[/itex]) and density ([itex]\rho[/itex]):
    [tex]\lambda \propto \frac {T^2}{n} [/tex]

    Thus, the higher the temperature of the gas, the longer the mean free path is. Also, the lower the density, the longer the mean free path is. The bulk motion of the gas does not affect the mean free path.

    -bombadil
     
  6. Jul 22, 2009 #5
    Hi bombadil,
    thank you very much, this changes everything!
    Where can I find the demonstration of this formula?

    Bye
     
  7. Jul 22, 2009 #6
    I can't find a good simple derivation online, so I'll get back to you with a simplified derivation in a day or two
     
  8. Jul 22, 2009 #7
    It's hard to measure thermal energy with a speedometer.

    Technically, thermal energy=kinetic energy. But thermal energy is kinda "useless" since it is harder to capture and use than normal kinetic energy.
     
  9. Jul 23, 2009 #8

    Chronos

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    Thermal and kinetic energy are interchangeable. A massive object streaking through space can be assigned a potential temperature, if so desired. It's the same priciple used to convert the kinetic energy of an asteroid striking earth into thermal energy.
     
  10. Jul 27, 2009 #9
    Bombadil, you can post the "real" demonstration, then I will try to understand it.
    Thank you all for your replies.
     
  11. Jul 27, 2009 #10
    The following is a very simplified discussion, and only intended to show scaling. I'll be omitting all numerical factors and calculate in cgs units.

    The strong interaction radius, [itex]r[/itex], is defined when the couloumb potential energy is equal to the average kinetic energy per particle:
    [tex] T\sim \frac{e^2}{r} [/tex]
    Where T is the temperature (I assume the electrons are in thermal equilibrium with the ions), and e is the fundamental charge. Thus,
    [tex]r=\frac{e^2}{T}[/tex]
    leading to,
    [tex]\sigma \sim r^2 \sim \frac{e^4}{T^2}[/tex]
    The above discussion only accounts for strong collisions, to account for weak collisions more analysis is needed and I'll just quote the result:
    [tex]\sigma \sim \frac{e^4}{T^2}\ln \Lambda[/tex]
    Where the quantity [itex]\ln \Lambda[/itex] is known as the http://en.wikipedia.org/wiki/Coulomb_collision" [Broken]. Since [itex]\Lambda[/itex] is contained in a logarithm its precise value is not that important. In fact, for most relevant parameters [itex]\ln \Lambda \sim 10-30[/itex], and is roughly constant. Thus the mean free path can now be written as:
    [tex]\lambda=\frac{1}{n \sigma}=\frac{T^2}{n e^4 \ln \Lambda}[/tex]
    where n is the particle number density. Don't forget that [itex]\lambda[/itex] is the distance it takes for a particle to change it's direction by 90 degrees

    For a very detailed discussion and exact calculations, look at http://ocw.mit.edu/NR/rdonlyres/Nuclear-Engineering/22-611JIntroduction-to-Plasma-Physics-IFall2003/15E93A3B-75CD-45C9-82AC-B9962044F499/0/chap3.pdf" [Broken] of a plasma MIT course. If you click on the link, look for the "90 degree" cross-section, [itex]\sigma_{90}[/itex], since that's what you need to calculate the mean free path as described above.
     
    Last edited by a moderator: May 4, 2017
  12. Jul 29, 2009 #11
    Bombadil, thank you very much!!
     
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